Research ArticleNumerical Solvability and Solution of an Inverse ProblemRelated to the Gibbs Phenomenon

Nassar H S Haidar

Center for Research in Applied Mathematics amp Statistics (CRAMS) AUL PO Box 14-6495 Cola Beirut Lebanon

Correspondence should be addressed to Nassar H S Haidar nhaidarsuffolkedu

Received 7 March 2015 Accepted 10 May 2015

Academic Editor Don Hong

Copyright copy 2015 Nassar H S HaidarThis is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

We report on the inverse problem for the truncated Fourier series representation of 119891(119909) isin BV(minus119871 119871) in a form with a quadraticdegeneracy revealing the existence of the Gibbs-Wilbraham phenomenon A new distribution-theoretic proof is proposed forthis phenomenon The paper studies moreover the iterative numerical solvability and solution of this inverse problem neardiscontinuities of 119891(119909)

1 Introduction

This paper reinvestigates the Fourier series [1 2] representa-tion

119892 (119909) =11988602+infin

sum119898=1

119886119898cos119898120587

119871119909+ 119887119898sin119898120587

119871119909 (1)

of a piecewise continuous 2119871 ndashperiodic signal 119891(119909) isinBV(minus119871 119871) with

1198860 =1119871int119871

minus119871

119891 (119909) 119889119909

119886119898=

1119871int119871

minus119871

119891 (119909) cos119898120587

119871119909119889119909

119887119898=

1119871int119871

minus119871

119891 (119909) sin119898120587

119871119909119889119909

(2)

The Gibbs phenomenon (see eg [3]) is a statement ofthe fact that the infinite series 119892(119909) tends to overshoot thepositive corner of a discontinuity of 119891(119909) by sim9 of thejump size and to undershoot the negative corner by the sameamount An overshootundershoot effect that is accompaniedwith spurious oscillations ldquoringingrdquo near the discontinuity (asdescribed in the appendix) when the series in (1) is truncatedat 119898 = 119873 However according to a theorem by Fejer [1] theinfinite series 119892(119909) of a BV(minus119871 119871) function converges to 119891(119909)at each point 119909 of continuity of 119891(119909)

This shortcoming in the infinite Fourier series represen-tation of piecewise continuous 119891(119909) was first observed byH Wilbraham in 1848 and then analyzed in detail [4] by JW Gibbs in 1898 The main reasons of the Gibbs-Wilbrahameffect are that (i) not all frequencies (only integer ones) areemployed in (1) (ii) 119886

119898and 119887119898happen to decay slowly with

increasing119898 and (iii) the global nature of the approximationof 119891(119909) the expansion coefficients are obtained via (2) byintegration over the entire period including the points ofdiscontinuity

What is unpleasant though with all of this is that theGibbs-Wilbraham effect is generic and is present for anyperiodic signal 119891(119909) isin BV(minus119871 119871) with isolated disconti-nuities The presence of this effect can in fact lead to quitenegative consequences when single infinite Fourier seriesmultiple infinite Fourier series or even infinite wavelet seriesare employed to approximate signals of various dimensionsin many fields such as radio engineering and signal transmis-sion

The Gibbs-Wilbraham effect can nevertheless have bothpositive and negative consequences in different applicationsThe negative consequences call for Gibbs effect reductionand this can in principle be achieved with the use [4] of avariety of filters This effect can also be reduced theoreticallyand for all purposes In addition to classical mathematicalfilters recently (in 2011) Rim and Yun [5] defined a kind ofspectral series to filter off completely the Gibbs effect near

Hindawi Publishing CorporationInternational Journal of Computational MathematicsVolume 2015 Article ID 212860 13 pageshttpdxdoiorg1011552015212860

2 International Journal of Computational Mathematics

a discontinuity The construction of this series is based onthe method of adding the Fourier coefficients of a Heavisidefunction to the given Fourier partial sums

This paper is organized as follows After this introductionwe propose in Section 2 a new distribution-theoretic prooffor the Gibbs-Wilbraham effect In Section 3 we advancethe inverse problem of the truncated Fourier series and thenew stereographic truncated Fourier series and its relationto resolution of the undershootovershoot pair associated tothe Gibbs-Wilbraham effect Here we illustrate how solvingthe inverse problem for the truncated representation (1)contains a quadratic degeneracy that indicates the existenceof theGibbs phenomenon Section 4 deals with the numericaliterative solution of the previous inverse problem and its con-vergence with a proof and demonstration that stereographicprojection does not affect such a numerical solution and itsposedness

2 Distribution-Theoretic Proof forthe Gibbs-Wilbraham Effect

Let 119891(119909) isin BV(minus119871 119871) be a 2119871 ndashperiodic piecewise continu-ously differentiable function on (minus119871 119871) and let 119909

119904be a point

in (minus119871 119871) at which 119891(119909) has a discontinuity of the first kindConsider 119909minus

119904= 119909119904minus 0 119909+

119904= 119909119904+ 0 and 119891minus

119904= 119891(119909minus

119904)

and 119891+119904

= 119891(119909+119904) to assume without loss of generality that

119891+119904gt 119891minus119904and define the discontinuity jump by

Δ119904= 119891+119904minus119891minus119904 (3)

Moreover

119891119888(119909) =

119891 (119909) 119909 = 119909119904

undefined 119909 = 119909119904

(4)

Theorem 1 Let a 2119871 ndashperiodic 119891(119909) isin BV(minus119871 119871) be discon-tinuous at 119909

119904 The Fourier series representation 119892(119909) of 119891(119909)

converges over (minus119871 119871) as119873 rarr infin so as

119892 (119909) =

119891minus119904minus 120588Δ119904 119909 = 119909minus

119904

12(119891minus119904+ 119891+119904) 119909 = 119909

119904

119891+119904+ 120588Δ119904 119909 = 119909+

119904

119891119888(119909) 119909 = 119909

119904119900119903 119909plusmn119904

(5)

where 120588 = 119906(0) an uncertainty

Proof According to distribution theory [6] when 119892(119909) coin-cides ae with 119891(119909) over (minus119871 119871) it may always be differenti-ated namely

1198921015840 (119909) = 1198911015840119888(119909) +Δ

119904120575 (119909 minus 119909

119904) (6)

with 120575 as Diracrsquos delta Cancellation of this differentiationis performed by sweeping the 119909-axis during a compensatingintegration from minus119871 to 119871 Indeed integration of (6) first fromminus119871 to 119909 leads to

int119909

minus119871

[1198921015840 (120591) minus1198911015840119888(120591)] 119889120591 = Δ

119904int119909

minus119871

120575 (120591 minus 119909119904) 119889120591 (7)

and since 119892(minus119871) = 119891119888(minus119871) then

119892 (119909) = 119891119888(119909) +Δ

119904119906 (119909 minus 119909

119904) (8)

with 119906 as Heavisidersquos unit step functionBy a theorem of Fejer [1] 119892(119909) should converge to 119891

119888(119909)

when119873 rarr infin forall119909 = 119909119904on (minus119871 119871) Hence the term Δ

119904119906(119909 minus

119909119904) in (8) can have a compact support only on an infinitesimal

interval 120590 = lim119873rarrinfin

120590119873= 0 to the right of 119909

119904 that is ending

at 119909+119904 Hence

119892 (119909) = 119891119888(119909) +Δ

119904119906 (119909+119904minus119909119904) (9)

Furthermore despite the fact that 119906(119909+119904minus 119909119904) = 1 119906(119909+

119904minus

119909119904) asymp 119906(0) which is an uncertainty to be denoted by 120588

Consequently relation (9) is representable as

119892 (119909) =

119891119888(119909) 119909 = 119909

119904

119891119888(119909) + 120588Δ

119904 119909 = 119909+

119904

(10)

Next complete the sweeping by integrating (6) from 119909 to119871 namely

int119871

119909

[1198921015840 (120591) minus1198911015840119888(120591)] 119889120591 = Δ

119904int119871

119909

120575 (120591 minus 119909119904) 119889120591 (11)

Here again since 119892(119871) = 119891119888(119871) then the left-hand side of

(11) becomes minus119892(119909) + 119891119888(119909) The right-hand side requires

standardization by means of the substitution 120591 = minus] leadingto 119889120591 = minus119889] [119909 119871] becoming [minus119909 minus119871] and

int119871

119909

120575 (120591 minus 119909119904) 119889120591 = minusint

minus119871

minus119909

120575 (minus]minus119909119904) 119889]

= minusintminus119871

minus119909

120575 (]+119909119904) 119889]

= intminus119909

minus119871

120575 (]+119909119904) 119889] = 119906 (minus119909+119909

119904)

(12)

It is obvious then that (11) is reducible to

119892 (119909) = 119891119888(119909) minusΔ

119904119906 (minus119909+119909

119904) (13)

Repeated application of the same arguments employed inthe derivation of (9)-(10) to (13) allows for writing it as

119892 (119909) = 119891119888(119909) minusΔ

119904119906 (119909minus119904+119909119904) (14)

and subsequently in the form

119892 (119909) =

119891119888(119909) 119909 = 119909

119904

119891119888(119909) minus 120588Δ

119904 119909 = 119909minus

119904

(15)

Taking into consideration that 119891119888(119909+119904) = 119891+119904and 119891

119888(119909minus119904) = 119891minus119904

in (10) and (15) respectively then combining these equationswith Dirichletrsquos theorem [1] leads to the required result

International Journal of Computational Mathematics 3

It should be pointed out that in the previous proofrelations (10) and (15) say nothing specifically about thebehavior of119892(119909) at119909minus

119904and119909+119904 respectively while asserting the

existence of a spiky behavior of 119892(119909) individually at 119909+119904then

at 119909minus119904 Moreover the uncertainty 120588 had been resolved in the

past (by Wilbraham see also the Appendix) computationallyto satisfy 120588 = 009 and long before the advance of the theoryof distributions (generalized functions) by S L Sobolev in1936 The number 009 for 120588 remains until now howevera mysterious theoretical puzzle that calls for a rigorousdistribution-theoretic justification

Clearly in the neighborhood of 119909119904 119892(119909) can only be a

rudimentary approximation to 119891(119909) even when 119873 rarr infinwith an added spike pair

119876119904(119909) =

+120588Δ119904 119909 = 119909+

119904

minus120588Δ119904 119909 = 119909minus

119904

0 119909 = 119909119904

(16)

to 119891plusmn119904at 119909plusmn119904 Relation (16) represents the Gibbs-Wilbraham

effect which due to the fact that |119909+119904

minus 119909minus119904| = 0 is

not practically computable This effect degenerates howeverwhen119873 is finite to an undershoot-overshoot pair119876

119904120590119873(119909) on

an interval 119868119904120590119873

= [119909119904minus 120590119873 119909119904+ 120590119873] splitted by Δ

119904at 119909119904 In

this case it is always possible to compute

119892 (119909) =

sim 119891 (119909) + 119876119904120590119873

(119909) 119909 isin 119868119904120590119873

119873 lt infin

sim 119891 (119909) 119909 notin 119868119904120590119873

(17)

Furthermore since lim119873rarrinfin

119876119904120590119873

(119909) = 119876119904(119909) then these

computations with (17) over 119868119904120590119873

can reveal only a feature119876119904120590119873

(119909) associated with the Gibbs-Wilbraham effect 119876119904(119909)

but not the effect itself

3 The Inverse Problem for TruncatedFourier Series

Substitution of any value of 119909 isin [minus119871 119871] in (1) yields a number119888 = 119892(119909) which may differ [2] from 119891(119909) by 119876

119904120590119873(119909) if 119909 is

in the neighborhood of a discontinuity of 119891(119909) Obviously 119888satisfies the nonlinear in 119909 trigonometric series equation

119873

sum119898=1

119866 (120574 119909119898) =119873

sum119898=1

119886119898cos119898120587

119871119909+ 119887119898sin119898120587

119871119909

= 119888 minus11988602

(18)

which is however linear in the data 120574 = 119886119898 119887119898infin119898=1

Definition 2 (the inverse problem for the truncated Fourierseries representation) Given a number 119888 = 119892(119909) satisfying(18) what is the corresponding 119909

This inverse problem happens to be a discretized versionof the similar deconvolution problem [4] of 119909 from a givennumber 119902 in a singular first kind integral equation of the form

intinfin

0119870(120574 119909 120591) 119889120591 = 119902 (19)

in which 120574 represents the data It is well known additionally(see eg [7 8]) that this deconvolution problem can becomputationally ill-posed if an arbitrarily small deviationof the 120574 data may cause an arbitrarily large deviation inthe solution 119909 This situation may further prevail when theoperator intinfin0 119870(120574 119909 120591)119889120591 does not continuously depend on120574

Moreover the inverse problem of the previous definitionhas a unique solution that satisfies (19) and is expected toaccept a double rootwhen the horizontal line ℎ(119909) = 119888 gt 119891(119909)intersects with a sharp peak (overshoot) of a truncated 119892(119909)We shall call this feature a duplicated (quadratic) degeneracyof the solution for this inverse problem

The existence of the Fourier series truncation peak (andthe corresponding two roots for 119909) is not directly evidentfrom the structure of 119866(120574 119909119898) and despite its linearity inthe data 120574 = 119886

119898 119887119898infin119898=1

31 The Tangent Half-Angle Stereographic Projection Thetangent half-angle substitution (the Euler-Weirstrass sub-stitution [9]) tan(1205792) is widely used in integral calculusfor finding antiderivatives of rational expressions of circularfunction pairs (cos 120579 sin 120579) Geometrically the constructiongoes like this draw the unit circle and let 119875 be the point(minus1 0) A line through 119875 (except the vertical line) passesthrough a point (cos 120579 sin 120579) on the circle and crosses the119910-axis at some point 119910 = 119905 Obviously 119905 determines theslope of this line Furthermore each of the lines (exceptthe vertical line) intersects the unit circle in exactly twopoints one of which is 119875 This determines a function frompoints on the unit circle to points on a straight line Thecircular functions determine therefore a map from points(cos 120579 sin 120579) on the unit circle to slopes 119905 In fact this isa stereographic projection [9] that expresses these pairs interms of one variable tan(1205792) = 119905

Numerical algorithms employing these trigonometricpairs like Fourier series (1) could in principle be influencedby such a projection In this context the tangent half-rangetransformation relates the angle 120579

119898= 119898(120587119871)119909 to the slope

119905119898= tan(120579

1198982) of a line intersecting the unit circle centered

around (0 0)Indeed as 120579

119898varies the point (cos 120579

119898 sin 120579119898) winds

repeatedly around the unit circle centered at (0 0) The point

(1 minus 1199052119898

1 + 1199052119898

2119905119898

1 + 1199052119898

) = (cos 120579119898 sin 120579119898) (20)

goes only once around the circle as 119905119898goes from minusinfin to infin

and never reaches the point (minus1 0) which is approached as alimit when 119905

119898rarr plusmninfin

Clearly the stereographic projection 119905119898

= tan(1205791198982) is

defined on the entire unit circle except at the projection point119875 (minus1 0) Where it is defined this mapping is smoothand bijective [9] It is conformal meaning that it preservesangles it is however not isometric that is it does not preservedistances as

119889120579119898=

2 119889119905119898

1 + 1199052119898

(21)

4 International Journal of Computational Mathematics

The rest of this section illustrates how stereographicprojection of Fourier series reveals a quadratic degeneracythat indicates the existence of the Gibbs-Wilbraham effect Italso addresses the question of solving the inverse problem forthe truncated Fourier series representations The impact ofstereographic projection on the ill-posedness of the numer-ical solution of this inverse problem is moreover a mainquestion that has been addressed in this paper

Our analysis starts however by illustrating how thetangent half-angle stereographic projection can provide anew insight into the behavior of truncated Fourier series neara discontinuity of 119891(119909)

Lemma 3 The solution 119909 of the inverse problem for thetruncated Fourier series representation satisfies

119873

sum119898=1

119867(120574 119909119898)

=119873

sum119898=1

(1198862119898+ 1198872119898) minus (119886

119898tan119898(1205872119871) 119909 minus 119887

119898)2

119886119898(tan2119898(1205872119871) 119909 + 1)

= 119888 minus11988602

(22)

Proof By studying solvability for 119909 of (18) Let 119905119898

=tan119898(1205872119871)119909 and then cos119898(120587119871)119909 = (1minus1199052

119898)(1+1199052

119898) and

sin119898(120587119871)119909 = 2119905119898(1 + 1199052

119898) Further consideration of these

facts in (18) leads to the tangent half-angle stereographicallytransformedS[ ] = [] form

minus119873

sum119898=1

S [119866 (120574 119909119898)] =119873

sum119898=1

(1198861198981199052119898minus 2119887119898119905119898minus 119886119898)

(1199052119898+ 1)

=11988602minus 119888

(23)

The harmonic trinomial in the left-hand side of (23) hastwo harmonic zeros 120585plusmn

119898 for each 119909 namely

120585plusmn119898=

2119871119898120587

tanminus1119905plusmn119898

119905plusmn119898=

119887119898

119886119898

plusmn120590119898

120590119898=

radic1198862119898+ 1198872119898

119886119898

(24)

It is worth noting at this point that these 120585plusmn119898zeros should

by no means be confused with the possible roots of the 119909variable Substitution then of (24) in (23) leads to

119873

sum119898=1

119886119898(119905119898minus 119905+119898) (119905119898minus 119905minus119898)

(1199052119898+ 1)

=11988602minus 119888 (25)

which is the same as

119873

sum119898=1

119886119898(119905119898minus 119887119898119886119898minus 120590119898) (119905119898minus 119887119898119886119898+ 120590119898)

(1199052119898+ 1)

=11988602minus 119888

(26)

The quadratic degeneracy of the harmonic zeros 120585plusmn119898of 119905119898

can disappear only when 120590119898= 0 for all119898 And this happens

to be equivalent with

119873

sum119898=1

1198862119898+ 1198872119898= 0 (27)

Obviously satisfaction of (27) by (26) reduces it to theequation

119873

sum119898=1

119886119898(119905119898minus 119887119898119886119898)2

(1199052119898+ 1)

=11988602minus 119888 (28)

which is the same as

119873

sum119898=1

(119886119898tan119898(1205872119871) 119909 minus 119887

119898)2

119886119898(tan2119898(1205872119871) 119909 + 1)

=11988602minus 119888 (29)

It is interesting to remark that condition (27) is entirelydifferent from

11988602+119873

sum119898=1

1198862119898+ 1198872119898gt 0 (30)

which can be derived from the Parseval identity [1] (com-pleteness relation)

11988602+infin

sum119898=1

1198862119898+ 1198872119898=

1119871int119871

minus119871

1198912(119909) 119889119909 (31)

Hence a possible nondegenerate solution for 119909 of the nonlin-ear equation (29) is on one hand fundamentally incorrectbecause it involves a violation of (30) and Parsevalrsquos identityOn the other hand a consistent solution 119909 in the senseof satisfaction of (31) is intrinsically degenerate because ofits emergence as a must from the nonlinear equation (26)which rewrites as

119873

sum119898=1

119886119898[(119905119898minus 119887119898119886119898)2minus 1205902119898]

(1199052119898+ 1)

=11988602minus 119888 (32)

Equation (32) is easily transformable to the requiredresult (23) here the proof completes

32 The Stereographic Fourier Series To further analyze thevariability of stereographically projected Fourier series wemake the substitution

120579119898= 120579119898(119909) = 119898

120587

119871119909 (33)

International Journal of Computational Mathematics 5

in 119866(120574 119909119898) to rewrite it as

119866 (120574 119909119898) = 119886119898cos 120579119898(119909) + 119887

119898sin 120579119898(119909) (34)

The tangent half-angle (stereographic) transformation119905119898= tan(120579

1198982) or 120579

119898= 2tanminus1119905

119898is conceived as

120579119898(119909) = 2tanminus1119905

119898(119909) = Φ [119905

119898(119909)] (35)

and subsequently

119866 = 119866 [120579119898(119909)] = 119866 Φ [119905

119898(119909)] = 119867 [119905

119898(119909)] (36)

It is clear then that this stereographic projection effectivelyleads to insertion of an additional bracket into the compo-sition of 119866 as a function of 119909 This paves the way to thedefinition

S 119866 [120579119898(119909)] = 119866 Φ [119905

119898(119909)] = 119867 [119905

119898(119909)]

= 119867 (120574 119909119898) (37)

Proposition 4 The following

S119889

119889119909

119873

sum119898=1

119866 [120579119898(119909)] =

119889

119889119909S119873

sum119898=1

119866 [120579119898(119909)] (38)

always holds

Proof Since 119873 is finite then the proof of S(119889119889119909)119866 =(119889119889119909)S119866 = (119889119889119909)119867 implies the correctness of thisproposition So let us differentiate 119866(119909) = 119866[120579

119898(119909)] with

respect to 119909 by applying the chain rule namely

119889119866

119889119909=

119889119866

119889Φ

119889Φ

119889119905119898

119889119905119898

119889119909=

119889119866

119889120579119898

119889120579119898

119889119905119898

119889119905119898

119889119909 (39)

which is the same as

119889119866

119889119909

= [minus119886119898sin 120579119898+ 119887119898cos 120579119898]

2(1 + 1199052119898)

120579119898

2sec2 (

120579119898

2)

(40)

Take then the S transformation of the previous relation toarrive after some algebra at

S119889

119889119909119866 = minus119898

120587

119871

(1198871198981199052119898+ 2119886119898119905119898minus 119887119898)

(1 + 1199052119898)

(41)

Invoke then

119867 = 119867(120574 119909119898) =(119886119898+ 2119887119898119905119898minus 1198861198981199052119898)

(1 + 1199052119898)

(42)

to differentiate it with respect to 119909 as

119889

119889119909119867 =

119889

119889119909S 119866 = minus119898

120587

119871

(1198871198981199052119898+ 2119886119898119905119898minus 119887119898)

(1 + 1199052119898)

(43)

which is the required result

Let us rewrite (1) in the form

120593 (119909) = 119892 (119909) minus 119888 = 0 (44)

then try to differentiate it term-by-term as

1205931015840 (119909) =119889

119889119909119892 (119909) =

119889

119889119909

infin

sum119898=1

119866 (120574 119909119898)

= minusinfin

sum119898=1

119898120587

119871(minus119887119898cos119898120587

119871119909+ 119886119898sin119898120587

119871119909)

= minusinfin

sum119898=1

119898120587

119871119866lowast (120574 119909119898)

(45)

where 119866lowast(120574 119909119898) is the harmonic conjugate of 119866(120574 119909119898)The trigonometric series 119892lowast(119909) = sum

infin

119898=1 119866lowast(120574 119909119898) is called

the trigonometric series conjugate (see eg [10]) to 119892(119909)In addition to the fact that the convergence of 119892lowast(119909)

can sometimes become worse than convergence of 119892(119909) (seeeg [10]) the presence of the 119898(120587119871) factor in (119889119889119909)119892(119909)can aggravate the convergence problem of (45) when thecoefficients 119886

119898and 119887119898do not decay fast enough

Next we may stereographically transform (120597120597119909)119892(119909)using the notationS[(119889119889119909)119892(119909)] = [(119889119889119909)119892(119909)] where

[119889

119889119909119892 (119909)] = minus

infin

sum119898=1

119898120587

119871119866lowast (120574 119909119898)

= minusinfin

sum119898=1

119898120587

119871

(1198871198981199052119898+ 2119886119898119905119898minus 119887119898)

(1 + 1199052119898)

(46)

Similarly for the stereographic Fourier series representation

119892 (119909) =11988602+infin

sum119898=1

119867(120574 119909119898) (47)

in which

119867(120574 119909119898) =(119886119898+ 2119887119898119905119898minus 1198861198981199052119898)

(1 + 1199052119898)

(48)

we may approximate (44) by

120595 (119909) = 119892 (119909) minus 119888 = 0 (49)

Differentiate then (49) term-by-term as

1205951015840 (119909) =119889

119889119909119892 (119909) =

119889

119889119909

infin

sum119898=1

119867(120574 119909119898)

= minusinfin

sum119898=1

119898120587

119871119867lowast (120574 119909119898)

= minusinfin

sum119898=1

119898120587

119871

(1198871198981199052119898+ 2119886119898119905119898minus 119887119898)

(1 + 1199052119898)

(50)

6 International Journal of Computational Mathematics

Note here that119867lowast (120574 119909119898)

=(1198871198981199052119898+ 2119886119898119905119898minus 119887119898)

(1 + 1199052119898)

=(119887119898tan119898(1205872119871) 119909 minus 119886

119898)2minus (1198862119898+ 1198872119898)

119887119898(tan2119898(1205872119871) 119909 + 1)

(51)

is the harmonic conjugate of119867(120574 119909119898) and that

119867lowast (120574 119909119898) = 119866lowast (120574 119909119898) (52)

This rather interesting fact leads to a remark that follows

Remark 5 Away from points of discontinuity of 119891(119909) thestereographically transformed differentiated Fourier series

[(119889119889119909)119892(119909)] which defines 119866lowast(120574 119909119898) is the same as thederivative (119889119889119909)119892(119909) which defines 119867lowast(120574 119909119898) of thestereographic Fourier series 119892(119909) Moreover even at pointsof discontinuity of 119891(119909) [(119889119889119909)119892(119909)] = (119889119889119909)119892(119909)

Theorem 6 (see [1]) Let 119891(119909) be (i) continuous on [minus119871 119871]such that (ii) 119891(minus119871) = 119891(119871) and let (iii) 1198911015840(119909) be piecewisecontinuous over [minus119871 119871] then the corresponding Fourier series119892(119909) of (1) or (47) is differentiable at each point where (iv)11989110158401015840(119909) exists

Theprevious remark can incidentally have useful applica-tions when differentiating certain slowly converging Fourierseries as illustrated by the following example

Example 7 Consider the 2120587-periodic odd signal 119891(119909)defined over one period by 119891(119909) = 119909 Its Fourier series isknown to be

119892 (119909) = 2infin

sum119898=1

(minus1)119898+1

119898sin119898119909 (53)

that is with 119886119898= 0 forall119898 119887

119898= 2((minus1)119898+1119898) and 119898(120587119871) =

119898Differentiating term-by-term we obtain

1198921015840 (119909) = 2infin

sum119898=1

(minus1)119898+1 cos119898119909 (54)

which is not uniformly convergent because 119891(119909) is not onlydiscontinuous at 119909 = plusmn120587 but 119891(minus120587) = 119891(120587) and 11989110158401015840(plusmn120587) doesnot exist In particular at 119909 = 0 the sum of the series 1198921015840(0) =2suminfin119898=1(minus1)

119898+1 oscillates between 0 and 1 that is divergent Itis also well known nevertheless that the seriessuminfin

119898=1(minus1)119898+1

is 119862 minus 1 summable [1 2] to 12 and that turns out to yieldthe right answer 1198921015840(0) = 1198911015840(0) = 1 Moreover at 119909 = 120587 thesum of the series 1198921015840(120587) = 2suminfin

119898=1(minus1)2119898+1 rarr minusinfin that is

divergentAlternatively by means of (47)

119892 (119909) =infin

sum119898=1

119887119898

2119905119898

1 + 1199052119898

= 2infin

sum119898=1

(minus1)119898+1

119898sin119898119909 (55)

and by means of (46)

1198921015840 (119909) =

[119889

119889119909119892 (119909)] = minus

infin

sum119898=1

119898119866lowast (120574 119909119898)

= minusinfin

sum119898=1

119898119887119898

1199052119898minus 1

1 + 1199052119898

= 2infin

sum119898=1

(minus1)119898+1(1 minus tan2119898(12) 119909)(1 + tan2119898(12) 119909)

= 2infin

sum119898=1

(minus1)119898+1 cos119898119909

(56)

Here at 119909 = 120587 the sum of the series 1198921015840(120587) =

2suminfin119898=1(minus1)

2119898+1 rarr minusinfin is also divergent

Note in general that for even signals that is when 119887119898= 0

forall119898 it follows from (22) that

119867(120574 119909119898)1003816100381610038161003816119887119898=0

= 119886119898

1 minus 1199052119898

1 + 1199052119898

= 119886119898cos119898120587

119871119909

= 119866 (120574 119909119898)1003816100381610038161003816119887119898=0

(57)

Moreover for odd signals that is when 119886119898

= 0 forall119898 thesituation is not of the same clarity

Indeed according to (22)

119867(120574 119909119898)1003816100381610038161003816119886119898=0

=00

119887119898

1 + 1199052119898

(58)

which is an uncertainty Luckily however this uncertainty isnot an essential one and can straightforwardly be resolved byresorting to the stereographic form (23) of (22) which yieldsthe correct result

119867(120574 119909119898)1003816100381610038161003816119886119898=0

= 119887119898

2119905119898

1 + 1199052119898

= 119887119898sin119898120587

119871119909

= 119866 (120574 119909119898)1003816100381610038161003816119886119898=0

(59)

Also when 119887119898= 0 forall119898

119867lowast (120574 119909119898)1003816100381610038161003816119887119898=0

=00

119886119898

1 + 1199052119898

(60)

and this is resolvable in a similar fashionBoth119866(120574 119909119898) of (2) and119867(120574 119909119898) of (22) equivalent

forms are nonlinear trigonometric functions of the 119909 solu-tion However while 119866(120574 119909119898) is linear in 120574 = 119886

119898 119887119898infin119898=1

thestereographic 119867(120574 119909119898) is nonlinear in this 120574 This canraise a question on a possibility for (120597120597119909)119892(119909) = (120597120597119909)119892(119909)at singular points of 119891(119909) A question that will be answerednegatively in the next section Furthermore since the data 120574 =119886119898 119887119898infin119898=1 is discrete then the operators sum

infin

119898=1 119866(120574 119909119898)and sum

infin

119898=1 119867(120574 119909119898) do not depend continuously on 120574 andthis is themain reason for ill-posedness of the present inverseproblem

International Journal of Computational Mathematics 7

In conclusion the quadratic degeneracy of the solution ofthe present inverse problem explicit when using 119867(120574 119909119898)is a nonlinear indicator of the existence of the peak 119876

119904120590119873(119909)

pertaining to the Gibbs phenomenon However the numer-ical ill-posedness associated with the relative values of atrial iterative variable in the 120574 set happens to remarkablyremain invariant under the tangent half-angle stereographicprojection

Following the same arguments of Wilbraham we maycancel the artificially induced differentiation by integrating(50) to represent the stereographic Fourier series 119892(119909) inintegral form as

119892 (119909) =11988602minusinfin

sum119898=1

int119909

0119898

sdot120587

119871

(119887119898tan119898(1205872119871) 120585 minus 119886

119898)2minus (1198862119898+ 1198872119898)

119887119898(tan2119898(1205872119871) 120585 + 1)

119889120585 =11988602

minusinfin

sum119898=1

int119909

0119898

sdot120587

119871

(119887119898tan2119898(1205872119871) 120585 minus 2119886

119898tan119898(1205872119871) 120585 minus 119887

119898)

(tan2119898(1205872119871) 120585 + 1)119889120585

(61)

4 Numerics of Solving the Inverse Problem

The solution of the present inverse problem is in fact aproblem of finding the roots of (44) when 119892(119909) is truncatedat 119898 = 119873 For that reason we shall focus on such rootsin the neighborhood of a step-like discontinuity of 119891(119909) at119909 = 119909119900 say where 119891(119909119900 minus 120576) lt 119891(119909119900 + 120576) 0 lt 120576 ≪ 1 Theassociated with the Gibbs phenomenon overshoot of 119892(119909) =11988602+sum

119873

119898=1 119866(120574 119909119898) or 11988602+sum119873

119898=1 119867(120574 119909119898) over119891(119909119900+120576)and undershoot of this 119892(119909) below 119891(119909119900minus120576)The intersectionof a horizontal line ℎ(119909) = 119888 ge 119891(119909119900 + 120576) corresponding to120593(119909) = 0 can be at two points 1205721 and 1205722 on both sides of thepeak location 1205720 representing two distinct roots of 120593(119909) = 0It can also be at one point 1205720 corresponding to themaximumof the truncated 119892(119909) that is a double root of 120593(119909) = 0

Obviously 1205721 lt 1205721 lt 1205720 lt 1205722 lt 1205722 where 1205721 and 1205722 arepossible points of inflection defined by

11989210158401015840 (1205721) = 12059310158401015840 (1205721) = 11989210158401015840 (1205722) = 12059310158401015840 (1205722) = 0 (62)

The existence of both 1205721 and 1205722 or at least one of themnamely 1205722 is however not mandatory Moreover for the 12057211205720 and 1205722 points we expect to have

1198921015840 (1205721) = 1205931015840 (1205721) gt 0

12059310158401015840 (1205721) gt 0

120593101584010158401015840 (1205721) lt 0

1198921015840 (1205720) = 1205931015840 (1205720) = 0

12059310158401015840 (1205720) lt 0

120593101584010158401015840 (1205720) = 0

1198921015840 (1205722) = 1205931015840 (1205722) lt 0

12059310158401015840 (1205722) gt 0

120593101584010158401015840 (1205722) gt 0(63)

Equation (44) can be solved for119909 120593(119909) = 0 by aNewton-Raphson (see eg [11 12]) iterative process

119909119899+1 = 119909

119899minus

120593 (119909119899)

1205931015840 (119909119899) (64)

which is repeated until a sufficiently accurate value is reachedThis process for each series truncation number119873 rewrites as

119909119899+1 = 119909

119899+[sum119873

119898=1 119866 (120574 119909119899 119898) + 11988602 minus 119888]

[sum119873

119898=1 119898(120587119871)119866lowast (120574 119909119899 119898)]

(65)

or in the stereographic form

119909119899+1 = 119909

119899+[sum119873

119898=1 119867(120574 119909119899 119898) + 11988602 minus 119888]

[sum119873

119898=1 119898(120587119871)119867lowast (120574 119909119899 119898)]

(66)

which is the same as

119909119899+1 = 119909

119899

+[sum119873

119898=1 ((119886119898 + 2119887119898119905119898minus 1198861198981199052119898) (1 + 1199052

119898)) + 11988602 minus 119888]

[sum119873

119898=1 119898(120587119871) ((1198871198981199052119898+ 2119886119898119905119898minus 119887119898) (1 + 1199052

119898))]

(67)

and is started off with some rather free initial value 1199090Obviously here 119905

119898= tan119898(1205872119871)119909

The sequence 119909119899will usually converge provided that 1199090

is close enough to the unknown zero 120572 and that

1205931015840 (1199090) = minusinfin

sum119898=1

119898120587

119871119866lowast (120574 1199090 119898) = 0

or 1205951015840 (1199090) = minusinfin

sum119898=1

119898120587

119871119867lowast (120574 1199090 119898) = 0

(68)

41 Computation of the Peak Location 1205720 Being a maximiza-tion problem for 120593(119909) = 0 the determination of 1205720 by theNewton-Raphson process [11] becomes

119909119899+1 = 119909

119899minus

1205931015840 (119909119899)

12059310158401015840 (119909119899)

(69)

or

119909119899+1 = 119909

119899minus

1205951015840 (119909119899)

12059510158401015840 (119909119899) (70)

in stereographic form

8 International Journal of Computational Mathematics

Straightforwardly this process can be expressed as

119909119899+1 = 119909

119899minus

[sum119873

119898=1 119898119866lowast (120574 119909119899 119898)]

[sum119873

119898=1 1198982 (120587119871) 119866 (120574 119909

119899 119898)]

(71)

or in the stereographic form

119909119899+1 = 119909

119899

minus[minussum119873

119898=1 119898(120587119871)119867lowast (120574 119909119899 119898)]

[sum119873

119898=1 1198982 (12058721198712) ((119886

1198981199052119898minus 2119887119898119905119898minus 119886119898) (1 + 1199052

119898))]

(72)

which is the same as119909119899+1 = 119909

119899

+[sum119873

119898=1 119898((1198871198981199052119898+ 2119886119898119905119898minus 119887119898) (1 + 1199052

119898))]

[sum119873

119898=1 1198982 (120587119871) ((119886

1198981199052119898minus 2119887119898119905119898minus 119886119898) (1 + 1199052

119898))]

(73)

It is remarkable how this algorithm is distinctively indepen-dent of 119888

Proposition 8 The ill-posedness of the inverse problem ofthe truncated Fourier series is invariant under stereographicprojection

Proof The tangent half-angle stereographic projectionthough not isometric (distance preserving) is neverthelessvariability preserving Indeed the main ingredients ofthis inverse problem are namely 120574 = 119886

119898 119887119898119873119898=1 119888 and

1199090 with (119889119889119909)119866[120579119898(119909)]1199090 Under this projection 120579

119898(1199090)

transforms to 119905119898(1199090) = tan[120579

119898(1199090)2] and 119866[120579

119898(1199090)]

transforms to 119867[119905119898(1199090)] But according to Proposition 4

(119889119889119909)119867[119905119898(119909)]1199090

= (119889119889119909)119866[120579119898(119909)]1199090 Hence the variation

of 119866(120574 1199090 119898) with 1199090 in the Fourier series inverse problemand the variation of119867(120574 1199090 119898) with 1199090 in the stereographicform are identical

Equivalently this makes the ill-posedness of this inverseproblem invariant under stereographic projection

Expected manifestations of the correctness of this propo-sition are as follows (i) the results of computations by the(71) and (73) algorithms for 1205720 with the same 1199090 must beidentical and (ii) the results of computations by the (65) and(67) algorithms for 1205721 (or 1205722) with the same 1199090 and 119888 mustalso be identical

42 Convergence Analysis Let us focus first on the con-vergence of the iterative process (65) or (67) In manysituationswhere term-by-termdifferentiability does not holdtheNewton-Raphsonmethod can always be replaced by somequasi-Newtonmethod [12 13] where 1205931015840(119909

119899) or1205951015840(119909

119899) in (65)

or (67) is replaced by a suitable approximation For examplein the chord method 1205931015840(119909

119899) is replaced by 1205931015840(1199090) for all

iterationsIf 120598119899= 120572minus119909

119899 120598119899+1 = 120572minus119909

119899+1 and120573119899 is some point between119909119899and 120572 it can be easily proved (see [11] or [12]) that

120598119899+1 = minus

12059310158401015840 (120573119899)

21205931015840 (119909119899)1205982119899 (74)

Taking absolute value of both sides gives

1003816100381610038161003816120598119899+11003816100381610038161003816 =

1003816100381610038161003816100381612059310158401015840 (120573119899)10038161003816100381610038161003816

2 10038161003816100381610038161205931015840 (119909119899)10038161003816100381610038161205982119899 (75)

implying quadratic convergence that is 120598119899is squared (the

number of correct digits roughly doubles) at each step Thesame arguments apply of course to the 120595(119909) = 0 equation

In actual fact for a distinct 120572 the convergence of119909119899 in (65)ndash(67) is at least quadratic in the neigh-

borhood of this 120572 The process may face a difficultywhen 1205931015840(119909

119899) = minussum

119873

119898=1 119898(120587119871)119866lowast(120574 119909119899 119898) or 1205951015840(119909

119899) =

minussum119873

119898=1 119898(120587119871)119867lowast(120574 119909119899 119898) diverges This is however a spe-

cial but not a general problem and (75) always holds if thefollowing conditions are satisfied

(i) 1205931015840(119909) = 0 forall119909 isin 119868 where 119868 = [120572 minus 120578 120572 + 120578] for some120578 ge |(120572 minus 1199090)|

(ii) 12059310158401015840(119909) is finite forall119909 isin 119868(iii) 1199090 is sufficiently close to 120572

The term sufficiently close in this context means the follow-ing

(a) Taylor approximation is accurate enough such that wecan ignore higher order terms

(b) (12)|12059310158401015840(119909119899)1205931015840(119909

119899)| lt 119870|12059310158401015840(120572)1205931015840(120572)| for some119870 lt

infin(c) 119870|12059310158401015840(120572)1205931015840(120572)|120598

119899lt 1 for 119899 isin N

Consequently (75) can be expressed in the following way1003816100381610038161003816120598119899+1

1003816100381610038161003816 = 1198721205982119899 (76)

where

119872 = sup119909isin119868

12

100381610038161003816100381610038161003816100381610038161003816

12059310158401015840 (119909)

1205931015840 (119909)

100381610038161003816100381610038161003816100381610038161003816 (77)

The initial point 1199090 has to be chosen such that conditions(i) through (iii) are satisfied where the third conditionrequires that 119872|1205980| lt 1 Moreover with increasing 119873 thedistance |1205721 minus 1205722| is known to decrease in a way makingthe satisfaction of these conditions very difficult and actuallyimpossible as119873 rarr infin Here it should be underlined that theiterative method (65)ndash(67) may fail to converge to the 120572 rootin the following situations

(i) If a point like 1205720 is included in 119868 the method willterminate due to division by zero

(ii) When the initial estimate 1199090 is poor the pertainingwrong 119868 can contribute to nonconvergence of thealgorithm

As for convergence of the Newton-Raphson processtowards 1205720 let 120576119899 = 1205720 minus 119909

119899 120576119899+1 = 1205720 minus 119909

119899+1 and 120579119899is some

point between 119909119899and 1205720 to arrive at

120576119899+1 = minus

120593101584010158401015840 (120579119899)

212059310158401015840 (119909119899)1205762119899 (78)

1003816100381610038161003816120576119899+11003816100381610038161003816 =

10038161003816100381610038161003816120593101584010158401015840 (120579119899)10038161003816100381610038161003816

2 100381610038161003816100381612059310158401015840 (119909119899)10038161003816100381610038161205762119899 (79)

International Journal of Computational Mathematics 9

The convergence of 119909119899 in (71)ndash(73) is at least quadratic

in the neighborhood of this 1205720 The process may face a diffi-culty however when 12059310158401015840(119909

119899) = minussum

infin

119898=1 1198982(120587119871)2119866(120574 119909

119899 119898)

or 12059510158401015840(119909119899) diverges Moreover (79) always holds if the follow-

ing conditions are satisfied

(i) 12059310158401015840(119909) = 0 forall119909 isin 119868 where 119868 = [1205720 minus120578 1205720 +120578] for some120578 ge |(1205720 minus 1199090)|

(ii) 120593101584010158401015840(119909) is finite forall119909 isin 119868

(iii) 1199090 is sufficiently close to 1205720

Sufficient closeness in this context means that

(a) Taylor approximation is accurate enough such that wecan ignore higher order terms

(b) (12)|120593101584010158401015840(119909119899)12059310158401015840(119909

119899)| lt 119870|120593101584010158401015840(1205720)120593

10158401015840(1205720)| for some119870 lt infin

(c) 119870|120593101584010158401015840(1205720)12059310158401015840(1205720)|120576119899 lt 1 for 119899 isin N

Consequently

1003816100381610038161003816120576119899+11003816100381610038161003816 = 1198721205762

119899 (80)

where

119872 = sup119909isin119868

12

100381610038161003816100381610038161003816100381610038161003816

120593101584010158401015840 (119909)

12059310158401015840 (119909)

100381610038161003816100381610038161003816100381610038161003816 (81)

43 Applications The standard iterative algorithms (71) or(73) for finding the location of overshoots in 119876

119904120590119873(119909) of

the Gibbs effect employ the fact that the first derivative of apeak in 119892(119909) or 119892(119909) respectively has a downward-going-zero-crossing at the peak maximum But the presence ofmany neighboring smaller peaks in the ringing close to theovershoot will cause many undesirable zero-crossings to beinvoked if the initial trial root 1199090 is not correctly chosenThis fact makes the numerical problem of peak finding in theGibbs effect of pathological difficulty A difficulty that aggra-vates with the increase in the number 119873 of truncated termsin the employed Fourier series Indeed with the increaseof 119873 the number of undesirable zero-crossings increasesin the neighborhood of 1205720 as illustrated in Section 1 Thiscan increasingly lead to oscillatory and even unstable valuesin sequence 119909

119899 the more the distance |1199090 minus 1205720| contains

unwanted zero-crossingsThese problems have been resolved in the two examples

that follow which simultaneously illustrate the applicabilityof Proposition 8

Example 9 Consider the 2120587-periodic odd signal 119891(119909) ofExample A1 defined over one period by

119891 (119909) =

minus1 minus120587 lt 119909 lt 0

1 0 le 119909 lt 120587(82)

which has for each truncation number 119873 the truncatedFourier series approximations

119892 (119909) ≃4120587

119873

sum119898=1

1(2119898 minus 1)

sin (2119898minus 1) 119909

119892 (119909) ≃8120587

119873

sum119898=1

119887119898119905119898

(1 + 1199052119898)=

8120587

119873

sum119898=1

1(2119898 minus 1)

119905119898

(1 + 1199052119898)

=8120587

119873

sum119898=1

1(2119898 minus 1)

tan ((2119898 minus 1) 2) 1199091 + [tan ((2119898 minus 1) 2) 119909]2

(83)

This signal has a jump discontinuity at 119909 = 0 ofmagnitude 2 We shall focus our computations on the Gibbsphenomenon overshoot near this discontinuity

Here is a printout list of results for the parameters of theposing inverse problem via computations by both the directand stereographic algorithms

It is remarkable how results by both algorithms arepractically identical and share the same following featuresthat are observed by results of the direct algorithm discussedbelow

Case I (119873 = 40) The computations of 1205720 for the previousexample using (71) with 1199090 = 002 lead to 1199091 = 0039 109this varies quickly with 119899 (only beyond the 4th decimal) tostabilize towards 1199093 = 1199094 = 1199095 = sdot sdot sdot = 0039 299 = 1205720Here using (1) gives 119892(1205720) = 119892(0039 299) = 1179 030 whichis quite close to the theoretically expected value by (A7) ofsim118

Then using (65) for 119888 = 109 and 1199090 = 0000381 leads to1199091 = 0021 697 this varies quickly with 119899 (only beyond the 3-d decimal) to stabilize quickly towards 1199095 = 1199096 = 1199097 = sdot sdot sdot =0028 368 = 1205721 Also using (65) for 119888 = 109 and 1199090 = 004leads to 1199091 = 0067 699 this varies quickly with 119899 (beyondthe 2nd decimal) to stabilize quickly towards 1199095 = 1199096 = 1199097 =sdot sdot sdot = 0052 873 = 1205722

For both1205721 and1205722 using (1) yields119892(1205721) = 119892(0028368) =119892(1205722) = 119892(0052873) = 109 = 119888 as expected

Case II (119873 = 100) Computations of 1205720 using (71) with1199090 = 002 lead to 1199091 = 0014 221 this varies quickly with119899 (only beyond the 4th decimal) to stabilize towards 1199093 =1199094 = 1199095 = sdot sdot sdot = 0015 708 = 1205720 Here using (1) gives119892(1205720) = 119892(0015708) = 1178 990 This result appears to beslightly excessive perhaps due to increased rounding off errorswith increasing 119873 to 100

Thenusing (65) for 119888 = 109 and1199090 = 0015 237 leads aftera negative value to 1199092 = 0014 462 this varies with 119899 (beyondthe 2nd decimal) to stabilize quickly towards 1199097 = 1199098 = 1199099 =sdot sdot sdot = 0011 3348 = 1205721 The used value of 1199090 is apparently notappropriate and this illustrates the generic ill-posedness of thisinverse problem

The pertaining computations of 1205722 with 1199090 = 0017000lead to 1199091 = 0020 434 this varies with 119899 (beyond the 3-ddecimal) to stabilize quickly towards 1199094 = 1199095 = 1199096 = sdot sdot sdot =0021 149 = 1205722

10 International Journal of Computational Mathematics

For both 1205721 and 1205722 using (1) yields 119892(1205721) =119892(00113348) = 119892(1205722) = 119892(0021149) = 109 = 119888 asexpected

Case III (119873 = 150) The computations of 1205720 using (71) with1199090 = 0005 lead to 1199091 = 0010 595 this varies insignificantlywith 119899 (only beyond the 4th decimal) to stabilize towards1199093 =1199094 = 1199095 = sdot sdot sdot = 0010 472 = 1205720 Here using (1) gives 119892(1205720) =119892(0010472) = 1178 980 which does not significantly differfrom 119892(1205720) when119873 = 100

Thenusing (65) for 119888 = 109 and1199090 = 0010 158 leads aftera negative value to 1199092 = 0009 643 and this varies (beyond the3-d decimal) to stabilize quickly towards1199097 = 1199098 = 1199099 = sdot sdot sdot =0007 565 = 1205721

The pertaining computations of 1205722 with 1199090 = 0010 800lead to 1199091 = 0025 998 this varies (beyond the 2nd decimal)to stabilize towards 11990914 = 11990915 = 11990916 = 11990917 = 11990918 = sdot sdot sdot =0014 099 = 1205722

For both 1205721 and 1205722 using (1) yields 119892(1205721) = 119892(007 565) =119892(1205722) = 119892(0014 099) = 109 = 119888

During the previous computations summarized inTable 1 it has been observed that varying 1199090 in the presentNewton-Raphson process for all119873 can lead to strong oscil-lations and even completely unstable values for 1205720 1205721 and 1205722of the Fourier series inverse problem which though ill-posed

is practically regularizible Apart from this drawback the(71) with (65) iterative algorithms (i) converge quickly to thetheoretically expected values and are (ii) of demonstratedstability of the iterations for properly chosen 1199090 (which act aseffective regularizing parameters) Moreover computationsby the stereographic algorithm invoking (73) and (67) sharethe samemagnitude and convergence rates with results by thedirect algorithm Finally it is demonstrated that the presentsignal exhibits an overshoot when119873 = 150 ofsim9of a jumpdiscontinuity of magnitude 2 that is located at a distance ofsim00105

To illustrate robustness of both numerical algorithms andtheir possible effective regularizibility we shall study also thenumerical inverse problem for the Fourier series representa-tion of an arbitrary (nonsymmetric) periodic signal

Example 10 Consider the 2120587-periodic signal 119891(119909) definedover one period by

119891 (119909) =

minus119909

120587minus 1 minus120587 lt 119909 lt 0

1 0 le 119909 lt 120587(84)

It has for each truncation number 119873 the truncated Fourierseries approximations

119892 (119909) ≃14+119873

sum119898=1

[(minus1)119898 minus 1]12058721198982 cos119898119909+

[2 minus (minus1)119898]120587119898

sin119898119909

119892 (119909) ≃14

+119873

sum119898=1

[(minus1)119898 minus 1] 12058721198982 + 2 ([2 minus (minus1)119898] 120587119898) tan ((2119898 minus 1) 2) 119909 minus ([(minus1)119898 minus 1] 12058721198982) [tan ((2119898 minus 1) 2) 119909]2

1 + [tan ((2119898 minus 1) 2) 119909]2

(85)

This signal has like the signal of Example 9 a jumpdiscontinuity at 119909 = 0 of magnitude 2 For the sake ofcomparison we shall also focus our computations on theGibbs phenomenon overshoot part of 119876

119904120590119873(119909) near this

discontinuityIt should be noted here that the results by both algorithms

summarized in Table 2 are also identical a feature indicatingthat the degree of ill-posedness of the posing inverse problemis not affected as predicted by Proposition 8 by the tangenthalf-angle stereographic projection

Case I (119873 = 40) The computations of 1205720 for this exampleusing (71) and (73) with 1199090 = 003 lead to an identicalsequence 119909

119899 that converges by both algorithms after the 1199094

iteration to 0078 139 with 119892(1205720) = 119892(1205720) = 1178 76

Case II (119873 = 100) The computations of 1205720 with 1199090 = 001lead to an identical sequence 119909

119899 that converges by both

algorithms after the 1199095 iteration to 0031 351 with 119892(1205720) =

119892(1205720) = 1178 87

Case III (119873 = 150) The computations of 1205720 with 1199090 = 0007lead also to an identical sequence 119909

119899 that converges by both

algorithms after the 1199095 iteration to 0020 915 with 119892(1205720) =

119892(1205720) = 1178 91

Therefore the present signal exhibits an overshoot when119873 = 150 of sim9 of a jump discontinuity of magnitude 2 thatis located at a distance of sim0020 9

Comparison of the computed Gibbs overshoot parame-ters of the two considered signals with jump discontinuitiesat the same 119909 = 0 of the same magnitude illustrates that therelative size of the overshoot depends essentially exclusivelyon the magnitude of the stimulating jump discontinuityFor instance based on the present numerical results onecan suggest here that nonsymmetric signals have equalovershoots with symmetric signals Their respective 1205720rsquos arehowever different Indeed the nonsymmetric signal has asignificantly larger 1205720 (0020 9) than the 1205720 (00105) of thesymmetric signal

International Journal of Computational Mathematics 11

Table 1 Results of computations for Example 9

119873 1205721 1205720 1205722

40 0028 368 0039 299 0052 873100 0011 335 0015 705 0021 149150 0007 565 0010 472 0014 099

Table 2 Results of computations for Example 10

119873 1205721 1205720 1205722

40 00564765 00781391 0105155100 00226546 00313516 00422024150 00151124 00209153 0028156

Table 3 Summary of computations of all overshoot parameters inExample 10

1205720results

119873 1199090

119909lowast

1205720

119892(1205720)

100 0030 1199093

0031 3516 1178 87200 0015 119909

30015 6919 1178 92

300 0008 1199093

0010 4648 1178 941205721results

119873 1199090

119909lowast

1205721

119888

0025 1199094

0022 2027 108100 0025 119909

30023 1344 110

0025 1199093

0024 2027 1120010 119909

30011 1117 108

200 0010 1199093

0011 5779 1100010 119909

40012 1123 112

0008 1199093

0007 410 08 108300 0008 119909

30007 720 95 110

0008 1199092

0008 077 32 1121205722results

119873 1199090

119909lowast

1205722

119888

100035 119909

40042 9439 108

035 1199094

0041 4445 110035 119909

40039 8513 112

2000018 119909

30021 4960 108

0018 1199093

0020 7456 1100018 119909

30019 9483 112

3000012 119909

30014 3360 108

0012 1199093

0013 8356 1100012 119909

40013 3039 112

A further analysis has also been performed of Example 10with 119873 = 100 200 300 to compute the corresponding1205720 with 1205721 1205722 that correspond simultaneously to 119888 =108 110 112 as illustrated in Figure 1 Results of thesecomputations are summarized in Table 3 and then used forthe partial 119876

119904120590119873(119909) plots of Figure 2 that interpolate 1205721 1205722

(for 119888 = 108 110 112) with 1205720 for each 119873 = 100 200 300In this table119909

lowaststands for the first convergence step of the 119909

119899

sequence of Newton Raphson iterations

002

05

10

004

g(x

)

xminus004 minus002

minus05

minus10

g(x) for N = 100

g(x) for N = 200

g(x) for N = 300

h(x) = c = 108

h(x) = c = 11

h(x) = c = 112

1205721 1205720 1205722

Figure 1 Plots of 119892(119909) near a discontinuity of 119891(119909) in Example 10

120

115

110

105

100

000 001 002 003 004 005

g(x

)

x

g(x) for N = 100

g(x) for N = 200

g(x) for N = 300

h(x) = c = 108

h(x) = c = 11

h(x) = c = 112

Figure 2 Plots of the 119892(119909) curves near a discontinuity of 119891(119909) inExample 10 interpolating 1205721 1205722 (for 119888 = 108 110 112) with 1205720 allfor119873 = 100 200 300

These results turn out to be extremely stable against varia-tions of the trial initial root1199090 Indeed in the computationallymost difficult situation namely when 119873 = 300 a changeof 1199090 from 0008 to 0020 leads to a variation of 119909

lowastfrom

1199092 to just 1199096 with the same asymptotic 1205720 = 00104648 and1205721 = 000807732 (for 119888 = 112)

5 Conclusions

This paper demonstrates that the quadratic degeneracy ofthe solution of the present inverse problem explicit whenusing 119867(120574 119909119898) is a nonlinear indicator of the existence ofGibbsrsquo phenomenon It provides also a distribution-theoreticproof for the existence of this phenomenon The reportedanalysis of numerical solvability of this problem illustrates thefollowing

(i) When suminfin

119898=1 119898(120587119871)119866lowast(120574 119909119898) = suminfin

119898=1 119898(120587119871)119867lowast(120574 119909119898) converge continuously in 119909 and are

12 International Journal of Computational Mathematics

nonzero at 1205721 (or 1205722) there exists a neighborhood(defined by 119868) of 120572 such that for all starting values 1199090in the neighborhood 119909

119899 will converge to 1205721 (or 1205722)

(ii) When suminfin

119898=1 119898(120587119871)119866lowast(120574 119909119898)=suminfin

119898=1 119898(120587119871)119867lowast(120574119909119898) converge continuously in 119909 and are nonzero at1205721 and 1205722 (but not at 1205720) and if 120593(119909) has a secondderivative which might be zero at 1205721 and 1205722 (and at1205720) then the convergence of 119909119899 is quadratic or fasterIf12059310158401015840(119909) = 0 at1205721 and1205722 then the convergence of 119909119899is merely quadratic

(iii) The convergence of the Newton-Raphson iterativeprocess to 1205720 can also be quadratic but restrictionson the derivatives of 120593(119909) have to be observed up tothe third order Otherwise the convergence of 119909

119899 to

1205720 can slow down to become only linear [12] with arate log210

(iv) The reported computations clearly illustrate that1205720 rarr 0 and |1205721 minus 1205722| rarr 0 when 119873 rarr infin thatis the infinite Fourier series will sum to 119891(119909) inthe neighborhood of a discontinuity except at thediscontinuity itself

Appendix

To illustrate the Gibbs-Wilbraham effect we shall use herethe same arguments employed by Wilbraham in 1848 inanalyzing a similar square wave

Example A1 Let 119891(119909) be a square wave periodic signalwhich is defined over one period by

119891 (119909) =

minus1 minus120587 lt 119909 le 0

1 0 le 119909 lt 120587(A1)

Its Fourier series representation is

119892 (119909) =4120587

infin

sum119898=1

1(2119898 minus 1)

sin (2119898minus 1) 119909 (A2)

This defines the truncated series

119878119873(119909) =

4120587

119873

sum119898=1

1(2119898 minus 1)

sin (2119898minus 1) 119909 (A3)

which may be differentiated as

119889

119889119909119878119873(119909) =

4120587

119873

sum119898=1

cos (2119898minus 1) 119909 (A4)

Since

2119873

sum119898=1

cos (2119899 minus 1) 119909 =sin 2119873119909

sin119909 (A5)

then119889

119889119909119878119873(119909) =

2120587

sin 2119873119909

sin119909 (A6)

It is possible to cancel the previous artificially induceddifferentiation by integrating the last result as

119878119873(119909) = int

119909

0

119889

119889120585119878119873(120585) 119889120585 =

2120587int119909

0

sin 2119873120585

sin 120585119889120585 (A7)

This integral starts at zero when 119909 = 0 in agreement with(A2) and increases until 2119873120585 = 120587 Then 120585 = 1205872119873 definesa turning point 1199091 The integral starts to decrease at whichpoint the numerator sin 2119873120585 goes negative For large 119873 thedenominator sin 120585 remains positive

The form (A7) distinctively from (A3) reveals that119878119873(119909) has turning points at the zeros of sin 2119873120585 These occur

when 2119873120585 = 119903120587 119903 = 1 2 3 4 5 that is at 120585 = 119903(1205872119873)Clearly then 119909

119903sim 120585119903= 119903(1205872119873) are turning points for 119878

119873(119909)

In particular 1199091 sim 1205851 = 1205872119873 defines the overshoot peaklocation 1205720 and 119878

119873(1199091) = (2120587) int12058721198730 (sin 2119873120585 sin 120585) 119889120585 is

its magnitude To evaluate 119878119873(1199091) = 119878

119873(1205720) numerically

let us change the variable of integration namely 2119873120585 = 119908or 120585 = 1199082119873 to rewrite the previous integral as 119878

119873(1199091) =

(2120587) int1205870 (sin119908(sin1199082119873)) 119889(1199082119873) Clearly when 119873 rarr

infin sin1199082119873 = 1199082119873 and

lim119873rarrinfin

119878119873(1199091) =

2120587int120587

0

sin119908119908

119889119908 =2120587Si (120587) = 118 (A8)

correct to two decimal pointsAs for 1199092 1199093 1199094 119909119903 lim119873rarrinfin[119878119873(119909119903) minus 119891(119909)]

decreases as we move away from the discontinuity butit should be remarked however that the location of theovershoot 1205720 = 1199091 = 1205872119873 moves towards the discontinuityaccording to

lim119873rarrinfin

1205720 = lim119873rarrinfin

120587

2119873= 0 (A9)

Conflict of Interests

The author declares that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

The numerical computations of Examples 9 and 10 have beenperformed using Mathematica 8 by Mounir Abou Yassineof the Department of CCE of AUL

References

[1] P L Butzer and R J Nessel Fourier Analysis and Approxima-tion Volume I One-Dimensional Theory Birkhauser 1971

[2] R E Edwards Fourier Series A Modern Introduction HoltRinehart amp Winston 1967

[3] W J Thompson ldquoFourier series and the Gibbs phenomenonrdquoAmerican Journal of Physics vol 60 no 5 pp 425ndash429 1992

[4] A J JerriThe Gibbs Phenomenon in Fourier Analysis Splines ampWavelet Approximations Kluwer Academic Publishers BostonMass USA 1990

International Journal of Computational Mathematics 13

[5] K S Rim and B I Yun ldquoGibbs phenomenon removal by addingHeaviside functionsrdquo Advances in Computational Mathematicsvol 38 no 4 pp 683ndash699 2013

[6] A H Zemanian Distribution Theory and Transform AnalysisMcGraw-Hill 1965

[7] G R Ayers and J C Dainty ldquoIterative blind deconvolutionmethod and its applicationsrdquo Optics Letters vol 13 no 7 pp547ndash549 1988

[8] N H S Haidar ldquoA nonlinear method for a severely ill-posedgeneralized moment problemrdquo Journal of Inverse amp Ill-PosedProblems vol 5 no 4 pp 323ndash335 1997

[9] G L Bradley and K J Smith Calculus Prentice Hall 1995[10] P Offner T Sonar and M Wirz ldquoDetecting strength and

location of jump discontinuities in numerical datardquo AppliedMathematics vol 4 no 12 pp 1ndash14 2013

[11] K E Atkinson An Introduction to Numerical Analysis JohnWiley amp Sons New York NY USA 1989

[12] E Suli and D F Mayers An Introduction to Numerical AnalysisCambridge University Press Cambridge UK 2003

[13] P Wang ldquoA third-order family of Newton-like iteration meth-ods for solving nonlinear equationsrdquo Journal of NumericalMathematics and Stochastics vol 3 no 1 pp 13ndash19 2011

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Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Discrete Dynamics in Nature and Society

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Stochastic AnalysisInternational Journal of

2 International Journal of Computational Mathematics

a discontinuity The construction of this series is based onthe method of adding the Fourier coefficients of a Heavisidefunction to the given Fourier partial sums

This paper is organized as follows After this introductionwe propose in Section 2 a new distribution-theoretic prooffor the Gibbs-Wilbraham effect In Section 3 we advancethe inverse problem of the truncated Fourier series and thenew stereographic truncated Fourier series and its relationto resolution of the undershootovershoot pair associated tothe Gibbs-Wilbraham effect Here we illustrate how solvingthe inverse problem for the truncated representation (1)contains a quadratic degeneracy that indicates the existenceof theGibbs phenomenon Section 4 deals with the numericaliterative solution of the previous inverse problem and its con-vergence with a proof and demonstration that stereographicprojection does not affect such a numerical solution and itsposedness

2 Distribution-Theoretic Proof forthe Gibbs-Wilbraham Effect

Let 119891(119909) isin BV(minus119871 119871) be a 2119871 ndashperiodic piecewise continu-ously differentiable function on (minus119871 119871) and let 119909

119904be a point

in (minus119871 119871) at which 119891(119909) has a discontinuity of the first kindConsider 119909minus

119904= 119909119904minus 0 119909+

119904= 119909119904+ 0 and 119891minus

119904= 119891(119909minus

119904)

and 119891+119904

= 119891(119909+119904) to assume without loss of generality that

119891+119904gt 119891minus119904and define the discontinuity jump by

Δ119904= 119891+119904minus119891minus119904 (3)

Moreover

119891119888(119909) =

119891 (119909) 119909 = 119909119904

undefined 119909 = 119909119904

(4)

Theorem 1 Let a 2119871 ndashperiodic 119891(119909) isin BV(minus119871 119871) be discon-tinuous at 119909

119904 The Fourier series representation 119892(119909) of 119891(119909)

converges over (minus119871 119871) as119873 rarr infin so as

119892 (119909) =

119891minus119904minus 120588Δ119904 119909 = 119909minus

119904

12(119891minus119904+ 119891+119904) 119909 = 119909

119904

119891+119904+ 120588Δ119904 119909 = 119909+

119904

119891119888(119909) 119909 = 119909

119904119900119903 119909plusmn119904

(5)

where 120588 = 119906(0) an uncertainty

Proof According to distribution theory [6] when 119892(119909) coin-cides ae with 119891(119909) over (minus119871 119871) it may always be differenti-ated namely

1198921015840 (119909) = 1198911015840119888(119909) +Δ

119904120575 (119909 minus 119909

119904) (6)

with 120575 as Diracrsquos delta Cancellation of this differentiationis performed by sweeping the 119909-axis during a compensatingintegration from minus119871 to 119871 Indeed integration of (6) first fromminus119871 to 119909 leads to

int119909

minus119871

[1198921015840 (120591) minus1198911015840119888(120591)] 119889120591 = Δ

119904int119909

minus119871

120575 (120591 minus 119909119904) 119889120591 (7)

and since 119892(minus119871) = 119891119888(minus119871) then

119892 (119909) = 119891119888(119909) +Δ

119904119906 (119909 minus 119909

119904) (8)

with 119906 as Heavisidersquos unit step functionBy a theorem of Fejer [1] 119892(119909) should converge to 119891

119888(119909)

when119873 rarr infin forall119909 = 119909119904on (minus119871 119871) Hence the term Δ

119904119906(119909 minus

119909119904) in (8) can have a compact support only on an infinitesimal

interval 120590 = lim119873rarrinfin

120590119873= 0 to the right of 119909

119904 that is ending

at 119909+119904 Hence

119892 (119909) = 119891119888(119909) +Δ

119904119906 (119909+119904minus119909119904) (9)

Furthermore despite the fact that 119906(119909+119904minus 119909119904) = 1 119906(119909+

119904minus

119909119904) asymp 119906(0) which is an uncertainty to be denoted by 120588

Consequently relation (9) is representable as

119892 (119909) =

119891119888(119909) 119909 = 119909

119904

119891119888(119909) + 120588Δ

119904 119909 = 119909+

119904

(10)

Next complete the sweeping by integrating (6) from 119909 to119871 namely

int119871

119909

[1198921015840 (120591) minus1198911015840119888(120591)] 119889120591 = Δ

119904int119871

119909

120575 (120591 minus 119909119904) 119889120591 (11)

Here again since 119892(119871) = 119891119888(119871) then the left-hand side of

(11) becomes minus119892(119909) + 119891119888(119909) The right-hand side requires

standardization by means of the substitution 120591 = minus] leadingto 119889120591 = minus119889] [119909 119871] becoming [minus119909 minus119871] and

int119871

119909

120575 (120591 minus 119909119904) 119889120591 = minusint

minus119871

minus119909

120575 (minus]minus119909119904) 119889]

= minusintminus119871

minus119909

120575 (]+119909119904) 119889]

= intminus119909

minus119871

120575 (]+119909119904) 119889] = 119906 (minus119909+119909

119904)

(12)

It is obvious then that (11) is reducible to

119892 (119909) = 119891119888(119909) minusΔ

119904119906 (minus119909+119909

119904) (13)

Repeated application of the same arguments employed inthe derivation of (9)-(10) to (13) allows for writing it as

119892 (119909) = 119891119888(119909) minusΔ

119904119906 (119909minus119904+119909119904) (14)

and subsequently in the form

119892 (119909) =

119891119888(119909) 119909 = 119909

119904

119891119888(119909) minus 120588Δ

119904 119909 = 119909minus

119904

(15)

Taking into consideration that 119891119888(119909+119904) = 119891+119904and 119891

119888(119909minus119904) = 119891minus119904

in (10) and (15) respectively then combining these equationswith Dirichletrsquos theorem [1] leads to the required result

International Journal of Computational Mathematics 3

It should be pointed out that in the previous proofrelations (10) and (15) say nothing specifically about thebehavior of119892(119909) at119909minus

119904and119909+119904 respectively while asserting the

existence of a spiky behavior of 119892(119909) individually at 119909+119904then

at 119909minus119904 Moreover the uncertainty 120588 had been resolved in the

past (by Wilbraham see also the Appendix) computationallyto satisfy 120588 = 009 and long before the advance of the theoryof distributions (generalized functions) by S L Sobolev in1936 The number 009 for 120588 remains until now howevera mysterious theoretical puzzle that calls for a rigorousdistribution-theoretic justification

Clearly in the neighborhood of 119909119904 119892(119909) can only be a

rudimentary approximation to 119891(119909) even when 119873 rarr infinwith an added spike pair

119876119904(119909) =

+120588Δ119904 119909 = 119909+

119904

minus120588Δ119904 119909 = 119909minus

119904

0 119909 = 119909119904

(16)

to 119891plusmn119904at 119909plusmn119904 Relation (16) represents the Gibbs-Wilbraham

effect which due to the fact that |119909+119904

minus 119909minus119904| = 0 is

not practically computable This effect degenerates howeverwhen119873 is finite to an undershoot-overshoot pair119876

119904120590119873(119909) on

an interval 119868119904120590119873

= [119909119904minus 120590119873 119909119904+ 120590119873] splitted by Δ

119904at 119909119904 In

this case it is always possible to compute

119892 (119909) =

sim 119891 (119909) + 119876119904120590119873

(119909) 119909 isin 119868119904120590119873

119873 lt infin

sim 119891 (119909) 119909 notin 119868119904120590119873

(17)

Furthermore since lim119873rarrinfin

119876119904120590119873

(119909) = 119876119904(119909) then these

computations with (17) over 119868119904120590119873

can reveal only a feature119876119904120590119873

(119909) associated with the Gibbs-Wilbraham effect 119876119904(119909)

but not the effect itself

3 The Inverse Problem for TruncatedFourier Series

Substitution of any value of 119909 isin [minus119871 119871] in (1) yields a number119888 = 119892(119909) which may differ [2] from 119891(119909) by 119876

119904120590119873(119909) if 119909 is

in the neighborhood of a discontinuity of 119891(119909) Obviously 119888satisfies the nonlinear in 119909 trigonometric series equation

119873

sum119898=1

119866 (120574 119909119898) =119873

sum119898=1

119886119898cos119898120587

119871119909+ 119887119898sin119898120587

119871119909

= 119888 minus11988602

(18)

which is however linear in the data 120574 = 119886119898 119887119898infin119898=1

Definition 2 (the inverse problem for the truncated Fourierseries representation) Given a number 119888 = 119892(119909) satisfying(18) what is the corresponding 119909

This inverse problem happens to be a discretized versionof the similar deconvolution problem [4] of 119909 from a givennumber 119902 in a singular first kind integral equation of the form

intinfin

0119870(120574 119909 120591) 119889120591 = 119902 (19)

in which 120574 represents the data It is well known additionally(see eg [7 8]) that this deconvolution problem can becomputationally ill-posed if an arbitrarily small deviationof the 120574 data may cause an arbitrarily large deviation inthe solution 119909 This situation may further prevail when theoperator intinfin0 119870(120574 119909 120591)119889120591 does not continuously depend on120574

Moreover the inverse problem of the previous definitionhas a unique solution that satisfies (19) and is expected toaccept a double rootwhen the horizontal line ℎ(119909) = 119888 gt 119891(119909)intersects with a sharp peak (overshoot) of a truncated 119892(119909)We shall call this feature a duplicated (quadratic) degeneracyof the solution for this inverse problem

The existence of the Fourier series truncation peak (andthe corresponding two roots for 119909) is not directly evidentfrom the structure of 119866(120574 119909119898) and despite its linearity inthe data 120574 = 119886

119898 119887119898infin119898=1

31 The Tangent Half-Angle Stereographic Projection Thetangent half-angle substitution (the Euler-Weirstrass sub-stitution [9]) tan(1205792) is widely used in integral calculusfor finding antiderivatives of rational expressions of circularfunction pairs (cos 120579 sin 120579) Geometrically the constructiongoes like this draw the unit circle and let 119875 be the point(minus1 0) A line through 119875 (except the vertical line) passesthrough a point (cos 120579 sin 120579) on the circle and crosses the119910-axis at some point 119910 = 119905 Obviously 119905 determines theslope of this line Furthermore each of the lines (exceptthe vertical line) intersects the unit circle in exactly twopoints one of which is 119875 This determines a function frompoints on the unit circle to points on a straight line Thecircular functions determine therefore a map from points(cos 120579 sin 120579) on the unit circle to slopes 119905 In fact this isa stereographic projection [9] that expresses these pairs interms of one variable tan(1205792) = 119905

Numerical algorithms employing these trigonometricpairs like Fourier series (1) could in principle be influencedby such a projection In this context the tangent half-rangetransformation relates the angle 120579

119898= 119898(120587119871)119909 to the slope

119905119898= tan(120579

1198982) of a line intersecting the unit circle centered

around (0 0)Indeed as 120579

119898varies the point (cos 120579

119898 sin 120579119898) winds

repeatedly around the unit circle centered at (0 0) The point

(1 minus 1199052119898

1 + 1199052119898

2119905119898

1 + 1199052119898

) = (cos 120579119898 sin 120579119898) (20)

goes only once around the circle as 119905119898goes from minusinfin to infin

and never reaches the point (minus1 0) which is approached as alimit when 119905

119898rarr plusmninfin

Clearly the stereographic projection 119905119898

= tan(1205791198982) is

defined on the entire unit circle except at the projection point119875 (minus1 0) Where it is defined this mapping is smoothand bijective [9] It is conformal meaning that it preservesangles it is however not isometric that is it does not preservedistances as

119889120579119898=

2 119889119905119898

1 + 1199052119898

(21)

4 International Journal of Computational Mathematics

The rest of this section illustrates how stereographicprojection of Fourier series reveals a quadratic degeneracythat indicates the existence of the Gibbs-Wilbraham effect Italso addresses the question of solving the inverse problem forthe truncated Fourier series representations The impact ofstereographic projection on the ill-posedness of the numer-ical solution of this inverse problem is moreover a mainquestion that has been addressed in this paper

Our analysis starts however by illustrating how thetangent half-angle stereographic projection can provide anew insight into the behavior of truncated Fourier series neara discontinuity of 119891(119909)

Lemma 3 The solution 119909 of the inverse problem for thetruncated Fourier series representation satisfies

119873

sum119898=1

119867(120574 119909119898)

=119873

sum119898=1

(1198862119898+ 1198872119898) minus (119886

119898tan119898(1205872119871) 119909 minus 119887

119898)2

119886119898(tan2119898(1205872119871) 119909 + 1)

= 119888 minus11988602

(22)

Proof By studying solvability for 119909 of (18) Let 119905119898

=tan119898(1205872119871)119909 and then cos119898(120587119871)119909 = (1minus1199052

119898)(1+1199052

119898) and

sin119898(120587119871)119909 = 2119905119898(1 + 1199052

119898) Further consideration of these

facts in (18) leads to the tangent half-angle stereographicallytransformedS[ ] = [] form

minus119873

sum119898=1

S [119866 (120574 119909119898)] =119873

sum119898=1

(1198861198981199052119898minus 2119887119898119905119898minus 119886119898)

(1199052119898+ 1)

=11988602minus 119888

(23)

The harmonic trinomial in the left-hand side of (23) hastwo harmonic zeros 120585plusmn

119898 for each 119909 namely

120585plusmn119898=

2119871119898120587

tanminus1119905plusmn119898

119905plusmn119898=

119887119898

119886119898

plusmn120590119898

120590119898=

radic1198862119898+ 1198872119898

119886119898

(24)

It is worth noting at this point that these 120585plusmn119898zeros should

by no means be confused with the possible roots of the 119909variable Substitution then of (24) in (23) leads to

119873

sum119898=1

119886119898(119905119898minus 119905+119898) (119905119898minus 119905minus119898)

(1199052119898+ 1)

=11988602minus 119888 (25)

which is the same as

119873

sum119898=1

119886119898(119905119898minus 119887119898119886119898minus 120590119898) (119905119898minus 119887119898119886119898+ 120590119898)

(1199052119898+ 1)

=11988602minus 119888

(26)

The quadratic degeneracy of the harmonic zeros 120585plusmn119898of 119905119898

can disappear only when 120590119898= 0 for all119898 And this happens

to be equivalent with

119873

sum119898=1

1198862119898+ 1198872119898= 0 (27)

Obviously satisfaction of (27) by (26) reduces it to theequation

119873

sum119898=1

119886119898(119905119898minus 119887119898119886119898)2

(1199052119898+ 1)

=11988602minus 119888 (28)

which is the same as

119873

sum119898=1

(119886119898tan119898(1205872119871) 119909 minus 119887

119898)2

119886119898(tan2119898(1205872119871) 119909 + 1)

=11988602minus 119888 (29)

It is interesting to remark that condition (27) is entirelydifferent from

11988602+119873

sum119898=1

1198862119898+ 1198872119898gt 0 (30)

which can be derived from the Parseval identity [1] (com-pleteness relation)

11988602+infin

sum119898=1

1198862119898+ 1198872119898=

1119871int119871

minus119871

1198912(119909) 119889119909 (31)

Hence a possible nondegenerate solution for 119909 of the nonlin-ear equation (29) is on one hand fundamentally incorrectbecause it involves a violation of (30) and Parsevalrsquos identityOn the other hand a consistent solution 119909 in the senseof satisfaction of (31) is intrinsically degenerate because ofits emergence as a must from the nonlinear equation (26)which rewrites as

119873

sum119898=1

119886119898[(119905119898minus 119887119898119886119898)2minus 1205902119898]

(1199052119898+ 1)

=11988602minus 119888 (32)

Equation (32) is easily transformable to the requiredresult (23) here the proof completes

32 The Stereographic Fourier Series To further analyze thevariability of stereographically projected Fourier series wemake the substitution

120579119898= 120579119898(119909) = 119898

120587

119871119909 (33)

International Journal of Computational Mathematics 5

in 119866(120574 119909119898) to rewrite it as

119866 (120574 119909119898) = 119886119898cos 120579119898(119909) + 119887

119898sin 120579119898(119909) (34)

The tangent half-angle (stereographic) transformation119905119898= tan(120579

1198982) or 120579

119898= 2tanminus1119905

119898is conceived as

120579119898(119909) = 2tanminus1119905

119898(119909) = Φ [119905

119898(119909)] (35)

and subsequently

119866 = 119866 [120579119898(119909)] = 119866 Φ [119905

119898(119909)] = 119867 [119905

119898(119909)] (36)

It is clear then that this stereographic projection effectivelyleads to insertion of an additional bracket into the compo-sition of 119866 as a function of 119909 This paves the way to thedefinition

S 119866 [120579119898(119909)] = 119866 Φ [119905

119898(119909)] = 119867 [119905

119898(119909)]

= 119867 (120574 119909119898) (37)

Proposition 4 The following

S119889

119889119909

119873

sum119898=1

119866 [120579119898(119909)] =

119889

119889119909S119873

sum119898=1

119866 [120579119898(119909)] (38)

always holds

Proof Since 119873 is finite then the proof of S(119889119889119909)119866 =(119889119889119909)S119866 = (119889119889119909)119867 implies the correctness of thisproposition So let us differentiate 119866(119909) = 119866[120579

119898(119909)] with

respect to 119909 by applying the chain rule namely

119889119866

119889119909=

119889119866

119889Φ

119889Φ

119889119905119898

119889119905119898

119889119909=

119889119866

119889120579119898

119889120579119898

119889119905119898

119889119905119898

119889119909 (39)

which is the same as

119889119866

119889119909

= [minus119886119898sin 120579119898+ 119887119898cos 120579119898]

2(1 + 1199052119898)

120579119898

2sec2 (

120579119898

2)

(40)

Take then the S transformation of the previous relation toarrive after some algebra at

S119889

119889119909119866 = minus119898

120587

119871

(1198871198981199052119898+ 2119886119898119905119898minus 119887119898)

(1 + 1199052119898)

(41)

Invoke then

119867 = 119867(120574 119909119898) =(119886119898+ 2119887119898119905119898minus 1198861198981199052119898)

(1 + 1199052119898)

(42)

to differentiate it with respect to 119909 as

119889

119889119909119867 =

119889

119889119909S 119866 = minus119898

120587

119871

(1198871198981199052119898+ 2119886119898119905119898minus 119887119898)

(1 + 1199052119898)

(43)

which is the required result

Let us rewrite (1) in the form

120593 (119909) = 119892 (119909) minus 119888 = 0 (44)

then try to differentiate it term-by-term as

1205931015840 (119909) =119889

119889119909119892 (119909) =

119889

119889119909

infin

sum119898=1

119866 (120574 119909119898)

= minusinfin

sum119898=1

119898120587

119871(minus119887119898cos119898120587

119871119909+ 119886119898sin119898120587

119871119909)

= minusinfin

sum119898=1

119898120587

119871119866lowast (120574 119909119898)

(45)

where 119866lowast(120574 119909119898) is the harmonic conjugate of 119866(120574 119909119898)The trigonometric series 119892lowast(119909) = sum

infin

119898=1 119866lowast(120574 119909119898) is called

the trigonometric series conjugate (see eg [10]) to 119892(119909)In addition to the fact that the convergence of 119892lowast(119909)

can sometimes become worse than convergence of 119892(119909) (seeeg [10]) the presence of the 119898(120587119871) factor in (119889119889119909)119892(119909)can aggravate the convergence problem of (45) when thecoefficients 119886

119898and 119887119898do not decay fast enough

Next we may stereographically transform (120597120597119909)119892(119909)using the notationS[(119889119889119909)119892(119909)] = [(119889119889119909)119892(119909)] where

[119889

119889119909119892 (119909)] = minus

infin

sum119898=1

119898120587

119871119866lowast (120574 119909119898)

= minusinfin

sum119898=1

119898120587

119871

(1198871198981199052119898+ 2119886119898119905119898minus 119887119898)

(1 + 1199052119898)

(46)

Similarly for the stereographic Fourier series representation

119892 (119909) =11988602+infin

sum119898=1

119867(120574 119909119898) (47)

in which

119867(120574 119909119898) =(119886119898+ 2119887119898119905119898minus 1198861198981199052119898)

(1 + 1199052119898)

(48)

we may approximate (44) by

120595 (119909) = 119892 (119909) minus 119888 = 0 (49)

Differentiate then (49) term-by-term as

1205951015840 (119909) =119889

119889119909119892 (119909) =

119889

119889119909

infin

sum119898=1

119867(120574 119909119898)

= minusinfin

sum119898=1

119898120587

119871119867lowast (120574 119909119898)

= minusinfin

sum119898=1

119898120587

119871

(1198871198981199052119898+ 2119886119898119905119898minus 119887119898)

(1 + 1199052119898)

(50)

6 International Journal of Computational Mathematics

Note here that119867lowast (120574 119909119898)

=(1198871198981199052119898+ 2119886119898119905119898minus 119887119898)

(1 + 1199052119898)

=(119887119898tan119898(1205872119871) 119909 minus 119886

119898)2minus (1198862119898+ 1198872119898)

119887119898(tan2119898(1205872119871) 119909 + 1)

(51)

is the harmonic conjugate of119867(120574 119909119898) and that

119867lowast (120574 119909119898) = 119866lowast (120574 119909119898) (52)

This rather interesting fact leads to a remark that follows

Remark 5 Away from points of discontinuity of 119891(119909) thestereographically transformed differentiated Fourier series

[(119889119889119909)119892(119909)] which defines 119866lowast(120574 119909119898) is the same as thederivative (119889119889119909)119892(119909) which defines 119867lowast(120574 119909119898) of thestereographic Fourier series 119892(119909) Moreover even at pointsof discontinuity of 119891(119909) [(119889119889119909)119892(119909)] = (119889119889119909)119892(119909)

Theorem 6 (see [1]) Let 119891(119909) be (i) continuous on [minus119871 119871]such that (ii) 119891(minus119871) = 119891(119871) and let (iii) 1198911015840(119909) be piecewisecontinuous over [minus119871 119871] then the corresponding Fourier series119892(119909) of (1) or (47) is differentiable at each point where (iv)11989110158401015840(119909) exists

Theprevious remark can incidentally have useful applica-tions when differentiating certain slowly converging Fourierseries as illustrated by the following example

Example 7 Consider the 2120587-periodic odd signal 119891(119909)defined over one period by 119891(119909) = 119909 Its Fourier series isknown to be

119892 (119909) = 2infin

sum119898=1

(minus1)119898+1

119898sin119898119909 (53)

that is with 119886119898= 0 forall119898 119887

119898= 2((minus1)119898+1119898) and 119898(120587119871) =

119898Differentiating term-by-term we obtain

1198921015840 (119909) = 2infin

sum119898=1

(minus1)119898+1 cos119898119909 (54)

which is not uniformly convergent because 119891(119909) is not onlydiscontinuous at 119909 = plusmn120587 but 119891(minus120587) = 119891(120587) and 11989110158401015840(plusmn120587) doesnot exist In particular at 119909 = 0 the sum of the series 1198921015840(0) =2suminfin119898=1(minus1)

119898+1 oscillates between 0 and 1 that is divergent Itis also well known nevertheless that the seriessuminfin

119898=1(minus1)119898+1

is 119862 minus 1 summable [1 2] to 12 and that turns out to yieldthe right answer 1198921015840(0) = 1198911015840(0) = 1 Moreover at 119909 = 120587 thesum of the series 1198921015840(120587) = 2suminfin

119898=1(minus1)2119898+1 rarr minusinfin that is

divergentAlternatively by means of (47)

119892 (119909) =infin

sum119898=1

119887119898

2119905119898

1 + 1199052119898

= 2infin

sum119898=1

(minus1)119898+1

119898sin119898119909 (55)

and by means of (46)

1198921015840 (119909) =

[119889

119889119909119892 (119909)] = minus

infin

sum119898=1

119898119866lowast (120574 119909119898)

= minusinfin

sum119898=1

119898119887119898

1199052119898minus 1

1 + 1199052119898

= 2infin

sum119898=1

(minus1)119898+1(1 minus tan2119898(12) 119909)(1 + tan2119898(12) 119909)

= 2infin

sum119898=1

(minus1)119898+1 cos119898119909

(56)

Here at 119909 = 120587 the sum of the series 1198921015840(120587) =

2suminfin119898=1(minus1)

2119898+1 rarr minusinfin is also divergent

Note in general that for even signals that is when 119887119898= 0

forall119898 it follows from (22) that

119867(120574 119909119898)1003816100381610038161003816119887119898=0

= 119886119898

1 minus 1199052119898

1 + 1199052119898

= 119886119898cos119898120587

119871119909

= 119866 (120574 119909119898)1003816100381610038161003816119887119898=0

(57)

Moreover for odd signals that is when 119886119898

= 0 forall119898 thesituation is not of the same clarity

Indeed according to (22)

119867(120574 119909119898)1003816100381610038161003816119886119898=0

=00

119887119898

1 + 1199052119898

(58)

which is an uncertainty Luckily however this uncertainty isnot an essential one and can straightforwardly be resolved byresorting to the stereographic form (23) of (22) which yieldsthe correct result

119867(120574 119909119898)1003816100381610038161003816119886119898=0

= 119887119898

2119905119898

1 + 1199052119898

= 119887119898sin119898120587

119871119909

= 119866 (120574 119909119898)1003816100381610038161003816119886119898=0

(59)

Also when 119887119898= 0 forall119898

119867lowast (120574 119909119898)1003816100381610038161003816119887119898=0

=00

119886119898

1 + 1199052119898

(60)

and this is resolvable in a similar fashionBoth119866(120574 119909119898) of (2) and119867(120574 119909119898) of (22) equivalent

forms are nonlinear trigonometric functions of the 119909 solu-tion However while 119866(120574 119909119898) is linear in 120574 = 119886

119898 119887119898infin119898=1

thestereographic 119867(120574 119909119898) is nonlinear in this 120574 This canraise a question on a possibility for (120597120597119909)119892(119909) = (120597120597119909)119892(119909)at singular points of 119891(119909) A question that will be answerednegatively in the next section Furthermore since the data 120574 =119886119898 119887119898infin119898=1 is discrete then the operators sum

infin

119898=1 119866(120574 119909119898)and sum

infin

119898=1 119867(120574 119909119898) do not depend continuously on 120574 andthis is themain reason for ill-posedness of the present inverseproblem

International Journal of Computational Mathematics 7

In conclusion the quadratic degeneracy of the solution ofthe present inverse problem explicit when using 119867(120574 119909119898)is a nonlinear indicator of the existence of the peak 119876

119904120590119873(119909)

pertaining to the Gibbs phenomenon However the numer-ical ill-posedness associated with the relative values of atrial iterative variable in the 120574 set happens to remarkablyremain invariant under the tangent half-angle stereographicprojection

Following the same arguments of Wilbraham we maycancel the artificially induced differentiation by integrating(50) to represent the stereographic Fourier series 119892(119909) inintegral form as

119892 (119909) =11988602minusinfin

sum119898=1

int119909

0119898

sdot120587

119871

(119887119898tan119898(1205872119871) 120585 minus 119886

119898)2minus (1198862119898+ 1198872119898)

119887119898(tan2119898(1205872119871) 120585 + 1)

119889120585 =11988602

minusinfin

sum119898=1

int119909

0119898

sdot120587

119871

(119887119898tan2119898(1205872119871) 120585 minus 2119886

119898tan119898(1205872119871) 120585 minus 119887

119898)

(tan2119898(1205872119871) 120585 + 1)119889120585

(61)

4 Numerics of Solving the Inverse Problem

The solution of the present inverse problem is in fact aproblem of finding the roots of (44) when 119892(119909) is truncatedat 119898 = 119873 For that reason we shall focus on such rootsin the neighborhood of a step-like discontinuity of 119891(119909) at119909 = 119909119900 say where 119891(119909119900 minus 120576) lt 119891(119909119900 + 120576) 0 lt 120576 ≪ 1 Theassociated with the Gibbs phenomenon overshoot of 119892(119909) =11988602+sum

119873

119898=1 119866(120574 119909119898) or 11988602+sum119873

119898=1 119867(120574 119909119898) over119891(119909119900+120576)and undershoot of this 119892(119909) below 119891(119909119900minus120576)The intersectionof a horizontal line ℎ(119909) = 119888 ge 119891(119909119900 + 120576) corresponding to120593(119909) = 0 can be at two points 1205721 and 1205722 on both sides of thepeak location 1205720 representing two distinct roots of 120593(119909) = 0It can also be at one point 1205720 corresponding to themaximumof the truncated 119892(119909) that is a double root of 120593(119909) = 0

Obviously 1205721 lt 1205721 lt 1205720 lt 1205722 lt 1205722 where 1205721 and 1205722 arepossible points of inflection defined by

11989210158401015840 (1205721) = 12059310158401015840 (1205721) = 11989210158401015840 (1205722) = 12059310158401015840 (1205722) = 0 (62)

The existence of both 1205721 and 1205722 or at least one of themnamely 1205722 is however not mandatory Moreover for the 12057211205720 and 1205722 points we expect to have

1198921015840 (1205721) = 1205931015840 (1205721) gt 0

12059310158401015840 (1205721) gt 0

120593101584010158401015840 (1205721) lt 0

1198921015840 (1205720) = 1205931015840 (1205720) = 0

12059310158401015840 (1205720) lt 0

120593101584010158401015840 (1205720) = 0

1198921015840 (1205722) = 1205931015840 (1205722) lt 0

12059310158401015840 (1205722) gt 0

120593101584010158401015840 (1205722) gt 0(63)

Equation (44) can be solved for119909 120593(119909) = 0 by aNewton-Raphson (see eg [11 12]) iterative process

119909119899+1 = 119909

119899minus

120593 (119909119899)

1205931015840 (119909119899) (64)

which is repeated until a sufficiently accurate value is reachedThis process for each series truncation number119873 rewrites as

119909119899+1 = 119909

119899+[sum119873

119898=1 119866 (120574 119909119899 119898) + 11988602 minus 119888]

[sum119873

119898=1 119898(120587119871)119866lowast (120574 119909119899 119898)]

(65)

or in the stereographic form

119909119899+1 = 119909

119899+[sum119873

119898=1 119867(120574 119909119899 119898) + 11988602 minus 119888]

[sum119873

119898=1 119898(120587119871)119867lowast (120574 119909119899 119898)]

(66)

which is the same as

119909119899+1 = 119909

119899

+[sum119873

119898=1 ((119886119898 + 2119887119898119905119898minus 1198861198981199052119898) (1 + 1199052

119898)) + 11988602 minus 119888]

[sum119873

119898=1 119898(120587119871) ((1198871198981199052119898+ 2119886119898119905119898minus 119887119898) (1 + 1199052

119898))]

(67)

and is started off with some rather free initial value 1199090Obviously here 119905

119898= tan119898(1205872119871)119909

The sequence 119909119899will usually converge provided that 1199090

is close enough to the unknown zero 120572 and that

1205931015840 (1199090) = minusinfin

sum119898=1

119898120587

119871119866lowast (120574 1199090 119898) = 0

or 1205951015840 (1199090) = minusinfin

sum119898=1

119898120587

119871119867lowast (120574 1199090 119898) = 0

(68)

41 Computation of the Peak Location 1205720 Being a maximiza-tion problem for 120593(119909) = 0 the determination of 1205720 by theNewton-Raphson process [11] becomes

119909119899+1 = 119909

119899minus

1205931015840 (119909119899)

12059310158401015840 (119909119899)

(69)

or

119909119899+1 = 119909

119899minus

1205951015840 (119909119899)

12059510158401015840 (119909119899) (70)

in stereographic form

8 International Journal of Computational Mathematics

Straightforwardly this process can be expressed as

119909119899+1 = 119909

119899minus

[sum119873

119898=1 119898119866lowast (120574 119909119899 119898)]

[sum119873

119898=1 1198982 (120587119871) 119866 (120574 119909

119899 119898)]

(71)

or in the stereographic form

119909119899+1 = 119909

119899

minus[minussum119873

119898=1 119898(120587119871)119867lowast (120574 119909119899 119898)]

[sum119873

119898=1 1198982 (12058721198712) ((119886

1198981199052119898minus 2119887119898119905119898minus 119886119898) (1 + 1199052

119898))]

(72)

which is the same as119909119899+1 = 119909

119899

+[sum119873

119898=1 119898((1198871198981199052119898+ 2119886119898119905119898minus 119887119898) (1 + 1199052

119898))]

[sum119873

119898=1 1198982 (120587119871) ((119886

1198981199052119898minus 2119887119898119905119898minus 119886119898) (1 + 1199052

119898))]

(73)

It is remarkable how this algorithm is distinctively indepen-dent of 119888

Proposition 8 The ill-posedness of the inverse problem ofthe truncated Fourier series is invariant under stereographicprojection

Proof The tangent half-angle stereographic projectionthough not isometric (distance preserving) is neverthelessvariability preserving Indeed the main ingredients ofthis inverse problem are namely 120574 = 119886

119898 119887119898119873119898=1 119888 and

1199090 with (119889119889119909)119866[120579119898(119909)]1199090 Under this projection 120579

119898(1199090)

transforms to 119905119898(1199090) = tan[120579

119898(1199090)2] and 119866[120579

119898(1199090)]

transforms to 119867[119905119898(1199090)] But according to Proposition 4

(119889119889119909)119867[119905119898(119909)]1199090

= (119889119889119909)119866[120579119898(119909)]1199090 Hence the variation

of 119866(120574 1199090 119898) with 1199090 in the Fourier series inverse problemand the variation of119867(120574 1199090 119898) with 1199090 in the stereographicform are identical

Equivalently this makes the ill-posedness of this inverseproblem invariant under stereographic projection

Expected manifestations of the correctness of this propo-sition are as follows (i) the results of computations by the(71) and (73) algorithms for 1205720 with the same 1199090 must beidentical and (ii) the results of computations by the (65) and(67) algorithms for 1205721 (or 1205722) with the same 1199090 and 119888 mustalso be identical

42 Convergence Analysis Let us focus first on the con-vergence of the iterative process (65) or (67) In manysituationswhere term-by-termdifferentiability does not holdtheNewton-Raphsonmethod can always be replaced by somequasi-Newtonmethod [12 13] where 1205931015840(119909

119899) or1205951015840(119909

119899) in (65)

or (67) is replaced by a suitable approximation For examplein the chord method 1205931015840(119909

119899) is replaced by 1205931015840(1199090) for all

iterationsIf 120598119899= 120572minus119909

119899 120598119899+1 = 120572minus119909

119899+1 and120573119899 is some point between119909119899and 120572 it can be easily proved (see [11] or [12]) that

120598119899+1 = minus

12059310158401015840 (120573119899)

21205931015840 (119909119899)1205982119899 (74)

Taking absolute value of both sides gives

1003816100381610038161003816120598119899+11003816100381610038161003816 =

1003816100381610038161003816100381612059310158401015840 (120573119899)10038161003816100381610038161003816

2 10038161003816100381610038161205931015840 (119909119899)10038161003816100381610038161205982119899 (75)

implying quadratic convergence that is 120598119899is squared (the

number of correct digits roughly doubles) at each step Thesame arguments apply of course to the 120595(119909) = 0 equation

In actual fact for a distinct 120572 the convergence of119909119899 in (65)ndash(67) is at least quadratic in the neigh-

borhood of this 120572 The process may face a difficultywhen 1205931015840(119909

119899) = minussum

119873

119898=1 119898(120587119871)119866lowast(120574 119909119899 119898) or 1205951015840(119909

119899) =

minussum119873

119898=1 119898(120587119871)119867lowast(120574 119909119899 119898) diverges This is however a spe-

cial but not a general problem and (75) always holds if thefollowing conditions are satisfied

(i) 1205931015840(119909) = 0 forall119909 isin 119868 where 119868 = [120572 minus 120578 120572 + 120578] for some120578 ge |(120572 minus 1199090)|

(ii) 12059310158401015840(119909) is finite forall119909 isin 119868(iii) 1199090 is sufficiently close to 120572

The term sufficiently close in this context means the follow-ing

(a) Taylor approximation is accurate enough such that wecan ignore higher order terms

(b) (12)|12059310158401015840(119909119899)1205931015840(119909

119899)| lt 119870|12059310158401015840(120572)1205931015840(120572)| for some119870 lt

infin(c) 119870|12059310158401015840(120572)1205931015840(120572)|120598

119899lt 1 for 119899 isin N

Consequently (75) can be expressed in the following way1003816100381610038161003816120598119899+1

1003816100381610038161003816 = 1198721205982119899 (76)

where

119872 = sup119909isin119868

12

100381610038161003816100381610038161003816100381610038161003816

12059310158401015840 (119909)

1205931015840 (119909)

100381610038161003816100381610038161003816100381610038161003816 (77)

The initial point 1199090 has to be chosen such that conditions(i) through (iii) are satisfied where the third conditionrequires that 119872|1205980| lt 1 Moreover with increasing 119873 thedistance |1205721 minus 1205722| is known to decrease in a way makingthe satisfaction of these conditions very difficult and actuallyimpossible as119873 rarr infin Here it should be underlined that theiterative method (65)ndash(67) may fail to converge to the 120572 rootin the following situations

(i) If a point like 1205720 is included in 119868 the method willterminate due to division by zero

(ii) When the initial estimate 1199090 is poor the pertainingwrong 119868 can contribute to nonconvergence of thealgorithm

As for convergence of the Newton-Raphson processtowards 1205720 let 120576119899 = 1205720 minus 119909

119899 120576119899+1 = 1205720 minus 119909

119899+1 and 120579119899is some

point between 119909119899and 1205720 to arrive at

120576119899+1 = minus

120593101584010158401015840 (120579119899)

212059310158401015840 (119909119899)1205762119899 (78)

1003816100381610038161003816120576119899+11003816100381610038161003816 =

10038161003816100381610038161003816120593101584010158401015840 (120579119899)10038161003816100381610038161003816

2 100381610038161003816100381612059310158401015840 (119909119899)10038161003816100381610038161205762119899 (79)

International Journal of Computational Mathematics 9

The convergence of 119909119899 in (71)ndash(73) is at least quadratic

in the neighborhood of this 1205720 The process may face a diffi-culty however when 12059310158401015840(119909

119899) = minussum

infin

119898=1 1198982(120587119871)2119866(120574 119909

119899 119898)

or 12059510158401015840(119909119899) diverges Moreover (79) always holds if the follow-

ing conditions are satisfied

(i) 12059310158401015840(119909) = 0 forall119909 isin 119868 where 119868 = [1205720 minus120578 1205720 +120578] for some120578 ge |(1205720 minus 1199090)|

(ii) 120593101584010158401015840(119909) is finite forall119909 isin 119868

(iii) 1199090 is sufficiently close to 1205720

Sufficient closeness in this context means that

(a) Taylor approximation is accurate enough such that wecan ignore higher order terms

(b) (12)|120593101584010158401015840(119909119899)12059310158401015840(119909

119899)| lt 119870|120593101584010158401015840(1205720)120593

10158401015840(1205720)| for some119870 lt infin

(c) 119870|120593101584010158401015840(1205720)12059310158401015840(1205720)|120576119899 lt 1 for 119899 isin N

Consequently

1003816100381610038161003816120576119899+11003816100381610038161003816 = 1198721205762

119899 (80)

where

119872 = sup119909isin119868

12

100381610038161003816100381610038161003816100381610038161003816

120593101584010158401015840 (119909)

12059310158401015840 (119909)

100381610038161003816100381610038161003816100381610038161003816 (81)

43 Applications The standard iterative algorithms (71) or(73) for finding the location of overshoots in 119876

119904120590119873(119909) of

the Gibbs effect employ the fact that the first derivative of apeak in 119892(119909) or 119892(119909) respectively has a downward-going-zero-crossing at the peak maximum But the presence ofmany neighboring smaller peaks in the ringing close to theovershoot will cause many undesirable zero-crossings to beinvoked if the initial trial root 1199090 is not correctly chosenThis fact makes the numerical problem of peak finding in theGibbs effect of pathological difficulty A difficulty that aggra-vates with the increase in the number 119873 of truncated termsin the employed Fourier series Indeed with the increaseof 119873 the number of undesirable zero-crossings increasesin the neighborhood of 1205720 as illustrated in Section 1 Thiscan increasingly lead to oscillatory and even unstable valuesin sequence 119909

119899 the more the distance |1199090 minus 1205720| contains

unwanted zero-crossingsThese problems have been resolved in the two examples

that follow which simultaneously illustrate the applicabilityof Proposition 8

Example 9 Consider the 2120587-periodic odd signal 119891(119909) ofExample A1 defined over one period by

119891 (119909) =

minus1 minus120587 lt 119909 lt 0

1 0 le 119909 lt 120587(82)

which has for each truncation number 119873 the truncatedFourier series approximations

119892 (119909) ≃4120587

119873

sum119898=1

1(2119898 minus 1)

sin (2119898minus 1) 119909

119892 (119909) ≃8120587

119873

sum119898=1

119887119898119905119898

(1 + 1199052119898)=

8120587

119873

sum119898=1

1(2119898 minus 1)

119905119898

(1 + 1199052119898)

=8120587

119873

sum119898=1

1(2119898 minus 1)

tan ((2119898 minus 1) 2) 1199091 + [tan ((2119898 minus 1) 2) 119909]2

(83)

This signal has a jump discontinuity at 119909 = 0 ofmagnitude 2 We shall focus our computations on the Gibbsphenomenon overshoot near this discontinuity

Here is a printout list of results for the parameters of theposing inverse problem via computations by both the directand stereographic algorithms

It is remarkable how results by both algorithms arepractically identical and share the same following featuresthat are observed by results of the direct algorithm discussedbelow

Case I (119873 = 40) The computations of 1205720 for the previousexample using (71) with 1199090 = 002 lead to 1199091 = 0039 109this varies quickly with 119899 (only beyond the 4th decimal) tostabilize towards 1199093 = 1199094 = 1199095 = sdot sdot sdot = 0039 299 = 1205720Here using (1) gives 119892(1205720) = 119892(0039 299) = 1179 030 whichis quite close to the theoretically expected value by (A7) ofsim118

Then using (65) for 119888 = 109 and 1199090 = 0000381 leads to1199091 = 0021 697 this varies quickly with 119899 (only beyond the 3-d decimal) to stabilize quickly towards 1199095 = 1199096 = 1199097 = sdot sdot sdot =0028 368 = 1205721 Also using (65) for 119888 = 109 and 1199090 = 004leads to 1199091 = 0067 699 this varies quickly with 119899 (beyondthe 2nd decimal) to stabilize quickly towards 1199095 = 1199096 = 1199097 =sdot sdot sdot = 0052 873 = 1205722

For both1205721 and1205722 using (1) yields119892(1205721) = 119892(0028368) =119892(1205722) = 119892(0052873) = 109 = 119888 as expected

Case II (119873 = 100) Computations of 1205720 using (71) with1199090 = 002 lead to 1199091 = 0014 221 this varies quickly with119899 (only beyond the 4th decimal) to stabilize towards 1199093 =1199094 = 1199095 = sdot sdot sdot = 0015 708 = 1205720 Here using (1) gives119892(1205720) = 119892(0015708) = 1178 990 This result appears to beslightly excessive perhaps due to increased rounding off errorswith increasing 119873 to 100

Thenusing (65) for 119888 = 109 and1199090 = 0015 237 leads aftera negative value to 1199092 = 0014 462 this varies with 119899 (beyondthe 2nd decimal) to stabilize quickly towards 1199097 = 1199098 = 1199099 =sdot sdot sdot = 0011 3348 = 1205721 The used value of 1199090 is apparently notappropriate and this illustrates the generic ill-posedness of thisinverse problem

The pertaining computations of 1205722 with 1199090 = 0017000lead to 1199091 = 0020 434 this varies with 119899 (beyond the 3-ddecimal) to stabilize quickly towards 1199094 = 1199095 = 1199096 = sdot sdot sdot =0021 149 = 1205722

10 International Journal of Computational Mathematics

For both 1205721 and 1205722 using (1) yields 119892(1205721) =119892(00113348) = 119892(1205722) = 119892(0021149) = 109 = 119888 asexpected

Case III (119873 = 150) The computations of 1205720 using (71) with1199090 = 0005 lead to 1199091 = 0010 595 this varies insignificantlywith 119899 (only beyond the 4th decimal) to stabilize towards1199093 =1199094 = 1199095 = sdot sdot sdot = 0010 472 = 1205720 Here using (1) gives 119892(1205720) =119892(0010472) = 1178 980 which does not significantly differfrom 119892(1205720) when119873 = 100

Thenusing (65) for 119888 = 109 and1199090 = 0010 158 leads aftera negative value to 1199092 = 0009 643 and this varies (beyond the3-d decimal) to stabilize quickly towards1199097 = 1199098 = 1199099 = sdot sdot sdot =0007 565 = 1205721

The pertaining computations of 1205722 with 1199090 = 0010 800lead to 1199091 = 0025 998 this varies (beyond the 2nd decimal)to stabilize towards 11990914 = 11990915 = 11990916 = 11990917 = 11990918 = sdot sdot sdot =0014 099 = 1205722

For both 1205721 and 1205722 using (1) yields 119892(1205721) = 119892(007 565) =119892(1205722) = 119892(0014 099) = 109 = 119888

During the previous computations summarized inTable 1 it has been observed that varying 1199090 in the presentNewton-Raphson process for all119873 can lead to strong oscil-lations and even completely unstable values for 1205720 1205721 and 1205722of the Fourier series inverse problem which though ill-posed

is practically regularizible Apart from this drawback the(71) with (65) iterative algorithms (i) converge quickly to thetheoretically expected values and are (ii) of demonstratedstability of the iterations for properly chosen 1199090 (which act aseffective regularizing parameters) Moreover computationsby the stereographic algorithm invoking (73) and (67) sharethe samemagnitude and convergence rates with results by thedirect algorithm Finally it is demonstrated that the presentsignal exhibits an overshoot when119873 = 150 ofsim9of a jumpdiscontinuity of magnitude 2 that is located at a distance ofsim00105

To illustrate robustness of both numerical algorithms andtheir possible effective regularizibility we shall study also thenumerical inverse problem for the Fourier series representa-tion of an arbitrary (nonsymmetric) periodic signal

Example 10 Consider the 2120587-periodic signal 119891(119909) definedover one period by

119891 (119909) =

minus119909

120587minus 1 minus120587 lt 119909 lt 0

1 0 le 119909 lt 120587(84)

It has for each truncation number 119873 the truncated Fourierseries approximations

119892 (119909) ≃14+119873

sum119898=1

[(minus1)119898 minus 1]12058721198982 cos119898119909+

[2 minus (minus1)119898]120587119898

sin119898119909

119892 (119909) ≃14

+119873

sum119898=1

[(minus1)119898 minus 1] 12058721198982 + 2 ([2 minus (minus1)119898] 120587119898) tan ((2119898 minus 1) 2) 119909 minus ([(minus1)119898 minus 1] 12058721198982) [tan ((2119898 minus 1) 2) 119909]2

1 + [tan ((2119898 minus 1) 2) 119909]2

(85)

This signal has like the signal of Example 9 a jumpdiscontinuity at 119909 = 0 of magnitude 2 For the sake ofcomparison we shall also focus our computations on theGibbs phenomenon overshoot part of 119876

119904120590119873(119909) near this

discontinuityIt should be noted here that the results by both algorithms

summarized in Table 2 are also identical a feature indicatingthat the degree of ill-posedness of the posing inverse problemis not affected as predicted by Proposition 8 by the tangenthalf-angle stereographic projection

Case I (119873 = 40) The computations of 1205720 for this exampleusing (71) and (73) with 1199090 = 003 lead to an identicalsequence 119909

119899 that converges by both algorithms after the 1199094

iteration to 0078 139 with 119892(1205720) = 119892(1205720) = 1178 76

Case II (119873 = 100) The computations of 1205720 with 1199090 = 001lead to an identical sequence 119909

119899 that converges by both

algorithms after the 1199095 iteration to 0031 351 with 119892(1205720) =

119892(1205720) = 1178 87

Case III (119873 = 150) The computations of 1205720 with 1199090 = 0007lead also to an identical sequence 119909

119899 that converges by both

algorithms after the 1199095 iteration to 0020 915 with 119892(1205720) =

119892(1205720) = 1178 91

Therefore the present signal exhibits an overshoot when119873 = 150 of sim9 of a jump discontinuity of magnitude 2 thatis located at a distance of sim0020 9

Comparison of the computed Gibbs overshoot parame-ters of the two considered signals with jump discontinuitiesat the same 119909 = 0 of the same magnitude illustrates that therelative size of the overshoot depends essentially exclusivelyon the magnitude of the stimulating jump discontinuityFor instance based on the present numerical results onecan suggest here that nonsymmetric signals have equalovershoots with symmetric signals Their respective 1205720rsquos arehowever different Indeed the nonsymmetric signal has asignificantly larger 1205720 (0020 9) than the 1205720 (00105) of thesymmetric signal

International Journal of Computational Mathematics 11

Table 1 Results of computations for Example 9

119873 1205721 1205720 1205722

40 0028 368 0039 299 0052 873100 0011 335 0015 705 0021 149150 0007 565 0010 472 0014 099

Table 2 Results of computations for Example 10

119873 1205721 1205720 1205722

40 00564765 00781391 0105155100 00226546 00313516 00422024150 00151124 00209153 0028156

Table 3 Summary of computations of all overshoot parameters inExample 10

1205720results

119873 1199090

119909lowast

1205720

119892(1205720)

100 0030 1199093

0031 3516 1178 87200 0015 119909

30015 6919 1178 92

300 0008 1199093

0010 4648 1178 941205721results

119873 1199090

119909lowast

1205721

119888

0025 1199094

0022 2027 108100 0025 119909

30023 1344 110

0025 1199093

0024 2027 1120010 119909

30011 1117 108

200 0010 1199093

0011 5779 1100010 119909

40012 1123 112

0008 1199093

0007 410 08 108300 0008 119909

30007 720 95 110

0008 1199092

0008 077 32 1121205722results

119873 1199090

119909lowast

1205722

119888

100035 119909

40042 9439 108

035 1199094

0041 4445 110035 119909

40039 8513 112

2000018 119909

30021 4960 108

0018 1199093

0020 7456 1100018 119909

30019 9483 112

3000012 119909

30014 3360 108

0012 1199093

0013 8356 1100012 119909

40013 3039 112

A further analysis has also been performed of Example 10with 119873 = 100 200 300 to compute the corresponding1205720 with 1205721 1205722 that correspond simultaneously to 119888 =108 110 112 as illustrated in Figure 1 Results of thesecomputations are summarized in Table 3 and then used forthe partial 119876

119904120590119873(119909) plots of Figure 2 that interpolate 1205721 1205722

(for 119888 = 108 110 112) with 1205720 for each 119873 = 100 200 300In this table119909

lowaststands for the first convergence step of the 119909

119899

sequence of Newton Raphson iterations

002

05

10

004

g(x

)

xminus004 minus002

minus05

minus10

g(x) for N = 100

g(x) for N = 200

g(x) for N = 300

h(x) = c = 108

h(x) = c = 11

h(x) = c = 112

1205721 1205720 1205722

Figure 1 Plots of 119892(119909) near a discontinuity of 119891(119909) in Example 10

120

115

110

105

100

000 001 002 003 004 005

g(x

)

x

g(x) for N = 100

g(x) for N = 200

g(x) for N = 300

h(x) = c = 108

h(x) = c = 11

h(x) = c = 112

Figure 2 Plots of the 119892(119909) curves near a discontinuity of 119891(119909) inExample 10 interpolating 1205721 1205722 (for 119888 = 108 110 112) with 1205720 allfor119873 = 100 200 300

These results turn out to be extremely stable against varia-tions of the trial initial root1199090 Indeed in the computationallymost difficult situation namely when 119873 = 300 a changeof 1199090 from 0008 to 0020 leads to a variation of 119909

lowastfrom

1199092 to just 1199096 with the same asymptotic 1205720 = 00104648 and1205721 = 000807732 (for 119888 = 112)

5 Conclusions

This paper demonstrates that the quadratic degeneracy ofthe solution of the present inverse problem explicit whenusing 119867(120574 119909119898) is a nonlinear indicator of the existence ofGibbsrsquo phenomenon It provides also a distribution-theoreticproof for the existence of this phenomenon The reportedanalysis of numerical solvability of this problem illustrates thefollowing

(i) When suminfin

119898=1 119898(120587119871)119866lowast(120574 119909119898) = suminfin

119898=1 119898(120587119871)119867lowast(120574 119909119898) converge continuously in 119909 and are

12 International Journal of Computational Mathematics

nonzero at 1205721 (or 1205722) there exists a neighborhood(defined by 119868) of 120572 such that for all starting values 1199090in the neighborhood 119909

119899 will converge to 1205721 (or 1205722)

(ii) When suminfin

119898=1 119898(120587119871)119866lowast(120574 119909119898)=suminfin

119898=1 119898(120587119871)119867lowast(120574119909119898) converge continuously in 119909 and are nonzero at1205721 and 1205722 (but not at 1205720) and if 120593(119909) has a secondderivative which might be zero at 1205721 and 1205722 (and at1205720) then the convergence of 119909119899 is quadratic or fasterIf12059310158401015840(119909) = 0 at1205721 and1205722 then the convergence of 119909119899is merely quadratic

(iii) The convergence of the Newton-Raphson iterativeprocess to 1205720 can also be quadratic but restrictionson the derivatives of 120593(119909) have to be observed up tothe third order Otherwise the convergence of 119909

119899 to

1205720 can slow down to become only linear [12] with arate log210

(iv) The reported computations clearly illustrate that1205720 rarr 0 and |1205721 minus 1205722| rarr 0 when 119873 rarr infin thatis the infinite Fourier series will sum to 119891(119909) inthe neighborhood of a discontinuity except at thediscontinuity itself

Appendix

To illustrate the Gibbs-Wilbraham effect we shall use herethe same arguments employed by Wilbraham in 1848 inanalyzing a similar square wave

Example A1 Let 119891(119909) be a square wave periodic signalwhich is defined over one period by

119891 (119909) =

minus1 minus120587 lt 119909 le 0

1 0 le 119909 lt 120587(A1)

Its Fourier series representation is

119892 (119909) =4120587

infin

sum119898=1

1(2119898 minus 1)

sin (2119898minus 1) 119909 (A2)

This defines the truncated series

119878119873(119909) =

4120587

119873

sum119898=1

1(2119898 minus 1)

sin (2119898minus 1) 119909 (A3)

which may be differentiated as

119889

119889119909119878119873(119909) =

4120587

119873

sum119898=1

cos (2119898minus 1) 119909 (A4)

Since

2119873

sum119898=1

cos (2119899 minus 1) 119909 =sin 2119873119909

sin119909 (A5)

then119889

119889119909119878119873(119909) =

2120587

sin 2119873119909

sin119909 (A6)

It is possible to cancel the previous artificially induceddifferentiation by integrating the last result as

119878119873(119909) = int

119909

0

119889

119889120585119878119873(120585) 119889120585 =

2120587int119909

0

sin 2119873120585

sin 120585119889120585 (A7)

This integral starts at zero when 119909 = 0 in agreement with(A2) and increases until 2119873120585 = 120587 Then 120585 = 1205872119873 definesa turning point 1199091 The integral starts to decrease at whichpoint the numerator sin 2119873120585 goes negative For large 119873 thedenominator sin 120585 remains positive

The form (A7) distinctively from (A3) reveals that119878119873(119909) has turning points at the zeros of sin 2119873120585 These occur

when 2119873120585 = 119903120587 119903 = 1 2 3 4 5 that is at 120585 = 119903(1205872119873)Clearly then 119909

119903sim 120585119903= 119903(1205872119873) are turning points for 119878

119873(119909)

In particular 1199091 sim 1205851 = 1205872119873 defines the overshoot peaklocation 1205720 and 119878

119873(1199091) = (2120587) int12058721198730 (sin 2119873120585 sin 120585) 119889120585 is

its magnitude To evaluate 119878119873(1199091) = 119878

119873(1205720) numerically

let us change the variable of integration namely 2119873120585 = 119908or 120585 = 1199082119873 to rewrite the previous integral as 119878

119873(1199091) =

(2120587) int1205870 (sin119908(sin1199082119873)) 119889(1199082119873) Clearly when 119873 rarr

infin sin1199082119873 = 1199082119873 and

lim119873rarrinfin

119878119873(1199091) =

2120587int120587

0

sin119908119908

119889119908 =2120587Si (120587) = 118 (A8)

correct to two decimal pointsAs for 1199092 1199093 1199094 119909119903 lim119873rarrinfin[119878119873(119909119903) minus 119891(119909)]

decreases as we move away from the discontinuity butit should be remarked however that the location of theovershoot 1205720 = 1199091 = 1205872119873 moves towards the discontinuityaccording to

lim119873rarrinfin

1205720 = lim119873rarrinfin

120587

2119873= 0 (A9)

Conflict of Interests

The author declares that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

The numerical computations of Examples 9 and 10 have beenperformed using Mathematica 8 by Mounir Abou Yassineof the Department of CCE of AUL

References

[1] P L Butzer and R J Nessel Fourier Analysis and Approxima-tion Volume I One-Dimensional Theory Birkhauser 1971

[2] R E Edwards Fourier Series A Modern Introduction HoltRinehart amp Winston 1967

[3] W J Thompson ldquoFourier series and the Gibbs phenomenonrdquoAmerican Journal of Physics vol 60 no 5 pp 425ndash429 1992

[4] A J JerriThe Gibbs Phenomenon in Fourier Analysis Splines ampWavelet Approximations Kluwer Academic Publishers BostonMass USA 1990

International Journal of Computational Mathematics 13

[5] K S Rim and B I Yun ldquoGibbs phenomenon removal by addingHeaviside functionsrdquo Advances in Computational Mathematicsvol 38 no 4 pp 683ndash699 2013

[6] A H Zemanian Distribution Theory and Transform AnalysisMcGraw-Hill 1965

[7] G R Ayers and J C Dainty ldquoIterative blind deconvolutionmethod and its applicationsrdquo Optics Letters vol 13 no 7 pp547ndash549 1988

[8] N H S Haidar ldquoA nonlinear method for a severely ill-posedgeneralized moment problemrdquo Journal of Inverse amp Ill-PosedProblems vol 5 no 4 pp 323ndash335 1997

[9] G L Bradley and K J Smith Calculus Prentice Hall 1995[10] P Offner T Sonar and M Wirz ldquoDetecting strength and

location of jump discontinuities in numerical datardquo AppliedMathematics vol 4 no 12 pp 1ndash14 2013

[11] K E Atkinson An Introduction to Numerical Analysis JohnWiley amp Sons New York NY USA 1989

[12] E Suli and D F Mayers An Introduction to Numerical AnalysisCambridge University Press Cambridge UK 2003

[13] P Wang ldquoA third-order family of Newton-like iteration meth-ods for solving nonlinear equationsrdquo Journal of NumericalMathematics and Stochastics vol 3 no 1 pp 13ndash19 2011

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Stochastic AnalysisInternational Journal of

International Journal of Computational Mathematics 3

It should be pointed out that in the previous proofrelations (10) and (15) say nothing specifically about thebehavior of119892(119909) at119909minus

119904and119909+119904 respectively while asserting the

existence of a spiky behavior of 119892(119909) individually at 119909+119904then

at 119909minus119904 Moreover the uncertainty 120588 had been resolved in the

past (by Wilbraham see also the Appendix) computationallyto satisfy 120588 = 009 and long before the advance of the theoryof distributions (generalized functions) by S L Sobolev in1936 The number 009 for 120588 remains until now howevera mysterious theoretical puzzle that calls for a rigorousdistribution-theoretic justification

Clearly in the neighborhood of 119909119904 119892(119909) can only be a

rudimentary approximation to 119891(119909) even when 119873 rarr infinwith an added spike pair

119876119904(119909) =

+120588Δ119904 119909 = 119909+

119904

minus120588Δ119904 119909 = 119909minus

119904

0 119909 = 119909119904

(16)

to 119891plusmn119904at 119909plusmn119904 Relation (16) represents the Gibbs-Wilbraham

effect which due to the fact that |119909+119904

minus 119909minus119904| = 0 is

not practically computable This effect degenerates howeverwhen119873 is finite to an undershoot-overshoot pair119876

119904120590119873(119909) on

an interval 119868119904120590119873

= [119909119904minus 120590119873 119909119904+ 120590119873] splitted by Δ

119904at 119909119904 In

this case it is always possible to compute

119892 (119909) =

sim 119891 (119909) + 119876119904120590119873

(119909) 119909 isin 119868119904120590119873

119873 lt infin

sim 119891 (119909) 119909 notin 119868119904120590119873

(17)

Furthermore since lim119873rarrinfin

119876119904120590119873

(119909) = 119876119904(119909) then these

computations with (17) over 119868119904120590119873

can reveal only a feature119876119904120590119873

(119909) associated with the Gibbs-Wilbraham effect 119876119904(119909)

but not the effect itself

3 The Inverse Problem for TruncatedFourier Series

Substitution of any value of 119909 isin [minus119871 119871] in (1) yields a number119888 = 119892(119909) which may differ [2] from 119891(119909) by 119876

119904120590119873(119909) if 119909 is

in the neighborhood of a discontinuity of 119891(119909) Obviously 119888satisfies the nonlinear in 119909 trigonometric series equation

119873

sum119898=1

119866 (120574 119909119898) =119873

sum119898=1

119886119898cos119898120587

119871119909+ 119887119898sin119898120587

119871119909

= 119888 minus11988602

(18)

which is however linear in the data 120574 = 119886119898 119887119898infin119898=1

Definition 2 (the inverse problem for the truncated Fourierseries representation) Given a number 119888 = 119892(119909) satisfying(18) what is the corresponding 119909

This inverse problem happens to be a discretized versionof the similar deconvolution problem [4] of 119909 from a givennumber 119902 in a singular first kind integral equation of the form

intinfin

0119870(120574 119909 120591) 119889120591 = 119902 (19)

in which 120574 represents the data It is well known additionally(see eg [7 8]) that this deconvolution problem can becomputationally ill-posed if an arbitrarily small deviationof the 120574 data may cause an arbitrarily large deviation inthe solution 119909 This situation may further prevail when theoperator intinfin0 119870(120574 119909 120591)119889120591 does not continuously depend on120574

Moreover the inverse problem of the previous definitionhas a unique solution that satisfies (19) and is expected toaccept a double rootwhen the horizontal line ℎ(119909) = 119888 gt 119891(119909)intersects with a sharp peak (overshoot) of a truncated 119892(119909)We shall call this feature a duplicated (quadratic) degeneracyof the solution for this inverse problem

The existence of the Fourier series truncation peak (andthe corresponding two roots for 119909) is not directly evidentfrom the structure of 119866(120574 119909119898) and despite its linearity inthe data 120574 = 119886

119898 119887119898infin119898=1

31 The Tangent Half-Angle Stereographic Projection Thetangent half-angle substitution (the Euler-Weirstrass sub-stitution [9]) tan(1205792) is widely used in integral calculusfor finding antiderivatives of rational expressions of circularfunction pairs (cos 120579 sin 120579) Geometrically the constructiongoes like this draw the unit circle and let 119875 be the point(minus1 0) A line through 119875 (except the vertical line) passesthrough a point (cos 120579 sin 120579) on the circle and crosses the119910-axis at some point 119910 = 119905 Obviously 119905 determines theslope of this line Furthermore each of the lines (exceptthe vertical line) intersects the unit circle in exactly twopoints one of which is 119875 This determines a function frompoints on the unit circle to points on a straight line Thecircular functions determine therefore a map from points(cos 120579 sin 120579) on the unit circle to slopes 119905 In fact this isa stereographic projection [9] that expresses these pairs interms of one variable tan(1205792) = 119905

Numerical algorithms employing these trigonometricpairs like Fourier series (1) could in principle be influencedby such a projection In this context the tangent half-rangetransformation relates the angle 120579

119898= 119898(120587119871)119909 to the slope

119905119898= tan(120579

1198982) of a line intersecting the unit circle centered

around (0 0)Indeed as 120579

119898varies the point (cos 120579

119898 sin 120579119898) winds

repeatedly around the unit circle centered at (0 0) The point

(1 minus 1199052119898

1 + 1199052119898

2119905119898

1 + 1199052119898

) = (cos 120579119898 sin 120579119898) (20)

goes only once around the circle as 119905119898goes from minusinfin to infin

and never reaches the point (minus1 0) which is approached as alimit when 119905

119898rarr plusmninfin

Clearly the stereographic projection 119905119898

= tan(1205791198982) is

defined on the entire unit circle except at the projection point119875 (minus1 0) Where it is defined this mapping is smoothand bijective [9] It is conformal meaning that it preservesangles it is however not isometric that is it does not preservedistances as

119889120579119898=

2 119889119905119898

1 + 1199052119898

(21)

4 International Journal of Computational Mathematics

The rest of this section illustrates how stereographicprojection of Fourier series reveals a quadratic degeneracythat indicates the existence of the Gibbs-Wilbraham effect Italso addresses the question of solving the inverse problem forthe truncated Fourier series representations The impact ofstereographic projection on the ill-posedness of the numer-ical solution of this inverse problem is moreover a mainquestion that has been addressed in this paper

Our analysis starts however by illustrating how thetangent half-angle stereographic projection can provide anew insight into the behavior of truncated Fourier series neara discontinuity of 119891(119909)

Lemma 3 The solution 119909 of the inverse problem for thetruncated Fourier series representation satisfies

119873

sum119898=1

119867(120574 119909119898)

=119873

sum119898=1

(1198862119898+ 1198872119898) minus (119886

119898tan119898(1205872119871) 119909 minus 119887

119898)2

119886119898(tan2119898(1205872119871) 119909 + 1)

= 119888 minus11988602

(22)

Proof By studying solvability for 119909 of (18) Let 119905119898

=tan119898(1205872119871)119909 and then cos119898(120587119871)119909 = (1minus1199052

119898)(1+1199052

119898) and

sin119898(120587119871)119909 = 2119905119898(1 + 1199052

119898) Further consideration of these

facts in (18) leads to the tangent half-angle stereographicallytransformedS[ ] = [] form

minus119873

sum119898=1

S [119866 (120574 119909119898)] =119873

sum119898=1

(1198861198981199052119898minus 2119887119898119905119898minus 119886119898)

(1199052119898+ 1)

=11988602minus 119888

(23)

The harmonic trinomial in the left-hand side of (23) hastwo harmonic zeros 120585plusmn

119898 for each 119909 namely

120585plusmn119898=

2119871119898120587

tanminus1119905plusmn119898

119905plusmn119898=

119887119898

119886119898

plusmn120590119898

120590119898=

radic1198862119898+ 1198872119898

119886119898

(24)

It is worth noting at this point that these 120585plusmn119898zeros should

by no means be confused with the possible roots of the 119909variable Substitution then of (24) in (23) leads to

119873

sum119898=1

119886119898(119905119898minus 119905+119898) (119905119898minus 119905minus119898)

(1199052119898+ 1)

=11988602minus 119888 (25)

which is the same as

119873

sum119898=1

119886119898(119905119898minus 119887119898119886119898minus 120590119898) (119905119898minus 119887119898119886119898+ 120590119898)

(1199052119898+ 1)

=11988602minus 119888

(26)

The quadratic degeneracy of the harmonic zeros 120585plusmn119898of 119905119898

can disappear only when 120590119898= 0 for all119898 And this happens

to be equivalent with

119873

sum119898=1

1198862119898+ 1198872119898= 0 (27)

Obviously satisfaction of (27) by (26) reduces it to theequation

119873

sum119898=1

119886119898(119905119898minus 119887119898119886119898)2

(1199052119898+ 1)

=11988602minus 119888 (28)

which is the same as

119873

sum119898=1

(119886119898tan119898(1205872119871) 119909 minus 119887

119898)2

119886119898(tan2119898(1205872119871) 119909 + 1)

=11988602minus 119888 (29)

It is interesting to remark that condition (27) is entirelydifferent from

11988602+119873

sum119898=1

1198862119898+ 1198872119898gt 0 (30)

which can be derived from the Parseval identity [1] (com-pleteness relation)

11988602+infin

sum119898=1

1198862119898+ 1198872119898=

1119871int119871

minus119871

1198912(119909) 119889119909 (31)

Hence a possible nondegenerate solution for 119909 of the nonlin-ear equation (29) is on one hand fundamentally incorrectbecause it involves a violation of (30) and Parsevalrsquos identityOn the other hand a consistent solution 119909 in the senseof satisfaction of (31) is intrinsically degenerate because ofits emergence as a must from the nonlinear equation (26)which rewrites as

119873

sum119898=1

119886119898[(119905119898minus 119887119898119886119898)2minus 1205902119898]

(1199052119898+ 1)

=11988602minus 119888 (32)

Equation (32) is easily transformable to the requiredresult (23) here the proof completes

32 The Stereographic Fourier Series To further analyze thevariability of stereographically projected Fourier series wemake the substitution

120579119898= 120579119898(119909) = 119898

120587

119871119909 (33)

International Journal of Computational Mathematics 5

in 119866(120574 119909119898) to rewrite it as

119866 (120574 119909119898) = 119886119898cos 120579119898(119909) + 119887

119898sin 120579119898(119909) (34)

The tangent half-angle (stereographic) transformation119905119898= tan(120579

1198982) or 120579

119898= 2tanminus1119905

119898is conceived as

120579119898(119909) = 2tanminus1119905

119898(119909) = Φ [119905

119898(119909)] (35)

and subsequently

119866 = 119866 [120579119898(119909)] = 119866 Φ [119905

119898(119909)] = 119867 [119905

119898(119909)] (36)

It is clear then that this stereographic projection effectivelyleads to insertion of an additional bracket into the compo-sition of 119866 as a function of 119909 This paves the way to thedefinition

S 119866 [120579119898(119909)] = 119866 Φ [119905

119898(119909)] = 119867 [119905

119898(119909)]

= 119867 (120574 119909119898) (37)

Proposition 4 The following

S119889

119889119909

119873

sum119898=1

119866 [120579119898(119909)] =

119889

119889119909S119873

sum119898=1

119866 [120579119898(119909)] (38)

always holds

Proof Since 119873 is finite then the proof of S(119889119889119909)119866 =(119889119889119909)S119866 = (119889119889119909)119867 implies the correctness of thisproposition So let us differentiate 119866(119909) = 119866[120579

119898(119909)] with

respect to 119909 by applying the chain rule namely

119889119866

119889119909=

119889119866

119889Φ

119889Φ

119889119905119898

119889119905119898

119889119909=

119889119866

119889120579119898

119889120579119898

119889119905119898

119889119905119898

119889119909 (39)

which is the same as

119889119866

119889119909

= [minus119886119898sin 120579119898+ 119887119898cos 120579119898]

2(1 + 1199052119898)

120579119898

2sec2 (

120579119898

2)

(40)

Take then the S transformation of the previous relation toarrive after some algebra at

S119889

119889119909119866 = minus119898

120587

119871

(1198871198981199052119898+ 2119886119898119905119898minus 119887119898)

(1 + 1199052119898)

(41)

Invoke then

119867 = 119867(120574 119909119898) =(119886119898+ 2119887119898119905119898minus 1198861198981199052119898)

(1 + 1199052119898)

(42)

to differentiate it with respect to 119909 as

119889

119889119909119867 =

119889

119889119909S 119866 = minus119898

120587

119871

(1198871198981199052119898+ 2119886119898119905119898minus 119887119898)

(1 + 1199052119898)

(43)

which is the required result

Let us rewrite (1) in the form

120593 (119909) = 119892 (119909) minus 119888 = 0 (44)

then try to differentiate it term-by-term as

1205931015840 (119909) =119889

119889119909119892 (119909) =

119889

119889119909

infin

sum119898=1

119866 (120574 119909119898)

= minusinfin

sum119898=1

119898120587

119871(minus119887119898cos119898120587

119871119909+ 119886119898sin119898120587

119871119909)

= minusinfin

sum119898=1

119898120587

119871119866lowast (120574 119909119898)

(45)

where 119866lowast(120574 119909119898) is the harmonic conjugate of 119866(120574 119909119898)The trigonometric series 119892lowast(119909) = sum

infin

119898=1 119866lowast(120574 119909119898) is called

the trigonometric series conjugate (see eg [10]) to 119892(119909)In addition to the fact that the convergence of 119892lowast(119909)

can sometimes become worse than convergence of 119892(119909) (seeeg [10]) the presence of the 119898(120587119871) factor in (119889119889119909)119892(119909)can aggravate the convergence problem of (45) when thecoefficients 119886

119898and 119887119898do not decay fast enough

Next we may stereographically transform (120597120597119909)119892(119909)using the notationS[(119889119889119909)119892(119909)] = [(119889119889119909)119892(119909)] where

[119889

119889119909119892 (119909)] = minus

infin

sum119898=1

119898120587

119871119866lowast (120574 119909119898)

= minusinfin

sum119898=1

119898120587

119871

(1198871198981199052119898+ 2119886119898119905119898minus 119887119898)

(1 + 1199052119898)

(46)

Similarly for the stereographic Fourier series representation

119892 (119909) =11988602+infin

sum119898=1

119867(120574 119909119898) (47)

in which

119867(120574 119909119898) =(119886119898+ 2119887119898119905119898minus 1198861198981199052119898)

(1 + 1199052119898)

(48)

we may approximate (44) by

120595 (119909) = 119892 (119909) minus 119888 = 0 (49)

Differentiate then (49) term-by-term as

1205951015840 (119909) =119889

119889119909119892 (119909) =

119889

119889119909

infin

sum119898=1

119867(120574 119909119898)

= minusinfin

sum119898=1

119898120587

119871119867lowast (120574 119909119898)

= minusinfin

sum119898=1

119898120587

119871

(1198871198981199052119898+ 2119886119898119905119898minus 119887119898)

(1 + 1199052119898)

(50)

6 International Journal of Computational Mathematics

Note here that119867lowast (120574 119909119898)

=(1198871198981199052119898+ 2119886119898119905119898minus 119887119898)

(1 + 1199052119898)

=(119887119898tan119898(1205872119871) 119909 minus 119886

119898)2minus (1198862119898+ 1198872119898)

119887119898(tan2119898(1205872119871) 119909 + 1)

(51)

is the harmonic conjugate of119867(120574 119909119898) and that

119867lowast (120574 119909119898) = 119866lowast (120574 119909119898) (52)

This rather interesting fact leads to a remark that follows

Remark 5 Away from points of discontinuity of 119891(119909) thestereographically transformed differentiated Fourier series

[(119889119889119909)119892(119909)] which defines 119866lowast(120574 119909119898) is the same as thederivative (119889119889119909)119892(119909) which defines 119867lowast(120574 119909119898) of thestereographic Fourier series 119892(119909) Moreover even at pointsof discontinuity of 119891(119909) [(119889119889119909)119892(119909)] = (119889119889119909)119892(119909)

Theorem 6 (see [1]) Let 119891(119909) be (i) continuous on [minus119871 119871]such that (ii) 119891(minus119871) = 119891(119871) and let (iii) 1198911015840(119909) be piecewisecontinuous over [minus119871 119871] then the corresponding Fourier series119892(119909) of (1) or (47) is differentiable at each point where (iv)11989110158401015840(119909) exists

Theprevious remark can incidentally have useful applica-tions when differentiating certain slowly converging Fourierseries as illustrated by the following example

Example 7 Consider the 2120587-periodic odd signal 119891(119909)defined over one period by 119891(119909) = 119909 Its Fourier series isknown to be

119892 (119909) = 2infin

sum119898=1

(minus1)119898+1

119898sin119898119909 (53)

that is with 119886119898= 0 forall119898 119887

119898= 2((minus1)119898+1119898) and 119898(120587119871) =

119898Differentiating term-by-term we obtain

1198921015840 (119909) = 2infin

sum119898=1

(minus1)119898+1 cos119898119909 (54)

which is not uniformly convergent because 119891(119909) is not onlydiscontinuous at 119909 = plusmn120587 but 119891(minus120587) = 119891(120587) and 11989110158401015840(plusmn120587) doesnot exist In particular at 119909 = 0 the sum of the series 1198921015840(0) =2suminfin119898=1(minus1)

119898+1 oscillates between 0 and 1 that is divergent Itis also well known nevertheless that the seriessuminfin

119898=1(minus1)119898+1

is 119862 minus 1 summable [1 2] to 12 and that turns out to yieldthe right answer 1198921015840(0) = 1198911015840(0) = 1 Moreover at 119909 = 120587 thesum of the series 1198921015840(120587) = 2suminfin

119898=1(minus1)2119898+1 rarr minusinfin that is

divergentAlternatively by means of (47)

119892 (119909) =infin

sum119898=1

119887119898

2119905119898

1 + 1199052119898

= 2infin

sum119898=1

(minus1)119898+1

119898sin119898119909 (55)

and by means of (46)

1198921015840 (119909) =

[119889

119889119909119892 (119909)] = minus

infin

sum119898=1

119898119866lowast (120574 119909119898)

= minusinfin

sum119898=1

119898119887119898

1199052119898minus 1

1 + 1199052119898

= 2infin

sum119898=1

(minus1)119898+1(1 minus tan2119898(12) 119909)(1 + tan2119898(12) 119909)

= 2infin

sum119898=1

(minus1)119898+1 cos119898119909

(56)

Here at 119909 = 120587 the sum of the series 1198921015840(120587) =

2suminfin119898=1(minus1)

2119898+1 rarr minusinfin is also divergent

Note in general that for even signals that is when 119887119898= 0

forall119898 it follows from (22) that

119867(120574 119909119898)1003816100381610038161003816119887119898=0

= 119886119898

1 minus 1199052119898

1 + 1199052119898

= 119886119898cos119898120587

119871119909

= 119866 (120574 119909119898)1003816100381610038161003816119887119898=0

(57)

Moreover for odd signals that is when 119886119898

= 0 forall119898 thesituation is not of the same clarity

Indeed according to (22)

119867(120574 119909119898)1003816100381610038161003816119886119898=0

=00

119887119898

1 + 1199052119898

(58)

which is an uncertainty Luckily however this uncertainty isnot an essential one and can straightforwardly be resolved byresorting to the stereographic form (23) of (22) which yieldsthe correct result

119867(120574 119909119898)1003816100381610038161003816119886119898=0

= 119887119898

2119905119898

1 + 1199052119898

= 119887119898sin119898120587

119871119909

= 119866 (120574 119909119898)1003816100381610038161003816119886119898=0

(59)

Also when 119887119898= 0 forall119898

119867lowast (120574 119909119898)1003816100381610038161003816119887119898=0

=00

119886119898

1 + 1199052119898

(60)

and this is resolvable in a similar fashionBoth119866(120574 119909119898) of (2) and119867(120574 119909119898) of (22) equivalent

forms are nonlinear trigonometric functions of the 119909 solu-tion However while 119866(120574 119909119898) is linear in 120574 = 119886

119898 119887119898infin119898=1

thestereographic 119867(120574 119909119898) is nonlinear in this 120574 This canraise a question on a possibility for (120597120597119909)119892(119909) = (120597120597119909)119892(119909)at singular points of 119891(119909) A question that will be answerednegatively in the next section Furthermore since the data 120574 =119886119898 119887119898infin119898=1 is discrete then the operators sum

infin

119898=1 119866(120574 119909119898)and sum

infin

119898=1 119867(120574 119909119898) do not depend continuously on 120574 andthis is themain reason for ill-posedness of the present inverseproblem

International Journal of Computational Mathematics 7

In conclusion the quadratic degeneracy of the solution ofthe present inverse problem explicit when using 119867(120574 119909119898)is a nonlinear indicator of the existence of the peak 119876

119904120590119873(119909)

pertaining to the Gibbs phenomenon However the numer-ical ill-posedness associated with the relative values of atrial iterative variable in the 120574 set happens to remarkablyremain invariant under the tangent half-angle stereographicprojection

Following the same arguments of Wilbraham we maycancel the artificially induced differentiation by integrating(50) to represent the stereographic Fourier series 119892(119909) inintegral form as

119892 (119909) =11988602minusinfin

sum119898=1

int119909

0119898

sdot120587

119871

(119887119898tan119898(1205872119871) 120585 minus 119886

119898)2minus (1198862119898+ 1198872119898)

119887119898(tan2119898(1205872119871) 120585 + 1)

119889120585 =11988602

minusinfin

sum119898=1

int119909

0119898

sdot120587

119871

(119887119898tan2119898(1205872119871) 120585 minus 2119886

119898tan119898(1205872119871) 120585 minus 119887

119898)

(tan2119898(1205872119871) 120585 + 1)119889120585

(61)

4 Numerics of Solving the Inverse Problem

The solution of the present inverse problem is in fact aproblem of finding the roots of (44) when 119892(119909) is truncatedat 119898 = 119873 For that reason we shall focus on such rootsin the neighborhood of a step-like discontinuity of 119891(119909) at119909 = 119909119900 say where 119891(119909119900 minus 120576) lt 119891(119909119900 + 120576) 0 lt 120576 ≪ 1 Theassociated with the Gibbs phenomenon overshoot of 119892(119909) =11988602+sum

119873

119898=1 119866(120574 119909119898) or 11988602+sum119873

119898=1 119867(120574 119909119898) over119891(119909119900+120576)and undershoot of this 119892(119909) below 119891(119909119900minus120576)The intersectionof a horizontal line ℎ(119909) = 119888 ge 119891(119909119900 + 120576) corresponding to120593(119909) = 0 can be at two points 1205721 and 1205722 on both sides of thepeak location 1205720 representing two distinct roots of 120593(119909) = 0It can also be at one point 1205720 corresponding to themaximumof the truncated 119892(119909) that is a double root of 120593(119909) = 0

Obviously 1205721 lt 1205721 lt 1205720 lt 1205722 lt 1205722 where 1205721 and 1205722 arepossible points of inflection defined by

11989210158401015840 (1205721) = 12059310158401015840 (1205721) = 11989210158401015840 (1205722) = 12059310158401015840 (1205722) = 0 (62)

The existence of both 1205721 and 1205722 or at least one of themnamely 1205722 is however not mandatory Moreover for the 12057211205720 and 1205722 points we expect to have

1198921015840 (1205721) = 1205931015840 (1205721) gt 0

12059310158401015840 (1205721) gt 0

120593101584010158401015840 (1205721) lt 0

1198921015840 (1205720) = 1205931015840 (1205720) = 0

12059310158401015840 (1205720) lt 0

120593101584010158401015840 (1205720) = 0

1198921015840 (1205722) = 1205931015840 (1205722) lt 0

12059310158401015840 (1205722) gt 0

120593101584010158401015840 (1205722) gt 0(63)

Equation (44) can be solved for119909 120593(119909) = 0 by aNewton-Raphson (see eg [11 12]) iterative process

119909119899+1 = 119909

119899minus

120593 (119909119899)

1205931015840 (119909119899) (64)

which is repeated until a sufficiently accurate value is reachedThis process for each series truncation number119873 rewrites as

119909119899+1 = 119909

119899+[sum119873

119898=1 119866 (120574 119909119899 119898) + 11988602 minus 119888]

[sum119873

119898=1 119898(120587119871)119866lowast (120574 119909119899 119898)]

(65)

or in the stereographic form

119909119899+1 = 119909

119899+[sum119873

119898=1 119867(120574 119909119899 119898) + 11988602 minus 119888]

[sum119873

119898=1 119898(120587119871)119867lowast (120574 119909119899 119898)]

(66)

which is the same as

119909119899+1 = 119909

119899

+[sum119873

119898=1 ((119886119898 + 2119887119898119905119898minus 1198861198981199052119898) (1 + 1199052

119898)) + 11988602 minus 119888]

[sum119873

119898=1 119898(120587119871) ((1198871198981199052119898+ 2119886119898119905119898minus 119887119898) (1 + 1199052

119898))]

(67)

and is started off with some rather free initial value 1199090Obviously here 119905

119898= tan119898(1205872119871)119909

The sequence 119909119899will usually converge provided that 1199090

is close enough to the unknown zero 120572 and that

1205931015840 (1199090) = minusinfin

sum119898=1

119898120587

119871119866lowast (120574 1199090 119898) = 0

or 1205951015840 (1199090) = minusinfin

sum119898=1

119898120587

119871119867lowast (120574 1199090 119898) = 0

(68)

41 Computation of the Peak Location 1205720 Being a maximiza-tion problem for 120593(119909) = 0 the determination of 1205720 by theNewton-Raphson process [11] becomes

119909119899+1 = 119909

119899minus

1205931015840 (119909119899)

12059310158401015840 (119909119899)

(69)

or

119909119899+1 = 119909

119899minus

1205951015840 (119909119899)

12059510158401015840 (119909119899) (70)

in stereographic form

8 International Journal of Computational Mathematics

Straightforwardly this process can be expressed as

119909119899+1 = 119909

119899minus

[sum119873

119898=1 119898119866lowast (120574 119909119899 119898)]

[sum119873

119898=1 1198982 (120587119871) 119866 (120574 119909

119899 119898)]

(71)

or in the stereographic form

119909119899+1 = 119909

119899

minus[minussum119873

119898=1 119898(120587119871)119867lowast (120574 119909119899 119898)]

[sum119873

119898=1 1198982 (12058721198712) ((119886

1198981199052119898minus 2119887119898119905119898minus 119886119898) (1 + 1199052

119898))]

(72)

which is the same as119909119899+1 = 119909

119899

+[sum119873

119898=1 119898((1198871198981199052119898+ 2119886119898119905119898minus 119887119898) (1 + 1199052

119898))]

[sum119873

119898=1 1198982 (120587119871) ((119886

1198981199052119898minus 2119887119898119905119898minus 119886119898) (1 + 1199052

119898))]

(73)

It is remarkable how this algorithm is distinctively indepen-dent of 119888

Proposition 8 The ill-posedness of the inverse problem ofthe truncated Fourier series is invariant under stereographicprojection

Proof The tangent half-angle stereographic projectionthough not isometric (distance preserving) is neverthelessvariability preserving Indeed the main ingredients ofthis inverse problem are namely 120574 = 119886

119898 119887119898119873119898=1 119888 and

1199090 with (119889119889119909)119866[120579119898(119909)]1199090 Under this projection 120579

119898(1199090)

transforms to 119905119898(1199090) = tan[120579

119898(1199090)2] and 119866[120579

119898(1199090)]

transforms to 119867[119905119898(1199090)] But according to Proposition 4

(119889119889119909)119867[119905119898(119909)]1199090

= (119889119889119909)119866[120579119898(119909)]1199090 Hence the variation

of 119866(120574 1199090 119898) with 1199090 in the Fourier series inverse problemand the variation of119867(120574 1199090 119898) with 1199090 in the stereographicform are identical

Equivalently this makes the ill-posedness of this inverseproblem invariant under stereographic projection

Expected manifestations of the correctness of this propo-sition are as follows (i) the results of computations by the(71) and (73) algorithms for 1205720 with the same 1199090 must beidentical and (ii) the results of computations by the (65) and(67) algorithms for 1205721 (or 1205722) with the same 1199090 and 119888 mustalso be identical

42 Convergence Analysis Let us focus first on the con-vergence of the iterative process (65) or (67) In manysituationswhere term-by-termdifferentiability does not holdtheNewton-Raphsonmethod can always be replaced by somequasi-Newtonmethod [12 13] where 1205931015840(119909

119899) or1205951015840(119909

119899) in (65)

or (67) is replaced by a suitable approximation For examplein the chord method 1205931015840(119909

119899) is replaced by 1205931015840(1199090) for all

iterationsIf 120598119899= 120572minus119909

119899 120598119899+1 = 120572minus119909

119899+1 and120573119899 is some point between119909119899and 120572 it can be easily proved (see [11] or [12]) that

120598119899+1 = minus

12059310158401015840 (120573119899)

21205931015840 (119909119899)1205982119899 (74)

Taking absolute value of both sides gives

1003816100381610038161003816120598119899+11003816100381610038161003816 =

1003816100381610038161003816100381612059310158401015840 (120573119899)10038161003816100381610038161003816

2 10038161003816100381610038161205931015840 (119909119899)10038161003816100381610038161205982119899 (75)

implying quadratic convergence that is 120598119899is squared (the

number of correct digits roughly doubles) at each step Thesame arguments apply of course to the 120595(119909) = 0 equation

In actual fact for a distinct 120572 the convergence of119909119899 in (65)ndash(67) is at least quadratic in the neigh-

borhood of this 120572 The process may face a difficultywhen 1205931015840(119909

119899) = minussum

119873

119898=1 119898(120587119871)119866lowast(120574 119909119899 119898) or 1205951015840(119909

119899) =

minussum119873

119898=1 119898(120587119871)119867lowast(120574 119909119899 119898) diverges This is however a spe-

cial but not a general problem and (75) always holds if thefollowing conditions are satisfied

(i) 1205931015840(119909) = 0 forall119909 isin 119868 where 119868 = [120572 minus 120578 120572 + 120578] for some120578 ge |(120572 minus 1199090)|

(ii) 12059310158401015840(119909) is finite forall119909 isin 119868(iii) 1199090 is sufficiently close to 120572

The term sufficiently close in this context means the follow-ing

(a) Taylor approximation is accurate enough such that wecan ignore higher order terms

(b) (12)|12059310158401015840(119909119899)1205931015840(119909

119899)| lt 119870|12059310158401015840(120572)1205931015840(120572)| for some119870 lt

infin(c) 119870|12059310158401015840(120572)1205931015840(120572)|120598

119899lt 1 for 119899 isin N

Consequently (75) can be expressed in the following way1003816100381610038161003816120598119899+1

1003816100381610038161003816 = 1198721205982119899 (76)

where

119872 = sup119909isin119868

12

100381610038161003816100381610038161003816100381610038161003816

12059310158401015840 (119909)

1205931015840 (119909)

100381610038161003816100381610038161003816100381610038161003816 (77)

The initial point 1199090 has to be chosen such that conditions(i) through (iii) are satisfied where the third conditionrequires that 119872|1205980| lt 1 Moreover with increasing 119873 thedistance |1205721 minus 1205722| is known to decrease in a way makingthe satisfaction of these conditions very difficult and actuallyimpossible as119873 rarr infin Here it should be underlined that theiterative method (65)ndash(67) may fail to converge to the 120572 rootin the following situations

(i) If a point like 1205720 is included in 119868 the method willterminate due to division by zero

(ii) When the initial estimate 1199090 is poor the pertainingwrong 119868 can contribute to nonconvergence of thealgorithm

As for convergence of the Newton-Raphson processtowards 1205720 let 120576119899 = 1205720 minus 119909

119899 120576119899+1 = 1205720 minus 119909

119899+1 and 120579119899is some

point between 119909119899and 1205720 to arrive at

120576119899+1 = minus

120593101584010158401015840 (120579119899)

212059310158401015840 (119909119899)1205762119899 (78)

1003816100381610038161003816120576119899+11003816100381610038161003816 =

10038161003816100381610038161003816120593101584010158401015840 (120579119899)10038161003816100381610038161003816

2 100381610038161003816100381612059310158401015840 (119909119899)10038161003816100381610038161205762119899 (79)

International Journal of Computational Mathematics 9

The convergence of 119909119899 in (71)ndash(73) is at least quadratic

in the neighborhood of this 1205720 The process may face a diffi-culty however when 12059310158401015840(119909

119899) = minussum

infin

119898=1 1198982(120587119871)2119866(120574 119909

119899 119898)

or 12059510158401015840(119909119899) diverges Moreover (79) always holds if the follow-

ing conditions are satisfied

(i) 12059310158401015840(119909) = 0 forall119909 isin 119868 where 119868 = [1205720 minus120578 1205720 +120578] for some120578 ge |(1205720 minus 1199090)|

(ii) 120593101584010158401015840(119909) is finite forall119909 isin 119868

(iii) 1199090 is sufficiently close to 1205720

Sufficient closeness in this context means that

(a) Taylor approximation is accurate enough such that wecan ignore higher order terms

(b) (12)|120593101584010158401015840(119909119899)12059310158401015840(119909

119899)| lt 119870|120593101584010158401015840(1205720)120593

10158401015840(1205720)| for some119870 lt infin

(c) 119870|120593101584010158401015840(1205720)12059310158401015840(1205720)|120576119899 lt 1 for 119899 isin N

Consequently

1003816100381610038161003816120576119899+11003816100381610038161003816 = 1198721205762

119899 (80)

where

119872 = sup119909isin119868

12

100381610038161003816100381610038161003816100381610038161003816

120593101584010158401015840 (119909)

12059310158401015840 (119909)

100381610038161003816100381610038161003816100381610038161003816 (81)

43 Applications The standard iterative algorithms (71) or(73) for finding the location of overshoots in 119876

119904120590119873(119909) of

the Gibbs effect employ the fact that the first derivative of apeak in 119892(119909) or 119892(119909) respectively has a downward-going-zero-crossing at the peak maximum But the presence ofmany neighboring smaller peaks in the ringing close to theovershoot will cause many undesirable zero-crossings to beinvoked if the initial trial root 1199090 is not correctly chosenThis fact makes the numerical problem of peak finding in theGibbs effect of pathological difficulty A difficulty that aggra-vates with the increase in the number 119873 of truncated termsin the employed Fourier series Indeed with the increaseof 119873 the number of undesirable zero-crossings increasesin the neighborhood of 1205720 as illustrated in Section 1 Thiscan increasingly lead to oscillatory and even unstable valuesin sequence 119909

119899 the more the distance |1199090 minus 1205720| contains

unwanted zero-crossingsThese problems have been resolved in the two examples

that follow which simultaneously illustrate the applicabilityof Proposition 8

Example 9 Consider the 2120587-periodic odd signal 119891(119909) ofExample A1 defined over one period by

119891 (119909) =

minus1 minus120587 lt 119909 lt 0

1 0 le 119909 lt 120587(82)

which has for each truncation number 119873 the truncatedFourier series approximations

119892 (119909) ≃4120587

119873

sum119898=1

1(2119898 minus 1)

sin (2119898minus 1) 119909

119892 (119909) ≃8120587

119873

sum119898=1

119887119898119905119898

(1 + 1199052119898)=

8120587

119873

sum119898=1

1(2119898 minus 1)

119905119898

(1 + 1199052119898)

=8120587

119873

sum119898=1

1(2119898 minus 1)

tan ((2119898 minus 1) 2) 1199091 + [tan ((2119898 minus 1) 2) 119909]2

(83)

This signal has a jump discontinuity at 119909 = 0 ofmagnitude 2 We shall focus our computations on the Gibbsphenomenon overshoot near this discontinuity

Here is a printout list of results for the parameters of theposing inverse problem via computations by both the directand stereographic algorithms

It is remarkable how results by both algorithms arepractically identical and share the same following featuresthat are observed by results of the direct algorithm discussedbelow

Case I (119873 = 40) The computations of 1205720 for the previousexample using (71) with 1199090 = 002 lead to 1199091 = 0039 109this varies quickly with 119899 (only beyond the 4th decimal) tostabilize towards 1199093 = 1199094 = 1199095 = sdot sdot sdot = 0039 299 = 1205720Here using (1) gives 119892(1205720) = 119892(0039 299) = 1179 030 whichis quite close to the theoretically expected value by (A7) ofsim118

Then using (65) for 119888 = 109 and 1199090 = 0000381 leads to1199091 = 0021 697 this varies quickly with 119899 (only beyond the 3-d decimal) to stabilize quickly towards 1199095 = 1199096 = 1199097 = sdot sdot sdot =0028 368 = 1205721 Also using (65) for 119888 = 109 and 1199090 = 004leads to 1199091 = 0067 699 this varies quickly with 119899 (beyondthe 2nd decimal) to stabilize quickly towards 1199095 = 1199096 = 1199097 =sdot sdot sdot = 0052 873 = 1205722

For both1205721 and1205722 using (1) yields119892(1205721) = 119892(0028368) =119892(1205722) = 119892(0052873) = 109 = 119888 as expected

Case II (119873 = 100) Computations of 1205720 using (71) with1199090 = 002 lead to 1199091 = 0014 221 this varies quickly with119899 (only beyond the 4th decimal) to stabilize towards 1199093 =1199094 = 1199095 = sdot sdot sdot = 0015 708 = 1205720 Here using (1) gives119892(1205720) = 119892(0015708) = 1178 990 This result appears to beslightly excessive perhaps due to increased rounding off errorswith increasing 119873 to 100

Thenusing (65) for 119888 = 109 and1199090 = 0015 237 leads aftera negative value to 1199092 = 0014 462 this varies with 119899 (beyondthe 2nd decimal) to stabilize quickly towards 1199097 = 1199098 = 1199099 =sdot sdot sdot = 0011 3348 = 1205721 The used value of 1199090 is apparently notappropriate and this illustrates the generic ill-posedness of thisinverse problem

The pertaining computations of 1205722 with 1199090 = 0017000lead to 1199091 = 0020 434 this varies with 119899 (beyond the 3-ddecimal) to stabilize quickly towards 1199094 = 1199095 = 1199096 = sdot sdot sdot =0021 149 = 1205722

10 International Journal of Computational Mathematics

For both 1205721 and 1205722 using (1) yields 119892(1205721) =119892(00113348) = 119892(1205722) = 119892(0021149) = 109 = 119888 asexpected

Case III (119873 = 150) The computations of 1205720 using (71) with1199090 = 0005 lead to 1199091 = 0010 595 this varies insignificantlywith 119899 (only beyond the 4th decimal) to stabilize towards1199093 =1199094 = 1199095 = sdot sdot sdot = 0010 472 = 1205720 Here using (1) gives 119892(1205720) =119892(0010472) = 1178 980 which does not significantly differfrom 119892(1205720) when119873 = 100

Thenusing (65) for 119888 = 109 and1199090 = 0010 158 leads aftera negative value to 1199092 = 0009 643 and this varies (beyond the3-d decimal) to stabilize quickly towards1199097 = 1199098 = 1199099 = sdot sdot sdot =0007 565 = 1205721

The pertaining computations of 1205722 with 1199090 = 0010 800lead to 1199091 = 0025 998 this varies (beyond the 2nd decimal)to stabilize towards 11990914 = 11990915 = 11990916 = 11990917 = 11990918 = sdot sdot sdot =0014 099 = 1205722

For both 1205721 and 1205722 using (1) yields 119892(1205721) = 119892(007 565) =119892(1205722) = 119892(0014 099) = 109 = 119888

During the previous computations summarized inTable 1 it has been observed that varying 1199090 in the presentNewton-Raphson process for all119873 can lead to strong oscil-lations and even completely unstable values for 1205720 1205721 and 1205722of the Fourier series inverse problem which though ill-posed

is practically regularizible Apart from this drawback the(71) with (65) iterative algorithms (i) converge quickly to thetheoretically expected values and are (ii) of demonstratedstability of the iterations for properly chosen 1199090 (which act aseffective regularizing parameters) Moreover computationsby the stereographic algorithm invoking (73) and (67) sharethe samemagnitude and convergence rates with results by thedirect algorithm Finally it is demonstrated that the presentsignal exhibits an overshoot when119873 = 150 ofsim9of a jumpdiscontinuity of magnitude 2 that is located at a distance ofsim00105

To illustrate robustness of both numerical algorithms andtheir possible effective regularizibility we shall study also thenumerical inverse problem for the Fourier series representa-tion of an arbitrary (nonsymmetric) periodic signal

Example 10 Consider the 2120587-periodic signal 119891(119909) definedover one period by

119891 (119909) =

minus119909

120587minus 1 minus120587 lt 119909 lt 0

1 0 le 119909 lt 120587(84)

It has for each truncation number 119873 the truncated Fourierseries approximations

119892 (119909) ≃14+119873

sum119898=1

[(minus1)119898 minus 1]12058721198982 cos119898119909+

[2 minus (minus1)119898]120587119898

sin119898119909

119892 (119909) ≃14

+119873

sum119898=1

[(minus1)119898 minus 1] 12058721198982 + 2 ([2 minus (minus1)119898] 120587119898) tan ((2119898 minus 1) 2) 119909 minus ([(minus1)119898 minus 1] 12058721198982) [tan ((2119898 minus 1) 2) 119909]2

1 + [tan ((2119898 minus 1) 2) 119909]2

(85)

This signal has like the signal of Example 9 a jumpdiscontinuity at 119909 = 0 of magnitude 2 For the sake ofcomparison we shall also focus our computations on theGibbs phenomenon overshoot part of 119876

119904120590119873(119909) near this

discontinuityIt should be noted here that the results by both algorithms

summarized in Table 2 are also identical a feature indicatingthat the degree of ill-posedness of the posing inverse problemis not affected as predicted by Proposition 8 by the tangenthalf-angle stereographic projection

Case I (119873 = 40) The computations of 1205720 for this exampleusing (71) and (73) with 1199090 = 003 lead to an identicalsequence 119909

119899 that converges by both algorithms after the 1199094

iteration to 0078 139 with 119892(1205720) = 119892(1205720) = 1178 76

Case II (119873 = 100) The computations of 1205720 with 1199090 = 001lead to an identical sequence 119909

119899 that converges by both

algorithms after the 1199095 iteration to 0031 351 with 119892(1205720) =

119892(1205720) = 1178 87

Case III (119873 = 150) The computations of 1205720 with 1199090 = 0007lead also to an identical sequence 119909

119899 that converges by both

algorithms after the 1199095 iteration to 0020 915 with 119892(1205720) =

119892(1205720) = 1178 91

Therefore the present signal exhibits an overshoot when119873 = 150 of sim9 of a jump discontinuity of magnitude 2 thatis located at a distance of sim0020 9

Comparison of the computed Gibbs overshoot parame-ters of the two considered signals with jump discontinuitiesat the same 119909 = 0 of the same magnitude illustrates that therelative size of the overshoot depends essentially exclusivelyon the magnitude of the stimulating jump discontinuityFor instance based on the present numerical results onecan suggest here that nonsymmetric signals have equalovershoots with symmetric signals Their respective 1205720rsquos arehowever different Indeed the nonsymmetric signal has asignificantly larger 1205720 (0020 9) than the 1205720 (00105) of thesymmetric signal

International Journal of Computational Mathematics 11

Table 1 Results of computations for Example 9

119873 1205721 1205720 1205722

40 0028 368 0039 299 0052 873100 0011 335 0015 705 0021 149150 0007 565 0010 472 0014 099

Table 2 Results of computations for Example 10

119873 1205721 1205720 1205722

40 00564765 00781391 0105155100 00226546 00313516 00422024150 00151124 00209153 0028156

Table 3 Summary of computations of all overshoot parameters inExample 10

1205720results

119873 1199090

119909lowast

1205720

119892(1205720)

100 0030 1199093

0031 3516 1178 87200 0015 119909

30015 6919 1178 92

300 0008 1199093

0010 4648 1178 941205721results

119873 1199090

119909lowast

1205721

119888

0025 1199094

0022 2027 108100 0025 119909

30023 1344 110

0025 1199093

0024 2027 1120010 119909

30011 1117 108

200 0010 1199093

0011 5779 1100010 119909

40012 1123 112

0008 1199093

0007 410 08 108300 0008 119909

30007 720 95 110

0008 1199092

0008 077 32 1121205722results

119873 1199090

119909lowast

1205722

119888

100035 119909

40042 9439 108

035 1199094

0041 4445 110035 119909

40039 8513 112

2000018 119909

30021 4960 108

0018 1199093

0020 7456 1100018 119909

30019 9483 112

3000012 119909

30014 3360 108

0012 1199093

0013 8356 1100012 119909

40013 3039 112

A further analysis has also been performed of Example 10with 119873 = 100 200 300 to compute the corresponding1205720 with 1205721 1205722 that correspond simultaneously to 119888 =108 110 112 as illustrated in Figure 1 Results of thesecomputations are summarized in Table 3 and then used forthe partial 119876

119904120590119873(119909) plots of Figure 2 that interpolate 1205721 1205722

(for 119888 = 108 110 112) with 1205720 for each 119873 = 100 200 300In this table119909

lowaststands for the first convergence step of the 119909

119899

sequence of Newton Raphson iterations

002

05

10

004

g(x

)

xminus004 minus002

minus05

minus10

g(x) for N = 100

g(x) for N = 200

g(x) for N = 300

h(x) = c = 108

h(x) = c = 11

h(x) = c = 112

1205721 1205720 1205722

Figure 1 Plots of 119892(119909) near a discontinuity of 119891(119909) in Example 10

120

115

110

105

100

000 001 002 003 004 005

g(x

)

x

g(x) for N = 100

g(x) for N = 200

g(x) for N = 300

h(x) = c = 108

h(x) = c = 11

h(x) = c = 112

Figure 2 Plots of the 119892(119909) curves near a discontinuity of 119891(119909) inExample 10 interpolating 1205721 1205722 (for 119888 = 108 110 112) with 1205720 allfor119873 = 100 200 300

These results turn out to be extremely stable against varia-tions of the trial initial root1199090 Indeed in the computationallymost difficult situation namely when 119873 = 300 a changeof 1199090 from 0008 to 0020 leads to a variation of 119909

lowastfrom

1199092 to just 1199096 with the same asymptotic 1205720 = 00104648 and1205721 = 000807732 (for 119888 = 112)

5 Conclusions

This paper demonstrates that the quadratic degeneracy ofthe solution of the present inverse problem explicit whenusing 119867(120574 119909119898) is a nonlinear indicator of the existence ofGibbsrsquo phenomenon It provides also a distribution-theoreticproof for the existence of this phenomenon The reportedanalysis of numerical solvability of this problem illustrates thefollowing

(i) When suminfin

119898=1 119898(120587119871)119866lowast(120574 119909119898) = suminfin

119898=1 119898(120587119871)119867lowast(120574 119909119898) converge continuously in 119909 and are

12 International Journal of Computational Mathematics

nonzero at 1205721 (or 1205722) there exists a neighborhood(defined by 119868) of 120572 such that for all starting values 1199090in the neighborhood 119909

119899 will converge to 1205721 (or 1205722)

(ii) When suminfin

119898=1 119898(120587119871)119866lowast(120574 119909119898)=suminfin

119898=1 119898(120587119871)119867lowast(120574119909119898) converge continuously in 119909 and are nonzero at1205721 and 1205722 (but not at 1205720) and if 120593(119909) has a secondderivative which might be zero at 1205721 and 1205722 (and at1205720) then the convergence of 119909119899 is quadratic or fasterIf12059310158401015840(119909) = 0 at1205721 and1205722 then the convergence of 119909119899is merely quadratic

(iii) The convergence of the Newton-Raphson iterativeprocess to 1205720 can also be quadratic but restrictionson the derivatives of 120593(119909) have to be observed up tothe third order Otherwise the convergence of 119909

119899 to

1205720 can slow down to become only linear [12] with arate log210

(iv) The reported computations clearly illustrate that1205720 rarr 0 and |1205721 minus 1205722| rarr 0 when 119873 rarr infin thatis the infinite Fourier series will sum to 119891(119909) inthe neighborhood of a discontinuity except at thediscontinuity itself

Appendix

To illustrate the Gibbs-Wilbraham effect we shall use herethe same arguments employed by Wilbraham in 1848 inanalyzing a similar square wave

Example A1 Let 119891(119909) be a square wave periodic signalwhich is defined over one period by

119891 (119909) =

minus1 minus120587 lt 119909 le 0

1 0 le 119909 lt 120587(A1)

Its Fourier series representation is

119892 (119909) =4120587

infin

sum119898=1

1(2119898 minus 1)

sin (2119898minus 1) 119909 (A2)

This defines the truncated series

119878119873(119909) =

4120587

119873

sum119898=1

1(2119898 minus 1)

sin (2119898minus 1) 119909 (A3)

which may be differentiated as

119889

119889119909119878119873(119909) =

4120587

119873

sum119898=1

cos (2119898minus 1) 119909 (A4)

Since

2119873

sum119898=1

cos (2119899 minus 1) 119909 =sin 2119873119909

sin119909 (A5)

then119889

119889119909119878119873(119909) =

2120587

sin 2119873119909

sin119909 (A6)

It is possible to cancel the previous artificially induceddifferentiation by integrating the last result as

119878119873(119909) = int

119909

0

119889

119889120585119878119873(120585) 119889120585 =

2120587int119909

0

sin 2119873120585

sin 120585119889120585 (A7)

This integral starts at zero when 119909 = 0 in agreement with(A2) and increases until 2119873120585 = 120587 Then 120585 = 1205872119873 definesa turning point 1199091 The integral starts to decrease at whichpoint the numerator sin 2119873120585 goes negative For large 119873 thedenominator sin 120585 remains positive

The form (A7) distinctively from (A3) reveals that119878119873(119909) has turning points at the zeros of sin 2119873120585 These occur

when 2119873120585 = 119903120587 119903 = 1 2 3 4 5 that is at 120585 = 119903(1205872119873)Clearly then 119909

119903sim 120585119903= 119903(1205872119873) are turning points for 119878

119873(119909)

In particular 1199091 sim 1205851 = 1205872119873 defines the overshoot peaklocation 1205720 and 119878

119873(1199091) = (2120587) int12058721198730 (sin 2119873120585 sin 120585) 119889120585 is

its magnitude To evaluate 119878119873(1199091) = 119878

119873(1205720) numerically

let us change the variable of integration namely 2119873120585 = 119908or 120585 = 1199082119873 to rewrite the previous integral as 119878

119873(1199091) =

(2120587) int1205870 (sin119908(sin1199082119873)) 119889(1199082119873) Clearly when 119873 rarr

infin sin1199082119873 = 1199082119873 and

lim119873rarrinfin

119878119873(1199091) =

2120587int120587

0

sin119908119908

119889119908 =2120587Si (120587) = 118 (A8)

correct to two decimal pointsAs for 1199092 1199093 1199094 119909119903 lim119873rarrinfin[119878119873(119909119903) minus 119891(119909)]

decreases as we move away from the discontinuity butit should be remarked however that the location of theovershoot 1205720 = 1199091 = 1205872119873 moves towards the discontinuityaccording to

lim119873rarrinfin

1205720 = lim119873rarrinfin

120587

2119873= 0 (A9)

Conflict of Interests

The author declares that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

The numerical computations of Examples 9 and 10 have beenperformed using Mathematica 8 by Mounir Abou Yassineof the Department of CCE of AUL

References

[1] P L Butzer and R J Nessel Fourier Analysis and Approxima-tion Volume I One-Dimensional Theory Birkhauser 1971

[2] R E Edwards Fourier Series A Modern Introduction HoltRinehart amp Winston 1967

[3] W J Thompson ldquoFourier series and the Gibbs phenomenonrdquoAmerican Journal of Physics vol 60 no 5 pp 425ndash429 1992

[4] A J JerriThe Gibbs Phenomenon in Fourier Analysis Splines ampWavelet Approximations Kluwer Academic Publishers BostonMass USA 1990

International Journal of Computational Mathematics 13

[5] K S Rim and B I Yun ldquoGibbs phenomenon removal by addingHeaviside functionsrdquo Advances in Computational Mathematicsvol 38 no 4 pp 683ndash699 2013

[6] A H Zemanian Distribution Theory and Transform AnalysisMcGraw-Hill 1965

[7] G R Ayers and J C Dainty ldquoIterative blind deconvolutionmethod and its applicationsrdquo Optics Letters vol 13 no 7 pp547ndash549 1988

[8] N H S Haidar ldquoA nonlinear method for a severely ill-posedgeneralized moment problemrdquo Journal of Inverse amp Ill-PosedProblems vol 5 no 4 pp 323ndash335 1997

[9] G L Bradley and K J Smith Calculus Prentice Hall 1995[10] P Offner T Sonar and M Wirz ldquoDetecting strength and

location of jump discontinuities in numerical datardquo AppliedMathematics vol 4 no 12 pp 1ndash14 2013

[11] K E Atkinson An Introduction to Numerical Analysis JohnWiley amp Sons New York NY USA 1989

[12] E Suli and D F Mayers An Introduction to Numerical AnalysisCambridge University Press Cambridge UK 2003

[13] P Wang ldquoA third-order family of Newton-like iteration meth-ods for solving nonlinear equationsrdquo Journal of NumericalMathematics and Stochastics vol 3 no 1 pp 13ndash19 2011

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

4 International Journal of Computational Mathematics

The rest of this section illustrates how stereographicprojection of Fourier series reveals a quadratic degeneracythat indicates the existence of the Gibbs-Wilbraham effect Italso addresses the question of solving the inverse problem forthe truncated Fourier series representations The impact ofstereographic projection on the ill-posedness of the numer-ical solution of this inverse problem is moreover a mainquestion that has been addressed in this paper

Our analysis starts however by illustrating how thetangent half-angle stereographic projection can provide anew insight into the behavior of truncated Fourier series neara discontinuity of 119891(119909)

Lemma 3 The solution 119909 of the inverse problem for thetruncated Fourier series representation satisfies

119873

sum119898=1

119867(120574 119909119898)

=119873

sum119898=1

(1198862119898+ 1198872119898) minus (119886

119898tan119898(1205872119871) 119909 minus 119887

119898)2

119886119898(tan2119898(1205872119871) 119909 + 1)

= 119888 minus11988602

(22)

Proof By studying solvability for 119909 of (18) Let 119905119898

=tan119898(1205872119871)119909 and then cos119898(120587119871)119909 = (1minus1199052

119898)(1+1199052

119898) and

sin119898(120587119871)119909 = 2119905119898(1 + 1199052

119898) Further consideration of these

facts in (18) leads to the tangent half-angle stereographicallytransformedS[ ] = [] form

minus119873

sum119898=1

S [119866 (120574 119909119898)] =119873

sum119898=1

(1198861198981199052119898minus 2119887119898119905119898minus 119886119898)

(1199052119898+ 1)

=11988602minus 119888

(23)

The harmonic trinomial in the left-hand side of (23) hastwo harmonic zeros 120585plusmn

119898 for each 119909 namely

120585plusmn119898=

2119871119898120587

tanminus1119905plusmn119898

119905plusmn119898=

119887119898

119886119898

plusmn120590119898

120590119898=

radic1198862119898+ 1198872119898

119886119898

(24)

It is worth noting at this point that these 120585plusmn119898zeros should

by no means be confused with the possible roots of the 119909variable Substitution then of (24) in (23) leads to

119873

sum119898=1

119886119898(119905119898minus 119905+119898) (119905119898minus 119905minus119898)

(1199052119898+ 1)

=11988602minus 119888 (25)

which is the same as

119873

sum119898=1

119886119898(119905119898minus 119887119898119886119898minus 120590119898) (119905119898minus 119887119898119886119898+ 120590119898)

(1199052119898+ 1)

=11988602minus 119888

(26)

The quadratic degeneracy of the harmonic zeros 120585plusmn119898of 119905119898

can disappear only when 120590119898= 0 for all119898 And this happens

to be equivalent with

119873

sum119898=1

1198862119898+ 1198872119898= 0 (27)

Obviously satisfaction of (27) by (26) reduces it to theequation

119873

sum119898=1

119886119898(119905119898minus 119887119898119886119898)2

(1199052119898+ 1)

=11988602minus 119888 (28)

which is the same as

119873

sum119898=1

(119886119898tan119898(1205872119871) 119909 minus 119887

119898)2

119886119898(tan2119898(1205872119871) 119909 + 1)

=11988602minus 119888 (29)

It is interesting to remark that condition (27) is entirelydifferent from

11988602+119873

sum119898=1

1198862119898+ 1198872119898gt 0 (30)

which can be derived from the Parseval identity [1] (com-pleteness relation)

11988602+infin

sum119898=1

1198862119898+ 1198872119898=

1119871int119871

minus119871

1198912(119909) 119889119909 (31)

Hence a possible nondegenerate solution for 119909 of the nonlin-ear equation (29) is on one hand fundamentally incorrectbecause it involves a violation of (30) and Parsevalrsquos identityOn the other hand a consistent solution 119909 in the senseof satisfaction of (31) is intrinsically degenerate because ofits emergence as a must from the nonlinear equation (26)which rewrites as

119873

sum119898=1

119886119898[(119905119898minus 119887119898119886119898)2minus 1205902119898]

(1199052119898+ 1)

=11988602minus 119888 (32)

Equation (32) is easily transformable to the requiredresult (23) here the proof completes

32 The Stereographic Fourier Series To further analyze thevariability of stereographically projected Fourier series wemake the substitution

120579119898= 120579119898(119909) = 119898

120587

119871119909 (33)

International Journal of Computational Mathematics 5

in 119866(120574 119909119898) to rewrite it as

119866 (120574 119909119898) = 119886119898cos 120579119898(119909) + 119887

119898sin 120579119898(119909) (34)

The tangent half-angle (stereographic) transformation119905119898= tan(120579

1198982) or 120579

119898= 2tanminus1119905

119898is conceived as

120579119898(119909) = 2tanminus1119905

119898(119909) = Φ [119905

119898(119909)] (35)

and subsequently

119866 = 119866 [120579119898(119909)] = 119866 Φ [119905

119898(119909)] = 119867 [119905

119898(119909)] (36)

It is clear then that this stereographic projection effectivelyleads to insertion of an additional bracket into the compo-sition of 119866 as a function of 119909 This paves the way to thedefinition

S 119866 [120579119898(119909)] = 119866 Φ [119905

119898(119909)] = 119867 [119905

119898(119909)]

= 119867 (120574 119909119898) (37)

Proposition 4 The following

S119889

119889119909

119873

sum119898=1

119866 [120579119898(119909)] =

119889

119889119909S119873

sum119898=1

119866 [120579119898(119909)] (38)

always holds

Proof Since 119873 is finite then the proof of S(119889119889119909)119866 =(119889119889119909)S119866 = (119889119889119909)119867 implies the correctness of thisproposition So let us differentiate 119866(119909) = 119866[120579

119898(119909)] with

respect to 119909 by applying the chain rule namely

119889119866

119889119909=

119889119866

119889Φ

119889Φ

119889119905119898

119889119905119898

119889119909=

119889119866

119889120579119898

119889120579119898

119889119905119898

119889119905119898

119889119909 (39)

which is the same as

119889119866

119889119909

= [minus119886119898sin 120579119898+ 119887119898cos 120579119898]

2(1 + 1199052119898)

120579119898

2sec2 (

120579119898

2)

(40)

Take then the S transformation of the previous relation toarrive after some algebra at

S119889

119889119909119866 = minus119898

120587

119871

(1198871198981199052119898+ 2119886119898119905119898minus 119887119898)

(1 + 1199052119898)

(41)

Invoke then

119867 = 119867(120574 119909119898) =(119886119898+ 2119887119898119905119898minus 1198861198981199052119898)

(1 + 1199052119898)

(42)

to differentiate it with respect to 119909 as

119889

119889119909119867 =

119889

119889119909S 119866 = minus119898

120587

119871

(1198871198981199052119898+ 2119886119898119905119898minus 119887119898)

(1 + 1199052119898)

(43)

which is the required result

Let us rewrite (1) in the form

120593 (119909) = 119892 (119909) minus 119888 = 0 (44)

then try to differentiate it term-by-term as

1205931015840 (119909) =119889

119889119909119892 (119909) =

119889

119889119909

infin

sum119898=1

119866 (120574 119909119898)

= minusinfin

sum119898=1

119898120587

119871(minus119887119898cos119898120587

119871119909+ 119886119898sin119898120587

119871119909)

= minusinfin

sum119898=1

119898120587

119871119866lowast (120574 119909119898)

(45)

where 119866lowast(120574 119909119898) is the harmonic conjugate of 119866(120574 119909119898)The trigonometric series 119892lowast(119909) = sum

infin

119898=1 119866lowast(120574 119909119898) is called

the trigonometric series conjugate (see eg [10]) to 119892(119909)In addition to the fact that the convergence of 119892lowast(119909)

can sometimes become worse than convergence of 119892(119909) (seeeg [10]) the presence of the 119898(120587119871) factor in (119889119889119909)119892(119909)can aggravate the convergence problem of (45) when thecoefficients 119886

119898and 119887119898do not decay fast enough

Next we may stereographically transform (120597120597119909)119892(119909)using the notationS[(119889119889119909)119892(119909)] = [(119889119889119909)119892(119909)] where

[119889

119889119909119892 (119909)] = minus

infin

sum119898=1

119898120587

119871119866lowast (120574 119909119898)

= minusinfin

sum119898=1

119898120587

119871

(1198871198981199052119898+ 2119886119898119905119898minus 119887119898)

(1 + 1199052119898)

(46)

Similarly for the stereographic Fourier series representation

119892 (119909) =11988602+infin

sum119898=1

119867(120574 119909119898) (47)

in which

119867(120574 119909119898) =(119886119898+ 2119887119898119905119898minus 1198861198981199052119898)

(1 + 1199052119898)

(48)

we may approximate (44) by

120595 (119909) = 119892 (119909) minus 119888 = 0 (49)

Differentiate then (49) term-by-term as

1205951015840 (119909) =119889

119889119909119892 (119909) =

119889

119889119909

infin

sum119898=1

119867(120574 119909119898)

= minusinfin

sum119898=1

119898120587

119871119867lowast (120574 119909119898)

= minusinfin

sum119898=1

119898120587

119871

(1198871198981199052119898+ 2119886119898119905119898minus 119887119898)

(1 + 1199052119898)

(50)

6 International Journal of Computational Mathematics

Note here that119867lowast (120574 119909119898)

=(1198871198981199052119898+ 2119886119898119905119898minus 119887119898)

(1 + 1199052119898)

=(119887119898tan119898(1205872119871) 119909 minus 119886

119898)2minus (1198862119898+ 1198872119898)

119887119898(tan2119898(1205872119871) 119909 + 1)

(51)

is the harmonic conjugate of119867(120574 119909119898) and that

119867lowast (120574 119909119898) = 119866lowast (120574 119909119898) (52)

This rather interesting fact leads to a remark that follows

Remark 5 Away from points of discontinuity of 119891(119909) thestereographically transformed differentiated Fourier series

[(119889119889119909)119892(119909)] which defines 119866lowast(120574 119909119898) is the same as thederivative (119889119889119909)119892(119909) which defines 119867lowast(120574 119909119898) of thestereographic Fourier series 119892(119909) Moreover even at pointsof discontinuity of 119891(119909) [(119889119889119909)119892(119909)] = (119889119889119909)119892(119909)

Theorem 6 (see [1]) Let 119891(119909) be (i) continuous on [minus119871 119871]such that (ii) 119891(minus119871) = 119891(119871) and let (iii) 1198911015840(119909) be piecewisecontinuous over [minus119871 119871] then the corresponding Fourier series119892(119909) of (1) or (47) is differentiable at each point where (iv)11989110158401015840(119909) exists

Theprevious remark can incidentally have useful applica-tions when differentiating certain slowly converging Fourierseries as illustrated by the following example

Example 7 Consider the 2120587-periodic odd signal 119891(119909)defined over one period by 119891(119909) = 119909 Its Fourier series isknown to be

119892 (119909) = 2infin

sum119898=1

(minus1)119898+1

119898sin119898119909 (53)

that is with 119886119898= 0 forall119898 119887

119898= 2((minus1)119898+1119898) and 119898(120587119871) =

119898Differentiating term-by-term we obtain

1198921015840 (119909) = 2infin

sum119898=1

(minus1)119898+1 cos119898119909 (54)

which is not uniformly convergent because 119891(119909) is not onlydiscontinuous at 119909 = plusmn120587 but 119891(minus120587) = 119891(120587) and 11989110158401015840(plusmn120587) doesnot exist In particular at 119909 = 0 the sum of the series 1198921015840(0) =2suminfin119898=1(minus1)

119898+1 oscillates between 0 and 1 that is divergent Itis also well known nevertheless that the seriessuminfin

119898=1(minus1)119898+1

is 119862 minus 1 summable [1 2] to 12 and that turns out to yieldthe right answer 1198921015840(0) = 1198911015840(0) = 1 Moreover at 119909 = 120587 thesum of the series 1198921015840(120587) = 2suminfin

119898=1(minus1)2119898+1 rarr minusinfin that is

divergentAlternatively by means of (47)

119892 (119909) =infin

sum119898=1

119887119898

2119905119898

1 + 1199052119898

= 2infin

sum119898=1

(minus1)119898+1

119898sin119898119909 (55)

and by means of (46)

1198921015840 (119909) =

[119889

119889119909119892 (119909)] = minus

infin

sum119898=1

119898119866lowast (120574 119909119898)

= minusinfin

sum119898=1

119898119887119898

1199052119898minus 1

1 + 1199052119898

= 2infin

sum119898=1

(minus1)119898+1(1 minus tan2119898(12) 119909)(1 + tan2119898(12) 119909)

= 2infin

sum119898=1

(minus1)119898+1 cos119898119909

(56)

Here at 119909 = 120587 the sum of the series 1198921015840(120587) =

2suminfin119898=1(minus1)

2119898+1 rarr minusinfin is also divergent

Note in general that for even signals that is when 119887119898= 0

forall119898 it follows from (22) that

119867(120574 119909119898)1003816100381610038161003816119887119898=0

= 119886119898

1 minus 1199052119898

1 + 1199052119898

= 119886119898cos119898120587

119871119909

= 119866 (120574 119909119898)1003816100381610038161003816119887119898=0

(57)

Moreover for odd signals that is when 119886119898

= 0 forall119898 thesituation is not of the same clarity

Indeed according to (22)

119867(120574 119909119898)1003816100381610038161003816119886119898=0

=00

119887119898

1 + 1199052119898

(58)

which is an uncertainty Luckily however this uncertainty isnot an essential one and can straightforwardly be resolved byresorting to the stereographic form (23) of (22) which yieldsthe correct result

119867(120574 119909119898)1003816100381610038161003816119886119898=0

= 119887119898

2119905119898

1 + 1199052119898

= 119887119898sin119898120587

119871119909

= 119866 (120574 119909119898)1003816100381610038161003816119886119898=0

(59)

Also when 119887119898= 0 forall119898

119867lowast (120574 119909119898)1003816100381610038161003816119887119898=0

=00

119886119898

1 + 1199052119898

(60)

and this is resolvable in a similar fashionBoth119866(120574 119909119898) of (2) and119867(120574 119909119898) of (22) equivalent

forms are nonlinear trigonometric functions of the 119909 solu-tion However while 119866(120574 119909119898) is linear in 120574 = 119886

119898 119887119898infin119898=1

thestereographic 119867(120574 119909119898) is nonlinear in this 120574 This canraise a question on a possibility for (120597120597119909)119892(119909) = (120597120597119909)119892(119909)at singular points of 119891(119909) A question that will be answerednegatively in the next section Furthermore since the data 120574 =119886119898 119887119898infin119898=1 is discrete then the operators sum

infin

119898=1 119866(120574 119909119898)and sum

infin

119898=1 119867(120574 119909119898) do not depend continuously on 120574 andthis is themain reason for ill-posedness of the present inverseproblem

International Journal of Computational Mathematics 7

In conclusion the quadratic degeneracy of the solution ofthe present inverse problem explicit when using 119867(120574 119909119898)is a nonlinear indicator of the existence of the peak 119876

119904120590119873(119909)

pertaining to the Gibbs phenomenon However the numer-ical ill-posedness associated with the relative values of atrial iterative variable in the 120574 set happens to remarkablyremain invariant under the tangent half-angle stereographicprojection

Following the same arguments of Wilbraham we maycancel the artificially induced differentiation by integrating(50) to represent the stereographic Fourier series 119892(119909) inintegral form as

119892 (119909) =11988602minusinfin

sum119898=1

int119909

0119898

sdot120587

119871

(119887119898tan119898(1205872119871) 120585 minus 119886

119898)2minus (1198862119898+ 1198872119898)

119887119898(tan2119898(1205872119871) 120585 + 1)

119889120585 =11988602

minusinfin

sum119898=1

int119909

0119898

sdot120587

119871

(119887119898tan2119898(1205872119871) 120585 minus 2119886

119898tan119898(1205872119871) 120585 minus 119887

119898)

(tan2119898(1205872119871) 120585 + 1)119889120585

(61)

4 Numerics of Solving the Inverse Problem

The solution of the present inverse problem is in fact aproblem of finding the roots of (44) when 119892(119909) is truncatedat 119898 = 119873 For that reason we shall focus on such rootsin the neighborhood of a step-like discontinuity of 119891(119909) at119909 = 119909119900 say where 119891(119909119900 minus 120576) lt 119891(119909119900 + 120576) 0 lt 120576 ≪ 1 Theassociated with the Gibbs phenomenon overshoot of 119892(119909) =11988602+sum

119873

119898=1 119866(120574 119909119898) or 11988602+sum119873

119898=1 119867(120574 119909119898) over119891(119909119900+120576)and undershoot of this 119892(119909) below 119891(119909119900minus120576)The intersectionof a horizontal line ℎ(119909) = 119888 ge 119891(119909119900 + 120576) corresponding to120593(119909) = 0 can be at two points 1205721 and 1205722 on both sides of thepeak location 1205720 representing two distinct roots of 120593(119909) = 0It can also be at one point 1205720 corresponding to themaximumof the truncated 119892(119909) that is a double root of 120593(119909) = 0

Obviously 1205721 lt 1205721 lt 1205720 lt 1205722 lt 1205722 where 1205721 and 1205722 arepossible points of inflection defined by

11989210158401015840 (1205721) = 12059310158401015840 (1205721) = 11989210158401015840 (1205722) = 12059310158401015840 (1205722) = 0 (62)

The existence of both 1205721 and 1205722 or at least one of themnamely 1205722 is however not mandatory Moreover for the 12057211205720 and 1205722 points we expect to have

1198921015840 (1205721) = 1205931015840 (1205721) gt 0

12059310158401015840 (1205721) gt 0

120593101584010158401015840 (1205721) lt 0

1198921015840 (1205720) = 1205931015840 (1205720) = 0

12059310158401015840 (1205720) lt 0

120593101584010158401015840 (1205720) = 0

1198921015840 (1205722) = 1205931015840 (1205722) lt 0

12059310158401015840 (1205722) gt 0

120593101584010158401015840 (1205722) gt 0(63)

Equation (44) can be solved for119909 120593(119909) = 0 by aNewton-Raphson (see eg [11 12]) iterative process

119909119899+1 = 119909

119899minus

120593 (119909119899)

1205931015840 (119909119899) (64)

which is repeated until a sufficiently accurate value is reachedThis process for each series truncation number119873 rewrites as

119909119899+1 = 119909

119899+[sum119873

119898=1 119866 (120574 119909119899 119898) + 11988602 minus 119888]

[sum119873

119898=1 119898(120587119871)119866lowast (120574 119909119899 119898)]

(65)

or in the stereographic form

119909119899+1 = 119909

119899+[sum119873

119898=1 119867(120574 119909119899 119898) + 11988602 minus 119888]

[sum119873

119898=1 119898(120587119871)119867lowast (120574 119909119899 119898)]

(66)

which is the same as

119909119899+1 = 119909

119899

+[sum119873

119898=1 ((119886119898 + 2119887119898119905119898minus 1198861198981199052119898) (1 + 1199052

119898)) + 11988602 minus 119888]

[sum119873

119898=1 119898(120587119871) ((1198871198981199052119898+ 2119886119898119905119898minus 119887119898) (1 + 1199052

119898))]

(67)

and is started off with some rather free initial value 1199090Obviously here 119905

119898= tan119898(1205872119871)119909

The sequence 119909119899will usually converge provided that 1199090

is close enough to the unknown zero 120572 and that

1205931015840 (1199090) = minusinfin

sum119898=1

119898120587

119871119866lowast (120574 1199090 119898) = 0

or 1205951015840 (1199090) = minusinfin

sum119898=1

119898120587

119871119867lowast (120574 1199090 119898) = 0

(68)

41 Computation of the Peak Location 1205720 Being a maximiza-tion problem for 120593(119909) = 0 the determination of 1205720 by theNewton-Raphson process [11] becomes

119909119899+1 = 119909

119899minus

1205931015840 (119909119899)

12059310158401015840 (119909119899)

(69)

or

119909119899+1 = 119909

119899minus

1205951015840 (119909119899)

12059510158401015840 (119909119899) (70)

in stereographic form

8 International Journal of Computational Mathematics

Straightforwardly this process can be expressed as

119909119899+1 = 119909

119899minus

[sum119873

119898=1 119898119866lowast (120574 119909119899 119898)]

[sum119873

119898=1 1198982 (120587119871) 119866 (120574 119909

119899 119898)]

(71)

or in the stereographic form

119909119899+1 = 119909

119899

minus[minussum119873

119898=1 119898(120587119871)119867lowast (120574 119909119899 119898)]

[sum119873

119898=1 1198982 (12058721198712) ((119886

1198981199052119898minus 2119887119898119905119898minus 119886119898) (1 + 1199052

119898))]

(72)

which is the same as119909119899+1 = 119909

119899

+[sum119873

119898=1 119898((1198871198981199052119898+ 2119886119898119905119898minus 119887119898) (1 + 1199052

119898))]

[sum119873

119898=1 1198982 (120587119871) ((119886

1198981199052119898minus 2119887119898119905119898minus 119886119898) (1 + 1199052

119898))]

(73)

It is remarkable how this algorithm is distinctively indepen-dent of 119888

Proposition 8 The ill-posedness of the inverse problem ofthe truncated Fourier series is invariant under stereographicprojection

Proof The tangent half-angle stereographic projectionthough not isometric (distance preserving) is neverthelessvariability preserving Indeed the main ingredients ofthis inverse problem are namely 120574 = 119886

119898 119887119898119873119898=1 119888 and

1199090 with (119889119889119909)119866[120579119898(119909)]1199090 Under this projection 120579

119898(1199090)

transforms to 119905119898(1199090) = tan[120579

119898(1199090)2] and 119866[120579

119898(1199090)]

transforms to 119867[119905119898(1199090)] But according to Proposition 4

(119889119889119909)119867[119905119898(119909)]1199090

= (119889119889119909)119866[120579119898(119909)]1199090 Hence the variation

of 119866(120574 1199090 119898) with 1199090 in the Fourier series inverse problemand the variation of119867(120574 1199090 119898) with 1199090 in the stereographicform are identical

Equivalently this makes the ill-posedness of this inverseproblem invariant under stereographic projection

Expected manifestations of the correctness of this propo-sition are as follows (i) the results of computations by the(71) and (73) algorithms for 1205720 with the same 1199090 must beidentical and (ii) the results of computations by the (65) and(67) algorithms for 1205721 (or 1205722) with the same 1199090 and 119888 mustalso be identical

42 Convergence Analysis Let us focus first on the con-vergence of the iterative process (65) or (67) In manysituationswhere term-by-termdifferentiability does not holdtheNewton-Raphsonmethod can always be replaced by somequasi-Newtonmethod [12 13] where 1205931015840(119909

119899) or1205951015840(119909

119899) in (65)

or (67) is replaced by a suitable approximation For examplein the chord method 1205931015840(119909

119899) is replaced by 1205931015840(1199090) for all

iterationsIf 120598119899= 120572minus119909

119899 120598119899+1 = 120572minus119909

119899+1 and120573119899 is some point between119909119899and 120572 it can be easily proved (see [11] or [12]) that

120598119899+1 = minus

12059310158401015840 (120573119899)

21205931015840 (119909119899)1205982119899 (74)

Taking absolute value of both sides gives

1003816100381610038161003816120598119899+11003816100381610038161003816 =

1003816100381610038161003816100381612059310158401015840 (120573119899)10038161003816100381610038161003816

2 10038161003816100381610038161205931015840 (119909119899)10038161003816100381610038161205982119899 (75)

implying quadratic convergence that is 120598119899is squared (the

number of correct digits roughly doubles) at each step Thesame arguments apply of course to the 120595(119909) = 0 equation

In actual fact for a distinct 120572 the convergence of119909119899 in (65)ndash(67) is at least quadratic in the neigh-

borhood of this 120572 The process may face a difficultywhen 1205931015840(119909

119899) = minussum

119873

119898=1 119898(120587119871)119866lowast(120574 119909119899 119898) or 1205951015840(119909

119899) =

minussum119873

119898=1 119898(120587119871)119867lowast(120574 119909119899 119898) diverges This is however a spe-

cial but not a general problem and (75) always holds if thefollowing conditions are satisfied

(i) 1205931015840(119909) = 0 forall119909 isin 119868 where 119868 = [120572 minus 120578 120572 + 120578] for some120578 ge |(120572 minus 1199090)|

(ii) 12059310158401015840(119909) is finite forall119909 isin 119868(iii) 1199090 is sufficiently close to 120572

The term sufficiently close in this context means the follow-ing

(a) Taylor approximation is accurate enough such that wecan ignore higher order terms

(b) (12)|12059310158401015840(119909119899)1205931015840(119909

119899)| lt 119870|12059310158401015840(120572)1205931015840(120572)| for some119870 lt

infin(c) 119870|12059310158401015840(120572)1205931015840(120572)|120598

119899lt 1 for 119899 isin N

Consequently (75) can be expressed in the following way1003816100381610038161003816120598119899+1

1003816100381610038161003816 = 1198721205982119899 (76)

where

119872 = sup119909isin119868

12

100381610038161003816100381610038161003816100381610038161003816

12059310158401015840 (119909)

1205931015840 (119909)

100381610038161003816100381610038161003816100381610038161003816 (77)

The initial point 1199090 has to be chosen such that conditions(i) through (iii) are satisfied where the third conditionrequires that 119872|1205980| lt 1 Moreover with increasing 119873 thedistance |1205721 minus 1205722| is known to decrease in a way makingthe satisfaction of these conditions very difficult and actuallyimpossible as119873 rarr infin Here it should be underlined that theiterative method (65)ndash(67) may fail to converge to the 120572 rootin the following situations

(i) If a point like 1205720 is included in 119868 the method willterminate due to division by zero

(ii) When the initial estimate 1199090 is poor the pertainingwrong 119868 can contribute to nonconvergence of thealgorithm

As for convergence of the Newton-Raphson processtowards 1205720 let 120576119899 = 1205720 minus 119909

119899 120576119899+1 = 1205720 minus 119909

119899+1 and 120579119899is some

point between 119909119899and 1205720 to arrive at

120576119899+1 = minus

120593101584010158401015840 (120579119899)

212059310158401015840 (119909119899)1205762119899 (78)

1003816100381610038161003816120576119899+11003816100381610038161003816 =

10038161003816100381610038161003816120593101584010158401015840 (120579119899)10038161003816100381610038161003816

2 100381610038161003816100381612059310158401015840 (119909119899)10038161003816100381610038161205762119899 (79)

International Journal of Computational Mathematics 9

The convergence of 119909119899 in (71)ndash(73) is at least quadratic

in the neighborhood of this 1205720 The process may face a diffi-culty however when 12059310158401015840(119909

119899) = minussum

infin

119898=1 1198982(120587119871)2119866(120574 119909

119899 119898)

or 12059510158401015840(119909119899) diverges Moreover (79) always holds if the follow-

ing conditions are satisfied

(i) 12059310158401015840(119909) = 0 forall119909 isin 119868 where 119868 = [1205720 minus120578 1205720 +120578] for some120578 ge |(1205720 minus 1199090)|

(ii) 120593101584010158401015840(119909) is finite forall119909 isin 119868

(iii) 1199090 is sufficiently close to 1205720

Sufficient closeness in this context means that

(a) Taylor approximation is accurate enough such that wecan ignore higher order terms

(b) (12)|120593101584010158401015840(119909119899)12059310158401015840(119909

119899)| lt 119870|120593101584010158401015840(1205720)120593

10158401015840(1205720)| for some119870 lt infin

(c) 119870|120593101584010158401015840(1205720)12059310158401015840(1205720)|120576119899 lt 1 for 119899 isin N

Consequently

1003816100381610038161003816120576119899+11003816100381610038161003816 = 1198721205762

119899 (80)

where

119872 = sup119909isin119868

12

100381610038161003816100381610038161003816100381610038161003816

120593101584010158401015840 (119909)

12059310158401015840 (119909)

100381610038161003816100381610038161003816100381610038161003816 (81)

43 Applications The standard iterative algorithms (71) or(73) for finding the location of overshoots in 119876

119904120590119873(119909) of

the Gibbs effect employ the fact that the first derivative of apeak in 119892(119909) or 119892(119909) respectively has a downward-going-zero-crossing at the peak maximum But the presence ofmany neighboring smaller peaks in the ringing close to theovershoot will cause many undesirable zero-crossings to beinvoked if the initial trial root 1199090 is not correctly chosenThis fact makes the numerical problem of peak finding in theGibbs effect of pathological difficulty A difficulty that aggra-vates with the increase in the number 119873 of truncated termsin the employed Fourier series Indeed with the increaseof 119873 the number of undesirable zero-crossings increasesin the neighborhood of 1205720 as illustrated in Section 1 Thiscan increasingly lead to oscillatory and even unstable valuesin sequence 119909

119899 the more the distance |1199090 minus 1205720| contains

unwanted zero-crossingsThese problems have been resolved in the two examples

that follow which simultaneously illustrate the applicabilityof Proposition 8

Example 9 Consider the 2120587-periodic odd signal 119891(119909) ofExample A1 defined over one period by

119891 (119909) =

minus1 minus120587 lt 119909 lt 0

1 0 le 119909 lt 120587(82)

which has for each truncation number 119873 the truncatedFourier series approximations

119892 (119909) ≃4120587

119873

sum119898=1

1(2119898 minus 1)

sin (2119898minus 1) 119909

119892 (119909) ≃8120587

119873

sum119898=1

119887119898119905119898

(1 + 1199052119898)=

8120587

119873

sum119898=1

1(2119898 minus 1)

119905119898

(1 + 1199052119898)

=8120587

119873

sum119898=1

1(2119898 minus 1)

tan ((2119898 minus 1) 2) 1199091 + [tan ((2119898 minus 1) 2) 119909]2

(83)

This signal has a jump discontinuity at 119909 = 0 ofmagnitude 2 We shall focus our computations on the Gibbsphenomenon overshoot near this discontinuity

Here is a printout list of results for the parameters of theposing inverse problem via computations by both the directand stereographic algorithms

It is remarkable how results by both algorithms arepractically identical and share the same following featuresthat are observed by results of the direct algorithm discussedbelow

Case I (119873 = 40) The computations of 1205720 for the previousexample using (71) with 1199090 = 002 lead to 1199091 = 0039 109this varies quickly with 119899 (only beyond the 4th decimal) tostabilize towards 1199093 = 1199094 = 1199095 = sdot sdot sdot = 0039 299 = 1205720Here using (1) gives 119892(1205720) = 119892(0039 299) = 1179 030 whichis quite close to the theoretically expected value by (A7) ofsim118

Then using (65) for 119888 = 109 and 1199090 = 0000381 leads to1199091 = 0021 697 this varies quickly with 119899 (only beyond the 3-d decimal) to stabilize quickly towards 1199095 = 1199096 = 1199097 = sdot sdot sdot =0028 368 = 1205721 Also using (65) for 119888 = 109 and 1199090 = 004leads to 1199091 = 0067 699 this varies quickly with 119899 (beyondthe 2nd decimal) to stabilize quickly towards 1199095 = 1199096 = 1199097 =sdot sdot sdot = 0052 873 = 1205722

For both1205721 and1205722 using (1) yields119892(1205721) = 119892(0028368) =119892(1205722) = 119892(0052873) = 109 = 119888 as expected

Case II (119873 = 100) Computations of 1205720 using (71) with1199090 = 002 lead to 1199091 = 0014 221 this varies quickly with119899 (only beyond the 4th decimal) to stabilize towards 1199093 =1199094 = 1199095 = sdot sdot sdot = 0015 708 = 1205720 Here using (1) gives119892(1205720) = 119892(0015708) = 1178 990 This result appears to beslightly excessive perhaps due to increased rounding off errorswith increasing 119873 to 100

Thenusing (65) for 119888 = 109 and1199090 = 0015 237 leads aftera negative value to 1199092 = 0014 462 this varies with 119899 (beyondthe 2nd decimal) to stabilize quickly towards 1199097 = 1199098 = 1199099 =sdot sdot sdot = 0011 3348 = 1205721 The used value of 1199090 is apparently notappropriate and this illustrates the generic ill-posedness of thisinverse problem

The pertaining computations of 1205722 with 1199090 = 0017000lead to 1199091 = 0020 434 this varies with 119899 (beyond the 3-ddecimal) to stabilize quickly towards 1199094 = 1199095 = 1199096 = sdot sdot sdot =0021 149 = 1205722

10 International Journal of Computational Mathematics

For both 1205721 and 1205722 using (1) yields 119892(1205721) =119892(00113348) = 119892(1205722) = 119892(0021149) = 109 = 119888 asexpected

Case III (119873 = 150) The computations of 1205720 using (71) with1199090 = 0005 lead to 1199091 = 0010 595 this varies insignificantlywith 119899 (only beyond the 4th decimal) to stabilize towards1199093 =1199094 = 1199095 = sdot sdot sdot = 0010 472 = 1205720 Here using (1) gives 119892(1205720) =119892(0010472) = 1178 980 which does not significantly differfrom 119892(1205720) when119873 = 100

Thenusing (65) for 119888 = 109 and1199090 = 0010 158 leads aftera negative value to 1199092 = 0009 643 and this varies (beyond the3-d decimal) to stabilize quickly towards1199097 = 1199098 = 1199099 = sdot sdot sdot =0007 565 = 1205721

The pertaining computations of 1205722 with 1199090 = 0010 800lead to 1199091 = 0025 998 this varies (beyond the 2nd decimal)to stabilize towards 11990914 = 11990915 = 11990916 = 11990917 = 11990918 = sdot sdot sdot =0014 099 = 1205722

For both 1205721 and 1205722 using (1) yields 119892(1205721) = 119892(007 565) =119892(1205722) = 119892(0014 099) = 109 = 119888

During the previous computations summarized inTable 1 it has been observed that varying 1199090 in the presentNewton-Raphson process for all119873 can lead to strong oscil-lations and even completely unstable values for 1205720 1205721 and 1205722of the Fourier series inverse problem which though ill-posed

is practically regularizible Apart from this drawback the(71) with (65) iterative algorithms (i) converge quickly to thetheoretically expected values and are (ii) of demonstratedstability of the iterations for properly chosen 1199090 (which act aseffective regularizing parameters) Moreover computationsby the stereographic algorithm invoking (73) and (67) sharethe samemagnitude and convergence rates with results by thedirect algorithm Finally it is demonstrated that the presentsignal exhibits an overshoot when119873 = 150 ofsim9of a jumpdiscontinuity of magnitude 2 that is located at a distance ofsim00105

To illustrate robustness of both numerical algorithms andtheir possible effective regularizibility we shall study also thenumerical inverse problem for the Fourier series representa-tion of an arbitrary (nonsymmetric) periodic signal

Example 10 Consider the 2120587-periodic signal 119891(119909) definedover one period by

119891 (119909) =

minus119909

120587minus 1 minus120587 lt 119909 lt 0

1 0 le 119909 lt 120587(84)

It has for each truncation number 119873 the truncated Fourierseries approximations

119892 (119909) ≃14+119873

sum119898=1

[(minus1)119898 minus 1]12058721198982 cos119898119909+

[2 minus (minus1)119898]120587119898

sin119898119909

119892 (119909) ≃14

+119873

sum119898=1

[(minus1)119898 minus 1] 12058721198982 + 2 ([2 minus (minus1)119898] 120587119898) tan ((2119898 minus 1) 2) 119909 minus ([(minus1)119898 minus 1] 12058721198982) [tan ((2119898 minus 1) 2) 119909]2

1 + [tan ((2119898 minus 1) 2) 119909]2

(85)

This signal has like the signal of Example 9 a jumpdiscontinuity at 119909 = 0 of magnitude 2 For the sake ofcomparison we shall also focus our computations on theGibbs phenomenon overshoot part of 119876

119904120590119873(119909) near this

discontinuityIt should be noted here that the results by both algorithms

summarized in Table 2 are also identical a feature indicatingthat the degree of ill-posedness of the posing inverse problemis not affected as predicted by Proposition 8 by the tangenthalf-angle stereographic projection

Case I (119873 = 40) The computations of 1205720 for this exampleusing (71) and (73) with 1199090 = 003 lead to an identicalsequence 119909

119899 that converges by both algorithms after the 1199094

iteration to 0078 139 with 119892(1205720) = 119892(1205720) = 1178 76

Case II (119873 = 100) The computations of 1205720 with 1199090 = 001lead to an identical sequence 119909

119899 that converges by both

algorithms after the 1199095 iteration to 0031 351 with 119892(1205720) =

119892(1205720) = 1178 87

Case III (119873 = 150) The computations of 1205720 with 1199090 = 0007lead also to an identical sequence 119909

119899 that converges by both

algorithms after the 1199095 iteration to 0020 915 with 119892(1205720) =

119892(1205720) = 1178 91

Therefore the present signal exhibits an overshoot when119873 = 150 of sim9 of a jump discontinuity of magnitude 2 thatis located at a distance of sim0020 9

Comparison of the computed Gibbs overshoot parame-ters of the two considered signals with jump discontinuitiesat the same 119909 = 0 of the same magnitude illustrates that therelative size of the overshoot depends essentially exclusivelyon the magnitude of the stimulating jump discontinuityFor instance based on the present numerical results onecan suggest here that nonsymmetric signals have equalovershoots with symmetric signals Their respective 1205720rsquos arehowever different Indeed the nonsymmetric signal has asignificantly larger 1205720 (0020 9) than the 1205720 (00105) of thesymmetric signal

International Journal of Computational Mathematics 11

Table 1 Results of computations for Example 9

119873 1205721 1205720 1205722

40 0028 368 0039 299 0052 873100 0011 335 0015 705 0021 149150 0007 565 0010 472 0014 099

Table 2 Results of computations for Example 10

119873 1205721 1205720 1205722

40 00564765 00781391 0105155100 00226546 00313516 00422024150 00151124 00209153 0028156

Table 3 Summary of computations of all overshoot parameters inExample 10

1205720results

119873 1199090

119909lowast

1205720

119892(1205720)

100 0030 1199093

0031 3516 1178 87200 0015 119909

30015 6919 1178 92

300 0008 1199093

0010 4648 1178 941205721results

119873 1199090

119909lowast

1205721

119888

0025 1199094

0022 2027 108100 0025 119909

30023 1344 110

0025 1199093

0024 2027 1120010 119909

30011 1117 108

200 0010 1199093

0011 5779 1100010 119909

40012 1123 112

0008 1199093

0007 410 08 108300 0008 119909

30007 720 95 110

0008 1199092

0008 077 32 1121205722results

119873 1199090

119909lowast

1205722

119888

100035 119909

40042 9439 108

035 1199094

0041 4445 110035 119909

40039 8513 112

2000018 119909

30021 4960 108

0018 1199093

0020 7456 1100018 119909

30019 9483 112

3000012 119909

30014 3360 108

0012 1199093

0013 8356 1100012 119909

40013 3039 112

A further analysis has also been performed of Example 10with 119873 = 100 200 300 to compute the corresponding1205720 with 1205721 1205722 that correspond simultaneously to 119888 =108 110 112 as illustrated in Figure 1 Results of thesecomputations are summarized in Table 3 and then used forthe partial 119876

119904120590119873(119909) plots of Figure 2 that interpolate 1205721 1205722

(for 119888 = 108 110 112) with 1205720 for each 119873 = 100 200 300In this table119909

lowaststands for the first convergence step of the 119909

119899

sequence of Newton Raphson iterations

002

05

10

004

g(x

)

xminus004 minus002

minus05

minus10

g(x) for N = 100

g(x) for N = 200

g(x) for N = 300

h(x) = c = 108

h(x) = c = 11

h(x) = c = 112

1205721 1205720 1205722

Figure 1 Plots of 119892(119909) near a discontinuity of 119891(119909) in Example 10

120

115

110

105

100

000 001 002 003 004 005

g(x

)

x

g(x) for N = 100

g(x) for N = 200

g(x) for N = 300

h(x) = c = 108

h(x) = c = 11

h(x) = c = 112

Figure 2 Plots of the 119892(119909) curves near a discontinuity of 119891(119909) inExample 10 interpolating 1205721 1205722 (for 119888 = 108 110 112) with 1205720 allfor119873 = 100 200 300

These results turn out to be extremely stable against varia-tions of the trial initial root1199090 Indeed in the computationallymost difficult situation namely when 119873 = 300 a changeof 1199090 from 0008 to 0020 leads to a variation of 119909

lowastfrom

1199092 to just 1199096 with the same asymptotic 1205720 = 00104648 and1205721 = 000807732 (for 119888 = 112)

5 Conclusions

This paper demonstrates that the quadratic degeneracy ofthe solution of the present inverse problem explicit whenusing 119867(120574 119909119898) is a nonlinear indicator of the existence ofGibbsrsquo phenomenon It provides also a distribution-theoreticproof for the existence of this phenomenon The reportedanalysis of numerical solvability of this problem illustrates thefollowing

(i) When suminfin

119898=1 119898(120587119871)119866lowast(120574 119909119898) = suminfin

119898=1 119898(120587119871)119867lowast(120574 119909119898) converge continuously in 119909 and are

12 International Journal of Computational Mathematics

nonzero at 1205721 (or 1205722) there exists a neighborhood(defined by 119868) of 120572 such that for all starting values 1199090in the neighborhood 119909

119899 will converge to 1205721 (or 1205722)

(ii) When suminfin

119898=1 119898(120587119871)119866lowast(120574 119909119898)=suminfin

119898=1 119898(120587119871)119867lowast(120574119909119898) converge continuously in 119909 and are nonzero at1205721 and 1205722 (but not at 1205720) and if 120593(119909) has a secondderivative which might be zero at 1205721 and 1205722 (and at1205720) then the convergence of 119909119899 is quadratic or fasterIf12059310158401015840(119909) = 0 at1205721 and1205722 then the convergence of 119909119899is merely quadratic

(iii) The convergence of the Newton-Raphson iterativeprocess to 1205720 can also be quadratic but restrictionson the derivatives of 120593(119909) have to be observed up tothe third order Otherwise the convergence of 119909

119899 to

1205720 can slow down to become only linear [12] with arate log210

(iv) The reported computations clearly illustrate that1205720 rarr 0 and |1205721 minus 1205722| rarr 0 when 119873 rarr infin thatis the infinite Fourier series will sum to 119891(119909) inthe neighborhood of a discontinuity except at thediscontinuity itself

Appendix

To illustrate the Gibbs-Wilbraham effect we shall use herethe same arguments employed by Wilbraham in 1848 inanalyzing a similar square wave

Example A1 Let 119891(119909) be a square wave periodic signalwhich is defined over one period by

119891 (119909) =

minus1 minus120587 lt 119909 le 0

1 0 le 119909 lt 120587(A1)

Its Fourier series representation is

119892 (119909) =4120587

infin

sum119898=1

1(2119898 minus 1)

sin (2119898minus 1) 119909 (A2)

This defines the truncated series

119878119873(119909) =

4120587

119873

sum119898=1

1(2119898 minus 1)

sin (2119898minus 1) 119909 (A3)

which may be differentiated as

119889

119889119909119878119873(119909) =

4120587

119873

sum119898=1

cos (2119898minus 1) 119909 (A4)

Since

2119873

sum119898=1

cos (2119899 minus 1) 119909 =sin 2119873119909

sin119909 (A5)

then119889

119889119909119878119873(119909) =

2120587

sin 2119873119909

sin119909 (A6)

It is possible to cancel the previous artificially induceddifferentiation by integrating the last result as

119878119873(119909) = int

119909

0

119889

119889120585119878119873(120585) 119889120585 =

2120587int119909

0

sin 2119873120585

sin 120585119889120585 (A7)

This integral starts at zero when 119909 = 0 in agreement with(A2) and increases until 2119873120585 = 120587 Then 120585 = 1205872119873 definesa turning point 1199091 The integral starts to decrease at whichpoint the numerator sin 2119873120585 goes negative For large 119873 thedenominator sin 120585 remains positive

The form (A7) distinctively from (A3) reveals that119878119873(119909) has turning points at the zeros of sin 2119873120585 These occur

when 2119873120585 = 119903120587 119903 = 1 2 3 4 5 that is at 120585 = 119903(1205872119873)Clearly then 119909

119903sim 120585119903= 119903(1205872119873) are turning points for 119878

119873(119909)

In particular 1199091 sim 1205851 = 1205872119873 defines the overshoot peaklocation 1205720 and 119878

119873(1199091) = (2120587) int12058721198730 (sin 2119873120585 sin 120585) 119889120585 is

its magnitude To evaluate 119878119873(1199091) = 119878

119873(1205720) numerically

let us change the variable of integration namely 2119873120585 = 119908or 120585 = 1199082119873 to rewrite the previous integral as 119878

119873(1199091) =

(2120587) int1205870 (sin119908(sin1199082119873)) 119889(1199082119873) Clearly when 119873 rarr

infin sin1199082119873 = 1199082119873 and

lim119873rarrinfin

119878119873(1199091) =

2120587int120587

0

sin119908119908

119889119908 =2120587Si (120587) = 118 (A8)

correct to two decimal pointsAs for 1199092 1199093 1199094 119909119903 lim119873rarrinfin[119878119873(119909119903) minus 119891(119909)]

decreases as we move away from the discontinuity butit should be remarked however that the location of theovershoot 1205720 = 1199091 = 1205872119873 moves towards the discontinuityaccording to

lim119873rarrinfin

1205720 = lim119873rarrinfin

120587

2119873= 0 (A9)

Conflict of Interests

The author declares that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

The numerical computations of Examples 9 and 10 have beenperformed using Mathematica 8 by Mounir Abou Yassineof the Department of CCE of AUL

References

[1] P L Butzer and R J Nessel Fourier Analysis and Approxima-tion Volume I One-Dimensional Theory Birkhauser 1971

[2] R E Edwards Fourier Series A Modern Introduction HoltRinehart amp Winston 1967

[3] W J Thompson ldquoFourier series and the Gibbs phenomenonrdquoAmerican Journal of Physics vol 60 no 5 pp 425ndash429 1992

[4] A J JerriThe Gibbs Phenomenon in Fourier Analysis Splines ampWavelet Approximations Kluwer Academic Publishers BostonMass USA 1990

International Journal of Computational Mathematics 13

[5] K S Rim and B I Yun ldquoGibbs phenomenon removal by addingHeaviside functionsrdquo Advances in Computational Mathematicsvol 38 no 4 pp 683ndash699 2013

[6] A H Zemanian Distribution Theory and Transform AnalysisMcGraw-Hill 1965

[7] G R Ayers and J C Dainty ldquoIterative blind deconvolutionmethod and its applicationsrdquo Optics Letters vol 13 no 7 pp547ndash549 1988

[8] N H S Haidar ldquoA nonlinear method for a severely ill-posedgeneralized moment problemrdquo Journal of Inverse amp Ill-PosedProblems vol 5 no 4 pp 323ndash335 1997

[9] G L Bradley and K J Smith Calculus Prentice Hall 1995[10] P Offner T Sonar and M Wirz ldquoDetecting strength and

location of jump discontinuities in numerical datardquo AppliedMathematics vol 4 no 12 pp 1ndash14 2013

[11] K E Atkinson An Introduction to Numerical Analysis JohnWiley amp Sons New York NY USA 1989

[12] E Suli and D F Mayers An Introduction to Numerical AnalysisCambridge University Press Cambridge UK 2003

[13] P Wang ldquoA third-order family of Newton-like iteration meth-ods for solving nonlinear equationsrdquo Journal of NumericalMathematics and Stochastics vol 3 no 1 pp 13ndash19 2011

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

International Journal of Computational Mathematics 5

in 119866(120574 119909119898) to rewrite it as

119866 (120574 119909119898) = 119886119898cos 120579119898(119909) + 119887

119898sin 120579119898(119909) (34)

The tangent half-angle (stereographic) transformation119905119898= tan(120579

1198982) or 120579

119898= 2tanminus1119905

119898is conceived as

120579119898(119909) = 2tanminus1119905

119898(119909) = Φ [119905

119898(119909)] (35)

and subsequently

119866 = 119866 [120579119898(119909)] = 119866 Φ [119905

119898(119909)] = 119867 [119905

119898(119909)] (36)

It is clear then that this stereographic projection effectivelyleads to insertion of an additional bracket into the compo-sition of 119866 as a function of 119909 This paves the way to thedefinition

S 119866 [120579119898(119909)] = 119866 Φ [119905

119898(119909)] = 119867 [119905

119898(119909)]

= 119867 (120574 119909119898) (37)

Proposition 4 The following

S119889

119889119909

119873

sum119898=1

119866 [120579119898(119909)] =

119889

119889119909S119873

sum119898=1

119866 [120579119898(119909)] (38)

always holds

Proof Since 119873 is finite then the proof of S(119889119889119909)119866 =(119889119889119909)S119866 = (119889119889119909)119867 implies the correctness of thisproposition So let us differentiate 119866(119909) = 119866[120579

119898(119909)] with

respect to 119909 by applying the chain rule namely

119889119866

119889119909=

119889119866

119889Φ

119889Φ

119889119905119898

119889119905119898

119889119909=

119889119866

119889120579119898

119889120579119898

119889119905119898

119889119905119898

119889119909 (39)

which is the same as

119889119866

119889119909

= [minus119886119898sin 120579119898+ 119887119898cos 120579119898]

2(1 + 1199052119898)

120579119898

2sec2 (

120579119898

2)

(40)

Take then the S transformation of the previous relation toarrive after some algebra at

S119889

119889119909119866 = minus119898

120587

119871

(1198871198981199052119898+ 2119886119898119905119898minus 119887119898)

(1 + 1199052119898)

(41)

Invoke then

119867 = 119867(120574 119909119898) =(119886119898+ 2119887119898119905119898minus 1198861198981199052119898)

(1 + 1199052119898)

(42)

to differentiate it with respect to 119909 as

119889

119889119909119867 =

119889

119889119909S 119866 = minus119898

120587

119871

(1198871198981199052119898+ 2119886119898119905119898minus 119887119898)

(1 + 1199052119898)

(43)

which is the required result

Let us rewrite (1) in the form

120593 (119909) = 119892 (119909) minus 119888 = 0 (44)

then try to differentiate it term-by-term as

1205931015840 (119909) =119889

119889119909119892 (119909) =

119889

119889119909

infin

sum119898=1

119866 (120574 119909119898)

= minusinfin

sum119898=1

119898120587

119871(minus119887119898cos119898120587

119871119909+ 119886119898sin119898120587

119871119909)

= minusinfin

sum119898=1

119898120587

119871119866lowast (120574 119909119898)

(45)

where 119866lowast(120574 119909119898) is the harmonic conjugate of 119866(120574 119909119898)The trigonometric series 119892lowast(119909) = sum

infin

119898=1 119866lowast(120574 119909119898) is called

the trigonometric series conjugate (see eg [10]) to 119892(119909)In addition to the fact that the convergence of 119892lowast(119909)

can sometimes become worse than convergence of 119892(119909) (seeeg [10]) the presence of the 119898(120587119871) factor in (119889119889119909)119892(119909)can aggravate the convergence problem of (45) when thecoefficients 119886

119898and 119887119898do not decay fast enough

Next we may stereographically transform (120597120597119909)119892(119909)using the notationS[(119889119889119909)119892(119909)] = [(119889119889119909)119892(119909)] where

[119889

119889119909119892 (119909)] = minus

infin

sum119898=1

119898120587

119871119866lowast (120574 119909119898)

= minusinfin

sum119898=1

119898120587

119871

(1198871198981199052119898+ 2119886119898119905119898minus 119887119898)

(1 + 1199052119898)

(46)

Similarly for the stereographic Fourier series representation

119892 (119909) =11988602+infin

sum119898=1

119867(120574 119909119898) (47)

in which

119867(120574 119909119898) =(119886119898+ 2119887119898119905119898minus 1198861198981199052119898)

(1 + 1199052119898)

(48)

we may approximate (44) by

120595 (119909) = 119892 (119909) minus 119888 = 0 (49)

Differentiate then (49) term-by-term as

1205951015840 (119909) =119889

119889119909119892 (119909) =

119889

119889119909

infin

sum119898=1

119867(120574 119909119898)

= minusinfin

sum119898=1

119898120587

119871119867lowast (120574 119909119898)

= minusinfin

sum119898=1

119898120587

119871

(1198871198981199052119898+ 2119886119898119905119898minus 119887119898)

(1 + 1199052119898)

(50)

6 International Journal of Computational Mathematics

Note here that119867lowast (120574 119909119898)

=(1198871198981199052119898+ 2119886119898119905119898minus 119887119898)

(1 + 1199052119898)

=(119887119898tan119898(1205872119871) 119909 minus 119886

119898)2minus (1198862119898+ 1198872119898)

119887119898(tan2119898(1205872119871) 119909 + 1)

(51)

is the harmonic conjugate of119867(120574 119909119898) and that

119867lowast (120574 119909119898) = 119866lowast (120574 119909119898) (52)

This rather interesting fact leads to a remark that follows

Remark 5 Away from points of discontinuity of 119891(119909) thestereographically transformed differentiated Fourier series

[(119889119889119909)119892(119909)] which defines 119866lowast(120574 119909119898) is the same as thederivative (119889119889119909)119892(119909) which defines 119867lowast(120574 119909119898) of thestereographic Fourier series 119892(119909) Moreover even at pointsof discontinuity of 119891(119909) [(119889119889119909)119892(119909)] = (119889119889119909)119892(119909)

Theorem 6 (see [1]) Let 119891(119909) be (i) continuous on [minus119871 119871]such that (ii) 119891(minus119871) = 119891(119871) and let (iii) 1198911015840(119909) be piecewisecontinuous over [minus119871 119871] then the corresponding Fourier series119892(119909) of (1) or (47) is differentiable at each point where (iv)11989110158401015840(119909) exists

Theprevious remark can incidentally have useful applica-tions when differentiating certain slowly converging Fourierseries as illustrated by the following example

Example 7 Consider the 2120587-periodic odd signal 119891(119909)defined over one period by 119891(119909) = 119909 Its Fourier series isknown to be

119892 (119909) = 2infin

sum119898=1

(minus1)119898+1

119898sin119898119909 (53)

that is with 119886119898= 0 forall119898 119887

119898= 2((minus1)119898+1119898) and 119898(120587119871) =

119898Differentiating term-by-term we obtain

1198921015840 (119909) = 2infin

sum119898=1

(minus1)119898+1 cos119898119909 (54)

which is not uniformly convergent because 119891(119909) is not onlydiscontinuous at 119909 = plusmn120587 but 119891(minus120587) = 119891(120587) and 11989110158401015840(plusmn120587) doesnot exist In particular at 119909 = 0 the sum of the series 1198921015840(0) =2suminfin119898=1(minus1)

119898+1 oscillates between 0 and 1 that is divergent Itis also well known nevertheless that the seriessuminfin

119898=1(minus1)119898+1

is 119862 minus 1 summable [1 2] to 12 and that turns out to yieldthe right answer 1198921015840(0) = 1198911015840(0) = 1 Moreover at 119909 = 120587 thesum of the series 1198921015840(120587) = 2suminfin

119898=1(minus1)2119898+1 rarr minusinfin that is

divergentAlternatively by means of (47)

119892 (119909) =infin

sum119898=1

119887119898

2119905119898

1 + 1199052119898

= 2infin

sum119898=1

(minus1)119898+1

119898sin119898119909 (55)

and by means of (46)

1198921015840 (119909) =

[119889

119889119909119892 (119909)] = minus

infin

sum119898=1

119898119866lowast (120574 119909119898)

= minusinfin

sum119898=1

119898119887119898

1199052119898minus 1

1 + 1199052119898

= 2infin

sum119898=1

(minus1)119898+1(1 minus tan2119898(12) 119909)(1 + tan2119898(12) 119909)

= 2infin

sum119898=1

(minus1)119898+1 cos119898119909

(56)

Here at 119909 = 120587 the sum of the series 1198921015840(120587) =

2suminfin119898=1(minus1)

2119898+1 rarr minusinfin is also divergent

Note in general that for even signals that is when 119887119898= 0

forall119898 it follows from (22) that

119867(120574 119909119898)1003816100381610038161003816119887119898=0

= 119886119898

1 minus 1199052119898

1 + 1199052119898

= 119886119898cos119898120587

119871119909

= 119866 (120574 119909119898)1003816100381610038161003816119887119898=0

(57)

Moreover for odd signals that is when 119886119898

= 0 forall119898 thesituation is not of the same clarity

Indeed according to (22)

119867(120574 119909119898)1003816100381610038161003816119886119898=0

=00

119887119898

1 + 1199052119898

(58)

which is an uncertainty Luckily however this uncertainty isnot an essential one and can straightforwardly be resolved byresorting to the stereographic form (23) of (22) which yieldsthe correct result

119867(120574 119909119898)1003816100381610038161003816119886119898=0

= 119887119898

2119905119898

1 + 1199052119898

= 119887119898sin119898120587

119871119909

= 119866 (120574 119909119898)1003816100381610038161003816119886119898=0

(59)

Also when 119887119898= 0 forall119898

119867lowast (120574 119909119898)1003816100381610038161003816119887119898=0

=00

119886119898

1 + 1199052119898

(60)

and this is resolvable in a similar fashionBoth119866(120574 119909119898) of (2) and119867(120574 119909119898) of (22) equivalent

forms are nonlinear trigonometric functions of the 119909 solu-tion However while 119866(120574 119909119898) is linear in 120574 = 119886

119898 119887119898infin119898=1

thestereographic 119867(120574 119909119898) is nonlinear in this 120574 This canraise a question on a possibility for (120597120597119909)119892(119909) = (120597120597119909)119892(119909)at singular points of 119891(119909) A question that will be answerednegatively in the next section Furthermore since the data 120574 =119886119898 119887119898infin119898=1 is discrete then the operators sum

infin

119898=1 119866(120574 119909119898)and sum

infin

119898=1 119867(120574 119909119898) do not depend continuously on 120574 andthis is themain reason for ill-posedness of the present inverseproblem

International Journal of Computational Mathematics 7

In conclusion the quadratic degeneracy of the solution ofthe present inverse problem explicit when using 119867(120574 119909119898)is a nonlinear indicator of the existence of the peak 119876

119904120590119873(119909)

pertaining to the Gibbs phenomenon However the numer-ical ill-posedness associated with the relative values of atrial iterative variable in the 120574 set happens to remarkablyremain invariant under the tangent half-angle stereographicprojection

Following the same arguments of Wilbraham we maycancel the artificially induced differentiation by integrating(50) to represent the stereographic Fourier series 119892(119909) inintegral form as

119892 (119909) =11988602minusinfin

sum119898=1

int119909

0119898

sdot120587

119871

(119887119898tan119898(1205872119871) 120585 minus 119886

119898)2minus (1198862119898+ 1198872119898)

119887119898(tan2119898(1205872119871) 120585 + 1)

119889120585 =11988602

minusinfin

sum119898=1

int119909

0119898

sdot120587

119871

(119887119898tan2119898(1205872119871) 120585 minus 2119886

119898tan119898(1205872119871) 120585 minus 119887

119898)

(tan2119898(1205872119871) 120585 + 1)119889120585

(61)

4 Numerics of Solving the Inverse Problem

The solution of the present inverse problem is in fact aproblem of finding the roots of (44) when 119892(119909) is truncatedat 119898 = 119873 For that reason we shall focus on such rootsin the neighborhood of a step-like discontinuity of 119891(119909) at119909 = 119909119900 say where 119891(119909119900 minus 120576) lt 119891(119909119900 + 120576) 0 lt 120576 ≪ 1 Theassociated with the Gibbs phenomenon overshoot of 119892(119909) =11988602+sum

119873

119898=1 119866(120574 119909119898) or 11988602+sum119873

119898=1 119867(120574 119909119898) over119891(119909119900+120576)and undershoot of this 119892(119909) below 119891(119909119900minus120576)The intersectionof a horizontal line ℎ(119909) = 119888 ge 119891(119909119900 + 120576) corresponding to120593(119909) = 0 can be at two points 1205721 and 1205722 on both sides of thepeak location 1205720 representing two distinct roots of 120593(119909) = 0It can also be at one point 1205720 corresponding to themaximumof the truncated 119892(119909) that is a double root of 120593(119909) = 0

Obviously 1205721 lt 1205721 lt 1205720 lt 1205722 lt 1205722 where 1205721 and 1205722 arepossible points of inflection defined by

11989210158401015840 (1205721) = 12059310158401015840 (1205721) = 11989210158401015840 (1205722) = 12059310158401015840 (1205722) = 0 (62)

The existence of both 1205721 and 1205722 or at least one of themnamely 1205722 is however not mandatory Moreover for the 12057211205720 and 1205722 points we expect to have

1198921015840 (1205721) = 1205931015840 (1205721) gt 0

12059310158401015840 (1205721) gt 0

120593101584010158401015840 (1205721) lt 0

1198921015840 (1205720) = 1205931015840 (1205720) = 0

12059310158401015840 (1205720) lt 0

120593101584010158401015840 (1205720) = 0

1198921015840 (1205722) = 1205931015840 (1205722) lt 0

12059310158401015840 (1205722) gt 0

120593101584010158401015840 (1205722) gt 0(63)

Equation (44) can be solved for119909 120593(119909) = 0 by aNewton-Raphson (see eg [11 12]) iterative process

119909119899+1 = 119909

119899minus

120593 (119909119899)

1205931015840 (119909119899) (64)

which is repeated until a sufficiently accurate value is reachedThis process for each series truncation number119873 rewrites as

119909119899+1 = 119909

119899+[sum119873

119898=1 119866 (120574 119909119899 119898) + 11988602 minus 119888]

[sum119873

119898=1 119898(120587119871)119866lowast (120574 119909119899 119898)]

(65)

or in the stereographic form

119909119899+1 = 119909

119899+[sum119873

119898=1 119867(120574 119909119899 119898) + 11988602 minus 119888]

[sum119873

119898=1 119898(120587119871)119867lowast (120574 119909119899 119898)]

(66)

which is the same as

119909119899+1 = 119909

119899

+[sum119873

119898=1 ((119886119898 + 2119887119898119905119898minus 1198861198981199052119898) (1 + 1199052

119898)) + 11988602 minus 119888]

[sum119873

119898=1 119898(120587119871) ((1198871198981199052119898+ 2119886119898119905119898minus 119887119898) (1 + 1199052

119898))]

(67)

and is started off with some rather free initial value 1199090Obviously here 119905

119898= tan119898(1205872119871)119909

The sequence 119909119899will usually converge provided that 1199090

is close enough to the unknown zero 120572 and that

1205931015840 (1199090) = minusinfin

sum119898=1

119898120587

119871119866lowast (120574 1199090 119898) = 0

or 1205951015840 (1199090) = minusinfin

sum119898=1

119898120587

119871119867lowast (120574 1199090 119898) = 0

(68)

41 Computation of the Peak Location 1205720 Being a maximiza-tion problem for 120593(119909) = 0 the determination of 1205720 by theNewton-Raphson process [11] becomes

119909119899+1 = 119909

119899minus

1205931015840 (119909119899)

12059310158401015840 (119909119899)

(69)

or

119909119899+1 = 119909

119899minus

1205951015840 (119909119899)

12059510158401015840 (119909119899) (70)

in stereographic form

8 International Journal of Computational Mathematics

Straightforwardly this process can be expressed as

119909119899+1 = 119909

119899minus

[sum119873

119898=1 119898119866lowast (120574 119909119899 119898)]

[sum119873

119898=1 1198982 (120587119871) 119866 (120574 119909

119899 119898)]

(71)

or in the stereographic form

119909119899+1 = 119909

119899

minus[minussum119873

119898=1 119898(120587119871)119867lowast (120574 119909119899 119898)]

[sum119873

119898=1 1198982 (12058721198712) ((119886

1198981199052119898minus 2119887119898119905119898minus 119886119898) (1 + 1199052

119898))]

(72)

which is the same as119909119899+1 = 119909

119899

+[sum119873

119898=1 119898((1198871198981199052119898+ 2119886119898119905119898minus 119887119898) (1 + 1199052

119898))]

[sum119873

119898=1 1198982 (120587119871) ((119886

1198981199052119898minus 2119887119898119905119898minus 119886119898) (1 + 1199052

119898))]

(73)

It is remarkable how this algorithm is distinctively indepen-dent of 119888

Proposition 8 The ill-posedness of the inverse problem ofthe truncated Fourier series is invariant under stereographicprojection

Proof The tangent half-angle stereographic projectionthough not isometric (distance preserving) is neverthelessvariability preserving Indeed the main ingredients ofthis inverse problem are namely 120574 = 119886

119898 119887119898119873119898=1 119888 and

1199090 with (119889119889119909)119866[120579119898(119909)]1199090 Under this projection 120579

119898(1199090)

transforms to 119905119898(1199090) = tan[120579

119898(1199090)2] and 119866[120579

119898(1199090)]

transforms to 119867[119905119898(1199090)] But according to Proposition 4

(119889119889119909)119867[119905119898(119909)]1199090

= (119889119889119909)119866[120579119898(119909)]1199090 Hence the variation

of 119866(120574 1199090 119898) with 1199090 in the Fourier series inverse problemand the variation of119867(120574 1199090 119898) with 1199090 in the stereographicform are identical

Equivalently this makes the ill-posedness of this inverseproblem invariant under stereographic projection

Expected manifestations of the correctness of this propo-sition are as follows (i) the results of computations by the(71) and (73) algorithms for 1205720 with the same 1199090 must beidentical and (ii) the results of computations by the (65) and(67) algorithms for 1205721 (or 1205722) with the same 1199090 and 119888 mustalso be identical

42 Convergence Analysis Let us focus first on the con-vergence of the iterative process (65) or (67) In manysituationswhere term-by-termdifferentiability does not holdtheNewton-Raphsonmethod can always be replaced by somequasi-Newtonmethod [12 13] where 1205931015840(119909

119899) or1205951015840(119909

119899) in (65)

or (67) is replaced by a suitable approximation For examplein the chord method 1205931015840(119909

119899) is replaced by 1205931015840(1199090) for all

iterationsIf 120598119899= 120572minus119909

119899 120598119899+1 = 120572minus119909

119899+1 and120573119899 is some point between119909119899and 120572 it can be easily proved (see [11] or [12]) that

120598119899+1 = minus

12059310158401015840 (120573119899)

21205931015840 (119909119899)1205982119899 (74)

Taking absolute value of both sides gives

1003816100381610038161003816120598119899+11003816100381610038161003816 =

1003816100381610038161003816100381612059310158401015840 (120573119899)10038161003816100381610038161003816

2 10038161003816100381610038161205931015840 (119909119899)10038161003816100381610038161205982119899 (75)

implying quadratic convergence that is 120598119899is squared (the

number of correct digits roughly doubles) at each step Thesame arguments apply of course to the 120595(119909) = 0 equation

In actual fact for a distinct 120572 the convergence of119909119899 in (65)ndash(67) is at least quadratic in the neigh-

borhood of this 120572 The process may face a difficultywhen 1205931015840(119909

119899) = minussum

119873

119898=1 119898(120587119871)119866lowast(120574 119909119899 119898) or 1205951015840(119909

119899) =

minussum119873

119898=1 119898(120587119871)119867lowast(120574 119909119899 119898) diverges This is however a spe-

cial but not a general problem and (75) always holds if thefollowing conditions are satisfied

(i) 1205931015840(119909) = 0 forall119909 isin 119868 where 119868 = [120572 minus 120578 120572 + 120578] for some120578 ge |(120572 minus 1199090)|

(ii) 12059310158401015840(119909) is finite forall119909 isin 119868(iii) 1199090 is sufficiently close to 120572

The term sufficiently close in this context means the follow-ing

(a) Taylor approximation is accurate enough such that wecan ignore higher order terms

(b) (12)|12059310158401015840(119909119899)1205931015840(119909

119899)| lt 119870|12059310158401015840(120572)1205931015840(120572)| for some119870 lt

infin(c) 119870|12059310158401015840(120572)1205931015840(120572)|120598

119899lt 1 for 119899 isin N

Consequently (75) can be expressed in the following way1003816100381610038161003816120598119899+1

1003816100381610038161003816 = 1198721205982119899 (76)

where

119872 = sup119909isin119868

12

100381610038161003816100381610038161003816100381610038161003816

12059310158401015840 (119909)

1205931015840 (119909)

100381610038161003816100381610038161003816100381610038161003816 (77)

The initial point 1199090 has to be chosen such that conditions(i) through (iii) are satisfied where the third conditionrequires that 119872|1205980| lt 1 Moreover with increasing 119873 thedistance |1205721 minus 1205722| is known to decrease in a way makingthe satisfaction of these conditions very difficult and actuallyimpossible as119873 rarr infin Here it should be underlined that theiterative method (65)ndash(67) may fail to converge to the 120572 rootin the following situations

(i) If a point like 1205720 is included in 119868 the method willterminate due to division by zero

(ii) When the initial estimate 1199090 is poor the pertainingwrong 119868 can contribute to nonconvergence of thealgorithm

As for convergence of the Newton-Raphson processtowards 1205720 let 120576119899 = 1205720 minus 119909

119899 120576119899+1 = 1205720 minus 119909

119899+1 and 120579119899is some

point between 119909119899and 1205720 to arrive at

120576119899+1 = minus

120593101584010158401015840 (120579119899)

212059310158401015840 (119909119899)1205762119899 (78)

1003816100381610038161003816120576119899+11003816100381610038161003816 =

10038161003816100381610038161003816120593101584010158401015840 (120579119899)10038161003816100381610038161003816

2 100381610038161003816100381612059310158401015840 (119909119899)10038161003816100381610038161205762119899 (79)

International Journal of Computational Mathematics 9

The convergence of 119909119899 in (71)ndash(73) is at least quadratic

in the neighborhood of this 1205720 The process may face a diffi-culty however when 12059310158401015840(119909

119899) = minussum

infin

119898=1 1198982(120587119871)2119866(120574 119909

119899 119898)

or 12059510158401015840(119909119899) diverges Moreover (79) always holds if the follow-

ing conditions are satisfied

(i) 12059310158401015840(119909) = 0 forall119909 isin 119868 where 119868 = [1205720 minus120578 1205720 +120578] for some120578 ge |(1205720 minus 1199090)|

(ii) 120593101584010158401015840(119909) is finite forall119909 isin 119868

(iii) 1199090 is sufficiently close to 1205720

Sufficient closeness in this context means that

(a) Taylor approximation is accurate enough such that wecan ignore higher order terms

(b) (12)|120593101584010158401015840(119909119899)12059310158401015840(119909

119899)| lt 119870|120593101584010158401015840(1205720)120593

10158401015840(1205720)| for some119870 lt infin

(c) 119870|120593101584010158401015840(1205720)12059310158401015840(1205720)|120576119899 lt 1 for 119899 isin N

Consequently

1003816100381610038161003816120576119899+11003816100381610038161003816 = 1198721205762

119899 (80)

where

119872 = sup119909isin119868

12

100381610038161003816100381610038161003816100381610038161003816

120593101584010158401015840 (119909)

12059310158401015840 (119909)

100381610038161003816100381610038161003816100381610038161003816 (81)

43 Applications The standard iterative algorithms (71) or(73) for finding the location of overshoots in 119876

119904120590119873(119909) of

the Gibbs effect employ the fact that the first derivative of apeak in 119892(119909) or 119892(119909) respectively has a downward-going-zero-crossing at the peak maximum But the presence ofmany neighboring smaller peaks in the ringing close to theovershoot will cause many undesirable zero-crossings to beinvoked if the initial trial root 1199090 is not correctly chosenThis fact makes the numerical problem of peak finding in theGibbs effect of pathological difficulty A difficulty that aggra-vates with the increase in the number 119873 of truncated termsin the employed Fourier series Indeed with the increaseof 119873 the number of undesirable zero-crossings increasesin the neighborhood of 1205720 as illustrated in Section 1 Thiscan increasingly lead to oscillatory and even unstable valuesin sequence 119909

119899 the more the distance |1199090 minus 1205720| contains

unwanted zero-crossingsThese problems have been resolved in the two examples

that follow which simultaneously illustrate the applicabilityof Proposition 8

Example 9 Consider the 2120587-periodic odd signal 119891(119909) ofExample A1 defined over one period by

119891 (119909) =

minus1 minus120587 lt 119909 lt 0

1 0 le 119909 lt 120587(82)

which has for each truncation number 119873 the truncatedFourier series approximations

119892 (119909) ≃4120587

119873

sum119898=1

1(2119898 minus 1)

sin (2119898minus 1) 119909

119892 (119909) ≃8120587

119873

sum119898=1

119887119898119905119898

(1 + 1199052119898)=

8120587

119873

sum119898=1

1(2119898 minus 1)

119905119898

(1 + 1199052119898)

=8120587

119873

sum119898=1

1(2119898 minus 1)

tan ((2119898 minus 1) 2) 1199091 + [tan ((2119898 minus 1) 2) 119909]2

(83)

This signal has a jump discontinuity at 119909 = 0 ofmagnitude 2 We shall focus our computations on the Gibbsphenomenon overshoot near this discontinuity

Here is a printout list of results for the parameters of theposing inverse problem via computations by both the directand stereographic algorithms

It is remarkable how results by both algorithms arepractically identical and share the same following featuresthat are observed by results of the direct algorithm discussedbelow

Case I (119873 = 40) The computations of 1205720 for the previousexample using (71) with 1199090 = 002 lead to 1199091 = 0039 109this varies quickly with 119899 (only beyond the 4th decimal) tostabilize towards 1199093 = 1199094 = 1199095 = sdot sdot sdot = 0039 299 = 1205720Here using (1) gives 119892(1205720) = 119892(0039 299) = 1179 030 whichis quite close to the theoretically expected value by (A7) ofsim118

Then using (65) for 119888 = 109 and 1199090 = 0000381 leads to1199091 = 0021 697 this varies quickly with 119899 (only beyond the 3-d decimal) to stabilize quickly towards 1199095 = 1199096 = 1199097 = sdot sdot sdot =0028 368 = 1205721 Also using (65) for 119888 = 109 and 1199090 = 004leads to 1199091 = 0067 699 this varies quickly with 119899 (beyondthe 2nd decimal) to stabilize quickly towards 1199095 = 1199096 = 1199097 =sdot sdot sdot = 0052 873 = 1205722

For both1205721 and1205722 using (1) yields119892(1205721) = 119892(0028368) =119892(1205722) = 119892(0052873) = 109 = 119888 as expected

Case II (119873 = 100) Computations of 1205720 using (71) with1199090 = 002 lead to 1199091 = 0014 221 this varies quickly with119899 (only beyond the 4th decimal) to stabilize towards 1199093 =1199094 = 1199095 = sdot sdot sdot = 0015 708 = 1205720 Here using (1) gives119892(1205720) = 119892(0015708) = 1178 990 This result appears to beslightly excessive perhaps due to increased rounding off errorswith increasing 119873 to 100

Thenusing (65) for 119888 = 109 and1199090 = 0015 237 leads aftera negative value to 1199092 = 0014 462 this varies with 119899 (beyondthe 2nd decimal) to stabilize quickly towards 1199097 = 1199098 = 1199099 =sdot sdot sdot = 0011 3348 = 1205721 The used value of 1199090 is apparently notappropriate and this illustrates the generic ill-posedness of thisinverse problem

The pertaining computations of 1205722 with 1199090 = 0017000lead to 1199091 = 0020 434 this varies with 119899 (beyond the 3-ddecimal) to stabilize quickly towards 1199094 = 1199095 = 1199096 = sdot sdot sdot =0021 149 = 1205722

10 International Journal of Computational Mathematics

For both 1205721 and 1205722 using (1) yields 119892(1205721) =119892(00113348) = 119892(1205722) = 119892(0021149) = 109 = 119888 asexpected

Case III (119873 = 150) The computations of 1205720 using (71) with1199090 = 0005 lead to 1199091 = 0010 595 this varies insignificantlywith 119899 (only beyond the 4th decimal) to stabilize towards1199093 =1199094 = 1199095 = sdot sdot sdot = 0010 472 = 1205720 Here using (1) gives 119892(1205720) =119892(0010472) = 1178 980 which does not significantly differfrom 119892(1205720) when119873 = 100

Thenusing (65) for 119888 = 109 and1199090 = 0010 158 leads aftera negative value to 1199092 = 0009 643 and this varies (beyond the3-d decimal) to stabilize quickly towards1199097 = 1199098 = 1199099 = sdot sdot sdot =0007 565 = 1205721

The pertaining computations of 1205722 with 1199090 = 0010 800lead to 1199091 = 0025 998 this varies (beyond the 2nd decimal)to stabilize towards 11990914 = 11990915 = 11990916 = 11990917 = 11990918 = sdot sdot sdot =0014 099 = 1205722

For both 1205721 and 1205722 using (1) yields 119892(1205721) = 119892(007 565) =119892(1205722) = 119892(0014 099) = 109 = 119888

During the previous computations summarized inTable 1 it has been observed that varying 1199090 in the presentNewton-Raphson process for all119873 can lead to strong oscil-lations and even completely unstable values for 1205720 1205721 and 1205722of the Fourier series inverse problem which though ill-posed

is practically regularizible Apart from this drawback the(71) with (65) iterative algorithms (i) converge quickly to thetheoretically expected values and are (ii) of demonstratedstability of the iterations for properly chosen 1199090 (which act aseffective regularizing parameters) Moreover computationsby the stereographic algorithm invoking (73) and (67) sharethe samemagnitude and convergence rates with results by thedirect algorithm Finally it is demonstrated that the presentsignal exhibits an overshoot when119873 = 150 ofsim9of a jumpdiscontinuity of magnitude 2 that is located at a distance ofsim00105

To illustrate robustness of both numerical algorithms andtheir possible effective regularizibility we shall study also thenumerical inverse problem for the Fourier series representa-tion of an arbitrary (nonsymmetric) periodic signal

Example 10 Consider the 2120587-periodic signal 119891(119909) definedover one period by

119891 (119909) =

minus119909

120587minus 1 minus120587 lt 119909 lt 0

1 0 le 119909 lt 120587(84)

It has for each truncation number 119873 the truncated Fourierseries approximations

119892 (119909) ≃14+119873

sum119898=1

[(minus1)119898 minus 1]12058721198982 cos119898119909+

[2 minus (minus1)119898]120587119898

sin119898119909

119892 (119909) ≃14

+119873

sum119898=1

[(minus1)119898 minus 1] 12058721198982 + 2 ([2 minus (minus1)119898] 120587119898) tan ((2119898 minus 1) 2) 119909 minus ([(minus1)119898 minus 1] 12058721198982) [tan ((2119898 minus 1) 2) 119909]2

1 + [tan ((2119898 minus 1) 2) 119909]2

(85)

This signal has like the signal of Example 9 a jumpdiscontinuity at 119909 = 0 of magnitude 2 For the sake ofcomparison we shall also focus our computations on theGibbs phenomenon overshoot part of 119876

119904120590119873(119909) near this

discontinuityIt should be noted here that the results by both algorithms

summarized in Table 2 are also identical a feature indicatingthat the degree of ill-posedness of the posing inverse problemis not affected as predicted by Proposition 8 by the tangenthalf-angle stereographic projection

Case I (119873 = 40) The computations of 1205720 for this exampleusing (71) and (73) with 1199090 = 003 lead to an identicalsequence 119909

119899 that converges by both algorithms after the 1199094

iteration to 0078 139 with 119892(1205720) = 119892(1205720) = 1178 76

Case II (119873 = 100) The computations of 1205720 with 1199090 = 001lead to an identical sequence 119909

119899 that converges by both

algorithms after the 1199095 iteration to 0031 351 with 119892(1205720) =

119892(1205720) = 1178 87

Case III (119873 = 150) The computations of 1205720 with 1199090 = 0007lead also to an identical sequence 119909

119899 that converges by both

algorithms after the 1199095 iteration to 0020 915 with 119892(1205720) =

119892(1205720) = 1178 91

Therefore the present signal exhibits an overshoot when119873 = 150 of sim9 of a jump discontinuity of magnitude 2 thatis located at a distance of sim0020 9

Comparison of the computed Gibbs overshoot parame-ters of the two considered signals with jump discontinuitiesat the same 119909 = 0 of the same magnitude illustrates that therelative size of the overshoot depends essentially exclusivelyon the magnitude of the stimulating jump discontinuityFor instance based on the present numerical results onecan suggest here that nonsymmetric signals have equalovershoots with symmetric signals Their respective 1205720rsquos arehowever different Indeed the nonsymmetric signal has asignificantly larger 1205720 (0020 9) than the 1205720 (00105) of thesymmetric signal

International Journal of Computational Mathematics 11

Table 1 Results of computations for Example 9

119873 1205721 1205720 1205722

40 0028 368 0039 299 0052 873100 0011 335 0015 705 0021 149150 0007 565 0010 472 0014 099

Table 2 Results of computations for Example 10

119873 1205721 1205720 1205722

40 00564765 00781391 0105155100 00226546 00313516 00422024150 00151124 00209153 0028156

Table 3 Summary of computations of all overshoot parameters inExample 10

1205720results

119873 1199090

119909lowast

1205720

119892(1205720)

100 0030 1199093

0031 3516 1178 87200 0015 119909

30015 6919 1178 92

300 0008 1199093

0010 4648 1178 941205721results

119873 1199090

119909lowast

1205721

119888

0025 1199094

0022 2027 108100 0025 119909

30023 1344 110

0025 1199093

0024 2027 1120010 119909

30011 1117 108

200 0010 1199093

0011 5779 1100010 119909

40012 1123 112

0008 1199093

0007 410 08 108300 0008 119909

30007 720 95 110

0008 1199092

0008 077 32 1121205722results

119873 1199090

119909lowast

1205722

119888

100035 119909

40042 9439 108

035 1199094

0041 4445 110035 119909

40039 8513 112

2000018 119909

30021 4960 108

0018 1199093

0020 7456 1100018 119909

30019 9483 112

3000012 119909

30014 3360 108

0012 1199093

0013 8356 1100012 119909

40013 3039 112

A further analysis has also been performed of Example 10with 119873 = 100 200 300 to compute the corresponding1205720 with 1205721 1205722 that correspond simultaneously to 119888 =108 110 112 as illustrated in Figure 1 Results of thesecomputations are summarized in Table 3 and then used forthe partial 119876

119904120590119873(119909) plots of Figure 2 that interpolate 1205721 1205722

(for 119888 = 108 110 112) with 1205720 for each 119873 = 100 200 300In this table119909

lowaststands for the first convergence step of the 119909

119899

sequence of Newton Raphson iterations

002

05

10

004

g(x

)

xminus004 minus002

minus05

minus10

g(x) for N = 100

g(x) for N = 200

g(x) for N = 300

h(x) = c = 108

h(x) = c = 11

h(x) = c = 112

1205721 1205720 1205722

Figure 1 Plots of 119892(119909) near a discontinuity of 119891(119909) in Example 10

120

115

110

105

100

000 001 002 003 004 005

g(x

)

x

g(x) for N = 100

g(x) for N = 200

g(x) for N = 300

h(x) = c = 108

h(x) = c = 11

h(x) = c = 112

Figure 2 Plots of the 119892(119909) curves near a discontinuity of 119891(119909) inExample 10 interpolating 1205721 1205722 (for 119888 = 108 110 112) with 1205720 allfor119873 = 100 200 300

These results turn out to be extremely stable against varia-tions of the trial initial root1199090 Indeed in the computationallymost difficult situation namely when 119873 = 300 a changeof 1199090 from 0008 to 0020 leads to a variation of 119909

lowastfrom

1199092 to just 1199096 with the same asymptotic 1205720 = 00104648 and1205721 = 000807732 (for 119888 = 112)

5 Conclusions

This paper demonstrates that the quadratic degeneracy ofthe solution of the present inverse problem explicit whenusing 119867(120574 119909119898) is a nonlinear indicator of the existence ofGibbsrsquo phenomenon It provides also a distribution-theoreticproof for the existence of this phenomenon The reportedanalysis of numerical solvability of this problem illustrates thefollowing

(i) When suminfin

119898=1 119898(120587119871)119866lowast(120574 119909119898) = suminfin

119898=1 119898(120587119871)119867lowast(120574 119909119898) converge continuously in 119909 and are

12 International Journal of Computational Mathematics

nonzero at 1205721 (or 1205722) there exists a neighborhood(defined by 119868) of 120572 such that for all starting values 1199090in the neighborhood 119909

119899 will converge to 1205721 (or 1205722)

(ii) When suminfin

119898=1 119898(120587119871)119866lowast(120574 119909119898)=suminfin

119898=1 119898(120587119871)119867lowast(120574119909119898) converge continuously in 119909 and are nonzero at1205721 and 1205722 (but not at 1205720) and if 120593(119909) has a secondderivative which might be zero at 1205721 and 1205722 (and at1205720) then the convergence of 119909119899 is quadratic or fasterIf12059310158401015840(119909) = 0 at1205721 and1205722 then the convergence of 119909119899is merely quadratic

(iii) The convergence of the Newton-Raphson iterativeprocess to 1205720 can also be quadratic but restrictionson the derivatives of 120593(119909) have to be observed up tothe third order Otherwise the convergence of 119909

119899 to

1205720 can slow down to become only linear [12] with arate log210

(iv) The reported computations clearly illustrate that1205720 rarr 0 and |1205721 minus 1205722| rarr 0 when 119873 rarr infin thatis the infinite Fourier series will sum to 119891(119909) inthe neighborhood of a discontinuity except at thediscontinuity itself

Appendix

To illustrate the Gibbs-Wilbraham effect we shall use herethe same arguments employed by Wilbraham in 1848 inanalyzing a similar square wave

Example A1 Let 119891(119909) be a square wave periodic signalwhich is defined over one period by

119891 (119909) =

minus1 minus120587 lt 119909 le 0

1 0 le 119909 lt 120587(A1)

Its Fourier series representation is

119892 (119909) =4120587

infin

sum119898=1

1(2119898 minus 1)

sin (2119898minus 1) 119909 (A2)

This defines the truncated series

119878119873(119909) =

4120587

119873

sum119898=1

1(2119898 minus 1)

sin (2119898minus 1) 119909 (A3)

which may be differentiated as

119889

119889119909119878119873(119909) =

4120587

119873

sum119898=1

cos (2119898minus 1) 119909 (A4)

Since

2119873

sum119898=1

cos (2119899 minus 1) 119909 =sin 2119873119909

sin119909 (A5)

then119889

119889119909119878119873(119909) =

2120587

sin 2119873119909

sin119909 (A6)

It is possible to cancel the previous artificially induceddifferentiation by integrating the last result as

119878119873(119909) = int

119909

0

119889

119889120585119878119873(120585) 119889120585 =

2120587int119909

0

sin 2119873120585

sin 120585119889120585 (A7)

This integral starts at zero when 119909 = 0 in agreement with(A2) and increases until 2119873120585 = 120587 Then 120585 = 1205872119873 definesa turning point 1199091 The integral starts to decrease at whichpoint the numerator sin 2119873120585 goes negative For large 119873 thedenominator sin 120585 remains positive

The form (A7) distinctively from (A3) reveals that119878119873(119909) has turning points at the zeros of sin 2119873120585 These occur

when 2119873120585 = 119903120587 119903 = 1 2 3 4 5 that is at 120585 = 119903(1205872119873)Clearly then 119909

119903sim 120585119903= 119903(1205872119873) are turning points for 119878

119873(119909)

In particular 1199091 sim 1205851 = 1205872119873 defines the overshoot peaklocation 1205720 and 119878

119873(1199091) = (2120587) int12058721198730 (sin 2119873120585 sin 120585) 119889120585 is

its magnitude To evaluate 119878119873(1199091) = 119878

119873(1205720) numerically

let us change the variable of integration namely 2119873120585 = 119908or 120585 = 1199082119873 to rewrite the previous integral as 119878

119873(1199091) =

(2120587) int1205870 (sin119908(sin1199082119873)) 119889(1199082119873) Clearly when 119873 rarr

infin sin1199082119873 = 1199082119873 and

lim119873rarrinfin

119878119873(1199091) =

2120587int120587

0

sin119908119908

119889119908 =2120587Si (120587) = 118 (A8)

correct to two decimal pointsAs for 1199092 1199093 1199094 119909119903 lim119873rarrinfin[119878119873(119909119903) minus 119891(119909)]

decreases as we move away from the discontinuity butit should be remarked however that the location of theovershoot 1205720 = 1199091 = 1205872119873 moves towards the discontinuityaccording to

lim119873rarrinfin

1205720 = lim119873rarrinfin

120587

2119873= 0 (A9)

Conflict of Interests

The author declares that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

The numerical computations of Examples 9 and 10 have beenperformed using Mathematica 8 by Mounir Abou Yassineof the Department of CCE of AUL

References

[1] P L Butzer and R J Nessel Fourier Analysis and Approxima-tion Volume I One-Dimensional Theory Birkhauser 1971

[2] R E Edwards Fourier Series A Modern Introduction HoltRinehart amp Winston 1967

[3] W J Thompson ldquoFourier series and the Gibbs phenomenonrdquoAmerican Journal of Physics vol 60 no 5 pp 425ndash429 1992

[4] A J JerriThe Gibbs Phenomenon in Fourier Analysis Splines ampWavelet Approximations Kluwer Academic Publishers BostonMass USA 1990

International Journal of Computational Mathematics 13

[5] K S Rim and B I Yun ldquoGibbs phenomenon removal by addingHeaviside functionsrdquo Advances in Computational Mathematicsvol 38 no 4 pp 683ndash699 2013

[6] A H Zemanian Distribution Theory and Transform AnalysisMcGraw-Hill 1965

[7] G R Ayers and J C Dainty ldquoIterative blind deconvolutionmethod and its applicationsrdquo Optics Letters vol 13 no 7 pp547ndash549 1988

[8] N H S Haidar ldquoA nonlinear method for a severely ill-posedgeneralized moment problemrdquo Journal of Inverse amp Ill-PosedProblems vol 5 no 4 pp 323ndash335 1997

[9] G L Bradley and K J Smith Calculus Prentice Hall 1995[10] P Offner T Sonar and M Wirz ldquoDetecting strength and

location of jump discontinuities in numerical datardquo AppliedMathematics vol 4 no 12 pp 1ndash14 2013

[11] K E Atkinson An Introduction to Numerical Analysis JohnWiley amp Sons New York NY USA 1989

[12] E Suli and D F Mayers An Introduction to Numerical AnalysisCambridge University Press Cambridge UK 2003

[13] P Wang ldquoA third-order family of Newton-like iteration meth-ods for solving nonlinear equationsrdquo Journal of NumericalMathematics and Stochastics vol 3 no 1 pp 13ndash19 2011

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Stochastic AnalysisInternational Journal of

6 International Journal of Computational Mathematics

Note here that119867lowast (120574 119909119898)

=(1198871198981199052119898+ 2119886119898119905119898minus 119887119898)

(1 + 1199052119898)

=(119887119898tan119898(1205872119871) 119909 minus 119886

119898)2minus (1198862119898+ 1198872119898)

119887119898(tan2119898(1205872119871) 119909 + 1)

(51)

is the harmonic conjugate of119867(120574 119909119898) and that

119867lowast (120574 119909119898) = 119866lowast (120574 119909119898) (52)

This rather interesting fact leads to a remark that follows

Remark 5 Away from points of discontinuity of 119891(119909) thestereographically transformed differentiated Fourier series

[(119889119889119909)119892(119909)] which defines 119866lowast(120574 119909119898) is the same as thederivative (119889119889119909)119892(119909) which defines 119867lowast(120574 119909119898) of thestereographic Fourier series 119892(119909) Moreover even at pointsof discontinuity of 119891(119909) [(119889119889119909)119892(119909)] = (119889119889119909)119892(119909)

Theorem 6 (see [1]) Let 119891(119909) be (i) continuous on [minus119871 119871]such that (ii) 119891(minus119871) = 119891(119871) and let (iii) 1198911015840(119909) be piecewisecontinuous over [minus119871 119871] then the corresponding Fourier series119892(119909) of (1) or (47) is differentiable at each point where (iv)11989110158401015840(119909) exists

Theprevious remark can incidentally have useful applica-tions when differentiating certain slowly converging Fourierseries as illustrated by the following example

Example 7 Consider the 2120587-periodic odd signal 119891(119909)defined over one period by 119891(119909) = 119909 Its Fourier series isknown to be

119892 (119909) = 2infin

sum119898=1

(minus1)119898+1

119898sin119898119909 (53)

that is with 119886119898= 0 forall119898 119887

119898= 2((minus1)119898+1119898) and 119898(120587119871) =

119898Differentiating term-by-term we obtain

1198921015840 (119909) = 2infin

sum119898=1

(minus1)119898+1 cos119898119909 (54)

which is not uniformly convergent because 119891(119909) is not onlydiscontinuous at 119909 = plusmn120587 but 119891(minus120587) = 119891(120587) and 11989110158401015840(plusmn120587) doesnot exist In particular at 119909 = 0 the sum of the series 1198921015840(0) =2suminfin119898=1(minus1)

119898+1 oscillates between 0 and 1 that is divergent Itis also well known nevertheless that the seriessuminfin

119898=1(minus1)119898+1

is 119862 minus 1 summable [1 2] to 12 and that turns out to yieldthe right answer 1198921015840(0) = 1198911015840(0) = 1 Moreover at 119909 = 120587 thesum of the series 1198921015840(120587) = 2suminfin

119898=1(minus1)2119898+1 rarr minusinfin that is

divergentAlternatively by means of (47)

119892 (119909) =infin

sum119898=1

119887119898

2119905119898

1 + 1199052119898

= 2infin

sum119898=1

(minus1)119898+1

119898sin119898119909 (55)

and by means of (46)

1198921015840 (119909) =

[119889

119889119909119892 (119909)] = minus

infin

sum119898=1

119898119866lowast (120574 119909119898)

= minusinfin

sum119898=1

119898119887119898

1199052119898minus 1

1 + 1199052119898

= 2infin

sum119898=1

(minus1)119898+1(1 minus tan2119898(12) 119909)(1 + tan2119898(12) 119909)

= 2infin

sum119898=1

(minus1)119898+1 cos119898119909

(56)

Here at 119909 = 120587 the sum of the series 1198921015840(120587) =

2suminfin119898=1(minus1)

2119898+1 rarr minusinfin is also divergent

Note in general that for even signals that is when 119887119898= 0

forall119898 it follows from (22) that

119867(120574 119909119898)1003816100381610038161003816119887119898=0

= 119886119898

1 minus 1199052119898

1 + 1199052119898

= 119886119898cos119898120587

119871119909

= 119866 (120574 119909119898)1003816100381610038161003816119887119898=0

(57)

Moreover for odd signals that is when 119886119898

= 0 forall119898 thesituation is not of the same clarity

Indeed according to (22)

119867(120574 119909119898)1003816100381610038161003816119886119898=0

=00

119887119898

1 + 1199052119898

(58)

which is an uncertainty Luckily however this uncertainty isnot an essential one and can straightforwardly be resolved byresorting to the stereographic form (23) of (22) which yieldsthe correct result

119867(120574 119909119898)1003816100381610038161003816119886119898=0

= 119887119898

2119905119898

1 + 1199052119898

= 119887119898sin119898120587

119871119909

= 119866 (120574 119909119898)1003816100381610038161003816119886119898=0

(59)

Also when 119887119898= 0 forall119898

119867lowast (120574 119909119898)1003816100381610038161003816119887119898=0

=00

119886119898

1 + 1199052119898

(60)

and this is resolvable in a similar fashionBoth119866(120574 119909119898) of (2) and119867(120574 119909119898) of (22) equivalent

forms are nonlinear trigonometric functions of the 119909 solu-tion However while 119866(120574 119909119898) is linear in 120574 = 119886

119898 119887119898infin119898=1

thestereographic 119867(120574 119909119898) is nonlinear in this 120574 This canraise a question on a possibility for (120597120597119909)119892(119909) = (120597120597119909)119892(119909)at singular points of 119891(119909) A question that will be answerednegatively in the next section Furthermore since the data 120574 =119886119898 119887119898infin119898=1 is discrete then the operators sum

infin

119898=1 119866(120574 119909119898)and sum

infin

119898=1 119867(120574 119909119898) do not depend continuously on 120574 andthis is themain reason for ill-posedness of the present inverseproblem

International Journal of Computational Mathematics 7

In conclusion the quadratic degeneracy of the solution ofthe present inverse problem explicit when using 119867(120574 119909119898)is a nonlinear indicator of the existence of the peak 119876

119904120590119873(119909)

pertaining to the Gibbs phenomenon However the numer-ical ill-posedness associated with the relative values of atrial iterative variable in the 120574 set happens to remarkablyremain invariant under the tangent half-angle stereographicprojection

Following the same arguments of Wilbraham we maycancel the artificially induced differentiation by integrating(50) to represent the stereographic Fourier series 119892(119909) inintegral form as

119892 (119909) =11988602minusinfin

sum119898=1

int119909

0119898

sdot120587

119871

(119887119898tan119898(1205872119871) 120585 minus 119886

119898)2minus (1198862119898+ 1198872119898)

119887119898(tan2119898(1205872119871) 120585 + 1)

119889120585 =11988602

minusinfin

sum119898=1

int119909

0119898

sdot120587

119871

(119887119898tan2119898(1205872119871) 120585 minus 2119886

119898tan119898(1205872119871) 120585 minus 119887

119898)

(tan2119898(1205872119871) 120585 + 1)119889120585

(61)

4 Numerics of Solving the Inverse Problem

The solution of the present inverse problem is in fact aproblem of finding the roots of (44) when 119892(119909) is truncatedat 119898 = 119873 For that reason we shall focus on such rootsin the neighborhood of a step-like discontinuity of 119891(119909) at119909 = 119909119900 say where 119891(119909119900 minus 120576) lt 119891(119909119900 + 120576) 0 lt 120576 ≪ 1 Theassociated with the Gibbs phenomenon overshoot of 119892(119909) =11988602+sum

119873

119898=1 119866(120574 119909119898) or 11988602+sum119873

119898=1 119867(120574 119909119898) over119891(119909119900+120576)and undershoot of this 119892(119909) below 119891(119909119900minus120576)The intersectionof a horizontal line ℎ(119909) = 119888 ge 119891(119909119900 + 120576) corresponding to120593(119909) = 0 can be at two points 1205721 and 1205722 on both sides of thepeak location 1205720 representing two distinct roots of 120593(119909) = 0It can also be at one point 1205720 corresponding to themaximumof the truncated 119892(119909) that is a double root of 120593(119909) = 0

Obviously 1205721 lt 1205721 lt 1205720 lt 1205722 lt 1205722 where 1205721 and 1205722 arepossible points of inflection defined by

11989210158401015840 (1205721) = 12059310158401015840 (1205721) = 11989210158401015840 (1205722) = 12059310158401015840 (1205722) = 0 (62)

The existence of both 1205721 and 1205722 or at least one of themnamely 1205722 is however not mandatory Moreover for the 12057211205720 and 1205722 points we expect to have

1198921015840 (1205721) = 1205931015840 (1205721) gt 0

12059310158401015840 (1205721) gt 0

120593101584010158401015840 (1205721) lt 0

1198921015840 (1205720) = 1205931015840 (1205720) = 0

12059310158401015840 (1205720) lt 0

120593101584010158401015840 (1205720) = 0

1198921015840 (1205722) = 1205931015840 (1205722) lt 0

12059310158401015840 (1205722) gt 0

120593101584010158401015840 (1205722) gt 0(63)

Equation (44) can be solved for119909 120593(119909) = 0 by aNewton-Raphson (see eg [11 12]) iterative process

119909119899+1 = 119909

119899minus

120593 (119909119899)

1205931015840 (119909119899) (64)

which is repeated until a sufficiently accurate value is reachedThis process for each series truncation number119873 rewrites as

119909119899+1 = 119909

119899+[sum119873

119898=1 119866 (120574 119909119899 119898) + 11988602 minus 119888]

[sum119873

119898=1 119898(120587119871)119866lowast (120574 119909119899 119898)]

(65)

or in the stereographic form

119909119899+1 = 119909

119899+[sum119873

119898=1 119867(120574 119909119899 119898) + 11988602 minus 119888]

[sum119873

119898=1 119898(120587119871)119867lowast (120574 119909119899 119898)]

(66)

which is the same as

119909119899+1 = 119909

119899

+[sum119873

119898=1 ((119886119898 + 2119887119898119905119898minus 1198861198981199052119898) (1 + 1199052

119898)) + 11988602 minus 119888]

[sum119873

119898=1 119898(120587119871) ((1198871198981199052119898+ 2119886119898119905119898minus 119887119898) (1 + 1199052

119898))]

(67)

and is started off with some rather free initial value 1199090Obviously here 119905

119898= tan119898(1205872119871)119909

The sequence 119909119899will usually converge provided that 1199090

is close enough to the unknown zero 120572 and that

1205931015840 (1199090) = minusinfin

sum119898=1

119898120587

119871119866lowast (120574 1199090 119898) = 0

or 1205951015840 (1199090) = minusinfin

sum119898=1

119898120587

119871119867lowast (120574 1199090 119898) = 0

(68)

41 Computation of the Peak Location 1205720 Being a maximiza-tion problem for 120593(119909) = 0 the determination of 1205720 by theNewton-Raphson process [11] becomes

119909119899+1 = 119909

119899minus

1205931015840 (119909119899)

12059310158401015840 (119909119899)

(69)

or

119909119899+1 = 119909

119899minus

1205951015840 (119909119899)

12059510158401015840 (119909119899) (70)

in stereographic form

8 International Journal of Computational Mathematics

Straightforwardly this process can be expressed as

119909119899+1 = 119909

119899minus

[sum119873

119898=1 119898119866lowast (120574 119909119899 119898)]

[sum119873

119898=1 1198982 (120587119871) 119866 (120574 119909

119899 119898)]

(71)

or in the stereographic form

119909119899+1 = 119909

119899

minus[minussum119873

119898=1 119898(120587119871)119867lowast (120574 119909119899 119898)]

[sum119873

119898=1 1198982 (12058721198712) ((119886

1198981199052119898minus 2119887119898119905119898minus 119886119898) (1 + 1199052

119898))]

(72)

which is the same as119909119899+1 = 119909

119899

+[sum119873

119898=1 119898((1198871198981199052119898+ 2119886119898119905119898minus 119887119898) (1 + 1199052

119898))]

[sum119873

119898=1 1198982 (120587119871) ((119886

1198981199052119898minus 2119887119898119905119898minus 119886119898) (1 + 1199052

119898))]

(73)

It is remarkable how this algorithm is distinctively indepen-dent of 119888

Proposition 8 The ill-posedness of the inverse problem ofthe truncated Fourier series is invariant under stereographicprojection

Proof The tangent half-angle stereographic projectionthough not isometric (distance preserving) is neverthelessvariability preserving Indeed the main ingredients ofthis inverse problem are namely 120574 = 119886

119898 119887119898119873119898=1 119888 and

1199090 with (119889119889119909)119866[120579119898(119909)]1199090 Under this projection 120579

119898(1199090)

transforms to 119905119898(1199090) = tan[120579

119898(1199090)2] and 119866[120579

119898(1199090)]

transforms to 119867[119905119898(1199090)] But according to Proposition 4

(119889119889119909)119867[119905119898(119909)]1199090

= (119889119889119909)119866[120579119898(119909)]1199090 Hence the variation

of 119866(120574 1199090 119898) with 1199090 in the Fourier series inverse problemand the variation of119867(120574 1199090 119898) with 1199090 in the stereographicform are identical

Equivalently this makes the ill-posedness of this inverseproblem invariant under stereographic projection

Expected manifestations of the correctness of this propo-sition are as follows (i) the results of computations by the(71) and (73) algorithms for 1205720 with the same 1199090 must beidentical and (ii) the results of computations by the (65) and(67) algorithms for 1205721 (or 1205722) with the same 1199090 and 119888 mustalso be identical

42 Convergence Analysis Let us focus first on the con-vergence of the iterative process (65) or (67) In manysituationswhere term-by-termdifferentiability does not holdtheNewton-Raphsonmethod can always be replaced by somequasi-Newtonmethod [12 13] where 1205931015840(119909

119899) or1205951015840(119909

119899) in (65)

or (67) is replaced by a suitable approximation For examplein the chord method 1205931015840(119909

119899) is replaced by 1205931015840(1199090) for all

iterationsIf 120598119899= 120572minus119909

119899 120598119899+1 = 120572minus119909

119899+1 and120573119899 is some point between119909119899and 120572 it can be easily proved (see [11] or [12]) that

120598119899+1 = minus

12059310158401015840 (120573119899)

21205931015840 (119909119899)1205982119899 (74)

Taking absolute value of both sides gives

1003816100381610038161003816120598119899+11003816100381610038161003816 =

1003816100381610038161003816100381612059310158401015840 (120573119899)10038161003816100381610038161003816

2 10038161003816100381610038161205931015840 (119909119899)10038161003816100381610038161205982119899 (75)

implying quadratic convergence that is 120598119899is squared (the

number of correct digits roughly doubles) at each step Thesame arguments apply of course to the 120595(119909) = 0 equation

In actual fact for a distinct 120572 the convergence of119909119899 in (65)ndash(67) is at least quadratic in the neigh-

borhood of this 120572 The process may face a difficultywhen 1205931015840(119909

119899) = minussum

119873

119898=1 119898(120587119871)119866lowast(120574 119909119899 119898) or 1205951015840(119909

119899) =

minussum119873

119898=1 119898(120587119871)119867lowast(120574 119909119899 119898) diverges This is however a spe-

cial but not a general problem and (75) always holds if thefollowing conditions are satisfied

(i) 1205931015840(119909) = 0 forall119909 isin 119868 where 119868 = [120572 minus 120578 120572 + 120578] for some120578 ge |(120572 minus 1199090)|

(ii) 12059310158401015840(119909) is finite forall119909 isin 119868(iii) 1199090 is sufficiently close to 120572

The term sufficiently close in this context means the follow-ing

(a) Taylor approximation is accurate enough such that wecan ignore higher order terms

(b) (12)|12059310158401015840(119909119899)1205931015840(119909

119899)| lt 119870|12059310158401015840(120572)1205931015840(120572)| for some119870 lt

infin(c) 119870|12059310158401015840(120572)1205931015840(120572)|120598

119899lt 1 for 119899 isin N

Consequently (75) can be expressed in the following way1003816100381610038161003816120598119899+1

1003816100381610038161003816 = 1198721205982119899 (76)

where

119872 = sup119909isin119868

12

100381610038161003816100381610038161003816100381610038161003816

12059310158401015840 (119909)

1205931015840 (119909)

100381610038161003816100381610038161003816100381610038161003816 (77)

The initial point 1199090 has to be chosen such that conditions(i) through (iii) are satisfied where the third conditionrequires that 119872|1205980| lt 1 Moreover with increasing 119873 thedistance |1205721 minus 1205722| is known to decrease in a way makingthe satisfaction of these conditions very difficult and actuallyimpossible as119873 rarr infin Here it should be underlined that theiterative method (65)ndash(67) may fail to converge to the 120572 rootin the following situations

(i) If a point like 1205720 is included in 119868 the method willterminate due to division by zero

(ii) When the initial estimate 1199090 is poor the pertainingwrong 119868 can contribute to nonconvergence of thealgorithm

As for convergence of the Newton-Raphson processtowards 1205720 let 120576119899 = 1205720 minus 119909

119899 120576119899+1 = 1205720 minus 119909

119899+1 and 120579119899is some

point between 119909119899and 1205720 to arrive at

120576119899+1 = minus

120593101584010158401015840 (120579119899)

212059310158401015840 (119909119899)1205762119899 (78)

1003816100381610038161003816120576119899+11003816100381610038161003816 =

10038161003816100381610038161003816120593101584010158401015840 (120579119899)10038161003816100381610038161003816

2 100381610038161003816100381612059310158401015840 (119909119899)10038161003816100381610038161205762119899 (79)

International Journal of Computational Mathematics 9

The convergence of 119909119899 in (71)ndash(73) is at least quadratic

in the neighborhood of this 1205720 The process may face a diffi-culty however when 12059310158401015840(119909

119899) = minussum

infin

119898=1 1198982(120587119871)2119866(120574 119909

119899 119898)

or 12059510158401015840(119909119899) diverges Moreover (79) always holds if the follow-

ing conditions are satisfied

(i) 12059310158401015840(119909) = 0 forall119909 isin 119868 where 119868 = [1205720 minus120578 1205720 +120578] for some120578 ge |(1205720 minus 1199090)|

(ii) 120593101584010158401015840(119909) is finite forall119909 isin 119868

(iii) 1199090 is sufficiently close to 1205720

Sufficient closeness in this context means that

(a) Taylor approximation is accurate enough such that wecan ignore higher order terms

(b) (12)|120593101584010158401015840(119909119899)12059310158401015840(119909

119899)| lt 119870|120593101584010158401015840(1205720)120593

10158401015840(1205720)| for some119870 lt infin

(c) 119870|120593101584010158401015840(1205720)12059310158401015840(1205720)|120576119899 lt 1 for 119899 isin N

Consequently

1003816100381610038161003816120576119899+11003816100381610038161003816 = 1198721205762

119899 (80)

where

119872 = sup119909isin119868

12

100381610038161003816100381610038161003816100381610038161003816

120593101584010158401015840 (119909)

12059310158401015840 (119909)

100381610038161003816100381610038161003816100381610038161003816 (81)

43 Applications The standard iterative algorithms (71) or(73) for finding the location of overshoots in 119876

119904120590119873(119909) of

the Gibbs effect employ the fact that the first derivative of apeak in 119892(119909) or 119892(119909) respectively has a downward-going-zero-crossing at the peak maximum But the presence ofmany neighboring smaller peaks in the ringing close to theovershoot will cause many undesirable zero-crossings to beinvoked if the initial trial root 1199090 is not correctly chosenThis fact makes the numerical problem of peak finding in theGibbs effect of pathological difficulty A difficulty that aggra-vates with the increase in the number 119873 of truncated termsin the employed Fourier series Indeed with the increaseof 119873 the number of undesirable zero-crossings increasesin the neighborhood of 1205720 as illustrated in Section 1 Thiscan increasingly lead to oscillatory and even unstable valuesin sequence 119909

119899 the more the distance |1199090 minus 1205720| contains

unwanted zero-crossingsThese problems have been resolved in the two examples

that follow which simultaneously illustrate the applicabilityof Proposition 8

Example 9 Consider the 2120587-periodic odd signal 119891(119909) ofExample A1 defined over one period by

119891 (119909) =

minus1 minus120587 lt 119909 lt 0

1 0 le 119909 lt 120587(82)

which has for each truncation number 119873 the truncatedFourier series approximations

119892 (119909) ≃4120587

119873

sum119898=1

1(2119898 minus 1)

sin (2119898minus 1) 119909

119892 (119909) ≃8120587

119873

sum119898=1

119887119898119905119898

(1 + 1199052119898)=

8120587

119873

sum119898=1

1(2119898 minus 1)

119905119898

(1 + 1199052119898)

=8120587

119873

sum119898=1

1(2119898 minus 1)

tan ((2119898 minus 1) 2) 1199091 + [tan ((2119898 minus 1) 2) 119909]2

(83)

This signal has a jump discontinuity at 119909 = 0 ofmagnitude 2 We shall focus our computations on the Gibbsphenomenon overshoot near this discontinuity

Here is a printout list of results for the parameters of theposing inverse problem via computations by both the directand stereographic algorithms

It is remarkable how results by both algorithms arepractically identical and share the same following featuresthat are observed by results of the direct algorithm discussedbelow

Case I (119873 = 40) The computations of 1205720 for the previousexample using (71) with 1199090 = 002 lead to 1199091 = 0039 109this varies quickly with 119899 (only beyond the 4th decimal) tostabilize towards 1199093 = 1199094 = 1199095 = sdot sdot sdot = 0039 299 = 1205720Here using (1) gives 119892(1205720) = 119892(0039 299) = 1179 030 whichis quite close to the theoretically expected value by (A7) ofsim118

Then using (65) for 119888 = 109 and 1199090 = 0000381 leads to1199091 = 0021 697 this varies quickly with 119899 (only beyond the 3-d decimal) to stabilize quickly towards 1199095 = 1199096 = 1199097 = sdot sdot sdot =0028 368 = 1205721 Also using (65) for 119888 = 109 and 1199090 = 004leads to 1199091 = 0067 699 this varies quickly with 119899 (beyondthe 2nd decimal) to stabilize quickly towards 1199095 = 1199096 = 1199097 =sdot sdot sdot = 0052 873 = 1205722

For both1205721 and1205722 using (1) yields119892(1205721) = 119892(0028368) =119892(1205722) = 119892(0052873) = 109 = 119888 as expected

Case II (119873 = 100) Computations of 1205720 using (71) with1199090 = 002 lead to 1199091 = 0014 221 this varies quickly with119899 (only beyond the 4th decimal) to stabilize towards 1199093 =1199094 = 1199095 = sdot sdot sdot = 0015 708 = 1205720 Here using (1) gives119892(1205720) = 119892(0015708) = 1178 990 This result appears to beslightly excessive perhaps due to increased rounding off errorswith increasing 119873 to 100

Thenusing (65) for 119888 = 109 and1199090 = 0015 237 leads aftera negative value to 1199092 = 0014 462 this varies with 119899 (beyondthe 2nd decimal) to stabilize quickly towards 1199097 = 1199098 = 1199099 =sdot sdot sdot = 0011 3348 = 1205721 The used value of 1199090 is apparently notappropriate and this illustrates the generic ill-posedness of thisinverse problem

The pertaining computations of 1205722 with 1199090 = 0017000lead to 1199091 = 0020 434 this varies with 119899 (beyond the 3-ddecimal) to stabilize quickly towards 1199094 = 1199095 = 1199096 = sdot sdot sdot =0021 149 = 1205722

10 International Journal of Computational Mathematics

For both 1205721 and 1205722 using (1) yields 119892(1205721) =119892(00113348) = 119892(1205722) = 119892(0021149) = 109 = 119888 asexpected

Case III (119873 = 150) The computations of 1205720 using (71) with1199090 = 0005 lead to 1199091 = 0010 595 this varies insignificantlywith 119899 (only beyond the 4th decimal) to stabilize towards1199093 =1199094 = 1199095 = sdot sdot sdot = 0010 472 = 1205720 Here using (1) gives 119892(1205720) =119892(0010472) = 1178 980 which does not significantly differfrom 119892(1205720) when119873 = 100

Thenusing (65) for 119888 = 109 and1199090 = 0010 158 leads aftera negative value to 1199092 = 0009 643 and this varies (beyond the3-d decimal) to stabilize quickly towards1199097 = 1199098 = 1199099 = sdot sdot sdot =0007 565 = 1205721

The pertaining computations of 1205722 with 1199090 = 0010 800lead to 1199091 = 0025 998 this varies (beyond the 2nd decimal)to stabilize towards 11990914 = 11990915 = 11990916 = 11990917 = 11990918 = sdot sdot sdot =0014 099 = 1205722

For both 1205721 and 1205722 using (1) yields 119892(1205721) = 119892(007 565) =119892(1205722) = 119892(0014 099) = 109 = 119888

During the previous computations summarized inTable 1 it has been observed that varying 1199090 in the presentNewton-Raphson process for all119873 can lead to strong oscil-lations and even completely unstable values for 1205720 1205721 and 1205722of the Fourier series inverse problem which though ill-posed

is practically regularizible Apart from this drawback the(71) with (65) iterative algorithms (i) converge quickly to thetheoretically expected values and are (ii) of demonstratedstability of the iterations for properly chosen 1199090 (which act aseffective regularizing parameters) Moreover computationsby the stereographic algorithm invoking (73) and (67) sharethe samemagnitude and convergence rates with results by thedirect algorithm Finally it is demonstrated that the presentsignal exhibits an overshoot when119873 = 150 ofsim9of a jumpdiscontinuity of magnitude 2 that is located at a distance ofsim00105

To illustrate robustness of both numerical algorithms andtheir possible effective regularizibility we shall study also thenumerical inverse problem for the Fourier series representa-tion of an arbitrary (nonsymmetric) periodic signal

Example 10 Consider the 2120587-periodic signal 119891(119909) definedover one period by

119891 (119909) =

minus119909

120587minus 1 minus120587 lt 119909 lt 0

1 0 le 119909 lt 120587(84)

It has for each truncation number 119873 the truncated Fourierseries approximations

119892 (119909) ≃14+119873

sum119898=1

[(minus1)119898 minus 1]12058721198982 cos119898119909+

[2 minus (minus1)119898]120587119898

sin119898119909

119892 (119909) ≃14

+119873

sum119898=1

[(minus1)119898 minus 1] 12058721198982 + 2 ([2 minus (minus1)119898] 120587119898) tan ((2119898 minus 1) 2) 119909 minus ([(minus1)119898 minus 1] 12058721198982) [tan ((2119898 minus 1) 2) 119909]2

1 + [tan ((2119898 minus 1) 2) 119909]2

(85)

This signal has like the signal of Example 9 a jumpdiscontinuity at 119909 = 0 of magnitude 2 For the sake ofcomparison we shall also focus our computations on theGibbs phenomenon overshoot part of 119876

119904120590119873(119909) near this

discontinuityIt should be noted here that the results by both algorithms

summarized in Table 2 are also identical a feature indicatingthat the degree of ill-posedness of the posing inverse problemis not affected as predicted by Proposition 8 by the tangenthalf-angle stereographic projection

Case I (119873 = 40) The computations of 1205720 for this exampleusing (71) and (73) with 1199090 = 003 lead to an identicalsequence 119909

119899 that converges by both algorithms after the 1199094

iteration to 0078 139 with 119892(1205720) = 119892(1205720) = 1178 76

Case II (119873 = 100) The computations of 1205720 with 1199090 = 001lead to an identical sequence 119909

119899 that converges by both

algorithms after the 1199095 iteration to 0031 351 with 119892(1205720) =

119892(1205720) = 1178 87

Case III (119873 = 150) The computations of 1205720 with 1199090 = 0007lead also to an identical sequence 119909

119899 that converges by both

algorithms after the 1199095 iteration to 0020 915 with 119892(1205720) =

119892(1205720) = 1178 91

Therefore the present signal exhibits an overshoot when119873 = 150 of sim9 of a jump discontinuity of magnitude 2 thatis located at a distance of sim0020 9

Comparison of the computed Gibbs overshoot parame-ters of the two considered signals with jump discontinuitiesat the same 119909 = 0 of the same magnitude illustrates that therelative size of the overshoot depends essentially exclusivelyon the magnitude of the stimulating jump discontinuityFor instance based on the present numerical results onecan suggest here that nonsymmetric signals have equalovershoots with symmetric signals Their respective 1205720rsquos arehowever different Indeed the nonsymmetric signal has asignificantly larger 1205720 (0020 9) than the 1205720 (00105) of thesymmetric signal

International Journal of Computational Mathematics 11

Table 1 Results of computations for Example 9

119873 1205721 1205720 1205722

40 0028 368 0039 299 0052 873100 0011 335 0015 705 0021 149150 0007 565 0010 472 0014 099

Table 2 Results of computations for Example 10

119873 1205721 1205720 1205722

40 00564765 00781391 0105155100 00226546 00313516 00422024150 00151124 00209153 0028156

Table 3 Summary of computations of all overshoot parameters inExample 10

1205720results

119873 1199090

119909lowast

1205720

119892(1205720)

100 0030 1199093

0031 3516 1178 87200 0015 119909

30015 6919 1178 92

300 0008 1199093

0010 4648 1178 941205721results

119873 1199090

119909lowast

1205721

119888

0025 1199094

0022 2027 108100 0025 119909

30023 1344 110

0025 1199093

0024 2027 1120010 119909

30011 1117 108

200 0010 1199093

0011 5779 1100010 119909

40012 1123 112

0008 1199093

0007 410 08 108300 0008 119909

30007 720 95 110

0008 1199092

0008 077 32 1121205722results

119873 1199090

119909lowast

1205722

119888

100035 119909

40042 9439 108

035 1199094

0041 4445 110035 119909

40039 8513 112

2000018 119909

30021 4960 108

0018 1199093

0020 7456 1100018 119909

30019 9483 112

3000012 119909

30014 3360 108

0012 1199093

0013 8356 1100012 119909

40013 3039 112

A further analysis has also been performed of Example 10with 119873 = 100 200 300 to compute the corresponding1205720 with 1205721 1205722 that correspond simultaneously to 119888 =108 110 112 as illustrated in Figure 1 Results of thesecomputations are summarized in Table 3 and then used forthe partial 119876

119904120590119873(119909) plots of Figure 2 that interpolate 1205721 1205722

(for 119888 = 108 110 112) with 1205720 for each 119873 = 100 200 300In this table119909

lowaststands for the first convergence step of the 119909

119899

sequence of Newton Raphson iterations

002

05

10

004

g(x

)

xminus004 minus002

minus05

minus10

g(x) for N = 100

g(x) for N = 200

g(x) for N = 300

h(x) = c = 108

h(x) = c = 11

h(x) = c = 112

1205721 1205720 1205722

Figure 1 Plots of 119892(119909) near a discontinuity of 119891(119909) in Example 10

120

115

110

105

100

000 001 002 003 004 005

g(x

)

x

g(x) for N = 100

g(x) for N = 200

g(x) for N = 300

h(x) = c = 108

h(x) = c = 11

h(x) = c = 112

Figure 2 Plots of the 119892(119909) curves near a discontinuity of 119891(119909) inExample 10 interpolating 1205721 1205722 (for 119888 = 108 110 112) with 1205720 allfor119873 = 100 200 300

These results turn out to be extremely stable against varia-tions of the trial initial root1199090 Indeed in the computationallymost difficult situation namely when 119873 = 300 a changeof 1199090 from 0008 to 0020 leads to a variation of 119909

lowastfrom

1199092 to just 1199096 with the same asymptotic 1205720 = 00104648 and1205721 = 000807732 (for 119888 = 112)

5 Conclusions

This paper demonstrates that the quadratic degeneracy ofthe solution of the present inverse problem explicit whenusing 119867(120574 119909119898) is a nonlinear indicator of the existence ofGibbsrsquo phenomenon It provides also a distribution-theoreticproof for the existence of this phenomenon The reportedanalysis of numerical solvability of this problem illustrates thefollowing

(i) When suminfin

119898=1 119898(120587119871)119866lowast(120574 119909119898) = suminfin

119898=1 119898(120587119871)119867lowast(120574 119909119898) converge continuously in 119909 and are

12 International Journal of Computational Mathematics

nonzero at 1205721 (or 1205722) there exists a neighborhood(defined by 119868) of 120572 such that for all starting values 1199090in the neighborhood 119909

119899 will converge to 1205721 (or 1205722)

(ii) When suminfin

119898=1 119898(120587119871)119866lowast(120574 119909119898)=suminfin

119898=1 119898(120587119871)119867lowast(120574119909119898) converge continuously in 119909 and are nonzero at1205721 and 1205722 (but not at 1205720) and if 120593(119909) has a secondderivative which might be zero at 1205721 and 1205722 (and at1205720) then the convergence of 119909119899 is quadratic or fasterIf12059310158401015840(119909) = 0 at1205721 and1205722 then the convergence of 119909119899is merely quadratic

(iii) The convergence of the Newton-Raphson iterativeprocess to 1205720 can also be quadratic but restrictionson the derivatives of 120593(119909) have to be observed up tothe third order Otherwise the convergence of 119909

119899 to

1205720 can slow down to become only linear [12] with arate log210

(iv) The reported computations clearly illustrate that1205720 rarr 0 and |1205721 minus 1205722| rarr 0 when 119873 rarr infin thatis the infinite Fourier series will sum to 119891(119909) inthe neighborhood of a discontinuity except at thediscontinuity itself

Appendix

To illustrate the Gibbs-Wilbraham effect we shall use herethe same arguments employed by Wilbraham in 1848 inanalyzing a similar square wave

Example A1 Let 119891(119909) be a square wave periodic signalwhich is defined over one period by

119891 (119909) =

minus1 minus120587 lt 119909 le 0

1 0 le 119909 lt 120587(A1)

Its Fourier series representation is

119892 (119909) =4120587

infin

sum119898=1

1(2119898 minus 1)

sin (2119898minus 1) 119909 (A2)

This defines the truncated series

119878119873(119909) =

4120587

119873

sum119898=1

1(2119898 minus 1)

sin (2119898minus 1) 119909 (A3)

which may be differentiated as

119889

119889119909119878119873(119909) =

4120587

119873

sum119898=1

cos (2119898minus 1) 119909 (A4)

Since

2119873

sum119898=1

cos (2119899 minus 1) 119909 =sin 2119873119909

sin119909 (A5)

then119889

119889119909119878119873(119909) =

2120587

sin 2119873119909

sin119909 (A6)

It is possible to cancel the previous artificially induceddifferentiation by integrating the last result as

119878119873(119909) = int

119909

0

119889

119889120585119878119873(120585) 119889120585 =

2120587int119909

0

sin 2119873120585

sin 120585119889120585 (A7)

This integral starts at zero when 119909 = 0 in agreement with(A2) and increases until 2119873120585 = 120587 Then 120585 = 1205872119873 definesa turning point 1199091 The integral starts to decrease at whichpoint the numerator sin 2119873120585 goes negative For large 119873 thedenominator sin 120585 remains positive

The form (A7) distinctively from (A3) reveals that119878119873(119909) has turning points at the zeros of sin 2119873120585 These occur

when 2119873120585 = 119903120587 119903 = 1 2 3 4 5 that is at 120585 = 119903(1205872119873)Clearly then 119909

119903sim 120585119903= 119903(1205872119873) are turning points for 119878

119873(119909)

In particular 1199091 sim 1205851 = 1205872119873 defines the overshoot peaklocation 1205720 and 119878

119873(1199091) = (2120587) int12058721198730 (sin 2119873120585 sin 120585) 119889120585 is

its magnitude To evaluate 119878119873(1199091) = 119878

119873(1205720) numerically

let us change the variable of integration namely 2119873120585 = 119908or 120585 = 1199082119873 to rewrite the previous integral as 119878

119873(1199091) =

(2120587) int1205870 (sin119908(sin1199082119873)) 119889(1199082119873) Clearly when 119873 rarr

infin sin1199082119873 = 1199082119873 and

lim119873rarrinfin

119878119873(1199091) =

2120587int120587

0

sin119908119908

119889119908 =2120587Si (120587) = 118 (A8)

correct to two decimal pointsAs for 1199092 1199093 1199094 119909119903 lim119873rarrinfin[119878119873(119909119903) minus 119891(119909)]

decreases as we move away from the discontinuity butit should be remarked however that the location of theovershoot 1205720 = 1199091 = 1205872119873 moves towards the discontinuityaccording to

lim119873rarrinfin

1205720 = lim119873rarrinfin

120587

2119873= 0 (A9)

Conflict of Interests

The author declares that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

The numerical computations of Examples 9 and 10 have beenperformed using Mathematica 8 by Mounir Abou Yassineof the Department of CCE of AUL

References

[1] P L Butzer and R J Nessel Fourier Analysis and Approxima-tion Volume I One-Dimensional Theory Birkhauser 1971

[2] R E Edwards Fourier Series A Modern Introduction HoltRinehart amp Winston 1967

[3] W J Thompson ldquoFourier series and the Gibbs phenomenonrdquoAmerican Journal of Physics vol 60 no 5 pp 425ndash429 1992

[4] A J JerriThe Gibbs Phenomenon in Fourier Analysis Splines ampWavelet Approximations Kluwer Academic Publishers BostonMass USA 1990

International Journal of Computational Mathematics 13

[5] K S Rim and B I Yun ldquoGibbs phenomenon removal by addingHeaviside functionsrdquo Advances in Computational Mathematicsvol 38 no 4 pp 683ndash699 2013

[6] A H Zemanian Distribution Theory and Transform AnalysisMcGraw-Hill 1965

[7] G R Ayers and J C Dainty ldquoIterative blind deconvolutionmethod and its applicationsrdquo Optics Letters vol 13 no 7 pp547ndash549 1988

[8] N H S Haidar ldquoA nonlinear method for a severely ill-posedgeneralized moment problemrdquo Journal of Inverse amp Ill-PosedProblems vol 5 no 4 pp 323ndash335 1997

[9] G L Bradley and K J Smith Calculus Prentice Hall 1995[10] P Offner T Sonar and M Wirz ldquoDetecting strength and

location of jump discontinuities in numerical datardquo AppliedMathematics vol 4 no 12 pp 1ndash14 2013

[11] K E Atkinson An Introduction to Numerical Analysis JohnWiley amp Sons New York NY USA 1989

[12] E Suli and D F Mayers An Introduction to Numerical AnalysisCambridge University Press Cambridge UK 2003

[13] P Wang ldquoA third-order family of Newton-like iteration meth-ods for solving nonlinear equationsrdquo Journal of NumericalMathematics and Stochastics vol 3 no 1 pp 13ndash19 2011

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Stochastic AnalysisInternational Journal of

International Journal of Computational Mathematics 7

In conclusion the quadratic degeneracy of the solution ofthe present inverse problem explicit when using 119867(120574 119909119898)is a nonlinear indicator of the existence of the peak 119876

119904120590119873(119909)

pertaining to the Gibbs phenomenon However the numer-ical ill-posedness associated with the relative values of atrial iterative variable in the 120574 set happens to remarkablyremain invariant under the tangent half-angle stereographicprojection

Following the same arguments of Wilbraham we maycancel the artificially induced differentiation by integrating(50) to represent the stereographic Fourier series 119892(119909) inintegral form as

119892 (119909) =11988602minusinfin

sum119898=1

int119909

0119898

sdot120587

119871

(119887119898tan119898(1205872119871) 120585 minus 119886

119898)2minus (1198862119898+ 1198872119898)

119887119898(tan2119898(1205872119871) 120585 + 1)

119889120585 =11988602

minusinfin

sum119898=1

int119909

0119898

sdot120587

119871

(119887119898tan2119898(1205872119871) 120585 minus 2119886

119898tan119898(1205872119871) 120585 minus 119887

119898)

(tan2119898(1205872119871) 120585 + 1)119889120585

(61)

4 Numerics of Solving the Inverse Problem

The solution of the present inverse problem is in fact aproblem of finding the roots of (44) when 119892(119909) is truncatedat 119898 = 119873 For that reason we shall focus on such rootsin the neighborhood of a step-like discontinuity of 119891(119909) at119909 = 119909119900 say where 119891(119909119900 minus 120576) lt 119891(119909119900 + 120576) 0 lt 120576 ≪ 1 Theassociated with the Gibbs phenomenon overshoot of 119892(119909) =11988602+sum

119873

119898=1 119866(120574 119909119898) or 11988602+sum119873

119898=1 119867(120574 119909119898) over119891(119909119900+120576)and undershoot of this 119892(119909) below 119891(119909119900minus120576)The intersectionof a horizontal line ℎ(119909) = 119888 ge 119891(119909119900 + 120576) corresponding to120593(119909) = 0 can be at two points 1205721 and 1205722 on both sides of thepeak location 1205720 representing two distinct roots of 120593(119909) = 0It can also be at one point 1205720 corresponding to themaximumof the truncated 119892(119909) that is a double root of 120593(119909) = 0

Obviously 1205721 lt 1205721 lt 1205720 lt 1205722 lt 1205722 where 1205721 and 1205722 arepossible points of inflection defined by

11989210158401015840 (1205721) = 12059310158401015840 (1205721) = 11989210158401015840 (1205722) = 12059310158401015840 (1205722) = 0 (62)

The existence of both 1205721 and 1205722 or at least one of themnamely 1205722 is however not mandatory Moreover for the 12057211205720 and 1205722 points we expect to have

1198921015840 (1205721) = 1205931015840 (1205721) gt 0

12059310158401015840 (1205721) gt 0

120593101584010158401015840 (1205721) lt 0

1198921015840 (1205720) = 1205931015840 (1205720) = 0

12059310158401015840 (1205720) lt 0

120593101584010158401015840 (1205720) = 0

1198921015840 (1205722) = 1205931015840 (1205722) lt 0

12059310158401015840 (1205722) gt 0

120593101584010158401015840 (1205722) gt 0(63)

Equation (44) can be solved for119909 120593(119909) = 0 by aNewton-Raphson (see eg [11 12]) iterative process

119909119899+1 = 119909

119899minus

120593 (119909119899)

1205931015840 (119909119899) (64)

which is repeated until a sufficiently accurate value is reachedThis process for each series truncation number119873 rewrites as

119909119899+1 = 119909

119899+[sum119873

119898=1 119866 (120574 119909119899 119898) + 11988602 minus 119888]

[sum119873

119898=1 119898(120587119871)119866lowast (120574 119909119899 119898)]

(65)

or in the stereographic form

119909119899+1 = 119909

119899+[sum119873

119898=1 119867(120574 119909119899 119898) + 11988602 minus 119888]

[sum119873

119898=1 119898(120587119871)119867lowast (120574 119909119899 119898)]

(66)

which is the same as

119909119899+1 = 119909

119899

+[sum119873

119898=1 ((119886119898 + 2119887119898119905119898minus 1198861198981199052119898) (1 + 1199052

119898)) + 11988602 minus 119888]

[sum119873

119898=1 119898(120587119871) ((1198871198981199052119898+ 2119886119898119905119898minus 119887119898) (1 + 1199052

119898))]

(67)

and is started off with some rather free initial value 1199090Obviously here 119905

119898= tan119898(1205872119871)119909

The sequence 119909119899will usually converge provided that 1199090

is close enough to the unknown zero 120572 and that

1205931015840 (1199090) = minusinfin

sum119898=1

119898120587

119871119866lowast (120574 1199090 119898) = 0

or 1205951015840 (1199090) = minusinfin

sum119898=1

119898120587

119871119867lowast (120574 1199090 119898) = 0

(68)

41 Computation of the Peak Location 1205720 Being a maximiza-tion problem for 120593(119909) = 0 the determination of 1205720 by theNewton-Raphson process [11] becomes

119909119899+1 = 119909

119899minus

1205931015840 (119909119899)

12059310158401015840 (119909119899)

(69)

or

119909119899+1 = 119909

119899minus

1205951015840 (119909119899)

12059510158401015840 (119909119899) (70)

in stereographic form

8 International Journal of Computational Mathematics

Straightforwardly this process can be expressed as

119909119899+1 = 119909

119899minus

[sum119873

119898=1 119898119866lowast (120574 119909119899 119898)]

[sum119873

119898=1 1198982 (120587119871) 119866 (120574 119909

119899 119898)]

(71)

or in the stereographic form

119909119899+1 = 119909

119899

minus[minussum119873

119898=1 119898(120587119871)119867lowast (120574 119909119899 119898)]

[sum119873

119898=1 1198982 (12058721198712) ((119886

1198981199052119898minus 2119887119898119905119898minus 119886119898) (1 + 1199052

119898))]

(72)

which is the same as119909119899+1 = 119909

119899

+[sum119873

119898=1 119898((1198871198981199052119898+ 2119886119898119905119898minus 119887119898) (1 + 1199052

119898))]

[sum119873

119898=1 1198982 (120587119871) ((119886

1198981199052119898minus 2119887119898119905119898minus 119886119898) (1 + 1199052

119898))]

(73)

It is remarkable how this algorithm is distinctively indepen-dent of 119888

Proposition 8 The ill-posedness of the inverse problem ofthe truncated Fourier series is invariant under stereographicprojection

Proof The tangent half-angle stereographic projectionthough not isometric (distance preserving) is neverthelessvariability preserving Indeed the main ingredients ofthis inverse problem are namely 120574 = 119886

119898 119887119898119873119898=1 119888 and

1199090 with (119889119889119909)119866[120579119898(119909)]1199090 Under this projection 120579

119898(1199090)

transforms to 119905119898(1199090) = tan[120579

119898(1199090)2] and 119866[120579

119898(1199090)]

transforms to 119867[119905119898(1199090)] But according to Proposition 4

(119889119889119909)119867[119905119898(119909)]1199090

= (119889119889119909)119866[120579119898(119909)]1199090 Hence the variation

of 119866(120574 1199090 119898) with 1199090 in the Fourier series inverse problemand the variation of119867(120574 1199090 119898) with 1199090 in the stereographicform are identical

Equivalently this makes the ill-posedness of this inverseproblem invariant under stereographic projection

Expected manifestations of the correctness of this propo-sition are as follows (i) the results of computations by the(71) and (73) algorithms for 1205720 with the same 1199090 must beidentical and (ii) the results of computations by the (65) and(67) algorithms for 1205721 (or 1205722) with the same 1199090 and 119888 mustalso be identical

42 Convergence Analysis Let us focus first on the con-vergence of the iterative process (65) or (67) In manysituationswhere term-by-termdifferentiability does not holdtheNewton-Raphsonmethod can always be replaced by somequasi-Newtonmethod [12 13] where 1205931015840(119909

119899) or1205951015840(119909

119899) in (65)

or (67) is replaced by a suitable approximation For examplein the chord method 1205931015840(119909

119899) is replaced by 1205931015840(1199090) for all

iterationsIf 120598119899= 120572minus119909

119899 120598119899+1 = 120572minus119909

119899+1 and120573119899 is some point between119909119899and 120572 it can be easily proved (see [11] or [12]) that

120598119899+1 = minus

12059310158401015840 (120573119899)

21205931015840 (119909119899)1205982119899 (74)

Taking absolute value of both sides gives

1003816100381610038161003816120598119899+11003816100381610038161003816 =

1003816100381610038161003816100381612059310158401015840 (120573119899)10038161003816100381610038161003816

2 10038161003816100381610038161205931015840 (119909119899)10038161003816100381610038161205982119899 (75)

implying quadratic convergence that is 120598119899is squared (the

number of correct digits roughly doubles) at each step Thesame arguments apply of course to the 120595(119909) = 0 equation

In actual fact for a distinct 120572 the convergence of119909119899 in (65)ndash(67) is at least quadratic in the neigh-

borhood of this 120572 The process may face a difficultywhen 1205931015840(119909

119899) = minussum

119873

119898=1 119898(120587119871)119866lowast(120574 119909119899 119898) or 1205951015840(119909

119899) =

minussum119873

119898=1 119898(120587119871)119867lowast(120574 119909119899 119898) diverges This is however a spe-

cial but not a general problem and (75) always holds if thefollowing conditions are satisfied

(i) 1205931015840(119909) = 0 forall119909 isin 119868 where 119868 = [120572 minus 120578 120572 + 120578] for some120578 ge |(120572 minus 1199090)|

(ii) 12059310158401015840(119909) is finite forall119909 isin 119868(iii) 1199090 is sufficiently close to 120572

The term sufficiently close in this context means the follow-ing

(a) Taylor approximation is accurate enough such that wecan ignore higher order terms

(b) (12)|12059310158401015840(119909119899)1205931015840(119909

119899)| lt 119870|12059310158401015840(120572)1205931015840(120572)| for some119870 lt

infin(c) 119870|12059310158401015840(120572)1205931015840(120572)|120598

119899lt 1 for 119899 isin N

Consequently (75) can be expressed in the following way1003816100381610038161003816120598119899+1

1003816100381610038161003816 = 1198721205982119899 (76)

where

119872 = sup119909isin119868

12

100381610038161003816100381610038161003816100381610038161003816

12059310158401015840 (119909)

1205931015840 (119909)

100381610038161003816100381610038161003816100381610038161003816 (77)

The initial point 1199090 has to be chosen such that conditions(i) through (iii) are satisfied where the third conditionrequires that 119872|1205980| lt 1 Moreover with increasing 119873 thedistance |1205721 minus 1205722| is known to decrease in a way makingthe satisfaction of these conditions very difficult and actuallyimpossible as119873 rarr infin Here it should be underlined that theiterative method (65)ndash(67) may fail to converge to the 120572 rootin the following situations

(i) If a point like 1205720 is included in 119868 the method willterminate due to division by zero

(ii) When the initial estimate 1199090 is poor the pertainingwrong 119868 can contribute to nonconvergence of thealgorithm

As for convergence of the Newton-Raphson processtowards 1205720 let 120576119899 = 1205720 minus 119909

119899 120576119899+1 = 1205720 minus 119909

119899+1 and 120579119899is some

point between 119909119899and 1205720 to arrive at

120576119899+1 = minus

120593101584010158401015840 (120579119899)

212059310158401015840 (119909119899)1205762119899 (78)

1003816100381610038161003816120576119899+11003816100381610038161003816 =

10038161003816100381610038161003816120593101584010158401015840 (120579119899)10038161003816100381610038161003816

2 100381610038161003816100381612059310158401015840 (119909119899)10038161003816100381610038161205762119899 (79)

International Journal of Computational Mathematics 9

The convergence of 119909119899 in (71)ndash(73) is at least quadratic

in the neighborhood of this 1205720 The process may face a diffi-culty however when 12059310158401015840(119909

119899) = minussum

infin

119898=1 1198982(120587119871)2119866(120574 119909

119899 119898)

or 12059510158401015840(119909119899) diverges Moreover (79) always holds if the follow-

ing conditions are satisfied

(i) 12059310158401015840(119909) = 0 forall119909 isin 119868 where 119868 = [1205720 minus120578 1205720 +120578] for some120578 ge |(1205720 minus 1199090)|

(ii) 120593101584010158401015840(119909) is finite forall119909 isin 119868

(iii) 1199090 is sufficiently close to 1205720

Sufficient closeness in this context means that

(a) Taylor approximation is accurate enough such that wecan ignore higher order terms

(b) (12)|120593101584010158401015840(119909119899)12059310158401015840(119909

119899)| lt 119870|120593101584010158401015840(1205720)120593

10158401015840(1205720)| for some119870 lt infin

(c) 119870|120593101584010158401015840(1205720)12059310158401015840(1205720)|120576119899 lt 1 for 119899 isin N

Consequently

1003816100381610038161003816120576119899+11003816100381610038161003816 = 1198721205762

119899 (80)

where

119872 = sup119909isin119868

12

100381610038161003816100381610038161003816100381610038161003816

120593101584010158401015840 (119909)

12059310158401015840 (119909)

100381610038161003816100381610038161003816100381610038161003816 (81)

43 Applications The standard iterative algorithms (71) or(73) for finding the location of overshoots in 119876

119904120590119873(119909) of

the Gibbs effect employ the fact that the first derivative of apeak in 119892(119909) or 119892(119909) respectively has a downward-going-zero-crossing at the peak maximum But the presence ofmany neighboring smaller peaks in the ringing close to theovershoot will cause many undesirable zero-crossings to beinvoked if the initial trial root 1199090 is not correctly chosenThis fact makes the numerical problem of peak finding in theGibbs effect of pathological difficulty A difficulty that aggra-vates with the increase in the number 119873 of truncated termsin the employed Fourier series Indeed with the increaseof 119873 the number of undesirable zero-crossings increasesin the neighborhood of 1205720 as illustrated in Section 1 Thiscan increasingly lead to oscillatory and even unstable valuesin sequence 119909

119899 the more the distance |1199090 minus 1205720| contains

unwanted zero-crossingsThese problems have been resolved in the two examples

that follow which simultaneously illustrate the applicabilityof Proposition 8

Example 9 Consider the 2120587-periodic odd signal 119891(119909) ofExample A1 defined over one period by

119891 (119909) =

minus1 minus120587 lt 119909 lt 0

1 0 le 119909 lt 120587(82)

which has for each truncation number 119873 the truncatedFourier series approximations

119892 (119909) ≃4120587

119873

sum119898=1

1(2119898 minus 1)

sin (2119898minus 1) 119909

119892 (119909) ≃8120587

119873

sum119898=1

119887119898119905119898

(1 + 1199052119898)=

8120587

119873

sum119898=1

1(2119898 minus 1)

119905119898

(1 + 1199052119898)

=8120587

119873

sum119898=1

1(2119898 minus 1)

tan ((2119898 minus 1) 2) 1199091 + [tan ((2119898 minus 1) 2) 119909]2

(83)

This signal has a jump discontinuity at 119909 = 0 ofmagnitude 2 We shall focus our computations on the Gibbsphenomenon overshoot near this discontinuity

Here is a printout list of results for the parameters of theposing inverse problem via computations by both the directand stereographic algorithms

It is remarkable how results by both algorithms arepractically identical and share the same following featuresthat are observed by results of the direct algorithm discussedbelow

Case I (119873 = 40) The computations of 1205720 for the previousexample using (71) with 1199090 = 002 lead to 1199091 = 0039 109this varies quickly with 119899 (only beyond the 4th decimal) tostabilize towards 1199093 = 1199094 = 1199095 = sdot sdot sdot = 0039 299 = 1205720Here using (1) gives 119892(1205720) = 119892(0039 299) = 1179 030 whichis quite close to the theoretically expected value by (A7) ofsim118

Then using (65) for 119888 = 109 and 1199090 = 0000381 leads to1199091 = 0021 697 this varies quickly with 119899 (only beyond the 3-d decimal) to stabilize quickly towards 1199095 = 1199096 = 1199097 = sdot sdot sdot =0028 368 = 1205721 Also using (65) for 119888 = 109 and 1199090 = 004leads to 1199091 = 0067 699 this varies quickly with 119899 (beyondthe 2nd decimal) to stabilize quickly towards 1199095 = 1199096 = 1199097 =sdot sdot sdot = 0052 873 = 1205722

For both1205721 and1205722 using (1) yields119892(1205721) = 119892(0028368) =119892(1205722) = 119892(0052873) = 109 = 119888 as expected

Case II (119873 = 100) Computations of 1205720 using (71) with1199090 = 002 lead to 1199091 = 0014 221 this varies quickly with119899 (only beyond the 4th decimal) to stabilize towards 1199093 =1199094 = 1199095 = sdot sdot sdot = 0015 708 = 1205720 Here using (1) gives119892(1205720) = 119892(0015708) = 1178 990 This result appears to beslightly excessive perhaps due to increased rounding off errorswith increasing 119873 to 100

Thenusing (65) for 119888 = 109 and1199090 = 0015 237 leads aftera negative value to 1199092 = 0014 462 this varies with 119899 (beyondthe 2nd decimal) to stabilize quickly towards 1199097 = 1199098 = 1199099 =sdot sdot sdot = 0011 3348 = 1205721 The used value of 1199090 is apparently notappropriate and this illustrates the generic ill-posedness of thisinverse problem

The pertaining computations of 1205722 with 1199090 = 0017000lead to 1199091 = 0020 434 this varies with 119899 (beyond the 3-ddecimal) to stabilize quickly towards 1199094 = 1199095 = 1199096 = sdot sdot sdot =0021 149 = 1205722

10 International Journal of Computational Mathematics

For both 1205721 and 1205722 using (1) yields 119892(1205721) =119892(00113348) = 119892(1205722) = 119892(0021149) = 109 = 119888 asexpected

Case III (119873 = 150) The computations of 1205720 using (71) with1199090 = 0005 lead to 1199091 = 0010 595 this varies insignificantlywith 119899 (only beyond the 4th decimal) to stabilize towards1199093 =1199094 = 1199095 = sdot sdot sdot = 0010 472 = 1205720 Here using (1) gives 119892(1205720) =119892(0010472) = 1178 980 which does not significantly differfrom 119892(1205720) when119873 = 100

Thenusing (65) for 119888 = 109 and1199090 = 0010 158 leads aftera negative value to 1199092 = 0009 643 and this varies (beyond the3-d decimal) to stabilize quickly towards1199097 = 1199098 = 1199099 = sdot sdot sdot =0007 565 = 1205721

The pertaining computations of 1205722 with 1199090 = 0010 800lead to 1199091 = 0025 998 this varies (beyond the 2nd decimal)to stabilize towards 11990914 = 11990915 = 11990916 = 11990917 = 11990918 = sdot sdot sdot =0014 099 = 1205722

For both 1205721 and 1205722 using (1) yields 119892(1205721) = 119892(007 565) =119892(1205722) = 119892(0014 099) = 109 = 119888

During the previous computations summarized inTable 1 it has been observed that varying 1199090 in the presentNewton-Raphson process for all119873 can lead to strong oscil-lations and even completely unstable values for 1205720 1205721 and 1205722of the Fourier series inverse problem which though ill-posed

is practically regularizible Apart from this drawback the(71) with (65) iterative algorithms (i) converge quickly to thetheoretically expected values and are (ii) of demonstratedstability of the iterations for properly chosen 1199090 (which act aseffective regularizing parameters) Moreover computationsby the stereographic algorithm invoking (73) and (67) sharethe samemagnitude and convergence rates with results by thedirect algorithm Finally it is demonstrated that the presentsignal exhibits an overshoot when119873 = 150 ofsim9of a jumpdiscontinuity of magnitude 2 that is located at a distance ofsim00105

To illustrate robustness of both numerical algorithms andtheir possible effective regularizibility we shall study also thenumerical inverse problem for the Fourier series representa-tion of an arbitrary (nonsymmetric) periodic signal

Example 10 Consider the 2120587-periodic signal 119891(119909) definedover one period by

119891 (119909) =

minus119909

120587minus 1 minus120587 lt 119909 lt 0

1 0 le 119909 lt 120587(84)

It has for each truncation number 119873 the truncated Fourierseries approximations

119892 (119909) ≃14+119873

sum119898=1

[(minus1)119898 minus 1]12058721198982 cos119898119909+

[2 minus (minus1)119898]120587119898

sin119898119909

119892 (119909) ≃14

+119873

sum119898=1

[(minus1)119898 minus 1] 12058721198982 + 2 ([2 minus (minus1)119898] 120587119898) tan ((2119898 minus 1) 2) 119909 minus ([(minus1)119898 minus 1] 12058721198982) [tan ((2119898 minus 1) 2) 119909]2

1 + [tan ((2119898 minus 1) 2) 119909]2

(85)

This signal has like the signal of Example 9 a jumpdiscontinuity at 119909 = 0 of magnitude 2 For the sake ofcomparison we shall also focus our computations on theGibbs phenomenon overshoot part of 119876

119904120590119873(119909) near this

discontinuityIt should be noted here that the results by both algorithms

summarized in Table 2 are also identical a feature indicatingthat the degree of ill-posedness of the posing inverse problemis not affected as predicted by Proposition 8 by the tangenthalf-angle stereographic projection

Case I (119873 = 40) The computations of 1205720 for this exampleusing (71) and (73) with 1199090 = 003 lead to an identicalsequence 119909

119899 that converges by both algorithms after the 1199094

iteration to 0078 139 with 119892(1205720) = 119892(1205720) = 1178 76

Case II (119873 = 100) The computations of 1205720 with 1199090 = 001lead to an identical sequence 119909

119899 that converges by both

algorithms after the 1199095 iteration to 0031 351 with 119892(1205720) =

119892(1205720) = 1178 87

Case III (119873 = 150) The computations of 1205720 with 1199090 = 0007lead also to an identical sequence 119909

119899 that converges by both

algorithms after the 1199095 iteration to 0020 915 with 119892(1205720) =

119892(1205720) = 1178 91

Therefore the present signal exhibits an overshoot when119873 = 150 of sim9 of a jump discontinuity of magnitude 2 thatis located at a distance of sim0020 9

Comparison of the computed Gibbs overshoot parame-ters of the two considered signals with jump discontinuitiesat the same 119909 = 0 of the same magnitude illustrates that therelative size of the overshoot depends essentially exclusivelyon the magnitude of the stimulating jump discontinuityFor instance based on the present numerical results onecan suggest here that nonsymmetric signals have equalovershoots with symmetric signals Their respective 1205720rsquos arehowever different Indeed the nonsymmetric signal has asignificantly larger 1205720 (0020 9) than the 1205720 (00105) of thesymmetric signal

International Journal of Computational Mathematics 11

Table 1 Results of computations for Example 9

119873 1205721 1205720 1205722

40 0028 368 0039 299 0052 873100 0011 335 0015 705 0021 149150 0007 565 0010 472 0014 099

Table 2 Results of computations for Example 10

119873 1205721 1205720 1205722

40 00564765 00781391 0105155100 00226546 00313516 00422024150 00151124 00209153 0028156

Table 3 Summary of computations of all overshoot parameters inExample 10

1205720results

119873 1199090

119909lowast

1205720

119892(1205720)

100 0030 1199093

0031 3516 1178 87200 0015 119909

30015 6919 1178 92

300 0008 1199093

0010 4648 1178 941205721results

119873 1199090

119909lowast

1205721

119888

0025 1199094

0022 2027 108100 0025 119909

30023 1344 110

0025 1199093

0024 2027 1120010 119909

30011 1117 108

200 0010 1199093

0011 5779 1100010 119909

40012 1123 112

0008 1199093

0007 410 08 108300 0008 119909

30007 720 95 110

0008 1199092

0008 077 32 1121205722results

119873 1199090

119909lowast

1205722

119888

100035 119909

40042 9439 108

035 1199094

0041 4445 110035 119909

40039 8513 112

2000018 119909

30021 4960 108

0018 1199093

0020 7456 1100018 119909

30019 9483 112

3000012 119909

30014 3360 108

0012 1199093

0013 8356 1100012 119909

40013 3039 112

A further analysis has also been performed of Example 10with 119873 = 100 200 300 to compute the corresponding1205720 with 1205721 1205722 that correspond simultaneously to 119888 =108 110 112 as illustrated in Figure 1 Results of thesecomputations are summarized in Table 3 and then used forthe partial 119876

119904120590119873(119909) plots of Figure 2 that interpolate 1205721 1205722

(for 119888 = 108 110 112) with 1205720 for each 119873 = 100 200 300In this table119909

lowaststands for the first convergence step of the 119909

119899

sequence of Newton Raphson iterations

002

05

10

004

g(x

)

xminus004 minus002

minus05

minus10

g(x) for N = 100

g(x) for N = 200

g(x) for N = 300

h(x) = c = 108

h(x) = c = 11

h(x) = c = 112

1205721 1205720 1205722

Figure 1 Plots of 119892(119909) near a discontinuity of 119891(119909) in Example 10

120

115

110

105

100

000 001 002 003 004 005

g(x

)

x

g(x) for N = 100

g(x) for N = 200

g(x) for N = 300

h(x) = c = 108

h(x) = c = 11

h(x) = c = 112

Figure 2 Plots of the 119892(119909) curves near a discontinuity of 119891(119909) inExample 10 interpolating 1205721 1205722 (for 119888 = 108 110 112) with 1205720 allfor119873 = 100 200 300

These results turn out to be extremely stable against varia-tions of the trial initial root1199090 Indeed in the computationallymost difficult situation namely when 119873 = 300 a changeof 1199090 from 0008 to 0020 leads to a variation of 119909

lowastfrom

1199092 to just 1199096 with the same asymptotic 1205720 = 00104648 and1205721 = 000807732 (for 119888 = 112)

5 Conclusions

This paper demonstrates that the quadratic degeneracy ofthe solution of the present inverse problem explicit whenusing 119867(120574 119909119898) is a nonlinear indicator of the existence ofGibbsrsquo phenomenon It provides also a distribution-theoreticproof for the existence of this phenomenon The reportedanalysis of numerical solvability of this problem illustrates thefollowing

(i) When suminfin

119898=1 119898(120587119871)119866lowast(120574 119909119898) = suminfin

119898=1 119898(120587119871)119867lowast(120574 119909119898) converge continuously in 119909 and are

12 International Journal of Computational Mathematics

nonzero at 1205721 (or 1205722) there exists a neighborhood(defined by 119868) of 120572 such that for all starting values 1199090in the neighborhood 119909

119899 will converge to 1205721 (or 1205722)

(ii) When suminfin

119898=1 119898(120587119871)119866lowast(120574 119909119898)=suminfin

119898=1 119898(120587119871)119867lowast(120574119909119898) converge continuously in 119909 and are nonzero at1205721 and 1205722 (but not at 1205720) and if 120593(119909) has a secondderivative which might be zero at 1205721 and 1205722 (and at1205720) then the convergence of 119909119899 is quadratic or fasterIf12059310158401015840(119909) = 0 at1205721 and1205722 then the convergence of 119909119899is merely quadratic

(iii) The convergence of the Newton-Raphson iterativeprocess to 1205720 can also be quadratic but restrictionson the derivatives of 120593(119909) have to be observed up tothe third order Otherwise the convergence of 119909

119899 to

1205720 can slow down to become only linear [12] with arate log210

(iv) The reported computations clearly illustrate that1205720 rarr 0 and |1205721 minus 1205722| rarr 0 when 119873 rarr infin thatis the infinite Fourier series will sum to 119891(119909) inthe neighborhood of a discontinuity except at thediscontinuity itself

Appendix

To illustrate the Gibbs-Wilbraham effect we shall use herethe same arguments employed by Wilbraham in 1848 inanalyzing a similar square wave

Example A1 Let 119891(119909) be a square wave periodic signalwhich is defined over one period by

119891 (119909) =

minus1 minus120587 lt 119909 le 0

1 0 le 119909 lt 120587(A1)

Its Fourier series representation is

119892 (119909) =4120587

infin

sum119898=1

1(2119898 minus 1)

sin (2119898minus 1) 119909 (A2)

This defines the truncated series

119878119873(119909) =

4120587

119873

sum119898=1

1(2119898 minus 1)

sin (2119898minus 1) 119909 (A3)

which may be differentiated as

119889

119889119909119878119873(119909) =

4120587

119873

sum119898=1

cos (2119898minus 1) 119909 (A4)

Since

2119873

sum119898=1

cos (2119899 minus 1) 119909 =sin 2119873119909

sin119909 (A5)

then119889

119889119909119878119873(119909) =

2120587

sin 2119873119909

sin119909 (A6)

It is possible to cancel the previous artificially induceddifferentiation by integrating the last result as

119878119873(119909) = int

119909

0

119889

119889120585119878119873(120585) 119889120585 =

2120587int119909

0

sin 2119873120585

sin 120585119889120585 (A7)

This integral starts at zero when 119909 = 0 in agreement with(A2) and increases until 2119873120585 = 120587 Then 120585 = 1205872119873 definesa turning point 1199091 The integral starts to decrease at whichpoint the numerator sin 2119873120585 goes negative For large 119873 thedenominator sin 120585 remains positive

The form (A7) distinctively from (A3) reveals that119878119873(119909) has turning points at the zeros of sin 2119873120585 These occur

when 2119873120585 = 119903120587 119903 = 1 2 3 4 5 that is at 120585 = 119903(1205872119873)Clearly then 119909

119903sim 120585119903= 119903(1205872119873) are turning points for 119878

119873(119909)

In particular 1199091 sim 1205851 = 1205872119873 defines the overshoot peaklocation 1205720 and 119878

119873(1199091) = (2120587) int12058721198730 (sin 2119873120585 sin 120585) 119889120585 is

its magnitude To evaluate 119878119873(1199091) = 119878

119873(1205720) numerically

let us change the variable of integration namely 2119873120585 = 119908or 120585 = 1199082119873 to rewrite the previous integral as 119878

119873(1199091) =

(2120587) int1205870 (sin119908(sin1199082119873)) 119889(1199082119873) Clearly when 119873 rarr

infin sin1199082119873 = 1199082119873 and

lim119873rarrinfin

119878119873(1199091) =

2120587int120587

0

sin119908119908

119889119908 =2120587Si (120587) = 118 (A8)

correct to two decimal pointsAs for 1199092 1199093 1199094 119909119903 lim119873rarrinfin[119878119873(119909119903) minus 119891(119909)]

decreases as we move away from the discontinuity butit should be remarked however that the location of theovershoot 1205720 = 1199091 = 1205872119873 moves towards the discontinuityaccording to

lim119873rarrinfin

1205720 = lim119873rarrinfin

120587

2119873= 0 (A9)

Conflict of Interests

The author declares that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

The numerical computations of Examples 9 and 10 have beenperformed using Mathematica 8 by Mounir Abou Yassineof the Department of CCE of AUL

References

[1] P L Butzer and R J Nessel Fourier Analysis and Approxima-tion Volume I One-Dimensional Theory Birkhauser 1971

[2] R E Edwards Fourier Series A Modern Introduction HoltRinehart amp Winston 1967

[3] W J Thompson ldquoFourier series and the Gibbs phenomenonrdquoAmerican Journal of Physics vol 60 no 5 pp 425ndash429 1992

[4] A J JerriThe Gibbs Phenomenon in Fourier Analysis Splines ampWavelet Approximations Kluwer Academic Publishers BostonMass USA 1990

International Journal of Computational Mathematics 13

[5] K S Rim and B I Yun ldquoGibbs phenomenon removal by addingHeaviside functionsrdquo Advances in Computational Mathematicsvol 38 no 4 pp 683ndash699 2013

[6] A H Zemanian Distribution Theory and Transform AnalysisMcGraw-Hill 1965

[7] G R Ayers and J C Dainty ldquoIterative blind deconvolutionmethod and its applicationsrdquo Optics Letters vol 13 no 7 pp547ndash549 1988

[8] N H S Haidar ldquoA nonlinear method for a severely ill-posedgeneralized moment problemrdquo Journal of Inverse amp Ill-PosedProblems vol 5 no 4 pp 323ndash335 1997

[9] G L Bradley and K J Smith Calculus Prentice Hall 1995[10] P Offner T Sonar and M Wirz ldquoDetecting strength and

location of jump discontinuities in numerical datardquo AppliedMathematics vol 4 no 12 pp 1ndash14 2013

[11] K E Atkinson An Introduction to Numerical Analysis JohnWiley amp Sons New York NY USA 1989

[12] E Suli and D F Mayers An Introduction to Numerical AnalysisCambridge University Press Cambridge UK 2003

[13] P Wang ldquoA third-order family of Newton-like iteration meth-ods for solving nonlinear equationsrdquo Journal of NumericalMathematics and Stochastics vol 3 no 1 pp 13ndash19 2011

Submit your manuscripts athttpwwwhindawicom

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

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Applied MathematicsJournal of

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Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Mathematical PhysicsAdvances in

Complex AnalysisJournal of

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Operations ResearchAdvances in

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Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Discrete Dynamics in Nature and Society

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

8 International Journal of Computational Mathematics

Straightforwardly this process can be expressed as

119909119899+1 = 119909

119899minus

[sum119873

119898=1 119898119866lowast (120574 119909119899 119898)]

[sum119873

119898=1 1198982 (120587119871) 119866 (120574 119909

119899 119898)]

(71)

or in the stereographic form

119909119899+1 = 119909

119899

minus[minussum119873

119898=1 119898(120587119871)119867lowast (120574 119909119899 119898)]

[sum119873

119898=1 1198982 (12058721198712) ((119886

1198981199052119898minus 2119887119898119905119898minus 119886119898) (1 + 1199052

119898))]

(72)

which is the same as119909119899+1 = 119909

119899

+[sum119873

119898=1 119898((1198871198981199052119898+ 2119886119898119905119898minus 119887119898) (1 + 1199052

119898))]

[sum119873

119898=1 1198982 (120587119871) ((119886

1198981199052119898minus 2119887119898119905119898minus 119886119898) (1 + 1199052

119898))]

(73)

It is remarkable how this algorithm is distinctively indepen-dent of 119888

Proposition 8 The ill-posedness of the inverse problem ofthe truncated Fourier series is invariant under stereographicprojection

Proof The tangent half-angle stereographic projectionthough not isometric (distance preserving) is neverthelessvariability preserving Indeed the main ingredients ofthis inverse problem are namely 120574 = 119886

119898 119887119898119873119898=1 119888 and

1199090 with (119889119889119909)119866[120579119898(119909)]1199090 Under this projection 120579

119898(1199090)

transforms to 119905119898(1199090) = tan[120579

119898(1199090)2] and 119866[120579

119898(1199090)]

transforms to 119867[119905119898(1199090)] But according to Proposition 4

(119889119889119909)119867[119905119898(119909)]1199090

= (119889119889119909)119866[120579119898(119909)]1199090 Hence the variation

of 119866(120574 1199090 119898) with 1199090 in the Fourier series inverse problemand the variation of119867(120574 1199090 119898) with 1199090 in the stereographicform are identical

Equivalently this makes the ill-posedness of this inverseproblem invariant under stereographic projection

Expected manifestations of the correctness of this propo-sition are as follows (i) the results of computations by the(71) and (73) algorithms for 1205720 with the same 1199090 must beidentical and (ii) the results of computations by the (65) and(67) algorithms for 1205721 (or 1205722) with the same 1199090 and 119888 mustalso be identical

42 Convergence Analysis Let us focus first on the con-vergence of the iterative process (65) or (67) In manysituationswhere term-by-termdifferentiability does not holdtheNewton-Raphsonmethod can always be replaced by somequasi-Newtonmethod [12 13] where 1205931015840(119909

119899) or1205951015840(119909

119899) in (65)

or (67) is replaced by a suitable approximation For examplein the chord method 1205931015840(119909

119899) is replaced by 1205931015840(1199090) for all

iterationsIf 120598119899= 120572minus119909

119899 120598119899+1 = 120572minus119909

119899+1 and120573119899 is some point between119909119899and 120572 it can be easily proved (see [11] or [12]) that

120598119899+1 = minus

12059310158401015840 (120573119899)

21205931015840 (119909119899)1205982119899 (74)

Taking absolute value of both sides gives

1003816100381610038161003816120598119899+11003816100381610038161003816 =

1003816100381610038161003816100381612059310158401015840 (120573119899)10038161003816100381610038161003816

2 10038161003816100381610038161205931015840 (119909119899)10038161003816100381610038161205982119899 (75)

implying quadratic convergence that is 120598119899is squared (the

number of correct digits roughly doubles) at each step Thesame arguments apply of course to the 120595(119909) = 0 equation

In actual fact for a distinct 120572 the convergence of119909119899 in (65)ndash(67) is at least quadratic in the neigh-

borhood of this 120572 The process may face a difficultywhen 1205931015840(119909

119899) = minussum

119873

119898=1 119898(120587119871)119866lowast(120574 119909119899 119898) or 1205951015840(119909

119899) =

minussum119873

119898=1 119898(120587119871)119867lowast(120574 119909119899 119898) diverges This is however a spe-

cial but not a general problem and (75) always holds if thefollowing conditions are satisfied

(i) 1205931015840(119909) = 0 forall119909 isin 119868 where 119868 = [120572 minus 120578 120572 + 120578] for some120578 ge |(120572 minus 1199090)|

(ii) 12059310158401015840(119909) is finite forall119909 isin 119868(iii) 1199090 is sufficiently close to 120572

The term sufficiently close in this context means the follow-ing

(a) Taylor approximation is accurate enough such that wecan ignore higher order terms

(b) (12)|12059310158401015840(119909119899)1205931015840(119909

119899)| lt 119870|12059310158401015840(120572)1205931015840(120572)| for some119870 lt

infin(c) 119870|12059310158401015840(120572)1205931015840(120572)|120598

119899lt 1 for 119899 isin N

Consequently (75) can be expressed in the following way1003816100381610038161003816120598119899+1

1003816100381610038161003816 = 1198721205982119899 (76)

where

119872 = sup119909isin119868

12

100381610038161003816100381610038161003816100381610038161003816

12059310158401015840 (119909)

1205931015840 (119909)

100381610038161003816100381610038161003816100381610038161003816 (77)

The initial point 1199090 has to be chosen such that conditions(i) through (iii) are satisfied where the third conditionrequires that 119872|1205980| lt 1 Moreover with increasing 119873 thedistance |1205721 minus 1205722| is known to decrease in a way makingthe satisfaction of these conditions very difficult and actuallyimpossible as119873 rarr infin Here it should be underlined that theiterative method (65)ndash(67) may fail to converge to the 120572 rootin the following situations

(i) If a point like 1205720 is included in 119868 the method willterminate due to division by zero

(ii) When the initial estimate 1199090 is poor the pertainingwrong 119868 can contribute to nonconvergence of thealgorithm

As for convergence of the Newton-Raphson processtowards 1205720 let 120576119899 = 1205720 minus 119909

119899 120576119899+1 = 1205720 minus 119909

119899+1 and 120579119899is some

point between 119909119899and 1205720 to arrive at

120576119899+1 = minus

120593101584010158401015840 (120579119899)

212059310158401015840 (119909119899)1205762119899 (78)

1003816100381610038161003816120576119899+11003816100381610038161003816 =

10038161003816100381610038161003816120593101584010158401015840 (120579119899)10038161003816100381610038161003816

2 100381610038161003816100381612059310158401015840 (119909119899)10038161003816100381610038161205762119899 (79)

International Journal of Computational Mathematics 9

The convergence of 119909119899 in (71)ndash(73) is at least quadratic

in the neighborhood of this 1205720 The process may face a diffi-culty however when 12059310158401015840(119909

119899) = minussum

infin

119898=1 1198982(120587119871)2119866(120574 119909

119899 119898)

or 12059510158401015840(119909119899) diverges Moreover (79) always holds if the follow-

ing conditions are satisfied

(i) 12059310158401015840(119909) = 0 forall119909 isin 119868 where 119868 = [1205720 minus120578 1205720 +120578] for some120578 ge |(1205720 minus 1199090)|

(ii) 120593101584010158401015840(119909) is finite forall119909 isin 119868

(iii) 1199090 is sufficiently close to 1205720

Sufficient closeness in this context means that

(a) Taylor approximation is accurate enough such that wecan ignore higher order terms

(b) (12)|120593101584010158401015840(119909119899)12059310158401015840(119909

119899)| lt 119870|120593101584010158401015840(1205720)120593

10158401015840(1205720)| for some119870 lt infin

(c) 119870|120593101584010158401015840(1205720)12059310158401015840(1205720)|120576119899 lt 1 for 119899 isin N

Consequently

1003816100381610038161003816120576119899+11003816100381610038161003816 = 1198721205762

119899 (80)

where

119872 = sup119909isin119868

12

100381610038161003816100381610038161003816100381610038161003816

120593101584010158401015840 (119909)

12059310158401015840 (119909)

100381610038161003816100381610038161003816100381610038161003816 (81)

43 Applications The standard iterative algorithms (71) or(73) for finding the location of overshoots in 119876

119904120590119873(119909) of

the Gibbs effect employ the fact that the first derivative of apeak in 119892(119909) or 119892(119909) respectively has a downward-going-zero-crossing at the peak maximum But the presence ofmany neighboring smaller peaks in the ringing close to theovershoot will cause many undesirable zero-crossings to beinvoked if the initial trial root 1199090 is not correctly chosenThis fact makes the numerical problem of peak finding in theGibbs effect of pathological difficulty A difficulty that aggra-vates with the increase in the number 119873 of truncated termsin the employed Fourier series Indeed with the increaseof 119873 the number of undesirable zero-crossings increasesin the neighborhood of 1205720 as illustrated in Section 1 Thiscan increasingly lead to oscillatory and even unstable valuesin sequence 119909

119899 the more the distance |1199090 minus 1205720| contains

unwanted zero-crossingsThese problems have been resolved in the two examples

that follow which simultaneously illustrate the applicabilityof Proposition 8

Example 9 Consider the 2120587-periodic odd signal 119891(119909) ofExample A1 defined over one period by

119891 (119909) =

minus1 minus120587 lt 119909 lt 0

1 0 le 119909 lt 120587(82)

which has for each truncation number 119873 the truncatedFourier series approximations

119892 (119909) ≃4120587

119873

sum119898=1

1(2119898 minus 1)

sin (2119898minus 1) 119909

119892 (119909) ≃8120587

119873

sum119898=1

119887119898119905119898

(1 + 1199052119898)=

8120587

119873

sum119898=1

1(2119898 minus 1)

119905119898

(1 + 1199052119898)

=8120587

119873

sum119898=1

1(2119898 minus 1)

tan ((2119898 minus 1) 2) 1199091 + [tan ((2119898 minus 1) 2) 119909]2

(83)

This signal has a jump discontinuity at 119909 = 0 ofmagnitude 2 We shall focus our computations on the Gibbsphenomenon overshoot near this discontinuity

Here is a printout list of results for the parameters of theposing inverse problem via computations by both the directand stereographic algorithms

It is remarkable how results by both algorithms arepractically identical and share the same following featuresthat are observed by results of the direct algorithm discussedbelow

Case I (119873 = 40) The computations of 1205720 for the previousexample using (71) with 1199090 = 002 lead to 1199091 = 0039 109this varies quickly with 119899 (only beyond the 4th decimal) tostabilize towards 1199093 = 1199094 = 1199095 = sdot sdot sdot = 0039 299 = 1205720Here using (1) gives 119892(1205720) = 119892(0039 299) = 1179 030 whichis quite close to the theoretically expected value by (A7) ofsim118

Then using (65) for 119888 = 109 and 1199090 = 0000381 leads to1199091 = 0021 697 this varies quickly with 119899 (only beyond the 3-d decimal) to stabilize quickly towards 1199095 = 1199096 = 1199097 = sdot sdot sdot =0028 368 = 1205721 Also using (65) for 119888 = 109 and 1199090 = 004leads to 1199091 = 0067 699 this varies quickly with 119899 (beyondthe 2nd decimal) to stabilize quickly towards 1199095 = 1199096 = 1199097 =sdot sdot sdot = 0052 873 = 1205722

For both1205721 and1205722 using (1) yields119892(1205721) = 119892(0028368) =119892(1205722) = 119892(0052873) = 109 = 119888 as expected

Case II (119873 = 100) Computations of 1205720 using (71) with1199090 = 002 lead to 1199091 = 0014 221 this varies quickly with119899 (only beyond the 4th decimal) to stabilize towards 1199093 =1199094 = 1199095 = sdot sdot sdot = 0015 708 = 1205720 Here using (1) gives119892(1205720) = 119892(0015708) = 1178 990 This result appears to beslightly excessive perhaps due to increased rounding off errorswith increasing 119873 to 100

Thenusing (65) for 119888 = 109 and1199090 = 0015 237 leads aftera negative value to 1199092 = 0014 462 this varies with 119899 (beyondthe 2nd decimal) to stabilize quickly towards 1199097 = 1199098 = 1199099 =sdot sdot sdot = 0011 3348 = 1205721 The used value of 1199090 is apparently notappropriate and this illustrates the generic ill-posedness of thisinverse problem

The pertaining computations of 1205722 with 1199090 = 0017000lead to 1199091 = 0020 434 this varies with 119899 (beyond the 3-ddecimal) to stabilize quickly towards 1199094 = 1199095 = 1199096 = sdot sdot sdot =0021 149 = 1205722

10 International Journal of Computational Mathematics

For both 1205721 and 1205722 using (1) yields 119892(1205721) =119892(00113348) = 119892(1205722) = 119892(0021149) = 109 = 119888 asexpected

Case III (119873 = 150) The computations of 1205720 using (71) with1199090 = 0005 lead to 1199091 = 0010 595 this varies insignificantlywith 119899 (only beyond the 4th decimal) to stabilize towards1199093 =1199094 = 1199095 = sdot sdot sdot = 0010 472 = 1205720 Here using (1) gives 119892(1205720) =119892(0010472) = 1178 980 which does not significantly differfrom 119892(1205720) when119873 = 100

Thenusing (65) for 119888 = 109 and1199090 = 0010 158 leads aftera negative value to 1199092 = 0009 643 and this varies (beyond the3-d decimal) to stabilize quickly towards1199097 = 1199098 = 1199099 = sdot sdot sdot =0007 565 = 1205721

The pertaining computations of 1205722 with 1199090 = 0010 800lead to 1199091 = 0025 998 this varies (beyond the 2nd decimal)to stabilize towards 11990914 = 11990915 = 11990916 = 11990917 = 11990918 = sdot sdot sdot =0014 099 = 1205722

For both 1205721 and 1205722 using (1) yields 119892(1205721) = 119892(007 565) =119892(1205722) = 119892(0014 099) = 109 = 119888

During the previous computations summarized inTable 1 it has been observed that varying 1199090 in the presentNewton-Raphson process for all119873 can lead to strong oscil-lations and even completely unstable values for 1205720 1205721 and 1205722of the Fourier series inverse problem which though ill-posed

is practically regularizible Apart from this drawback the(71) with (65) iterative algorithms (i) converge quickly to thetheoretically expected values and are (ii) of demonstratedstability of the iterations for properly chosen 1199090 (which act aseffective regularizing parameters) Moreover computationsby the stereographic algorithm invoking (73) and (67) sharethe samemagnitude and convergence rates with results by thedirect algorithm Finally it is demonstrated that the presentsignal exhibits an overshoot when119873 = 150 ofsim9of a jumpdiscontinuity of magnitude 2 that is located at a distance ofsim00105

To illustrate robustness of both numerical algorithms andtheir possible effective regularizibility we shall study also thenumerical inverse problem for the Fourier series representa-tion of an arbitrary (nonsymmetric) periodic signal

Example 10 Consider the 2120587-periodic signal 119891(119909) definedover one period by

119891 (119909) =

minus119909

120587minus 1 minus120587 lt 119909 lt 0

1 0 le 119909 lt 120587(84)

It has for each truncation number 119873 the truncated Fourierseries approximations

119892 (119909) ≃14+119873

sum119898=1

[(minus1)119898 minus 1]12058721198982 cos119898119909+

[2 minus (minus1)119898]120587119898

sin119898119909

119892 (119909) ≃14

+119873

sum119898=1

[(minus1)119898 minus 1] 12058721198982 + 2 ([2 minus (minus1)119898] 120587119898) tan ((2119898 minus 1) 2) 119909 minus ([(minus1)119898 minus 1] 12058721198982) [tan ((2119898 minus 1) 2) 119909]2

1 + [tan ((2119898 minus 1) 2) 119909]2

(85)

This signal has like the signal of Example 9 a jumpdiscontinuity at 119909 = 0 of magnitude 2 For the sake ofcomparison we shall also focus our computations on theGibbs phenomenon overshoot part of 119876

119904120590119873(119909) near this

discontinuityIt should be noted here that the results by both algorithms

summarized in Table 2 are also identical a feature indicatingthat the degree of ill-posedness of the posing inverse problemis not affected as predicted by Proposition 8 by the tangenthalf-angle stereographic projection

Case I (119873 = 40) The computations of 1205720 for this exampleusing (71) and (73) with 1199090 = 003 lead to an identicalsequence 119909

119899 that converges by both algorithms after the 1199094

iteration to 0078 139 with 119892(1205720) = 119892(1205720) = 1178 76

Case II (119873 = 100) The computations of 1205720 with 1199090 = 001lead to an identical sequence 119909

119899 that converges by both

algorithms after the 1199095 iteration to 0031 351 with 119892(1205720) =

119892(1205720) = 1178 87

Case III (119873 = 150) The computations of 1205720 with 1199090 = 0007lead also to an identical sequence 119909

119899 that converges by both

algorithms after the 1199095 iteration to 0020 915 with 119892(1205720) =

119892(1205720) = 1178 91

Therefore the present signal exhibits an overshoot when119873 = 150 of sim9 of a jump discontinuity of magnitude 2 thatis located at a distance of sim0020 9

Comparison of the computed Gibbs overshoot parame-ters of the two considered signals with jump discontinuitiesat the same 119909 = 0 of the same magnitude illustrates that therelative size of the overshoot depends essentially exclusivelyon the magnitude of the stimulating jump discontinuityFor instance based on the present numerical results onecan suggest here that nonsymmetric signals have equalovershoots with symmetric signals Their respective 1205720rsquos arehowever different Indeed the nonsymmetric signal has asignificantly larger 1205720 (0020 9) than the 1205720 (00105) of thesymmetric signal

International Journal of Computational Mathematics 11

Table 1 Results of computations for Example 9

119873 1205721 1205720 1205722

40 0028 368 0039 299 0052 873100 0011 335 0015 705 0021 149150 0007 565 0010 472 0014 099

Table 2 Results of computations for Example 10

119873 1205721 1205720 1205722

40 00564765 00781391 0105155100 00226546 00313516 00422024150 00151124 00209153 0028156

Table 3 Summary of computations of all overshoot parameters inExample 10

1205720results

119873 1199090

119909lowast

1205720

119892(1205720)

100 0030 1199093

0031 3516 1178 87200 0015 119909

30015 6919 1178 92

300 0008 1199093

0010 4648 1178 941205721results

119873 1199090

119909lowast

1205721

119888

0025 1199094

0022 2027 108100 0025 119909

30023 1344 110

0025 1199093

0024 2027 1120010 119909

30011 1117 108

200 0010 1199093

0011 5779 1100010 119909

40012 1123 112

0008 1199093

0007 410 08 108300 0008 119909

30007 720 95 110

0008 1199092

0008 077 32 1121205722results

119873 1199090

119909lowast

1205722

119888

100035 119909

40042 9439 108

035 1199094

0041 4445 110035 119909

40039 8513 112

2000018 119909

30021 4960 108

0018 1199093

0020 7456 1100018 119909

30019 9483 112

3000012 119909

30014 3360 108

0012 1199093

0013 8356 1100012 119909

40013 3039 112

A further analysis has also been performed of Example 10with 119873 = 100 200 300 to compute the corresponding1205720 with 1205721 1205722 that correspond simultaneously to 119888 =108 110 112 as illustrated in Figure 1 Results of thesecomputations are summarized in Table 3 and then used forthe partial 119876

119904120590119873(119909) plots of Figure 2 that interpolate 1205721 1205722

(for 119888 = 108 110 112) with 1205720 for each 119873 = 100 200 300In this table119909

lowaststands for the first convergence step of the 119909

119899

sequence of Newton Raphson iterations

002

05

10

004

g(x

)

xminus004 minus002

minus05

minus10

g(x) for N = 100

g(x) for N = 200

g(x) for N = 300

h(x) = c = 108

h(x) = c = 11

h(x) = c = 112

1205721 1205720 1205722

Figure 1 Plots of 119892(119909) near a discontinuity of 119891(119909) in Example 10

120

115

110

105

100

000 001 002 003 004 005

g(x

)

x

g(x) for N = 100

g(x) for N = 200

g(x) for N = 300

h(x) = c = 108

h(x) = c = 11

h(x) = c = 112

Figure 2 Plots of the 119892(119909) curves near a discontinuity of 119891(119909) inExample 10 interpolating 1205721 1205722 (for 119888 = 108 110 112) with 1205720 allfor119873 = 100 200 300

These results turn out to be extremely stable against varia-tions of the trial initial root1199090 Indeed in the computationallymost difficult situation namely when 119873 = 300 a changeof 1199090 from 0008 to 0020 leads to a variation of 119909

lowastfrom

1199092 to just 1199096 with the same asymptotic 1205720 = 00104648 and1205721 = 000807732 (for 119888 = 112)

5 Conclusions

This paper demonstrates that the quadratic degeneracy ofthe solution of the present inverse problem explicit whenusing 119867(120574 119909119898) is a nonlinear indicator of the existence ofGibbsrsquo phenomenon It provides also a distribution-theoreticproof for the existence of this phenomenon The reportedanalysis of numerical solvability of this problem illustrates thefollowing

(i) When suminfin

119898=1 119898(120587119871)119866lowast(120574 119909119898) = suminfin

119898=1 119898(120587119871)119867lowast(120574 119909119898) converge continuously in 119909 and are

12 International Journal of Computational Mathematics

nonzero at 1205721 (or 1205722) there exists a neighborhood(defined by 119868) of 120572 such that for all starting values 1199090in the neighborhood 119909

119899 will converge to 1205721 (or 1205722)

(ii) When suminfin

119898=1 119898(120587119871)119866lowast(120574 119909119898)=suminfin

119898=1 119898(120587119871)119867lowast(120574119909119898) converge continuously in 119909 and are nonzero at1205721 and 1205722 (but not at 1205720) and if 120593(119909) has a secondderivative which might be zero at 1205721 and 1205722 (and at1205720) then the convergence of 119909119899 is quadratic or fasterIf12059310158401015840(119909) = 0 at1205721 and1205722 then the convergence of 119909119899is merely quadratic

(iii) The convergence of the Newton-Raphson iterativeprocess to 1205720 can also be quadratic but restrictionson the derivatives of 120593(119909) have to be observed up tothe third order Otherwise the convergence of 119909

119899 to

1205720 can slow down to become only linear [12] with arate log210

(iv) The reported computations clearly illustrate that1205720 rarr 0 and |1205721 minus 1205722| rarr 0 when 119873 rarr infin thatis the infinite Fourier series will sum to 119891(119909) inthe neighborhood of a discontinuity except at thediscontinuity itself

Appendix

To illustrate the Gibbs-Wilbraham effect we shall use herethe same arguments employed by Wilbraham in 1848 inanalyzing a similar square wave

Example A1 Let 119891(119909) be a square wave periodic signalwhich is defined over one period by

119891 (119909) =

minus1 minus120587 lt 119909 le 0

1 0 le 119909 lt 120587(A1)

Its Fourier series representation is

119892 (119909) =4120587

infin

sum119898=1

1(2119898 minus 1)

sin (2119898minus 1) 119909 (A2)

This defines the truncated series

119878119873(119909) =

4120587

119873

sum119898=1

1(2119898 minus 1)

sin (2119898minus 1) 119909 (A3)

which may be differentiated as

119889

119889119909119878119873(119909) =

4120587

119873

sum119898=1

cos (2119898minus 1) 119909 (A4)

Since

2119873

sum119898=1

cos (2119899 minus 1) 119909 =sin 2119873119909

sin119909 (A5)

then119889

119889119909119878119873(119909) =

2120587

sin 2119873119909

sin119909 (A6)

It is possible to cancel the previous artificially induceddifferentiation by integrating the last result as

119878119873(119909) = int

119909

0

119889

119889120585119878119873(120585) 119889120585 =

2120587int119909

0

sin 2119873120585

sin 120585119889120585 (A7)

This integral starts at zero when 119909 = 0 in agreement with(A2) and increases until 2119873120585 = 120587 Then 120585 = 1205872119873 definesa turning point 1199091 The integral starts to decrease at whichpoint the numerator sin 2119873120585 goes negative For large 119873 thedenominator sin 120585 remains positive

The form (A7) distinctively from (A3) reveals that119878119873(119909) has turning points at the zeros of sin 2119873120585 These occur

when 2119873120585 = 119903120587 119903 = 1 2 3 4 5 that is at 120585 = 119903(1205872119873)Clearly then 119909

119903sim 120585119903= 119903(1205872119873) are turning points for 119878

119873(119909)

In particular 1199091 sim 1205851 = 1205872119873 defines the overshoot peaklocation 1205720 and 119878

119873(1199091) = (2120587) int12058721198730 (sin 2119873120585 sin 120585) 119889120585 is

its magnitude To evaluate 119878119873(1199091) = 119878

119873(1205720) numerically

let us change the variable of integration namely 2119873120585 = 119908or 120585 = 1199082119873 to rewrite the previous integral as 119878

119873(1199091) =

(2120587) int1205870 (sin119908(sin1199082119873)) 119889(1199082119873) Clearly when 119873 rarr

infin sin1199082119873 = 1199082119873 and

lim119873rarrinfin

119878119873(1199091) =

2120587int120587

0

sin119908119908

119889119908 =2120587Si (120587) = 118 (A8)

correct to two decimal pointsAs for 1199092 1199093 1199094 119909119903 lim119873rarrinfin[119878119873(119909119903) minus 119891(119909)]

decreases as we move away from the discontinuity butit should be remarked however that the location of theovershoot 1205720 = 1199091 = 1205872119873 moves towards the discontinuityaccording to

lim119873rarrinfin

1205720 = lim119873rarrinfin

120587

2119873= 0 (A9)

Conflict of Interests

The author declares that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

The numerical computations of Examples 9 and 10 have beenperformed using Mathematica 8 by Mounir Abou Yassineof the Department of CCE of AUL

References

[1] P L Butzer and R J Nessel Fourier Analysis and Approxima-tion Volume I One-Dimensional Theory Birkhauser 1971

[2] R E Edwards Fourier Series A Modern Introduction HoltRinehart amp Winston 1967

[3] W J Thompson ldquoFourier series and the Gibbs phenomenonrdquoAmerican Journal of Physics vol 60 no 5 pp 425ndash429 1992

[4] A J JerriThe Gibbs Phenomenon in Fourier Analysis Splines ampWavelet Approximations Kluwer Academic Publishers BostonMass USA 1990

International Journal of Computational Mathematics 13

[5] K S Rim and B I Yun ldquoGibbs phenomenon removal by addingHeaviside functionsrdquo Advances in Computational Mathematicsvol 38 no 4 pp 683ndash699 2013

[6] A H Zemanian Distribution Theory and Transform AnalysisMcGraw-Hill 1965

[7] G R Ayers and J C Dainty ldquoIterative blind deconvolutionmethod and its applicationsrdquo Optics Letters vol 13 no 7 pp547ndash549 1988

[8] N H S Haidar ldquoA nonlinear method for a severely ill-posedgeneralized moment problemrdquo Journal of Inverse amp Ill-PosedProblems vol 5 no 4 pp 323ndash335 1997

[9] G L Bradley and K J Smith Calculus Prentice Hall 1995[10] P Offner T Sonar and M Wirz ldquoDetecting strength and

location of jump discontinuities in numerical datardquo AppliedMathematics vol 4 no 12 pp 1ndash14 2013

[11] K E Atkinson An Introduction to Numerical Analysis JohnWiley amp Sons New York NY USA 1989

[12] E Suli and D F Mayers An Introduction to Numerical AnalysisCambridge University Press Cambridge UK 2003

[13] P Wang ldquoA third-order family of Newton-like iteration meth-ods for solving nonlinear equationsrdquo Journal of NumericalMathematics and Stochastics vol 3 no 1 pp 13ndash19 2011

Submit your manuscripts athttpwwwhindawicom

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Mathematical Problems in Engineering

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Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Mathematical PhysicsAdvances in

Complex AnalysisJournal of

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Discrete Dynamics in Nature and Society

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

International Journal of Computational Mathematics 9

The convergence of 119909119899 in (71)ndash(73) is at least quadratic

in the neighborhood of this 1205720 The process may face a diffi-culty however when 12059310158401015840(119909

119899) = minussum

infin

119898=1 1198982(120587119871)2119866(120574 119909

119899 119898)

or 12059510158401015840(119909119899) diverges Moreover (79) always holds if the follow-

ing conditions are satisfied

(i) 12059310158401015840(119909) = 0 forall119909 isin 119868 where 119868 = [1205720 minus120578 1205720 +120578] for some120578 ge |(1205720 minus 1199090)|

(ii) 120593101584010158401015840(119909) is finite forall119909 isin 119868

(iii) 1199090 is sufficiently close to 1205720

Sufficient closeness in this context means that

(a) Taylor approximation is accurate enough such that wecan ignore higher order terms

(b) (12)|120593101584010158401015840(119909119899)12059310158401015840(119909

119899)| lt 119870|120593101584010158401015840(1205720)120593

10158401015840(1205720)| for some119870 lt infin

(c) 119870|120593101584010158401015840(1205720)12059310158401015840(1205720)|120576119899 lt 1 for 119899 isin N

Consequently

1003816100381610038161003816120576119899+11003816100381610038161003816 = 1198721205762

119899 (80)

where

119872 = sup119909isin119868

12

100381610038161003816100381610038161003816100381610038161003816

120593101584010158401015840 (119909)

12059310158401015840 (119909)

100381610038161003816100381610038161003816100381610038161003816 (81)

43 Applications The standard iterative algorithms (71) or(73) for finding the location of overshoots in 119876

119904120590119873(119909) of

the Gibbs effect employ the fact that the first derivative of apeak in 119892(119909) or 119892(119909) respectively has a downward-going-zero-crossing at the peak maximum But the presence ofmany neighboring smaller peaks in the ringing close to theovershoot will cause many undesirable zero-crossings to beinvoked if the initial trial root 1199090 is not correctly chosenThis fact makes the numerical problem of peak finding in theGibbs effect of pathological difficulty A difficulty that aggra-vates with the increase in the number 119873 of truncated termsin the employed Fourier series Indeed with the increaseof 119873 the number of undesirable zero-crossings increasesin the neighborhood of 1205720 as illustrated in Section 1 Thiscan increasingly lead to oscillatory and even unstable valuesin sequence 119909

119899 the more the distance |1199090 minus 1205720| contains

unwanted zero-crossingsThese problems have been resolved in the two examples

that follow which simultaneously illustrate the applicabilityof Proposition 8

Example 9 Consider the 2120587-periodic odd signal 119891(119909) ofExample A1 defined over one period by

119891 (119909) =

minus1 minus120587 lt 119909 lt 0

1 0 le 119909 lt 120587(82)

which has for each truncation number 119873 the truncatedFourier series approximations

119892 (119909) ≃4120587

119873

sum119898=1

1(2119898 minus 1)

sin (2119898minus 1) 119909

119892 (119909) ≃8120587

119873

sum119898=1

119887119898119905119898

(1 + 1199052119898)=

8120587

119873

sum119898=1

1(2119898 minus 1)

119905119898

(1 + 1199052119898)

=8120587

119873

sum119898=1

1(2119898 minus 1)

tan ((2119898 minus 1) 2) 1199091 + [tan ((2119898 minus 1) 2) 119909]2

(83)

This signal has a jump discontinuity at 119909 = 0 ofmagnitude 2 We shall focus our computations on the Gibbsphenomenon overshoot near this discontinuity

Here is a printout list of results for the parameters of theposing inverse problem via computations by both the directand stereographic algorithms

It is remarkable how results by both algorithms arepractically identical and share the same following featuresthat are observed by results of the direct algorithm discussedbelow

Case I (119873 = 40) The computations of 1205720 for the previousexample using (71) with 1199090 = 002 lead to 1199091 = 0039 109this varies quickly with 119899 (only beyond the 4th decimal) tostabilize towards 1199093 = 1199094 = 1199095 = sdot sdot sdot = 0039 299 = 1205720Here using (1) gives 119892(1205720) = 119892(0039 299) = 1179 030 whichis quite close to the theoretically expected value by (A7) ofsim118

Then using (65) for 119888 = 109 and 1199090 = 0000381 leads to1199091 = 0021 697 this varies quickly with 119899 (only beyond the 3-d decimal) to stabilize quickly towards 1199095 = 1199096 = 1199097 = sdot sdot sdot =0028 368 = 1205721 Also using (65) for 119888 = 109 and 1199090 = 004leads to 1199091 = 0067 699 this varies quickly with 119899 (beyondthe 2nd decimal) to stabilize quickly towards 1199095 = 1199096 = 1199097 =sdot sdot sdot = 0052 873 = 1205722

For both1205721 and1205722 using (1) yields119892(1205721) = 119892(0028368) =119892(1205722) = 119892(0052873) = 109 = 119888 as expected

Case II (119873 = 100) Computations of 1205720 using (71) with1199090 = 002 lead to 1199091 = 0014 221 this varies quickly with119899 (only beyond the 4th decimal) to stabilize towards 1199093 =1199094 = 1199095 = sdot sdot sdot = 0015 708 = 1205720 Here using (1) gives119892(1205720) = 119892(0015708) = 1178 990 This result appears to beslightly excessive perhaps due to increased rounding off errorswith increasing 119873 to 100

Thenusing (65) for 119888 = 109 and1199090 = 0015 237 leads aftera negative value to 1199092 = 0014 462 this varies with 119899 (beyondthe 2nd decimal) to stabilize quickly towards 1199097 = 1199098 = 1199099 =sdot sdot sdot = 0011 3348 = 1205721 The used value of 1199090 is apparently notappropriate and this illustrates the generic ill-posedness of thisinverse problem

The pertaining computations of 1205722 with 1199090 = 0017000lead to 1199091 = 0020 434 this varies with 119899 (beyond the 3-ddecimal) to stabilize quickly towards 1199094 = 1199095 = 1199096 = sdot sdot sdot =0021 149 = 1205722

10 International Journal of Computational Mathematics

For both 1205721 and 1205722 using (1) yields 119892(1205721) =119892(00113348) = 119892(1205722) = 119892(0021149) = 109 = 119888 asexpected

Case III (119873 = 150) The computations of 1205720 using (71) with1199090 = 0005 lead to 1199091 = 0010 595 this varies insignificantlywith 119899 (only beyond the 4th decimal) to stabilize towards1199093 =1199094 = 1199095 = sdot sdot sdot = 0010 472 = 1205720 Here using (1) gives 119892(1205720) =119892(0010472) = 1178 980 which does not significantly differfrom 119892(1205720) when119873 = 100

Thenusing (65) for 119888 = 109 and1199090 = 0010 158 leads aftera negative value to 1199092 = 0009 643 and this varies (beyond the3-d decimal) to stabilize quickly towards1199097 = 1199098 = 1199099 = sdot sdot sdot =0007 565 = 1205721

The pertaining computations of 1205722 with 1199090 = 0010 800lead to 1199091 = 0025 998 this varies (beyond the 2nd decimal)to stabilize towards 11990914 = 11990915 = 11990916 = 11990917 = 11990918 = sdot sdot sdot =0014 099 = 1205722

For both 1205721 and 1205722 using (1) yields 119892(1205721) = 119892(007 565) =119892(1205722) = 119892(0014 099) = 109 = 119888

During the previous computations summarized inTable 1 it has been observed that varying 1199090 in the presentNewton-Raphson process for all119873 can lead to strong oscil-lations and even completely unstable values for 1205720 1205721 and 1205722of the Fourier series inverse problem which though ill-posed

is practically regularizible Apart from this drawback the(71) with (65) iterative algorithms (i) converge quickly to thetheoretically expected values and are (ii) of demonstratedstability of the iterations for properly chosen 1199090 (which act aseffective regularizing parameters) Moreover computationsby the stereographic algorithm invoking (73) and (67) sharethe samemagnitude and convergence rates with results by thedirect algorithm Finally it is demonstrated that the presentsignal exhibits an overshoot when119873 = 150 ofsim9of a jumpdiscontinuity of magnitude 2 that is located at a distance ofsim00105

To illustrate robustness of both numerical algorithms andtheir possible effective regularizibility we shall study also thenumerical inverse problem for the Fourier series representa-tion of an arbitrary (nonsymmetric) periodic signal

Example 10 Consider the 2120587-periodic signal 119891(119909) definedover one period by

119891 (119909) =

minus119909

120587minus 1 minus120587 lt 119909 lt 0

1 0 le 119909 lt 120587(84)

It has for each truncation number 119873 the truncated Fourierseries approximations

119892 (119909) ≃14+119873

sum119898=1

[(minus1)119898 minus 1]12058721198982 cos119898119909+

[2 minus (minus1)119898]120587119898

sin119898119909

119892 (119909) ≃14

+119873

sum119898=1

[(minus1)119898 minus 1] 12058721198982 + 2 ([2 minus (minus1)119898] 120587119898) tan ((2119898 minus 1) 2) 119909 minus ([(minus1)119898 minus 1] 12058721198982) [tan ((2119898 minus 1) 2) 119909]2

1 + [tan ((2119898 minus 1) 2) 119909]2

(85)

This signal has like the signal of Example 9 a jumpdiscontinuity at 119909 = 0 of magnitude 2 For the sake ofcomparison we shall also focus our computations on theGibbs phenomenon overshoot part of 119876

119904120590119873(119909) near this

discontinuityIt should be noted here that the results by both algorithms

summarized in Table 2 are also identical a feature indicatingthat the degree of ill-posedness of the posing inverse problemis not affected as predicted by Proposition 8 by the tangenthalf-angle stereographic projection

Case I (119873 = 40) The computations of 1205720 for this exampleusing (71) and (73) with 1199090 = 003 lead to an identicalsequence 119909

119899 that converges by both algorithms after the 1199094

iteration to 0078 139 with 119892(1205720) = 119892(1205720) = 1178 76

Case II (119873 = 100) The computations of 1205720 with 1199090 = 001lead to an identical sequence 119909

119899 that converges by both

algorithms after the 1199095 iteration to 0031 351 with 119892(1205720) =

119892(1205720) = 1178 87

Case III (119873 = 150) The computations of 1205720 with 1199090 = 0007lead also to an identical sequence 119909

119899 that converges by both

algorithms after the 1199095 iteration to 0020 915 with 119892(1205720) =

119892(1205720) = 1178 91

Therefore the present signal exhibits an overshoot when119873 = 150 of sim9 of a jump discontinuity of magnitude 2 thatis located at a distance of sim0020 9

Comparison of the computed Gibbs overshoot parame-ters of the two considered signals with jump discontinuitiesat the same 119909 = 0 of the same magnitude illustrates that therelative size of the overshoot depends essentially exclusivelyon the magnitude of the stimulating jump discontinuityFor instance based on the present numerical results onecan suggest here that nonsymmetric signals have equalovershoots with symmetric signals Their respective 1205720rsquos arehowever different Indeed the nonsymmetric signal has asignificantly larger 1205720 (0020 9) than the 1205720 (00105) of thesymmetric signal

International Journal of Computational Mathematics 11

Table 1 Results of computations for Example 9

119873 1205721 1205720 1205722

40 0028 368 0039 299 0052 873100 0011 335 0015 705 0021 149150 0007 565 0010 472 0014 099

Table 2 Results of computations for Example 10

119873 1205721 1205720 1205722

40 00564765 00781391 0105155100 00226546 00313516 00422024150 00151124 00209153 0028156

Table 3 Summary of computations of all overshoot parameters inExample 10

1205720results

119873 1199090

119909lowast

1205720

119892(1205720)

100 0030 1199093

0031 3516 1178 87200 0015 119909

30015 6919 1178 92

300 0008 1199093

0010 4648 1178 941205721results

119873 1199090

119909lowast

1205721

119888

0025 1199094

0022 2027 108100 0025 119909

30023 1344 110

0025 1199093

0024 2027 1120010 119909

30011 1117 108

200 0010 1199093

0011 5779 1100010 119909

40012 1123 112

0008 1199093

0007 410 08 108300 0008 119909

30007 720 95 110

0008 1199092

0008 077 32 1121205722results

119873 1199090

119909lowast

1205722

119888

100035 119909

40042 9439 108

035 1199094

0041 4445 110035 119909

40039 8513 112

2000018 119909

30021 4960 108

0018 1199093

0020 7456 1100018 119909

30019 9483 112

3000012 119909

30014 3360 108

0012 1199093

0013 8356 1100012 119909

40013 3039 112

A further analysis has also been performed of Example 10with 119873 = 100 200 300 to compute the corresponding1205720 with 1205721 1205722 that correspond simultaneously to 119888 =108 110 112 as illustrated in Figure 1 Results of thesecomputations are summarized in Table 3 and then used forthe partial 119876

119904120590119873(119909) plots of Figure 2 that interpolate 1205721 1205722

(for 119888 = 108 110 112) with 1205720 for each 119873 = 100 200 300In this table119909

lowaststands for the first convergence step of the 119909

119899

sequence of Newton Raphson iterations

002

05

10

004

g(x

)

xminus004 minus002

minus05

minus10

g(x) for N = 100

g(x) for N = 200

g(x) for N = 300

h(x) = c = 108

h(x) = c = 11

h(x) = c = 112

1205721 1205720 1205722

Figure 1 Plots of 119892(119909) near a discontinuity of 119891(119909) in Example 10

120

115

110

105

100

000 001 002 003 004 005

g(x

)

x

g(x) for N = 100

g(x) for N = 200

g(x) for N = 300

h(x) = c = 108

h(x) = c = 11

h(x) = c = 112

Figure 2 Plots of the 119892(119909) curves near a discontinuity of 119891(119909) inExample 10 interpolating 1205721 1205722 (for 119888 = 108 110 112) with 1205720 allfor119873 = 100 200 300

These results turn out to be extremely stable against varia-tions of the trial initial root1199090 Indeed in the computationallymost difficult situation namely when 119873 = 300 a changeof 1199090 from 0008 to 0020 leads to a variation of 119909

lowastfrom

1199092 to just 1199096 with the same asymptotic 1205720 = 00104648 and1205721 = 000807732 (for 119888 = 112)

5 Conclusions

This paper demonstrates that the quadratic degeneracy ofthe solution of the present inverse problem explicit whenusing 119867(120574 119909119898) is a nonlinear indicator of the existence ofGibbsrsquo phenomenon It provides also a distribution-theoreticproof for the existence of this phenomenon The reportedanalysis of numerical solvability of this problem illustrates thefollowing

(i) When suminfin

119898=1 119898(120587119871)119866lowast(120574 119909119898) = suminfin

119898=1 119898(120587119871)119867lowast(120574 119909119898) converge continuously in 119909 and are

12 International Journal of Computational Mathematics

nonzero at 1205721 (or 1205722) there exists a neighborhood(defined by 119868) of 120572 such that for all starting values 1199090in the neighborhood 119909

119899 will converge to 1205721 (or 1205722)

(ii) When suminfin

119898=1 119898(120587119871)119866lowast(120574 119909119898)=suminfin

119898=1 119898(120587119871)119867lowast(120574119909119898) converge continuously in 119909 and are nonzero at1205721 and 1205722 (but not at 1205720) and if 120593(119909) has a secondderivative which might be zero at 1205721 and 1205722 (and at1205720) then the convergence of 119909119899 is quadratic or fasterIf12059310158401015840(119909) = 0 at1205721 and1205722 then the convergence of 119909119899is merely quadratic

(iii) The convergence of the Newton-Raphson iterativeprocess to 1205720 can also be quadratic but restrictionson the derivatives of 120593(119909) have to be observed up tothe third order Otherwise the convergence of 119909

119899 to

1205720 can slow down to become only linear [12] with arate log210

(iv) The reported computations clearly illustrate that1205720 rarr 0 and |1205721 minus 1205722| rarr 0 when 119873 rarr infin thatis the infinite Fourier series will sum to 119891(119909) inthe neighborhood of a discontinuity except at thediscontinuity itself

Appendix

To illustrate the Gibbs-Wilbraham effect we shall use herethe same arguments employed by Wilbraham in 1848 inanalyzing a similar square wave

Example A1 Let 119891(119909) be a square wave periodic signalwhich is defined over one period by

119891 (119909) =

minus1 minus120587 lt 119909 le 0

1 0 le 119909 lt 120587(A1)

Its Fourier series representation is

119892 (119909) =4120587

infin

sum119898=1

1(2119898 minus 1)

sin (2119898minus 1) 119909 (A2)

This defines the truncated series

119878119873(119909) =

4120587

119873

sum119898=1

1(2119898 minus 1)

sin (2119898minus 1) 119909 (A3)

which may be differentiated as

119889

119889119909119878119873(119909) =

4120587

119873

sum119898=1

cos (2119898minus 1) 119909 (A4)

Since

2119873

sum119898=1

cos (2119899 minus 1) 119909 =sin 2119873119909

sin119909 (A5)

then119889

119889119909119878119873(119909) =

2120587

sin 2119873119909

sin119909 (A6)

It is possible to cancel the previous artificially induceddifferentiation by integrating the last result as

119878119873(119909) = int

119909

0

119889

119889120585119878119873(120585) 119889120585 =

2120587int119909

0

sin 2119873120585

sin 120585119889120585 (A7)

This integral starts at zero when 119909 = 0 in agreement with(A2) and increases until 2119873120585 = 120587 Then 120585 = 1205872119873 definesa turning point 1199091 The integral starts to decrease at whichpoint the numerator sin 2119873120585 goes negative For large 119873 thedenominator sin 120585 remains positive

The form (A7) distinctively from (A3) reveals that119878119873(119909) has turning points at the zeros of sin 2119873120585 These occur

when 2119873120585 = 119903120587 119903 = 1 2 3 4 5 that is at 120585 = 119903(1205872119873)Clearly then 119909

119903sim 120585119903= 119903(1205872119873) are turning points for 119878

119873(119909)

In particular 1199091 sim 1205851 = 1205872119873 defines the overshoot peaklocation 1205720 and 119878

119873(1199091) = (2120587) int12058721198730 (sin 2119873120585 sin 120585) 119889120585 is

its magnitude To evaluate 119878119873(1199091) = 119878

119873(1205720) numerically

let us change the variable of integration namely 2119873120585 = 119908or 120585 = 1199082119873 to rewrite the previous integral as 119878

119873(1199091) =

(2120587) int1205870 (sin119908(sin1199082119873)) 119889(1199082119873) Clearly when 119873 rarr

infin sin1199082119873 = 1199082119873 and

lim119873rarrinfin

119878119873(1199091) =

2120587int120587

0

sin119908119908

119889119908 =2120587Si (120587) = 118 (A8)

correct to two decimal pointsAs for 1199092 1199093 1199094 119909119903 lim119873rarrinfin[119878119873(119909119903) minus 119891(119909)]

decreases as we move away from the discontinuity butit should be remarked however that the location of theovershoot 1205720 = 1199091 = 1205872119873 moves towards the discontinuityaccording to

lim119873rarrinfin

1205720 = lim119873rarrinfin

120587

2119873= 0 (A9)

Conflict of Interests

The author declares that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

The numerical computations of Examples 9 and 10 have beenperformed using Mathematica 8 by Mounir Abou Yassineof the Department of CCE of AUL

References

[1] P L Butzer and R J Nessel Fourier Analysis and Approxima-tion Volume I One-Dimensional Theory Birkhauser 1971

[2] R E Edwards Fourier Series A Modern Introduction HoltRinehart amp Winston 1967

[3] W J Thompson ldquoFourier series and the Gibbs phenomenonrdquoAmerican Journal of Physics vol 60 no 5 pp 425ndash429 1992

[4] A J JerriThe Gibbs Phenomenon in Fourier Analysis Splines ampWavelet Approximations Kluwer Academic Publishers BostonMass USA 1990

International Journal of Computational Mathematics 13

[5] K S Rim and B I Yun ldquoGibbs phenomenon removal by addingHeaviside functionsrdquo Advances in Computational Mathematicsvol 38 no 4 pp 683ndash699 2013

[6] A H Zemanian Distribution Theory and Transform AnalysisMcGraw-Hill 1965

[7] G R Ayers and J C Dainty ldquoIterative blind deconvolutionmethod and its applicationsrdquo Optics Letters vol 13 no 7 pp547ndash549 1988

[8] N H S Haidar ldquoA nonlinear method for a severely ill-posedgeneralized moment problemrdquo Journal of Inverse amp Ill-PosedProblems vol 5 no 4 pp 323ndash335 1997

[9] G L Bradley and K J Smith Calculus Prentice Hall 1995[10] P Offner T Sonar and M Wirz ldquoDetecting strength and

location of jump discontinuities in numerical datardquo AppliedMathematics vol 4 no 12 pp 1ndash14 2013

[11] K E Atkinson An Introduction to Numerical Analysis JohnWiley amp Sons New York NY USA 1989

[12] E Suli and D F Mayers An Introduction to Numerical AnalysisCambridge University Press Cambridge UK 2003

[13] P Wang ldquoA third-order family of Newton-like iteration meth-ods for solving nonlinear equationsrdquo Journal of NumericalMathematics and Stochastics vol 3 no 1 pp 13ndash19 2011

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

10 International Journal of Computational Mathematics

For both 1205721 and 1205722 using (1) yields 119892(1205721) =119892(00113348) = 119892(1205722) = 119892(0021149) = 109 = 119888 asexpected

Case III (119873 = 150) The computations of 1205720 using (71) with1199090 = 0005 lead to 1199091 = 0010 595 this varies insignificantlywith 119899 (only beyond the 4th decimal) to stabilize towards1199093 =1199094 = 1199095 = sdot sdot sdot = 0010 472 = 1205720 Here using (1) gives 119892(1205720) =119892(0010472) = 1178 980 which does not significantly differfrom 119892(1205720) when119873 = 100

Thenusing (65) for 119888 = 109 and1199090 = 0010 158 leads aftera negative value to 1199092 = 0009 643 and this varies (beyond the3-d decimal) to stabilize quickly towards1199097 = 1199098 = 1199099 = sdot sdot sdot =0007 565 = 1205721

The pertaining computations of 1205722 with 1199090 = 0010 800lead to 1199091 = 0025 998 this varies (beyond the 2nd decimal)to stabilize towards 11990914 = 11990915 = 11990916 = 11990917 = 11990918 = sdot sdot sdot =0014 099 = 1205722

For both 1205721 and 1205722 using (1) yields 119892(1205721) = 119892(007 565) =119892(1205722) = 119892(0014 099) = 109 = 119888

During the previous computations summarized inTable 1 it has been observed that varying 1199090 in the presentNewton-Raphson process for all119873 can lead to strong oscil-lations and even completely unstable values for 1205720 1205721 and 1205722of the Fourier series inverse problem which though ill-posed

is practically regularizible Apart from this drawback the(71) with (65) iterative algorithms (i) converge quickly to thetheoretically expected values and are (ii) of demonstratedstability of the iterations for properly chosen 1199090 (which act aseffective regularizing parameters) Moreover computationsby the stereographic algorithm invoking (73) and (67) sharethe samemagnitude and convergence rates with results by thedirect algorithm Finally it is demonstrated that the presentsignal exhibits an overshoot when119873 = 150 ofsim9of a jumpdiscontinuity of magnitude 2 that is located at a distance ofsim00105

To illustrate robustness of both numerical algorithms andtheir possible effective regularizibility we shall study also thenumerical inverse problem for the Fourier series representa-tion of an arbitrary (nonsymmetric) periodic signal

Example 10 Consider the 2120587-periodic signal 119891(119909) definedover one period by

119891 (119909) =

minus119909

120587minus 1 minus120587 lt 119909 lt 0

1 0 le 119909 lt 120587(84)

It has for each truncation number 119873 the truncated Fourierseries approximations

119892 (119909) ≃14+119873

sum119898=1

[(minus1)119898 minus 1]12058721198982 cos119898119909+

[2 minus (minus1)119898]120587119898

sin119898119909

119892 (119909) ≃14

+119873

sum119898=1

[(minus1)119898 minus 1] 12058721198982 + 2 ([2 minus (minus1)119898] 120587119898) tan ((2119898 minus 1) 2) 119909 minus ([(minus1)119898 minus 1] 12058721198982) [tan ((2119898 minus 1) 2) 119909]2

1 + [tan ((2119898 minus 1) 2) 119909]2

(85)

This signal has like the signal of Example 9 a jumpdiscontinuity at 119909 = 0 of magnitude 2 For the sake ofcomparison we shall also focus our computations on theGibbs phenomenon overshoot part of 119876

119904120590119873(119909) near this

discontinuityIt should be noted here that the results by both algorithms

summarized in Table 2 are also identical a feature indicatingthat the degree of ill-posedness of the posing inverse problemis not affected as predicted by Proposition 8 by the tangenthalf-angle stereographic projection

Case I (119873 = 40) The computations of 1205720 for this exampleusing (71) and (73) with 1199090 = 003 lead to an identicalsequence 119909

119899 that converges by both algorithms after the 1199094

iteration to 0078 139 with 119892(1205720) = 119892(1205720) = 1178 76

Case II (119873 = 100) The computations of 1205720 with 1199090 = 001lead to an identical sequence 119909

119899 that converges by both

algorithms after the 1199095 iteration to 0031 351 with 119892(1205720) =

119892(1205720) = 1178 87

Case III (119873 = 150) The computations of 1205720 with 1199090 = 0007lead also to an identical sequence 119909

119899 that converges by both

algorithms after the 1199095 iteration to 0020 915 with 119892(1205720) =

119892(1205720) = 1178 91

Therefore the present signal exhibits an overshoot when119873 = 150 of sim9 of a jump discontinuity of magnitude 2 thatis located at a distance of sim0020 9

Comparison of the computed Gibbs overshoot parame-ters of the two considered signals with jump discontinuitiesat the same 119909 = 0 of the same magnitude illustrates that therelative size of the overshoot depends essentially exclusivelyon the magnitude of the stimulating jump discontinuityFor instance based on the present numerical results onecan suggest here that nonsymmetric signals have equalovershoots with symmetric signals Their respective 1205720rsquos arehowever different Indeed the nonsymmetric signal has asignificantly larger 1205720 (0020 9) than the 1205720 (00105) of thesymmetric signal

International Journal of Computational Mathematics 11

Table 1 Results of computations for Example 9

119873 1205721 1205720 1205722

40 0028 368 0039 299 0052 873100 0011 335 0015 705 0021 149150 0007 565 0010 472 0014 099

Table 2 Results of computations for Example 10

119873 1205721 1205720 1205722

40 00564765 00781391 0105155100 00226546 00313516 00422024150 00151124 00209153 0028156

Table 3 Summary of computations of all overshoot parameters inExample 10

1205720results

119873 1199090

119909lowast

1205720

119892(1205720)

100 0030 1199093

0031 3516 1178 87200 0015 119909

30015 6919 1178 92

300 0008 1199093

0010 4648 1178 941205721results

119873 1199090

119909lowast

1205721

119888

0025 1199094

0022 2027 108100 0025 119909

30023 1344 110

0025 1199093

0024 2027 1120010 119909

30011 1117 108

200 0010 1199093

0011 5779 1100010 119909

40012 1123 112

0008 1199093

0007 410 08 108300 0008 119909

30007 720 95 110

0008 1199092

0008 077 32 1121205722results

119873 1199090

119909lowast

1205722

119888

100035 119909

40042 9439 108

035 1199094

0041 4445 110035 119909

40039 8513 112

2000018 119909

30021 4960 108

0018 1199093

0020 7456 1100018 119909

30019 9483 112

3000012 119909

30014 3360 108

0012 1199093

0013 8356 1100012 119909

40013 3039 112

A further analysis has also been performed of Example 10with 119873 = 100 200 300 to compute the corresponding1205720 with 1205721 1205722 that correspond simultaneously to 119888 =108 110 112 as illustrated in Figure 1 Results of thesecomputations are summarized in Table 3 and then used forthe partial 119876

119904120590119873(119909) plots of Figure 2 that interpolate 1205721 1205722

(for 119888 = 108 110 112) with 1205720 for each 119873 = 100 200 300In this table119909

lowaststands for the first convergence step of the 119909

119899

sequence of Newton Raphson iterations

002

05

10

004

g(x

)

xminus004 minus002

minus05

minus10

g(x) for N = 100

g(x) for N = 200

g(x) for N = 300

h(x) = c = 108

h(x) = c = 11

h(x) = c = 112

1205721 1205720 1205722

Figure 1 Plots of 119892(119909) near a discontinuity of 119891(119909) in Example 10

120

115

110

105

100

000 001 002 003 004 005

g(x

)

x

g(x) for N = 100

g(x) for N = 200

g(x) for N = 300

h(x) = c = 108

h(x) = c = 11

h(x) = c = 112

Figure 2 Plots of the 119892(119909) curves near a discontinuity of 119891(119909) inExample 10 interpolating 1205721 1205722 (for 119888 = 108 110 112) with 1205720 allfor119873 = 100 200 300

These results turn out to be extremely stable against varia-tions of the trial initial root1199090 Indeed in the computationallymost difficult situation namely when 119873 = 300 a changeof 1199090 from 0008 to 0020 leads to a variation of 119909

lowastfrom

1199092 to just 1199096 with the same asymptotic 1205720 = 00104648 and1205721 = 000807732 (for 119888 = 112)

5 Conclusions

This paper demonstrates that the quadratic degeneracy ofthe solution of the present inverse problem explicit whenusing 119867(120574 119909119898) is a nonlinear indicator of the existence ofGibbsrsquo phenomenon It provides also a distribution-theoreticproof for the existence of this phenomenon The reportedanalysis of numerical solvability of this problem illustrates thefollowing

(i) When suminfin

119898=1 119898(120587119871)119866lowast(120574 119909119898) = suminfin

119898=1 119898(120587119871)119867lowast(120574 119909119898) converge continuously in 119909 and are

12 International Journal of Computational Mathematics

nonzero at 1205721 (or 1205722) there exists a neighborhood(defined by 119868) of 120572 such that for all starting values 1199090in the neighborhood 119909

119899 will converge to 1205721 (or 1205722)

(ii) When suminfin

119898=1 119898(120587119871)119866lowast(120574 119909119898)=suminfin

119898=1 119898(120587119871)119867lowast(120574119909119898) converge continuously in 119909 and are nonzero at1205721 and 1205722 (but not at 1205720) and if 120593(119909) has a secondderivative which might be zero at 1205721 and 1205722 (and at1205720) then the convergence of 119909119899 is quadratic or fasterIf12059310158401015840(119909) = 0 at1205721 and1205722 then the convergence of 119909119899is merely quadratic

(iii) The convergence of the Newton-Raphson iterativeprocess to 1205720 can also be quadratic but restrictionson the derivatives of 120593(119909) have to be observed up tothe third order Otherwise the convergence of 119909

119899 to

1205720 can slow down to become only linear [12] with arate log210

(iv) The reported computations clearly illustrate that1205720 rarr 0 and |1205721 minus 1205722| rarr 0 when 119873 rarr infin thatis the infinite Fourier series will sum to 119891(119909) inthe neighborhood of a discontinuity except at thediscontinuity itself

Appendix

To illustrate the Gibbs-Wilbraham effect we shall use herethe same arguments employed by Wilbraham in 1848 inanalyzing a similar square wave

Example A1 Let 119891(119909) be a square wave periodic signalwhich is defined over one period by

119891 (119909) =

minus1 minus120587 lt 119909 le 0

1 0 le 119909 lt 120587(A1)

Its Fourier series representation is

119892 (119909) =4120587

infin

sum119898=1

1(2119898 minus 1)

sin (2119898minus 1) 119909 (A2)

This defines the truncated series

119878119873(119909) =

4120587

119873

sum119898=1

1(2119898 minus 1)

sin (2119898minus 1) 119909 (A3)

which may be differentiated as

119889

119889119909119878119873(119909) =

4120587

119873

sum119898=1

cos (2119898minus 1) 119909 (A4)

Since

2119873

sum119898=1

cos (2119899 minus 1) 119909 =sin 2119873119909

sin119909 (A5)

then119889

119889119909119878119873(119909) =

2120587

sin 2119873119909

sin119909 (A6)

It is possible to cancel the previous artificially induceddifferentiation by integrating the last result as

119878119873(119909) = int

119909

0

119889

119889120585119878119873(120585) 119889120585 =

2120587int119909

0

sin 2119873120585

sin 120585119889120585 (A7)

This integral starts at zero when 119909 = 0 in agreement with(A2) and increases until 2119873120585 = 120587 Then 120585 = 1205872119873 definesa turning point 1199091 The integral starts to decrease at whichpoint the numerator sin 2119873120585 goes negative For large 119873 thedenominator sin 120585 remains positive

The form (A7) distinctively from (A3) reveals that119878119873(119909) has turning points at the zeros of sin 2119873120585 These occur

when 2119873120585 = 119903120587 119903 = 1 2 3 4 5 that is at 120585 = 119903(1205872119873)Clearly then 119909

119903sim 120585119903= 119903(1205872119873) are turning points for 119878

119873(119909)

In particular 1199091 sim 1205851 = 1205872119873 defines the overshoot peaklocation 1205720 and 119878

119873(1199091) = (2120587) int12058721198730 (sin 2119873120585 sin 120585) 119889120585 is

its magnitude To evaluate 119878119873(1199091) = 119878

119873(1205720) numerically

let us change the variable of integration namely 2119873120585 = 119908or 120585 = 1199082119873 to rewrite the previous integral as 119878

119873(1199091) =

(2120587) int1205870 (sin119908(sin1199082119873)) 119889(1199082119873) Clearly when 119873 rarr

infin sin1199082119873 = 1199082119873 and

lim119873rarrinfin

119878119873(1199091) =

2120587int120587

0

sin119908119908

119889119908 =2120587Si (120587) = 118 (A8)

correct to two decimal pointsAs for 1199092 1199093 1199094 119909119903 lim119873rarrinfin[119878119873(119909119903) minus 119891(119909)]

decreases as we move away from the discontinuity butit should be remarked however that the location of theovershoot 1205720 = 1199091 = 1205872119873 moves towards the discontinuityaccording to

lim119873rarrinfin

1205720 = lim119873rarrinfin

120587

2119873= 0 (A9)

Conflict of Interests

The author declares that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

The numerical computations of Examples 9 and 10 have beenperformed using Mathematica 8 by Mounir Abou Yassineof the Department of CCE of AUL

References

[1] P L Butzer and R J Nessel Fourier Analysis and Approxima-tion Volume I One-Dimensional Theory Birkhauser 1971

[2] R E Edwards Fourier Series A Modern Introduction HoltRinehart amp Winston 1967

[3] W J Thompson ldquoFourier series and the Gibbs phenomenonrdquoAmerican Journal of Physics vol 60 no 5 pp 425ndash429 1992

[4] A J JerriThe Gibbs Phenomenon in Fourier Analysis Splines ampWavelet Approximations Kluwer Academic Publishers BostonMass USA 1990

International Journal of Computational Mathematics 13

[5] K S Rim and B I Yun ldquoGibbs phenomenon removal by addingHeaviside functionsrdquo Advances in Computational Mathematicsvol 38 no 4 pp 683ndash699 2013

[6] A H Zemanian Distribution Theory and Transform AnalysisMcGraw-Hill 1965

[7] G R Ayers and J C Dainty ldquoIterative blind deconvolutionmethod and its applicationsrdquo Optics Letters vol 13 no 7 pp547ndash549 1988

[8] N H S Haidar ldquoA nonlinear method for a severely ill-posedgeneralized moment problemrdquo Journal of Inverse amp Ill-PosedProblems vol 5 no 4 pp 323ndash335 1997

[9] G L Bradley and K J Smith Calculus Prentice Hall 1995[10] P Offner T Sonar and M Wirz ldquoDetecting strength and

location of jump discontinuities in numerical datardquo AppliedMathematics vol 4 no 12 pp 1ndash14 2013

[11] K E Atkinson An Introduction to Numerical Analysis JohnWiley amp Sons New York NY USA 1989

[12] E Suli and D F Mayers An Introduction to Numerical AnalysisCambridge University Press Cambridge UK 2003

[13] P Wang ldquoA third-order family of Newton-like iteration meth-ods for solving nonlinear equationsrdquo Journal of NumericalMathematics and Stochastics vol 3 no 1 pp 13ndash19 2011

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

International Journal of Computational Mathematics 11

Table 1 Results of computations for Example 9

119873 1205721 1205720 1205722

40 0028 368 0039 299 0052 873100 0011 335 0015 705 0021 149150 0007 565 0010 472 0014 099

Table 2 Results of computations for Example 10

119873 1205721 1205720 1205722

40 00564765 00781391 0105155100 00226546 00313516 00422024150 00151124 00209153 0028156

Table 3 Summary of computations of all overshoot parameters inExample 10

1205720results

119873 1199090

119909lowast

1205720

119892(1205720)

100 0030 1199093

0031 3516 1178 87200 0015 119909

30015 6919 1178 92

300 0008 1199093

0010 4648 1178 941205721results

119873 1199090

119909lowast

1205721

119888

0025 1199094

0022 2027 108100 0025 119909

30023 1344 110

0025 1199093

0024 2027 1120010 119909

30011 1117 108

200 0010 1199093

0011 5779 1100010 119909

40012 1123 112

0008 1199093

0007 410 08 108300 0008 119909

30007 720 95 110

0008 1199092

0008 077 32 1121205722results

119873 1199090

119909lowast

1205722

119888

100035 119909

40042 9439 108

035 1199094

0041 4445 110035 119909

40039 8513 112

2000018 119909

30021 4960 108

0018 1199093

0020 7456 1100018 119909

30019 9483 112

3000012 119909

30014 3360 108

0012 1199093

0013 8356 1100012 119909

40013 3039 112

A further analysis has also been performed of Example 10with 119873 = 100 200 300 to compute the corresponding1205720 with 1205721 1205722 that correspond simultaneously to 119888 =108 110 112 as illustrated in Figure 1 Results of thesecomputations are summarized in Table 3 and then used forthe partial 119876

119904120590119873(119909) plots of Figure 2 that interpolate 1205721 1205722

(for 119888 = 108 110 112) with 1205720 for each 119873 = 100 200 300In this table119909

lowaststands for the first convergence step of the 119909

119899

sequence of Newton Raphson iterations

002

05

10

004

g(x

)

xminus004 minus002

minus05

minus10

g(x) for N = 100

g(x) for N = 200

g(x) for N = 300

h(x) = c = 108

h(x) = c = 11

h(x) = c = 112

1205721 1205720 1205722

Figure 1 Plots of 119892(119909) near a discontinuity of 119891(119909) in Example 10

120

115

110

105

100

000 001 002 003 004 005

g(x

)

x

g(x) for N = 100

g(x) for N = 200

g(x) for N = 300

h(x) = c = 108

h(x) = c = 11

h(x) = c = 112

Figure 2 Plots of the 119892(119909) curves near a discontinuity of 119891(119909) inExample 10 interpolating 1205721 1205722 (for 119888 = 108 110 112) with 1205720 allfor119873 = 100 200 300

These results turn out to be extremely stable against varia-tions of the trial initial root1199090 Indeed in the computationallymost difficult situation namely when 119873 = 300 a changeof 1199090 from 0008 to 0020 leads to a variation of 119909

lowastfrom

1199092 to just 1199096 with the same asymptotic 1205720 = 00104648 and1205721 = 000807732 (for 119888 = 112)

5 Conclusions

This paper demonstrates that the quadratic degeneracy ofthe solution of the present inverse problem explicit whenusing 119867(120574 119909119898) is a nonlinear indicator of the existence ofGibbsrsquo phenomenon It provides also a distribution-theoreticproof for the existence of this phenomenon The reportedanalysis of numerical solvability of this problem illustrates thefollowing

(i) When suminfin

119898=1 119898(120587119871)119866lowast(120574 119909119898) = suminfin

119898=1 119898(120587119871)119867lowast(120574 119909119898) converge continuously in 119909 and are

12 International Journal of Computational Mathematics

nonzero at 1205721 (or 1205722) there exists a neighborhood(defined by 119868) of 120572 such that for all starting values 1199090in the neighborhood 119909

119899 will converge to 1205721 (or 1205722)

(ii) When suminfin

119898=1 119898(120587119871)119866lowast(120574 119909119898)=suminfin

119898=1 119898(120587119871)119867lowast(120574119909119898) converge continuously in 119909 and are nonzero at1205721 and 1205722 (but not at 1205720) and if 120593(119909) has a secondderivative which might be zero at 1205721 and 1205722 (and at1205720) then the convergence of 119909119899 is quadratic or fasterIf12059310158401015840(119909) = 0 at1205721 and1205722 then the convergence of 119909119899is merely quadratic

(iii) The convergence of the Newton-Raphson iterativeprocess to 1205720 can also be quadratic but restrictionson the derivatives of 120593(119909) have to be observed up tothe third order Otherwise the convergence of 119909

119899 to

1205720 can slow down to become only linear [12] with arate log210

(iv) The reported computations clearly illustrate that1205720 rarr 0 and |1205721 minus 1205722| rarr 0 when 119873 rarr infin thatis the infinite Fourier series will sum to 119891(119909) inthe neighborhood of a discontinuity except at thediscontinuity itself

Appendix

To illustrate the Gibbs-Wilbraham effect we shall use herethe same arguments employed by Wilbraham in 1848 inanalyzing a similar square wave

Example A1 Let 119891(119909) be a square wave periodic signalwhich is defined over one period by

119891 (119909) =

minus1 minus120587 lt 119909 le 0

1 0 le 119909 lt 120587(A1)

Its Fourier series representation is

119892 (119909) =4120587

infin

sum119898=1

1(2119898 minus 1)

sin (2119898minus 1) 119909 (A2)

This defines the truncated series

119878119873(119909) =

4120587

119873

sum119898=1

1(2119898 minus 1)

sin (2119898minus 1) 119909 (A3)

which may be differentiated as

119889

119889119909119878119873(119909) =

4120587

119873

sum119898=1

cos (2119898minus 1) 119909 (A4)

Since

2119873

sum119898=1

cos (2119899 minus 1) 119909 =sin 2119873119909

sin119909 (A5)

then119889

119889119909119878119873(119909) =

2120587

sin 2119873119909

sin119909 (A6)

It is possible to cancel the previous artificially induceddifferentiation by integrating the last result as

119878119873(119909) = int

119909

0

119889

119889120585119878119873(120585) 119889120585 =

2120587int119909

0

sin 2119873120585

sin 120585119889120585 (A7)

This integral starts at zero when 119909 = 0 in agreement with(A2) and increases until 2119873120585 = 120587 Then 120585 = 1205872119873 definesa turning point 1199091 The integral starts to decrease at whichpoint the numerator sin 2119873120585 goes negative For large 119873 thedenominator sin 120585 remains positive

The form (A7) distinctively from (A3) reveals that119878119873(119909) has turning points at the zeros of sin 2119873120585 These occur

when 2119873120585 = 119903120587 119903 = 1 2 3 4 5 that is at 120585 = 119903(1205872119873)Clearly then 119909

119903sim 120585119903= 119903(1205872119873) are turning points for 119878

119873(119909)

In particular 1199091 sim 1205851 = 1205872119873 defines the overshoot peaklocation 1205720 and 119878

119873(1199091) = (2120587) int12058721198730 (sin 2119873120585 sin 120585) 119889120585 is

its magnitude To evaluate 119878119873(1199091) = 119878

119873(1205720) numerically

let us change the variable of integration namely 2119873120585 = 119908or 120585 = 1199082119873 to rewrite the previous integral as 119878

119873(1199091) =

(2120587) int1205870 (sin119908(sin1199082119873)) 119889(1199082119873) Clearly when 119873 rarr

infin sin1199082119873 = 1199082119873 and

lim119873rarrinfin

119878119873(1199091) =

2120587int120587

0

sin119908119908

119889119908 =2120587Si (120587) = 118 (A8)

correct to two decimal pointsAs for 1199092 1199093 1199094 119909119903 lim119873rarrinfin[119878119873(119909119903) minus 119891(119909)]

decreases as we move away from the discontinuity butit should be remarked however that the location of theovershoot 1205720 = 1199091 = 1205872119873 moves towards the discontinuityaccording to

lim119873rarrinfin

1205720 = lim119873rarrinfin

120587

2119873= 0 (A9)

Conflict of Interests

The author declares that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

The numerical computations of Examples 9 and 10 have beenperformed using Mathematica 8 by Mounir Abou Yassineof the Department of CCE of AUL

References

[1] P L Butzer and R J Nessel Fourier Analysis and Approxima-tion Volume I One-Dimensional Theory Birkhauser 1971

[2] R E Edwards Fourier Series A Modern Introduction HoltRinehart amp Winston 1967

[3] W J Thompson ldquoFourier series and the Gibbs phenomenonrdquoAmerican Journal of Physics vol 60 no 5 pp 425ndash429 1992

[4] A J JerriThe Gibbs Phenomenon in Fourier Analysis Splines ampWavelet Approximations Kluwer Academic Publishers BostonMass USA 1990

International Journal of Computational Mathematics 13

[5] K S Rim and B I Yun ldquoGibbs phenomenon removal by addingHeaviside functionsrdquo Advances in Computational Mathematicsvol 38 no 4 pp 683ndash699 2013

[6] A H Zemanian Distribution Theory and Transform AnalysisMcGraw-Hill 1965

[7] G R Ayers and J C Dainty ldquoIterative blind deconvolutionmethod and its applicationsrdquo Optics Letters vol 13 no 7 pp547ndash549 1988

[8] N H S Haidar ldquoA nonlinear method for a severely ill-posedgeneralized moment problemrdquo Journal of Inverse amp Ill-PosedProblems vol 5 no 4 pp 323ndash335 1997

[9] G L Bradley and K J Smith Calculus Prentice Hall 1995[10] P Offner T Sonar and M Wirz ldquoDetecting strength and

location of jump discontinuities in numerical datardquo AppliedMathematics vol 4 no 12 pp 1ndash14 2013

[11] K E Atkinson An Introduction to Numerical Analysis JohnWiley amp Sons New York NY USA 1989

[12] E Suli and D F Mayers An Introduction to Numerical AnalysisCambridge University Press Cambridge UK 2003

[13] P Wang ldquoA third-order family of Newton-like iteration meth-ods for solving nonlinear equationsrdquo Journal of NumericalMathematics and Stochastics vol 3 no 1 pp 13ndash19 2011

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

12 International Journal of Computational Mathematics

nonzero at 1205721 (or 1205722) there exists a neighborhood(defined by 119868) of 120572 such that for all starting values 1199090in the neighborhood 119909

119899 will converge to 1205721 (or 1205722)

(ii) When suminfin

119898=1 119898(120587119871)119866lowast(120574 119909119898)=suminfin

119898=1 119898(120587119871)119867lowast(120574119909119898) converge continuously in 119909 and are nonzero at1205721 and 1205722 (but not at 1205720) and if 120593(119909) has a secondderivative which might be zero at 1205721 and 1205722 (and at1205720) then the convergence of 119909119899 is quadratic or fasterIf12059310158401015840(119909) = 0 at1205721 and1205722 then the convergence of 119909119899is merely quadratic

(iii) The convergence of the Newton-Raphson iterativeprocess to 1205720 can also be quadratic but restrictionson the derivatives of 120593(119909) have to be observed up tothe third order Otherwise the convergence of 119909

119899 to

1205720 can slow down to become only linear [12] with arate log210

(iv) The reported computations clearly illustrate that1205720 rarr 0 and |1205721 minus 1205722| rarr 0 when 119873 rarr infin thatis the infinite Fourier series will sum to 119891(119909) inthe neighborhood of a discontinuity except at thediscontinuity itself

Appendix

To illustrate the Gibbs-Wilbraham effect we shall use herethe same arguments employed by Wilbraham in 1848 inanalyzing a similar square wave

Example A1 Let 119891(119909) be a square wave periodic signalwhich is defined over one period by

119891 (119909) =

minus1 minus120587 lt 119909 le 0

1 0 le 119909 lt 120587(A1)

Its Fourier series representation is

119892 (119909) =4120587

infin

sum119898=1

1(2119898 minus 1)

sin (2119898minus 1) 119909 (A2)

This defines the truncated series

119878119873(119909) =

4120587

119873

sum119898=1

1(2119898 minus 1)

sin (2119898minus 1) 119909 (A3)

which may be differentiated as

119889

119889119909119878119873(119909) =

4120587

119873

sum119898=1

cos (2119898minus 1) 119909 (A4)

Since

2119873

sum119898=1

cos (2119899 minus 1) 119909 =sin 2119873119909

sin119909 (A5)

then119889

119889119909119878119873(119909) =

2120587

sin 2119873119909

sin119909 (A6)

It is possible to cancel the previous artificially induceddifferentiation by integrating the last result as

119878119873(119909) = int

119909

0

119889

119889120585119878119873(120585) 119889120585 =

2120587int119909

0

sin 2119873120585

sin 120585119889120585 (A7)

This integral starts at zero when 119909 = 0 in agreement with(A2) and increases until 2119873120585 = 120587 Then 120585 = 1205872119873 definesa turning point 1199091 The integral starts to decrease at whichpoint the numerator sin 2119873120585 goes negative For large 119873 thedenominator sin 120585 remains positive

The form (A7) distinctively from (A3) reveals that119878119873(119909) has turning points at the zeros of sin 2119873120585 These occur

when 2119873120585 = 119903120587 119903 = 1 2 3 4 5 that is at 120585 = 119903(1205872119873)Clearly then 119909

119903sim 120585119903= 119903(1205872119873) are turning points for 119878

119873(119909)

In particular 1199091 sim 1205851 = 1205872119873 defines the overshoot peaklocation 1205720 and 119878

119873(1199091) = (2120587) int12058721198730 (sin 2119873120585 sin 120585) 119889120585 is

its magnitude To evaluate 119878119873(1199091) = 119878

119873(1205720) numerically

let us change the variable of integration namely 2119873120585 = 119908or 120585 = 1199082119873 to rewrite the previous integral as 119878

119873(1199091) =

(2120587) int1205870 (sin119908(sin1199082119873)) 119889(1199082119873) Clearly when 119873 rarr

infin sin1199082119873 = 1199082119873 and

lim119873rarrinfin

119878119873(1199091) =

2120587int120587

0

sin119908119908

119889119908 =2120587Si (120587) = 118 (A8)

correct to two decimal pointsAs for 1199092 1199093 1199094 119909119903 lim119873rarrinfin[119878119873(119909119903) minus 119891(119909)]

decreases as we move away from the discontinuity butit should be remarked however that the location of theovershoot 1205720 = 1199091 = 1205872119873 moves towards the discontinuityaccording to

lim119873rarrinfin

1205720 = lim119873rarrinfin

120587

2119873= 0 (A9)

Conflict of Interests

The author declares that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

The numerical computations of Examples 9 and 10 have beenperformed using Mathematica 8 by Mounir Abou Yassineof the Department of CCE of AUL

References

[1] P L Butzer and R J Nessel Fourier Analysis and Approxima-tion Volume I One-Dimensional Theory Birkhauser 1971

[2] R E Edwards Fourier Series A Modern Introduction HoltRinehart amp Winston 1967

[3] W J Thompson ldquoFourier series and the Gibbs phenomenonrdquoAmerican Journal of Physics vol 60 no 5 pp 425ndash429 1992

[4] A J JerriThe Gibbs Phenomenon in Fourier Analysis Splines ampWavelet Approximations Kluwer Academic Publishers BostonMass USA 1990

International Journal of Computational Mathematics 13

[5] K S Rim and B I Yun ldquoGibbs phenomenon removal by addingHeaviside functionsrdquo Advances in Computational Mathematicsvol 38 no 4 pp 683ndash699 2013

[6] A H Zemanian Distribution Theory and Transform AnalysisMcGraw-Hill 1965

[7] G R Ayers and J C Dainty ldquoIterative blind deconvolutionmethod and its applicationsrdquo Optics Letters vol 13 no 7 pp547ndash549 1988

[8] N H S Haidar ldquoA nonlinear method for a severely ill-posedgeneralized moment problemrdquo Journal of Inverse amp Ill-PosedProblems vol 5 no 4 pp 323ndash335 1997

[9] G L Bradley and K J Smith Calculus Prentice Hall 1995[10] P Offner T Sonar and M Wirz ldquoDetecting strength and

location of jump discontinuities in numerical datardquo AppliedMathematics vol 4 no 12 pp 1ndash14 2013

[11] K E Atkinson An Introduction to Numerical Analysis JohnWiley amp Sons New York NY USA 1989

[12] E Suli and D F Mayers An Introduction to Numerical AnalysisCambridge University Press Cambridge UK 2003

[13] P Wang ldquoA third-order family of Newton-like iteration meth-ods for solving nonlinear equationsrdquo Journal of NumericalMathematics and Stochastics vol 3 no 1 pp 13ndash19 2011

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

International Journal of Computational Mathematics 13

[5] K S Rim and B I Yun ldquoGibbs phenomenon removal by addingHeaviside functionsrdquo Advances in Computational Mathematicsvol 38 no 4 pp 683ndash699 2013

[6] A H Zemanian Distribution Theory and Transform AnalysisMcGraw-Hill 1965

[7] G R Ayers and J C Dainty ldquoIterative blind deconvolutionmethod and its applicationsrdquo Optics Letters vol 13 no 7 pp547ndash549 1988

[8] N H S Haidar ldquoA nonlinear method for a severely ill-posedgeneralized moment problemrdquo Journal of Inverse amp Ill-PosedProblems vol 5 no 4 pp 323ndash335 1997

[9] G L Bradley and K J Smith Calculus Prentice Hall 1995[10] P Offner T Sonar and M Wirz ldquoDetecting strength and

location of jump discontinuities in numerical datardquo AppliedMathematics vol 4 no 12 pp 1ndash14 2013

[11] K E Atkinson An Introduction to Numerical Analysis JohnWiley amp Sons New York NY USA 1989

[12] E Suli and D F Mayers An Introduction to Numerical AnalysisCambridge University Press Cambridge UK 2003

[13] P Wang ldquoA third-order family of Newton-like iteration meth-ods for solving nonlinear equationsrdquo Journal of NumericalMathematics and Stochastics vol 3 no 1 pp 13ndash19 2011

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of