Examples of Fourier series - Kenyatta University

1 Leif Mejlbro Examples of Fourier series Download free books at Download free ebooks at MejlbroExamples of Fourier seriesCalculus 4c-1 Download free ebooks at of Fourier series Calculus 4c-1 2008 Leif Mejlbro & Ventus Publishing ApSISBN 978-87-7681-380-2 Download free ebooks at of Fourier series 4 ContentsContents Introduction 1. Sum function of Fourier series 2. Fourier series and uniform convergence 3. Parseval s equation 4. Fourier series in the theory of beams 5 662101115 Stand out from the crowdDesigned for graduates with less than one year of full-time postgraduate work experience, London Business School s Masters in Management will expand your thinking and provide you with the foundations for a successful career in programme is developed in consultation with recruiters to provide you with the key skills that top employers demand.

2 Through 11 months of full-time study, you will gain the business knowledge and capabilities to increase your career choices and stand out from the are now open for entry in September more information visit email or call +44 (0)20 7000 7573 Masters in ManagementLondon Business SchoolRegent s ParkLondon NW1 4 SAUnited KingdomTel +44 (0)20 7000 7573 Email your careerPlease click the advertDownload free ebooks at of Fourier series 5 IntroductionIntroductionHere we present a collection of Examples of applications of the theory of Fourier series . The reader isalso referred toCalculus 4bas well as toCalculus should no longer be necessary rigourously to use the ADIC-model, described inCalculus 1candCalculus 2c, because we now assume that the reader can do this if I have tried to be careful about this text, it is impossible to avoid errors, in particular in thefirst edition.

3 It is my hope that the reader will show some understanding of my Mejlbro20th May 2008 Download free ebooks at of Fourier series 6 1 Sum function of Fourier seriesA general remark. In some textbooks the formulation of the main theorem also includes theunnecessary assumption that the graph of the function does not have vertical half tangents. It shouldbe replaced by the claim thatf L2over the given interval of period. However, since most peopleonly know the old version, I have checked in all Examples that the graph of the function does not havehalf tangents. Just in example thatcosn =( 1)n,n N0. Find and prove an analogous expression forcosn 2and forsinn 2.

4 (Hint: check the expressions forn=2p,p N0,andforn=2p 1,p N).0 3/2*Pi-PiPi/2(cos(t),sin(t)) 1 1 may interpret (cost,sint) as a point on the unit unit circle has the length 2 , so by winding an axis round the unit circle we see thatn alwayslies in ( 1,0) [rectangular coordinates] fornodd, and in (1,0) follows immediately from the geometric interpretation thatcosn =( 1) get in the same way that atcosn 2= 0fornulige,( 1)n/2fornlige,andsinn 2= ( 1)(n 1)/2fornulige, function of Fourier seriesDownload free ebooks at of Fourier series 7 example the Fourier series for the functionf K2 , which is given in the interval] , ]byf(t)

5 = 0for <t 0,1for0<t ,and find the sum of the series fort= 4 224xObviously,f(t) is piecewiseC1without vertical half tangents, sof K 2 . Then the adjusted functionf (t) is defined byf (t)= f(t)fort =p , p Z,1/2fort=p , p Fourier series is pointwise convergent everywhere with the sum functionf (t). In particular, thesum of the Fourier series att=0isf (0) =12,(the last question).Sum function of Fourier series UBS 2010. All rights for a career where your ideas could really make a difference? UBS s Graduate Programme and internships are a chance for you to experience for yourself what it s like to be part of a global team that rewards your input and believes in succeeding you are in your academic career, make your future a part of ours by visiting You re full of energyand ideas.

6 And that s just what we are looking click the advertDownload free ebooks at of Fourier series 8 The Fourier coefficients are thena0=1 f(t)dt=1 0dt=1,an=1 f(t)cosnt dt=1 0cosnt dt=1n [sinnt] 0=0,n 1,bn=1 f(t)sinnt dt=1 0sinnt dt= 1n [cosnt] 0=1 ( 1)nn ,henceb2n=0ogb2n+1=2 12n+ Fourier series is (with = instead of )f (t)=12a0+ n=1{ancosnt+bnsinnt}=12+2 n=012n+1sin(2n+1) the Fourier series for the functionf K2 , given in the interval] , ]byf(t)= 0for <t 0,sintfor0<t ,and find the sum of the series fort=p ,p 4 224xThe functionfis piecewiseC1without any vertical half tangents, hencef K 2.

7 Sincefis contin-uous, we even havef (t)=f(t), so the symbol can be replaced by the equality sign =,f(t)=12a0+ n=1{ancosnt+bnsinnt}.It follows immediately ( the last question) that the sum of the Fourier series att=p ,p Z,isgiven byf(p ) = 0, (cf. the graph).The Fourier coefficients area0=1 f(t)dt=1 0sintdt=1 [ cost] 0=2 ,a1=1 0sint costdt=12 sin2t 0=0,Sum function of Fourier seriesDownload free ebooks at of Fourier series 9 Sum function of Fourier seriesan=1 0sint cosnt dt=12 0{sin(n+1)t sin(n 1)t}dt=12 1n 1cos(n 1)t 1n+1cos(n+1)t 0=12 1n 1 ( 1)n 1 1 1n+1 ( 1)n+1 1 = 1 1+( 1)nn2 1forn> ,1+( 1)n= 2forneven,0fornodd,hencea2n+1=0forn 1, anda2n= 2 14n2 1,n N,(replacenby 2n).

8 Analogously,b1=1 0sin2tdt=1 12 0{cos2t+sin2t}dt=12,and forn>1wegetbn=1 0sint sinnt dt=12 0{cos(n 1)t cos(n+1)t}dt= up we get the Fourier series (with =, cf. above)f(t)=12a0+ n=1{ancosnt+bnsinnt}=1 +12sint 2 n=114n2 1cos of the last ,p Z,f(p )=0=1 2 n=114n2 1,hence by a rearrangement n=114n2 1= can also prove this result by a decomposition and then consider the sectional sequence,sN=N n=114n2 1=N n=11(2n 1)(2n+1)=12N n=1 12n 1 12n+1 =12 1 12N+1 12 Download free ebooks at of Fourier series 10 forN , hence n=114n2 1= limN sN= the periodic functionf:R R,ofperiod2 , be given in the interval] , ]byf(t)= 0,fort ] , /2[,sint,fort [ /2, /2],0fort ] /2, ].

9 Find the Fourier series of the function and its sum function. 1 3 2 1123xThe functionfis piecewiseC1without vertical half tangents, hencef K 2 . According to the maintheorem, the Fourier theorem is thenpointwise convergenteverywhere, and its sum function isf (t)= 1/2fort= 2+2p , p Z,1/2fort= 2+2p , p Z,f(t) (t) is discontinuous, the Fourier seriescannotbe uniformly ,f( t)= f(t), so the function is odd, and thusan=0foreveryn N0,andbn=2 0f(t)sinnt dt=2 /20sint sinnt dt=1 /20{cos((n 1)t) cos((n+1)t)} the exceptional casen= 1 we get insteadb1=1 /20(1 cos 2t)dt=1 t 12sin 2t /20=12,and forn N\{1}we getbn=1 1n 1sin((n 1)t) 1n+1sin((n+1)t) /20=1 1n 1sin n 12 1n+1sin n+12.

10 Sum function of Fourier seriesDownload free ebooks at of Fourier series 11 It follows immediately that ifn>1 is odd,n=2p+1,p 1, thenb2p+1=0(notethatb1=12hasbeen calculated separately) and that (forn=2peven)b2p=1 12p 1sin p 2 12p+1sin p + 2 =1 12p 1 cos(p ) sin 2 12p+1 cosp sin 2 =1 ( 1)p+1 12p 1+12p+1 =1 ( 1)p+1 4p4p2 changing variablep n, it follows thatfhas the Fourier seriesf 12sint+ n=11 ( 1)n 1 4n4n2 1sin 2nt=f (t),where we already have proved that the series is pointwise convergent with the adjusted functionf (t)as its sum function of Fourier seriesPlease click the advertDownload free ebooks at of Fourier series 12 example the Fourier series for the periodic functionf K2 , given in the interval] , ]byf(t)=|sint|.