Mah 80-00 Mo Ar 0 Chaer 9 Fourier Series ad alicaios o differeial equaios (ad arial differeial equaios) 9.-9. 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.
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 K5 0 5 0
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 ()
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.
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.
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
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
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.)
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.
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
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?