NUS - MA1505 (2012) - Chapter 6

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2 MA1505 Chapter 6. Fourier Series

For example, sine and cosine functions are periodic

2.

f(x) =c,cconstant, is a periodic function of period

pfor every positive number p.

x, x2, x3, , ex, ln x are not periodic.

6.1.2 Some algebraic properties of periodic

functions

From (1),

f(x+ 2p) =f((x+p) +p) =f(x+p) =f(x).

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Thus (by induction) for any positive integer n,

f(x+np) =f(x), for allx.

Hence 2p, 3p, are also periods off.

Further, iff andg have periodp, then the function

h(x) = a f(x) +bg(x) with a, b constants also has

period p.

6.1.3 Trigonometric series

Our aim is to represent various periodic functions of

period 2 in terms of simple functions

1, cos x, sin x, cos2x, sin2x, , cos nx, sin nx, (2)

which have period 2.

The series that arises in this connection will be of the

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form

a0+a1cos x+b1sin x+a2cos 2x+b2sin 2x+

=a0+

n=1(ancos nx+bnsin nx) (3)

where a0, a1, a2, , b1, b2, are real constants.

Series (3) is called a trigonometric series, and an

andbnare called coefficientsof the series.

The set of functions (2) is often called atrigonomet-

ricsystem.

We note that each term of the series (3) has period

2. Hence if the series converges, its sum will be a

periodic function of period 2.

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6.2 Fourier Series

Assume that f(x) is a periodic function of period

2and that it can be represented by a trigonometric

series

f(x) =a0+

n=1

(ancos nx+bnsin nx). (4)

That is, we assume that the series on the right con-

verges and hasf(x) as its sum.

We say the right hand side of (4) is the Fourier series

off(x).

Givenf(x), our task now is to determine the coeffi-

cientsanandbn.

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6.2.1 Determine a0

We integrate both sides of (4) from to :

f(x)dx=

(a0+

n=1(ancos nx+bnsin nx)) dx.

Assuming that term by term integration is allowed,

we obtain

f(x)dx

= a0

dx+

n=1

(an

cos nxdx+bn

sin nxdx)

= 2a0+

n=1

ansin nx

n

+

bn

cos nx

n

= 2a0

Soa0= 1

2

f(x)dx.

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6.2.2 Determine am, m >0

We multiply both sides of (4) by cos mx and inte-

grate term by term from to :

f(x)cos mx dx

=a0

cos mx dx+

n=1

an

cos nx cos mx dx

+bn

sin nx cos mx dx (5)Computing the three integrations on the right hand

sides of (5):

(i) cos mx dx=

sin mxm

= 0.

(ii)

sin nx cos mx dx= 0,

since sin nx is odd and cos mx is even

and

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(iii)

cos nx cos mx dx

= 1

2

(cos(m+n)x+ cos(m n)x) dx

=

1

2sin(m+n)x

m+n +sin(m n)x

m n

m =n12m

[mx+ sin mx cos mx]

m=n

=

0 m =n m=n

Substituting the above results back in (5), we get

am= 1

f(x)cos mxdx, m= 1, 2,

6.2.3 Determine bm, m >0

We multiply (4) by sinmx and integrate from

to :

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f(x)sinmx dx

=a0

sin mx dx+

n=1

an

cos nx sin mx dx

+bn

sin nx sin mx dx

(6)

=

n=1

bn

sin nx sin mx dx

as the first two integrands on the right hand side of

(6) are odd functions.

Now

sin nx sin mxdx

= 1

2

(cos(n m)x cos(n+m)x)dx

=

12

sin(nm)x

nm

sin(n+m)xn+m

m =n12m

[mx sin mx cos mx]

m=n

=

0 m =n m=n

Thus bm= 1

f(x)sin mxdx, m= 1, 2, .

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6.2.4 Euler formulas

Given a periodic function f(x) of period 2 with

Fourier series

a0+

n=1

(ancos nx+bnsin nx).

Its coefficients are known asFourier coefficientsand

are given by

a0 = 1

2

f(x)dx

an = 1

f(x)cos nxdx, n= 1, 2, (7)

bn =

1

f(x)sin nxdx, n= 1, 2,

(7) are known as Euler formulas.

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6.2.5 Representation by a Fourier series

If a periodic functionf(x) with period 2is piecewise

continuous in the interval x and has

a left hand derivative and right hand derivative at

each point of the interval, then the Fourier series with

coefficients (7) is convergent. Its sum is f(x) except

at a pointx0at whichf(x) is discontinuous and the

sum of the series is the average of the left hand and

right hand limits off atx0.

6.2.6 Example

Find the Fourier series off(x) given by

f(x) = k, if < x

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andf(x) =f(x+ 2).

Functions of this kind occur as external forces act-

ing on mechanical systems, electromotive forces in

electric circuits, etc.

f(x) is piecewise continuous and the value off at

a single point does not affect the integral. We can

therefore leavefundefined at x= 0,x= .

Solution. We observe that over the interval (, ),

fis an odd function. Thusf(x)cos nxis also an odd

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function. Thus by (7), an = 0 for n = 0, 1, 2, ,

and

bn = 1

f(x)sin nxdx

= 2

0

k sin nxdx= 2k

n(1 cos n)

= 2k

n(1 (1)n).

b1 = 4k

, b2= 0, b3=

4k

3,

b4 = 0, b5=4k5

, .

The Fourier series for the square wave is, therefore,

4k

(sin x+1

3sin 3x+

1

5sin 5x+ ).

6.2.7 An approximation for

From the previous section, the series converges to

f(x) in (0, ).

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Settingx= 2 , we get

k=4k

(1

1

3+

1

5 )

i.e. 4 = 1 13+

15 (Leibniz).

Note that at all the points of discontinuity (0,,etc)

off, the sum of the series is equal to 0, which is the

average of the left hand and right hand limits off

(e.g. they are k andk respectively atx= 0).

6.2.8 Periodic functions of period p= 2L

Letf(x) be a periodic function of periodp= 2L.

We set v = xL . Then x = v L

and at x = L, v =

.

We now view fas a function ofv and put f(x) =

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g(v). Then g becomes a periodic function of period

2. Iff(x) has a fourier series, then so hasg(v). We

have

g(v) =a0+

n=1

(ancos nv+bnsin nv)

with

a0 = 1

2

g(v)dv= 1

2 L

L

g(v)

Ldx

= 1

2L

L

L

f(x)dx

and for n= 1, 2, 3,

an = 1

g(v)cos nv dv

= 1

L

LL

f(x)cosnx

L dx,

bn = 1

L

LL

f(x)sinnx

L dx.

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Sinceg(v) =f(x), we get

f(x) =a0+

n=1

ancos

n

Lx+bnsin

n

Lx

witha0, an andbn as given above.

The interval of integration in the above formula can

be replaced by any interval of length p = 2L, for

example, by 0 x 2L or L x 3L.

6.2.9 Example

Letfbe a periodic square wave of periodp= 2L= 4

defined as follows :

f(x) =

0, if 2< x < 1k, if 1< x

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To find the Fourier series off, we compute

a0 = 1

4

22

f(x)dx

an = 1

2

22

f(x)cosnx

2 dx

and sincef is even,

bn= 0 for n= 1, 2, .

a0 = 1

4

2

2

f(x)dx=1

4

1

1

kdx =k

2

an = 1

2

22

f(x)cosnx

2 dx=

1

2

11

kcosnx

2 dx

= 2k

nsin

n

2

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Hencean= 0 ifnis even and

an=

2k

n ifn= 1, 5, 9,

2k

n ifn= 3, 7, 11,

Hence

f(x) =k

2+

2k

(cos

2x

1

3cos

3

2x+

1

5cos

5

2x ).

6.2.10 Fourier cosine and sine series

Using LL

f(x)dx=

0 iff is odd

2L0 f(x)dx iff is even.

we obtain the following two series.

The Fourier series of an even functionf(x) of period

2Lis the Fourier cosine series

f(x) =a0+

n=1

ancosnx

L ,

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with

a0 = 1

L

L0

f(x)dx,

an = 2

L

L0

f(x)cosnx

L dx, n= 1, 2, .

The Fourier series of an odd function f(x) of period

2Lis a Fourier sine series

f(x) =

n=1

bnsinnx

L ,

with bn= 2

L

L0

f(x)sinnx

L dx.

6.2.11 Sum and Scalar multiplication

The Fourier coefficients of f1+f2 are the sums of

corresponding Fourier coefficients off1andf2.

The Fourier coefficients of cf (c a constant) are c

times the corresponding Fourier coefficients off.

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6.2.12 Example

Saw tooth function

f(x) =x+, < x < ,

f(x) =f(x+ 2).

We note thatf=f1+f2, wheref1=x, f2=.

The Fourier coefficients forf2area0= and

an= 0 =bn,n 1.

The functionf1=xis odd.

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Thusan= 0 for alln, and

bn = 2

0

x sin nxdx

= 2

x

cos nx

n

0

0

cos nx

n dx

= 2

(1)n

n sin nx

n20

=

(1)n+12

n

So

f(x) = f1(x) +f2(x)

=

n=1

(1)n+12

n sin

nx

+

= + 2

n=1

(1)n+1

n sin nx

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6.3 Half-range Expansions

In various applications there is a practical need to use

Fourier series in connection with functionsfthat are

given on some interval only, say, 0 x L.

Figure (a)

6.3.1 Extension off(x)

We could extend f(x) as a periodic function with

period L and then represent the extended function

by a Fourier series, which in general would involve

both sine and cosine terms. We can do better and

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always get a cosine series by first extendingf(x) from

0 x Las an even function on the interval L

x L as in figure (b) and then extend this new

function as a periodic function of period 2L, and since

it is even, we can represent it by a Fourier cosine

series.

Figure (b)

Also, we can extend f(x) from 0 x L as an

odd function on L x L as in figure (c) and

then extend this new function as a periodic function

of period 2L, and since it is odd, it is represented by

a Fourier sine series.

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Figure (c)

6.3.2 Half range expansion

Such two series are called the two half range expan-

sions of the function fwhich is given only on half

the range.

The cosine half range expansion is

f(x) =a0+

1

an

cosnx

L

with

a0 = 1

L

L0

f(x)dx,

an = 2L

L0

f(x)cosnxL

dx, n= 1, 2,

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The sine half range expansion is

f(x) =1

bnsinn

Lx

with

bn= 2L

L

0

f(x)sinnxL

dx, n= 1, 2, .

6.3.3 Example

Find the two half range expansions for

f(x) =

0, 0< x < 2

1, 2 < x < .

For the cosine half range expansion, we have

a0= 1

/20

0dx+

/2

1dx

=

1

2

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and

an

= 2

/20

0cos nxdx+

/2

1cos nxdx

= 2

[sin n sin1

2nn

]

Thusansimplifies to

an= 2

n sinn

2

Indeed,

a1=2

, a3=

2

3, a5=

2

5, ,

andan= 0 ifn 1 andnis even.

The cosine half range expansion is

f(x) = 1

2+ 2

m=1

(1)m

(2m 1)cos(2m 1)x

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For the sine half range expansion, we have

bn

= 2

/20

0sin nxdx+

/2

1sin nxdx

= 2

[ cos n+ cos12n

n ]

Thusbnsimplifies to

bn= 2

n(1)n+1 + cosn

2

Indeed,

b1= 2 , b2=

42 , b3=

23, b4= 0,

b5= 25, b6=

46 , b7=

27, b8= 0, .

The sine half range expansion is

f(x) =

2

m=1sin(4m3)x

4m3 2 sin(4m2)x

4m2 +sin(4m1)x

4m1

NUS - MA1505 (2012) - Chapter 6

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