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Kap2: EULERS FORMULA: e i = cos + i sin . Orwritting in complex numbers: (e i )n= cos n + i sin n

Z = x+ iy = r (cos + i sin )= r e i , a b= e bln a

z 1/n= n r cos n

+ i sin n

. e z = 1+ z + z 2

2 !+ z

3

3 !...

Kap7:FOURIER TRANSFORM:Average of f(x on (a, b

a

b

f ( x)dx

b a."#e average value of sin$2 nx an% cos$2 nx

12 &&

&

sin 2 nx dx= 12 pi &

&

cos 2 nxdx= 1 /2

#e average value of sin mx cos nx over a perio% is:1

2 &&

&

sin mx cos nxdx = '

12 &&

&

sin mx sin nxdx =' m n

1/2 ' m= n '' m= n= '

12 &&

&

cos mxcos nxdx =' m n

1/2 ' m= n '1 m= n= '

. "#e fourier coeff.

1: f ( x)= 12

a ' + a 1 cos x+ a 2 cos 2x... b1 sin x+ b 2sin 2x... An%

olving a n a n=1&&

&

f ( x)cos nxdx an% bn is

bn=1&&

&

f ( x)sin nx dx . )ere is an example for it. * is ' from

pi +x+ ' an% 1 from '+x+pi.an% %o t#e same for bn an% put

#em toget#er in 1:- a n=1&['

&

' cos nxdx +&

'

1cos nx dx]COMPLEX FORM OF FOURIER SERIES:

f ( x)= c' + c1 eix + c 1 e

ix + c 2 e2i x ... =

n=

n=

cn ein x . or in ot#er

wa c n= 12 &&

&

f ( x)e inx dx %o t#e same t#is as %one in

xample 1: On OTHER INTERVALS t#e coefficients , t#e can

e written a n=1&&

&

f ( x)cos nxdx bn=1&'

2 &

f ( x)sin nxdx

n% t#e c n=1

2 &'2 &

f ( x)e inx dx . or u can write t#em in form

a n=1l l

l

f ( x)cos n & xl

dx an% same for bn an% c n ./f

function is even li0e x$2 or cos x, w#ose grap# for negativ x isust a reflection in axis of its grap# for positive x

f ( x)= f ( x) an% t#e opposite for o%% func. f ( x)= f ( x) .

l

l f ( x)dx={

' if f ( x)isodd 2

'

l

f ( x)dx if f ( x)iseven an% again if f(x is

%%, {b n= 2l 'l

f ( x)sin n & xl

dx

a n= 'if f(x is even t#en.

{a n= 2l 'l

f ( x)cos n & xl

dx

b n= '. PARSEVAL'S THEOREM: A

elation between t#e average of t#e s uare of f(x an% t#eoefficients in t#e fourier series for f(x t#at t#e s uare is finite.

f ( x)= 12

a ' +1

a n cos nx+1

bnsin nx . EFINITION OF

FOURIER TRANSFORM. f ( x)=

g ( )ei x d !

g ( )= 12 &

f ( x)e i x

dx . FOURIER SINETRANSFORMS.4e %efine f s ( x) an% g s( ) a pair ofourier sine transforms representing o%% functions.

f s( x)=2&''

g s( )sin xd ,

g s ( )= 2&''

f s( s)sin xdx . FOURIER COSINE

TRANSFORM.4e %efine f x ( x) an% g s( ) a pair ofourier cosine transform representing even function.

f s ( x)=2&''

g s ( )sin xd

g c( )=2&''

f c( s)cos x dx .

KAP ". OR INAR# IFFERENTIAL E$UATIONS./67A8 */89" O8 78 7;?57: 4e #ave y ' ' + A ' + B y= ' ,

D = d /dx . Dy= dyd x

= y' , D 2 y= d dx

dydx

= d 2 y

d x2= y ' '

D12 y+ y+ B = ' ( D2+ D+ B) y= ' t#is is x$2xpression

( D+ 1)( D+ B) y= ' ( D+ B) y= ' C ( D+ 1) y= ' y= c1 e

B x , y= c2 e x , y= c1 e

B x+ c 2 e x . /n general it is:

y= c1 ea x + c 2 e

b x solution of ( D a )( D b) y= ' /f we #ave

a = b= a , y = ( Ax+ B)e a x . /f t#e roots are + i D t#en t#esolution is y= e x( Ae i D x+ B ei D x) . or into forms li0e

y= e a x(c1 sin D x+ c 2 cos D x)= y= c eax

sin (D x+ E) . 97@O6O8 78 5/67A8 7;9 O* 78/IA"/I79 O* : ! ( y ' )= p& y '

! ( y' ' )= p2& p y' y' ' ! ( y ' ' ' )= p3 & p2 y' p y ' ' y ' ' '

KAP . CALCULUS OF VARIATIONS:

7 1: /f f ( z )= u ( x , y)+ iv ( x , y) is anal tic in a

region, t#en in t#at regionu x

= v x

=u y . "#e e uations are

calle% t#e @AA66 on con%itions.")7O87> 2: /f u(x, an% v(x, an% t#eir partial %erivativeswit# respect to x an% are continuous an% satisf t#e @auc#

8ienmann con%itions in a region, t#en f( is anal tic at all pointsinsi%e t#e region.9O>7 7*/6/"/O69: A regular point of f( is a point at w#ic#f( is anal tic. A singular point or singularity of f( is a point atw#ic# f( is not anal tic. /t is calle% an isolate% singular point iff( is anal tic ever w#ere else insi%e some small circle about t#esingular point.")7O87> 3. /f f( is anal tic in a region 8, t#en it #as%erivatives of all or%ers at points insi%e t#e region an% can beexpan%e% in a ta lor series about an point S' insi%e t#e region."#e power series converges insi%e t#e circle about S' t#at exten%s

to t#e nearest singular point.")7O87> : @A T: @Aultipl f( b ( S' $m, w#ere m is aninteger greater t#at or e ual to t#e or%er n of t#e pole, %ifferentiatet#e result m 1 times, %ivi%e b (m 1 ! An% evaluate t#e resultingexpression at S'.

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CAUCH#'S FORMULA:

f ' ( z )= 12 &i

f (( )d((( z )2

f ' (a )= 12 &i

f ( z )dz ( z a )2

f (n)( z )= n +

2 &i f (( )d((( z )n+ 1

f (n )(a )= n+

2 &i f ( z )dz ( z a )n+ 1

LAURENT SERIES EXAMPLE: f ( z )=12

z (2 z )(1+ z )#is function #as singular points at ', 2, an% 1. 4e write itn a form.

FOURIER TRANSFORM TIL 9 VISE:

'

cos ( x

a 2+ ( 2 = &2 a e

a

( x) .

LAPLACE TRANSFORM OF ERIVATIVES:t is for t#e first %erivative ! [ f ' (t ) ]= s ! [ f (t )] f ( ' ) *or#e secon% %erivative ! [ f ' ' (t )]= s 2 ! [ f (t )] s f ( ' ) f ' ( ' )SOME OTHER METHO S FOR FIN IN3 RESI UES:

='

2 &

d t 1+ b cos2 t

=12

'

2 &

d t 1+ b cos 2 t

. dz = i ei t d t = i zd t or

dt = 1i z dz we set z = e

it

, cos t =12 ( z + z

1

) an% get

= 12 i

d t bB

z B+( b2

+ 1) z 2+ bB

#ere is not so muc# place so ..

2i b

z d z

( z 2 z W2 )( z 2 z (

2 ) z W

2 = ( b 2 2 b+ 1)/b . *iren0le poler, bare polene / z W er innenfor @ og begge

esi% ene er1

2( z W2 z 2 )=

b

X b+ 1 . ette gir =&

1+ blir integran%en 1, og = &

RESI UE THEOREM SIMPLER 5A#:

dx

1+ x2. 4e

onsi%er

dz

1+ z 2

4#ere @ is close%oun%ar of t#e semiircle s#own #ere. *orn pY1, t#e semicirclencloses t#e singular point

i an% no ot#ersH t#eesi%ues of t#e integran% at i is :

* ( i )= lim z i ( z i )1

( z i )( z + i )=

12 i

. "#en t#e value of

#e contour integral is 2 &i (1 /2i )= & . /ntegral in two parts.(1n integral along t#e x axis from p to pH for t#is part xH (2 anntegral along t#e semicircle, w#ere z = p e i . "#en we #ave

C

dz

1+ z 2=

p

p

dx

1+ x2+

'

&

p i ei

d

1+ p2 e 2i 4e 0now t#e value of

t#e contour integral is & no matter #ow large p becomes sincet#ere are no ot#er singular points besi%es i in t#e upper #alf

plane.5et p - t#en t#e secon% integral on t#e rig#t of t#ee uation above ten%s to ero since t#e numerator contains p an%t#e %enominator p$2. "#us t#e first term on t#e rig#t ten%s to

& (t#e value of t#e contour integral as p - an% we

#ave

dx1+ x2

= & . .$ / 0 .$2D CA3 B 4/ D .2

5A!4A. A3& 3. #*A! 2" .$ "*20)

P ( x)Q( x)

dx

) /f ?(x an% ;(x are pol nomials wit# t#e %egree of ; at least twogreater t#an t#e %egree of ?, an% if ;( #as no real eros. /f t#eintegran% ?(x U;(x is an even function, t#en we can also fin% t#eintegral from ' to infinit . COS X; IS AN EVEN FUNCTIONAN SIN X; IS AN O FUNCTION.

SOME IMPORTANT IFFERENTIAL RELATIONS:4#en we #ave ( D 2+ ( 2) y= ' 4e get D= i ( an% t#esolution can be written in two forms: y= Ae i(t + B e i(t or

y= c1 sin ( t + c 2 cos ( t .A SECON OR ER IFFERENTIAL EXAMPLE:

5e

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USIN3 LAPLACE TRANSFORM TO O6TAIN SOLUTIONSFOR COUPLE E$UATIONS:

x(t )+ 2 n x(t )+ n2 x (t )= ' y(t )+ 2n y(t )+ n2 y(t )= [ x(t )

we will use laplace transfrom

o obtain t#e solution of x(t an% (t . /nitial con%itionsx(' (' '. x(t )= \ 4e call 5( R an% 5(x =.

x ' ' + 2 nx ' + n2 x= ' y ' ' + 2 ny ' + n2 y= [ x'

"ransforms of %erivatives og .

! ( y' )= p& y' , ! ( y ' ' )= p2 p y ' y ' ' we insert t#is into e

6 ( p2+ 2 n p + n2) \ = ' 6 ( p+ n)2= \ & ( p 2+ 2 n p + n2 ) [ p 6 = ' & ( p + n)2= [ p 6

& = \ p [

( p+ n)B , 6 = \

( p+ n )2 ! 1[ \ ( p+ n )2 ]= \ ! 1[ 1( p+ n)2 ]

! 1

[ 1

( p+ n)

1

( p+ n) ]= ! 1 [# ( p ) $ ( p )]= g % b 5S3B

# ( p )= 1 p+ n

! [ g (t )]= 1 p+ n

g (t )= e n t g (t J )= e n( t J )

$ ( p )= 1 p+ n

! [ $ ( p )]= 1( p + n)

e nt %( J)= e n J

'

t

\ en (t J )e n Jd J ='

t

\ e nt d J x(t )= \ t e n t x( ' )= \

& = [ p \

p+ nB[\ ! 1[ p( p+ n )2 1( p+ n)2 ]

! 1 [# ( p ) $ ( p )]= g %# ( p )= p( p+ n )2

(1 n t )e nt = g (t )

$ ( p )= 1( p+ n)2

%(t )= t e nt y( t )= [\ en t [ t 22 n t 3T ]THIS IS VER# IMPORTANT FOR IFFERENTIAL E$.

y ' ' + y' 2 y= Bsin 2 x . instea% of tac0ling t#is problem%irectl , we are first going to solve t#e e uation . 9ince exp(2ix

& ' ' + & ' 2 & = Be 2i x cos 2x W / sin 2x is complex, t#e solutionR ma be complex also. "#en & = & *+ i & is e uivalent to

wo e uations& * ' ' + & * ' ' 2 & *=Be

2i x= Bcos 2x& ' ' + . ' 2& = Be

2i x= Bsin 2x9ince t#e

econ% e uation above is t#e same as t#e uestion, we see t#at t#eolution of t#e uestion is t#e imaginar part of R. "#us to fin%

y p for t#e uestion we fin% & p for t#e e uation aboven% ta0e its imaginar part. 4e observe t#at 2i is not e ual toit#er of t#e roots of t#e auxiliar e uation in R e uation.ollowing t#e met#o% of t#e last paragrap#, we assume a solutionf t#e form & p= C e

2 ix an% subsitute it into R e uation to get:

( B+ 2i 2)C e 2i x= Be 2i xC = B2 i T

=B( 2i T)

B' = 1

A (i+ 3)

& p= 1A

(i+ 3)e2 i x ta0ing t#e imaginar part of RSp we fin%

y p

= 1

A cos 2x

3

A sin 2x

.

THE CAUCH# EULER E$UATION. x2 d

2 yd x 2

+ a x d yd x

+ b y = ' 4e assume a trial solution given b

y= xm , %ifferentiating, we #ave.d yd x

= m xm 1 An%

d 2 y

d x 2= m (m 1) xm 2 . 9ubstituting into t#e original e uation

x2 (m(m 1) xm 2)+ a x (m xm1 )+ b( xm)= '

8earranging gives : m2+ (a 1)m+ b= ' we t#en can solve form. "#ere are t#ree particular cases of interest.@ase ]1: "wo %istinct roots, m1 an% m2@ase ]2: One real repeate% root, m@ase ]3: @omplex roots, a Di ./n @ase]1, t#e solution is given b : y= c1 x

m1+ c2 xm2

/n @ase]2: t#e solution is given b y= c1 xm

ln( x)+ c 2 xm

/n @ase]3: t#e solution is given b y= c1 x

acos(Dln( x))+ c 2 x

asin (Dln ( x))

)ere is an example for Freen function.

EN INHOMO3EN LI3NIN3EN. y ' ' + P x y ' + Q x y = * x Ii mer0er oss at #vis y%( x) er en l^sning av %en #omogene ligningen 8(x ', og

Sp (x en eller annen l^sning av %en in#omogene (parti0ul_rl^sning , sZ er ogsZ: y( x)= y%( x)+ y p( x) en l^sning av %enin#omogene ligningen. 9p^rsmZlet er %a om %enne l^sningen

passer me% gitte grensever%ier for y( x' ) og y ' ( x ' ) ,sli0 at vi fZr %en ^ns0e%e ent %ige l^sningen. >e%

y%( x)= c1 y1( x)+ c2 y2( x) fZr vi betingelsenec1 y1( x' )+ c2 y2( x' )= y( x' ) y p( x' )

c 1 y1( x' )+ c2 y ' 2( ' )= y' ( x' ) y p( x' )0rav til l^sning er

som 0 ent at %eterminanten til ligningssettet mZ v_re uli0 null, %vs.

7 = y1( x' ) y2 ( x ' ) y ' 1( x' ) y' 2 ( x) ' . Og %ette vet vi fra f^r er o0

si%en y1( x , y2( x er line_rt uav#engige. ()vis #^ re si%en /ligningsettet ovenfor s0ulle v_re li0 null, #ar vi %ire0te at

y p( x) er en l^sning som tilfresstiller 0ravene, og er %erme%%en s^0te ent %ige l^sningen erme% er problemet formelt l^st.

en generelle l^sningen av %en in#omogene %iff ligningen #arformen y( x)= c1 y1( x)+ c2 y2( x)+ y p ( x) 6este problem blirimi%lerti% Z finne en parti0ul_r l^sning y p( x) . )er er %etflere meto%er som 0an bru0es. 7t vanlig tilfelle er %iff ligninger#vor venstre si%en #ar 0onstante 0oeffis%ienter.

y ' ' + ay ' + by= * ( x )er 0an vi 0omme langt me% lit

strategis0 g etting.1. * ( x)= A e rx , pr8v y p( x)= B er x

2. * ( x)= A sin rx+ B cos rx , pr8v y p ( x)= C sin rx + D cos rx3.

* ( x)= polynomav grad 3 , pr8v y p( x)= polynom avgrad 3 B

* ( x)= e x ( A sin rx+ B cos rx ) , pr8v y p ( x)= e x (C sin rx + D cos rx )

70semple: y ' ' 2 ' + y= A e rx . 5^sning av %en #omogen lig y%( x)= c1 e

x+ c 2 x e x vi pr^ver y p( x)= B e

rx og finner

B= A

(r 1)2 og %erme% y p( x)=

A(r 1)2 .

Anot#er example is: y ' ' + B y ' + B = cos# x 4e want to fin%t#e general solution to t#e %ifferential e uation, t#at is we want tofin% t#e solution to t#e #omogeneous %ifferential e uation.

y ' ' + B y' + B = ' *rom t#e c#aracteristic e uation\ 2+ B\ +B=(\ +2 )2= ' \ = 2,2 9ince we #ave repeate%

root, we #ave to intro%uce a factor of x for one solution to ensure

linear in%epen%ence. 9o we obtain u1= e 2 x

,u 2= x e 2 x

"#ewrons0ian of t#ese two functions is

e 2 x x e 2 x

2 e 2 x e 2 x(2 x 1)= e 2x e 2 x(2 x 1)+ 2x e 2 xe 2 x

eBx (2 x1)+2 xe B x=(2 x+1 +2x )e Bx=e Bx Necause t#e

wrons0ian is non ero, t#e two functions are linearl in%epen%ent,so t#is is in face t#e general solution for t#e #omogeneous %iffe uation. 4e see0 functions A(x an% N(x so

A( x)u 1+ B( x)u2 is a general solution of t#e non#omogeneous e uation. 4e nee% onl calculate t#e integrals

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A( x)= 17 u 2 ( x) B( x)dx , B ( x)= 17

u 1 b ( x)dx i.e.

A( x)= 1e Bx

x e 2x cos# x dx= xe 2 x cos# x dx A( x)= 1

1Xe x((V( x 1)+ e2 x (3 x 1))+ C )

B( x)= 1e B x

e 2x cos# xdx = e 2 xcos# x dx= 1T e x(3+ e2 x)+ C

FROM CHAPTER : EULER E$UATION EXERCISES.

= x1

x2

x 1+ y ' 2 dx 4e #ave " ( x , y , y ' )= x 1+ y' 2 dx

" y

= ', " y'

= ( x 1+ y' 2) y '

u = y' 2+ 1d udu

= 12 u

x 1+ y' 2

y '

x d ( y ' 2+ 1)

dy '

2 y ' 2+ 1=

x d ( y ' 2)

dy+ d

dy(1)

2 y ' 2+ 1 "

partia y' = x y '

y' 2+ 1 from 7uler e uation we #ave

x

" y '

" y

= ' x x y'

y ' 2+ 1= ' - "

y= '

x y ' y' 2+ 1

= c (constant ) ( x y ' )2= c2 ( y' 2+ 1)

x2 y ' 2= y ' 2 c2+ c 2 y ' = c

x2 c 2 , y' =

dyd x

= c

x2 c2dx

dy= c x2 c 2 dx y= c

x2c 2 1dx

arc cos# ( x)= 1

x2 1 y = c cos# 1( x/c ) x

c= cos#( y c )ax = cos# (ay+ b )

a = 1c

! b= c

constants c an% 0 are %etermine% from given

oints x1! x2 .HERE IS ANOTHER $UESTION SAME LIKE A6OVE:

= x1

x2

(1+ y y ' )2 dx " ( x , y , y ' )= (1+ y y ' )2 t#en

" y

= 2 ' (1+ y y' ) ! " y '

= 2 y(1+ y y' )

d d x

" y '

" y

= ' d d x

(2 y (1+ y y ' )) 2 y ' (1+ y y' )= '

#e %erivation ca n #appen wit# pro%uct t#ing.d

d x(u v )= v d u

d x+ u

d vd x

An% at last we get.

2( y2 y '' + 2 y y' 2+ y ' ) 2 y ' (1+ y y ' )= ' 2 y( y ' ' y2)= 'n% we #ave to fin% t#e x w#ic# is e ual to x=

12 c

y2 c' 2 c

.

RESI UE EXAMPLE: C

cos pi x

( 1 B 2 )( 1 + B z 2 )dx %en blir til

C

ei pi x

(1 B2)(1+ B z 2)dz er integras onsveien @ er en lu00et

urve ra 8 til 8 pZ %en reelle a0sen og en #alvsir0el me% ra%ius / vre #alvplan. 7tter or%ans lemma vil integralet over

#alvsir0elen gZ mot null nZr * . 9i%en integran%en #aroler pZ %en reelle a0sen for 1U2 og , mZ vi ogsZ leggemZ #alvsir0ler run%t %isse / vre #alvplan. Ii%ere er %et / vre

#alvplan en pol for iU2. ?rinsipal ver%ien av integralet er %a gitt

ve% P

cos & x

( 1 Bx 2 )( 1 + Bx 2 ) dx som er li0

2 & i *es ( z = i /2 )+ & i [ *es ( z = 1 / 2 )+ *es ( z = 1 / 2 ) ] Alle

olene er en0le, og vi #ar *es ( z = i /2)= 1X

i e&2 %en an%re

*es ( z = 1/2)= *es ( z = 1/2)= iX resultatet blir %a.

C

cos pi x(1 B2)(1+ B z 2)

dx = &X

(1+ e&2 )

FOURIER TRANSFROM EXAMPLE:

f ( x)=2x + a a x'

2x + 2a ' xa' ellers

en fourier transformerte er %a.

ANOTHER FRO6ENIBS METHO EXAMPLE.

T