Ae2 Mathematics : Fourier Series

Transcription

1 Ae Mthemtics : Fourier Series J. D. Gibbon (Professor J. D Gibbon, Dept of Mthemtics j.d.gibbon@ic.c.uk jdg These notes re not identicl word-for-word with my lectures which will be given on WB. Some of these notes my contin more emples thn the corresponding lecture while in other cses the lecture my contin more detiled working. I will not be hnding out copies of these notes you re therefore dvised to ttend lectures nd tke your own. Contents Introduction Derivtion of the Fourier coefficients m nd b m. Proof of (., (. nd ( Prsevl s equlity Emple Odd nd even functions & periodic etension 5 3. Odd nd even functions Emples Periodic etension nd hlf-rnge series An emple of periodic etension Fourier Trnsforms s the limit of Fourier series Do not confuse me with Dr J. Gibbons who is lso in the Mthemtics Dept.

2 5th// (emfs.te Introduction et us consider functions f( tht re periodic on the -is with period p f( + p f(. (. If f( is plotted over severl periods its repetitive nture becomes cler. Two obvious emples re sin nd cos ech of which hve period. (. lso generlizes to f( + np f( for integer n. (. Another emple is the swtooth function sketched below which is defined s f( on nd is lso periodic of period p. This function is, however, discontinuous t (n +. f( Yet nother emple seen below is squre wve of period p : f(, The swtooth nd squre-wve functions re typicl emples of rough functions in the sense tht they contin discontinuities. The question we wish to consider is whether such rough periodic functions cn be represented by n infinite series, not of polynomils, but of sines nd cosines?

3 5th// (emfs.te Derivtion of the Fourier coefficients m nd b m The generl convention for wht re clled Fourier series is to consider periodic functions of period over the domin [, ] on the -is. Then : The Fourier series for periodic function f(, periodic on [, ] hs Fourier series representtion f( + m m cos where the Fourier coefficients m nd b m re given by m while f( cos d, } + b m sin b m (.3 is consistent with putting m in m defined in (.. (. f( sin d, (. f( d. (.3. Proof of (., (. nd (.3 Split the cosine nd sine in (. into the combintions to turn (. into where f( + c m + m cos θ [ep(iθ + ep( iθ] sin θ [ep(iθ ep( iθ] (.4 i m (m ib m e im/ + ( m + ib m e im/} c m e im/ (.5 ( m ib m m >, m + ib m m <, c. (.6 We cn invert this to find the c m by multiplying by e in/ (n is nother integer nd integrting cross the epnsion in (.5 f(e in/ d + m c m e i(m n/ d. (.7

4 5th// (emfs.te 3 When m n the integrl on the RHS of (.7 is e i(m n/ d (ei(m n e i(m n i(m n sin(m n m n (m n m n (.8 Thus only one term is non-zero in the infinite series on the RHS of (.7, leving us with c m Tking rel & imginry prts & using (.6 we obtin (.. f(e im/ d, for m,,,... (.9 Under wht circumstnces is the epnsion (. relistic representtion of our periodic function f(? There is proof too complicted for this syllbus becuse it requires use of the Riemnn-ebsegue emm tht shows the following version of Fourier s Theorem : Provided f( is finite t its discontinuities in [, ] nd it lso hs finite number of mim & minim on this intervl then the Fourier series (. converges to f( t points where f( is continuous nd t discontinuities d (sy, f( must converge to lim [f( d δ + f( d + δ]. (. δ This, of course, is the verge vlue of the function t d.. Prsevl s equlity Consider the squre of the infinite series from m to m in (.5 nd form the integrl f( d c n c m e i(n m/ d. (. m n The integrl on the RHS hs lredy been found in (.8 so (. becomes f( m n d m c m m n Now use the definition of c m in (.6 to show tht m c m c + c m 4 + m Thus we hve Prsevl s equlity (see formul sheet : f( d + m m (. ( m + b m. (.3 ( m + b m. (.4

5 5th// (emfs.te 4.3 Emple f( Consider the function f( which is lso periodic over the whole -is. In this cse nd we hve (.5 m f( cos (m d, b m f( sin (m d, (.6 nd, for m, f( d d, (.7 in which cse /. For m m cos (m d d[sin (m] m [ ] sin(m m sin (m m [ ] cos(m m [ ( m ]. m (.8 Thus we hve Now we clculte b m : m even m m odd m (.9 b m sin (m d d[cos (m] m [ ] cos(m m + cos (m d m cos(m m. (.

6 5th// (emfs.te 5 Thus b m ( m+ m. (. Together the complete series is : f( 4 } } cos cos 3 sin sin sin ( Odd nd even functions & periodic etension 3. Odd nd even functions (i A function f( is sid to be even if it hs the property f( f(. (i A function f( is sid to be odd if it hs the property f( f(. Typicl emples of even functions re cos or n for integer vlues of n. Typicl emples of odd functions re sin or n+ for integer vlues of n. Note tht mny functions re neither odd nor even : for emple ep. However they cn be split down into even nd odd prts An emple is f( [f( + f( ] }} even [ e e + e ] }} cosh Why re these relevnt to Fourier series? + + [f( f( ]. (3. }} odd [ e e ]. (3. }} sinh Consider first b m for n even function f( : we split the domin down into [, ] nd [, ]. On [, ] we chnge vribles from y nd chnge the sign on the integrl when switching limits b m f( sin d + f( sin d ( f( sin d + f( y sin my d( y ( my f( sin d f( y sin d(y for n even f( f( sin ( m (3.3 d for n odd f( Thus if f( is n even function then the two integrls cncel. For m the story is the Note tht while the dummy vribles within the two integrls in (3.3 re different, this does not mtter becuse the limits in the two re identicl nd finite.

7 5th// (emfs.te 6 opposite m f( cos d + f( cos d ( f( cos d + f( y cos my d( y ( my f( cos d + f( y cos d(y f( cos ( m d for n even f( for n odd f( (3.4 Thus we hve the generl result : (i If f( is n even function then m f( cos d, b m. (3.5 (ii If f( is n odd function then m, b m f( sin d. (3.6 Note tht the non-zero integrls re over [, ] with fctor of outside & thus double-up. 3. Emples Emple : (even Consider the function f( (3.7 nd periodic over the whole -is. f( Now nd is n even function so b m. d, (3.8

8 5th// (emfs.te 7 so. Moreover, Therefore m The first few terms in the series re 3 f( 4 Now pply Prsevl s equlity. Firstly cos(m d d[sin(m] d m [ ] sin(m m sin(m d m [ ] cos(m m m [ ( m ]. (3.9 nd so from (.4, using nd m, m even m m odd 4 m (3. } cos 3 cos 5 cos ( f( d r or ( 9.87 nd nd 4 /96.5 d 3 (3. (r + 4 ( (3.4 Emple : (odd Consider the function f( (3.5 nd periodic over the whole -is. This is n odd function with. 3 By definition, plots cn be mde of only finite number of terms : these cpture the stright lines well but oscilltions re seen to occur round the pices of the tringles : these re clled Gibbs phenomen.

9 5th// (emfs.te 8 f(, Thus m including nd b m m sin(m d [ ] cos(m m [ ( m ] (3.6 Therefore m even b m 4 m odd (3.7 m The first few terms in the series re f( 4 } sin 3 sin 5 sin ( Oscilltory Gibbs phenomen re seen round the discontinuities in plots of finite number of terms. 3.3 Periodic etension nd hlf-rnge series If we re given function f( defined only over [, ] then, by definition, this is not periodic function so Fourier s Theorem is not pplicble. To circumvent this problem, construct globl function F ( tht is equl to f( in [, ] but is etended over the whole -is s periodic function. Consider f( below : f( The globl function F ( could be constructed two wys : the first is to repet it d infinitum s reflection in the y-is to mke it n even function : we sy tht F ( is the even etension of f(. Clerly b m.

10 5th// (emfs.te 9 F ( The second is to repet it d infinitum s n nti-reflection in the y-is to mke it n odd function : we sy tht F ( is the odd etension of f(. Clerly m. F ( In both cses the etension F ( is periodic function to which Fourier s Theorem pplies. The even nd odd etensions re mtter of choice : the resultnt Fourier series re clled hlf-rnge series for obvious resons. 3.4 An emple of periodic etension Define the tent-function f( (3.9 : The even etension of f( defined in (3.9 with is sketched below : Becuse this etension is even by construction, b m. To clculte m : nd so. Moreover f(d m f( cos m d / / d + / cos m d + ( d 4 (3. / ( cos m d. (3.

11 5th// (emfs.te Integrtion by prts gives b b cos m d cos m d m [sin m]b (3. [ m sin m + ] b cos m m (3.3 m m [sin m] / + [ sin m]/ + [cos m]/ m m m [ sin m] / m [cos m] / m [ cos( m cos m] (3.4 Using the fct tht cos m ( m : we hve cos m ( r when m r (even nd cos m when m r + (odd. Thus m m r + (odd [cos r ] r m r (even (3.5 The r-even terms re zero leving only r-odd terms with m /r (m r but r odd : f( 4 } cos cos 6 cos (3.6 : The odd etension of f( defined in (3.9 with is sketched below : Becuse this etension is odd by construction, m. b m Integrtion by prts gives b / b sin m d sin m d + / ( sin m d (3.7 sin m d m [cos m]b (3.8 [ m cos m + m sin m ] b (3.9

12 5th// (emfs.te b m m [cos m] / [ cos m]/ + [sin m]/ m m + m [ cos m] / m [sin m] / m [ sin( m sin m] m sin( m m r (even sin(r+ m r + (odd (r+ Now sin(r + cos r ( r. Thus m r (even b m 4( r m r + (odd (r+ Thus the odd-rnge epnsion is f( 4 } sin sin 3 sin 5 sin (3.3 (3.3 (3.3 Note: put in (3.6 or put / in (3.3 & end up with the sme epnsion for ( Fourier Trnsforms s the limit of Fourier series The Fourier Trnsform 4 of function f( nd its inverse on the infinite domin [, ] is defined by (FT f(k e ik f( d, (inverse FT f( e ik f(k dk. (4. To prove f( is the inverse of f(k s stted in (4., we use (.5 f( lim + m ( f( e im / d e im/ Dirichlet integrl (4. et m /m nd κ m / m m/. Define δκ m κ m+ κ m /. Then f( lim m 4 Not eminble but useful in other courses. ( δκ m e iκm e iκm f( d. (4.3

13 5th// (emfs.te We define nd in the limit k δκ m f (k e ik f( d (4.4 f( lim lim κ m m e iκm f (κ m δκ m (4.5 e ik f(k dk (4.6 which is the inverse Fourier Trnsform bsed on the Fourier Trnsform s the limit of (4.4 f(k lim f(k Emple : To find f(k when f( e where > : e ik f( d. (4.7 (4.8 f(k e ik e d + e ik e d. (4.9 The integrls re simple nd the resulting functions vnish t ± but not t. f(k ik + + ik + k. (4. Note finlly tht when the prmeter is very lrge the function f( hs very shrp cusp t which gets shrper s is incresed wheres f(k flttens out.