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1 Mah Mo Ar 0 Chaer 9 Fourier Series ad alicaios o differeial equaios (ad arial differeial equaios) Fourier series defiiio ad covergece. The idea of Fourier series is relaed o he liear algebra coces of do roduc, orm, ad rojecio. We'll review his coecio afer he defiiio of Fourier series: Le f : K, /= be a iecewise coiuous fucio, or equivalely, exed o f : =/= as a Keriodic fucio. Examle oe could cosider he -eriodic exesio of f =, iiially defied o he Kierval K,, o all of =. Is grah is he so-called "e fucio", e 3 K3 K K 0 3 The Fourier coeffices of a Keriodic fucio f are comued via he defiiios Ad he Fourier series for f is give by a d K a 0 d K b d K f d f cos d, ; f si d, ; f w a 0 C a cos C = b si. = The idea is ha he arial sums of he Fourier series of f should acually coverge o f. The reasos why his should be rue combie liear algebra ideas relaed o orhoormal basis vecors ad rojecio, wih aalysis ideas relaed o covergece. Le's do a examle o illusrae he magic, before discussig (ars of) why he covergece acually haes.

2 Exercise Cosider he eve fucio f = o he ierval K % %, exeded o be he Keriodic "e fucio" e of age. Fid he Fourier coefficies a 0, a, b ad he Fourier series for e. The aswer is below, alog wih a grah of arial sum of he Fourier series. 3 K3 K K 0 3 soluio: e w K 4 odd cos f d / K 4 $ j = 0 4 $j C cos j C $ : lo f, =K0..0, color = black ; 3 K0 K

3 Usig echology o comue Fourier coefficies: f d / ; f := / a0 d $ f d; K () assume, ieger ; # his will le Male aem o evaluae he iegrals a d / $ f $cos $ d : K b d / $ f $si $ d : K a ; b ; a0 := K ~ K ~ 0 ()

4 So wha's goig o? Recall he ideas of do roduc, agle, orhoormal basis ad rojecio oo subsaces, i =, from liear algebra: For x, y =, he do roduc x, y d k x k y k saisfies for all vecors x, y, z = ad scalars s =: = a) x, x R 0 ad = 0 if ad oly if x = 0 b) x, y = y, x c) x, y C z = x, y C x, z d) s x, y = s x, y = x, s y From hese four roeries oe ca defie he orm or magiude of a vecor by x = x, x ad he disace bewee wo vecors x, y by dis x, y d x K y. Oe ca check wih algebra ha he Cauchy-Schwarz iequaliy holds: x, y % x y, wih equaliy if ad oly if x, y are scalar muliles of each oher. Oe cosequece of he Cauchy- Schwarz iequaliy is he riagle iequaliy x C y % x C y, wih equaliy if ad oly if x, y are o-egaive scalar muliles of each oher. Equivalely, i erms of Euclidea disace, dis x, z % dis x, y C dis y, z. Aoher cosequece of he Cauchy-Schwarz iequaliy is ha oe ca defie he agle q bewee x, y via cos q d x, y, x y for 0 % q %, because K % x, y % holds so ha q exiss. I aricular wo vecors x, y are x y eredicular, or orhogoal if ad oly if x, y = 0. If oe has a K dimesioal subsace W 4 = a orhoormal basis u, u,... u for W is a collecio of ui vecors (ormalized o legh ), which are also muually orhogoal. (Oe ca fid such bases via he Gram-Schmid algorihm.) For such a orho-ormal basis he eares oi rojecio of a vecor x = oo W is give by roj W x = x, u u C x, u u C... C x, u u = k = For ay x (already) i W, roj W x = x. x, u k u k.

5 The eire algebraic/geomeric develome o he revious age jus deeded o he four algebraic roeries a,b,c,d for he do roduc. So i ca be geeralized: Defiiio Le V is ay real-scalar vecor sace. we call V a ier roduc sace if here is a ier roduc f, g for which he ier roduc saisfies c f, g, h V ad scalars s =: a) f, f R 0. f, f = 0 if ad oly if f = 0. b) f, g = g, f. c) f, g C h = f, g C f, h d) s f, g = s f, g = f, s g. I his case oe ca defie f = f, f, dis f, g = f K g ; rove he Cauchy-Schwarz iequaliy ad he riagle iequaliies; defie agles bewee vecors, ad i aricular, orhogoaliy bewee vecors; fid orho-ormal bases u, u,... u for fiie-dimesioal subsaces W, ad rove ha for ay f V he eares eleme i W o f is give by roj W f = f, u u C f, u u C... C f, u u = k! f, u k O u k. = Theorem Le V = f : =/= s.. f is iecewise coiuous ad Keriodic. Defie f, g d f g d. K ) The V,, is a ier roduc sace. ) Le V d sa, cos, cos,..., cos, si, si,... si. The he C fucios lised i his collecio are a orhoormal basis for he C dimesioal subsace V. I aricular, for ay f V he eares fucio i V o f is give by roj V f =! f, O C! f, cos O cos C =! f, si O si = = a 0 C a cos C = b si = where a 0, a, b are he Fourier coefficies defied o age.

6 Exercise ) Check ha, cos, cos,..., cos, si, si,... si are orhoormal wih resec o he ier roduc f, g d f g d K so Hi: cos m C k = cos m cos k K si m si k si m C k = si m cos k C cos m si k cos m cos k = cos m C k C cos mkk (eve if m = k si m si k = cos mkk Kcos m C k (eve if m = k cos m si k = si m C k C si Km C k

7 Exercise 3) Cosider he K eriodic odd fucio saw defie by exedig f =, K! % as a K eriodic fucio. sawooh fucio 3 K3 K K K 3 K3 Fid he Fourier series for saw. Hi: you oiced ha for he eve e fucio i Exercise he sie Fourier coefficies were all zero. Which oes will be zero for ay odd fucio? Why? soluio: saw w = K C si 0 f d /$ = K C $si $ : lo f, =K0..0, color = black ; 3 K0 K5 K 5 0

8 Covergece Theorems (These require some careful mahemaical aalysis o rove - hey are ofe discussed i Mah 50, for examle.) Theorem Le f : =/= be Keriodic ad iecewise coiuous. Le f = roj V f = a 0 C a cos C = b si = be he Fourier series rucaed a. The lim / fkf = lim / K f Kf I oher words, he disace bewee f ad f coverges o zero, where we are usig he disace fucio ha we ge from he ier roduc, dis f, g = f K g = f K g, f K g = K d = 0. f K g d. Theorem If f is as i Theorem, ad is (also) iecewise differeiable wih a mos jum discoiuiies, he (i) for ay 0 such ha f is differeiable a 0 lim / f = f 0 0 (oiwise covergece). (ii) for ay 0 where f is o differeiable (bu is eiher coiuous or has a jum discoiuiy), he where lim / f 0 = f K 0 = lim / 0 K f, f C 0 f K 0 C f C 0 = lim / 0 C f Examles: ) The rucaed Fourier series for he e fucio, e coverge o e for all. I fac, i ca be show ha he covergece is uiform, i.e. c e O 0d s.. R 0 e Ke oce. ) The rucaed Fourier series for he sawooh fucio, saw coverge o saw for all! e for all a s C k, k Z (i.e. everywhere exce a he jum ois). A hese jum ois he Fourier series coverges o he average of he lef ad righ had limis of saw, which is 0. (I fac, each arial sum evaluaes o 0 a hose ois.) The covergece a he oher values is oiwise, bu o uiform, as he covergece akes loger earer he jum ois.)

9 Exercise 4) We ca derive "magic" summaio formulas usig Fourier series. (See your homework for some more.) From Theorem we kow ha he Fourier series for e coverges for all. I aricular 4a) Deduce 4b) Verify ad use o show 0 = e 0 = K 4 odd C 3 C 5 C... = odd = cos $0. = odd C eve = odd C 4 = = = 6. = 8.

10 Differeiaig Fourier Series: Theorem 3 Le f be Keriodic, iecewise differeiable ad coiuous, ad wih f# iecewise coiuous. Le f have Fourier series f w a 0 C a cos C = b si. = The f# has he Fourier series you'd exec by differeiaig erm by erm: roof: Le f# have Fourier series The f#w K a si C = b cos = f#w A 0 C A cos C = B si. = A = f# cos d, ;. K Iegrae by ars wih u = cos, dv = f# d, du =K si d, v = f : K Similarly, A 0 = 0, B =K a. f# cos d = f K si K K = 0 C K K f si d = b. f K si d Remark: This is aalogous o wha haeed wih Lalace rasform. I ha case, he rasform of he derivaive mulilied he rasform of he origial fucio by s (ad here were correcio erms for he iiial values). I his case he raformed variables are he a, b which deed o. Ad he Fourier series "rasform" of he derivaive of a fucio mulilies hese coefficies by (ad ermues hem). A

11 Exercise 5a Use he differeiaio heorem ad he Fourier series for e o fid he Fourier series for he square wave, square, which is he Keriodic exesio of f = K K!! 0 0!! square() K3 K 3 K (You will fid he series direcly from he defiiio i your homework.) 5b) Deduce he magic formula K 3 C 5 K 7 C...= K k k = = 0 k C 4. soluio: square w 4 odd si 0 f3 d / 4 $ = 0 $ C $si $ C $ : lo f3, =K0..0, color = black ; K0 K5 5 0 K Could you check he Fourier coefficies wih echology?