Fourier Series. x 2 = 2L2. n 2. x 2

Transcription

1 Fourier Series Fourier series strted life s method to solve problems bout the flow of het through ordinry mterils. It hs grown so fr tht if you serch our librry s ctlog for the keyword Fourier you will find 68 entries s of this dte. It is tool in bstrct nlysis nd electromgnetism nd sttistics nd rdio communiction nd.... People hve even tried to use it to nlyze the stock mrket. (It didn t help.) The representtion of musicl sounds s sums of wves of vrious frequencies is n udible exmple. It provides n indispensible tool in solving prtil differentil equtions, nd lter chpter will show some of these tools t work. 5. Exmples The power series or Tylor series is bsed on the ide tht you cn write generl function s n infinite series of powers. The ide of Fourier series is tht you cn write function s n infinite series of sines nd cosines. You cn lso use functions other thn trigonometric ones, but I ll leve tht generliztion side for now, except to sy tht egendre polynomils re n importnt exmple of functions used for such more generl expnsions. An exmple: On the intervl < x < the function x 2 vries from to 2. It cn be written s the series of cosines x 2 = π 2 = π 2 ( ) n n 2 cos nπx [ cos πx 2πx cos 4 + 3πx cos 9 ] (5.) To see if this is even plusible, exmine successive prtil sums of the series, tking one term, then two terms, etc. Sketch the grphs of these prtil sums to see if they strt to look like the function they re supposed to represent (left grph). The grphs of the series, using terms up to n = 5 do pretty well t representing the prbol. x x The sme function cn be written in terms of sines with nother series: x 2 = 22 π [ ( ) n+ n 2 ( ( ) n π 2 n 3 ) ) ] sin nπx (5.2) nd gin you cn see how the series behves by tking one to severl terms of the series. (right grph) The grphs show the prbol y = x 2 nd prtil sums of the two series with terms up to n =, 3, 5. Jmes Nering, University of Mimi

2 5 Fourier Series 2 The second form doesn t work s well s the first one, nd there s reson for tht. The sine functions ll go to zero t x = nd x 2 doesn t, mking it hrd for the sum of sines to pproximte the desired function. They cn do it, but it tkes lot more terms in the series to get stisfctory result. The series Eq. (5.) hs terms tht go to zero s /n 2, while the terms in the series Eq. (5.2) go to zero only s /n.* 5.2 Computing Fourier Series How do you determine the detils of these series strting from the originl function? For the Tylor series, the trick ws to ssume series to be n infinitely long polynomil nd then to evlute it (nd its successive derivtives) t point. You require tht ll of these vlues mtch those of the desired function t tht one point. Tht method won t work in this cse. (Actully I ve red tht it cn work here too, but with ridiculous mount of lbor nd some mthemticlly suspect procedures.) The ide of Fourier s procedure is like one tht you cn use to determine the components of vector in three dimensions. You write such vector s A = A x ˆx + A y ŷ + A z ẑ And then use the orthonormlity of the bsis vectors, ˆx. ŷ = etc. Tke the sclr product of the preceding eqution with ˆx. ˆx. A = ˆx. ( A x ˆx + A y ŷ + A z ẑ ) = A x nd ŷ. A = Ay nd ẑ. A = Az (5.3) This lets you get ll the components of A. For exmple, ẑ γ β A ŷ ˆx. A = Ax = A cos α ŷ. A = Ay = A cos β (5.4) ˆx α ẑ. A = Az = A cos γ This shows the three direction cosines for the vector A. You will occsionlly see these numbers used to describe vectors in three dimensions, nd it s esy to see tht cos 2 α + cos 2 β + cos 2 γ =. In order to stress the close nlogy between this sclr product nd wht you do in Fourier series, I will introduce nother nottion for the sclr product. You don t typiclly see it in introductory courses for the simple reson tht it isn t needed there. Here however it will turn out to be very useful, nd in the next chpter you will see nothing but this nottion. Insted of ˆx. A or A. B you use ˆx, A or A, B. The ngle brcket nottion will mke it very esy to generlize the ide of dot product to cover other things. In this nottion the bove equtions will pper s ˆx, A = A cos α, ŷ, A = A cos β, ẑ, A = A cos γ nd they men exctly the sme thing s Eq. (5.4). * For nimted sequences showing the convergence of some of these series, see

3 5 Fourier Series 3 There re orthogonlity reltions similr to the ones for ˆx, ŷ, nd ẑ, but for sines nd cosines. et n nd m represent integers, then dx sin ( nπx ) sin ( mπx ) { n m = /2 n = m (5.5) This is sort of like ˆx. ẑ = nd ŷ. ŷ =, where the nlog of ˆx is sin πx/ nd the nlog of ŷ is sin 2πx/. The biggest difference is tht it doesn t stop with three vectors in the bsis; it keeps on with n infinite number of vlues of n nd the corresponding different sines. There re n infinite number of very different possible functions, so you need n infinite number of bsis functions in order to express generl function s sum of them. The integrl Eq. (5.5) is continuous nlog of the coordinte representtion of the common dot product. The sum over three terms A x B x +A y B y +A z B z becomes sum (integrl) over continuous index, the integrtion vrible. By using this integrl s generliztion of the ordinry sclr product, you cn sy tht sin(πx/) nd sin(2πx/) re orthogonl. et i be n index tking on the vlues x, y, nd z, then the nottion A i is function of the vrible i. In this cse the independent vrible tkes on just three possible vlues insted of the infinite number in Eq. (5.5). How do you derive n identity such s Eq. (5.5)? The first method is just stright integrtion, using the right trigonometric identities. The esier (nd more generl) method cn wit for few pges. cos(x ± y) = cos x cos y sin x sin y, subtrct: cos(x y) cos(x + y) = 2 sin x sin y Use this in the integrl. 2 dx sin ( nπx ) sin ( mπx ) = [ ( ) ( )] (n m)πx (n + m)πx dx cos cos Now do the integrl, ssuming n m nd tht n nd m re positive integers. ( ) ( ) (n m)πx = (n m)π sin (n + m)πx (n + m)π sin = (5.6) Why ssume tht the integers re positive? Aren t the negtive integers llowed too? Yes, but they ren t needed. Put n = into sin(nπx/) nd you get the sme function s for n = +, but turned upside down. It isn t n independent function, just times wht you lredy hve. Using it would be sort of like using for your bsis not only ˆx, ŷ, nd ẑ but ˆx, ŷ, nd ẑ too. Do the n = m cse of the integrl yourself. Computing n Exmple For simple exmple, tke the function f(x) =, the constnt on the intervl < x <, nd ssume tht there is series representtion for f on this intervl. = ( nπx ) n sin ( < x < ) (5.7) Multiply both sides by the sine of mπx/ nd integrte from to. dx sin ( mπx ) = dx sin ( mπx ) n sin n= ( nπx ) (5.8)

4 5 Fourier Series 4 Interchnge the order of the sum nd the integrl, nd the integrl tht shows up is the orthogonlity integrl derived just bove. When you use the orthogonlity of the sines, only one term in the infinite series survives. dx sin ( mπx ) = = n dx sin n= n= ( mπx ) sin { n. (n m) /2 (n = m) = m /2. Now ll you hve to do is to evlute the integrl on the left. dx sin ( mπx ) = [ mπ cos mπx This is zero for even m, nd when you equte it to (5.9) you get m = 4 mπ for m odd ] ( nπx ) = [ ( ) m ] mπ You cn relbel the indices so tht the sum shows only odd integers m = 2k + nd the Fourier series is 4 mπx sin π m = 4 (2k + )πx sin =, ( < x < ) (5.) π 2k + m odd > k= (5.9) highest hrmonic: 5 highest hrmonic: 2 highest hrmonic: The grphs show the sum of the series up to 2k + = 5, 9, 99 respectively. It is not very rpidly converging series, but it s strt. You cn see from the grphs tht ner the end of the intervl, where the function is discontinuous, the series hs hrd time hndling the jump. The resulting overshoot is clled the Gibbs phenomenon, nd it is nlyzed in section 5.7. Nottion The point of introducing tht other nottion for the sclr product comes right here. The sme nottion is used for these integrls. In this context define f, g = dx f(x) * g(x) (5.) nd it will behve just the sme wy tht A. B does. Eq. (5.5) then becomes un, u m = { n m /2 n = m precisely nlogous to ( nπx ) where u n (x) = sin ŷ, ˆx, ˆx = nd ẑ = (5.2) These u n re orthogonl to ech other even though they ren t normlized to one the wy tht ˆx nd ŷ re, but tht turns out not to mtter. u n, u n = /2 insted of =, so you simply keep trck of it.

5 5 Fourier Series 5 (Wht hppens to the series Eq. (5.7) if you multiply every u n by 2? Nothing, becuse the coefficients n get multiplied by /2.) The Fourier series mnipultions, Eqs. (5.7), (5.8), (5.9), become = n u n then um, = u m, n u n = n um, u n = m um, u m n= (5.3) This is fr more compct thn you see in the steps between Eq. (5.7) nd Eq. (5.). You still hve to evlute the integrls u m, nd u m, u m, but when you mster this nottion you ll likely mke fewer mistkes in figuring out wht integrl you hve to do. Agin, you cn think of Eq. (5.) s continuous nlog of the discrete sum of three terms, A, B = Ax B x + A y B y + A z B z. The nlogy between the vectors such s ˆx nd functions such s sine is relly fr deeper, nd it is centrl to the subject of the next chpter. In order not to get confused by the nottion, you hve to distinguish between whole function f, nd the vlue of tht function t point, f(x). The former is the whole grph of the function, nd the ltter is one point of the grph, nlogous to sying tht A is the whole vector nd A y is one of its components. The sclr product nottion defined in Eq. (5.) is not necessrily restricted to the intervl < x <. Depending on context it cn be over ny intervl tht you hppen to be considering t the time. In Eq. (5.) there is complex conjugtion symbol. The functions here hve been rel, so this mde no difference, but you will often del with complex functions nd then the fct tht the nottion f, g includes conjugtion is importnt. This nottion is specil cse of the generl development tht will come in section 6.6. The bsis vectors such s ˆx re conventionlly normlized to one, ˆx. ˆx =, but you don t hve to require it even there, nd in the context of Fourier series it would clutter up the nottion to require u n, u n =, so I don t bother. Some Exmples To get used to this nottion, try showing tht these pirs of functions re orthogonl on the intervl < x <. Sketch grphs of both functions in every cse. x, 3 2 x = sin πx/, cos πx/ = sin 3πx/, 2x = The nottion hs complex conjugtion built into it, but these exmples re ll rel. Wht if they ren t? Show tht these re orthogonl too. How do you grph these? Not esily.* e 2iπx/, e 2iπx/ = 4 (7 + i)x, ix = Extending the function In Equtions (5.) nd (5.2) the originl function ws specified on the intervl < x <. The two Fourier series tht represent it cn be evluted for ny x. Do they equl x 2 everywhere? No. The first series involves only cosines, so it is n even function of x, but it s periodic: f(x + 2) = f(x). The second series hs only sines, so it s odd, nd it too is periodic with period 2. * but see if you cn find copy of the book by Jhnke nd Emde, published long before computers. They show exmples. Also check out bnchoff/script/cfgind.html or bennett/jomcg/

6 5 Fourier Series 6 Here the discontinuity in the sine series is more obvious, fct relted to its slower convergence. 5.3 Choice of Bsis When you work with components of vectors in two or three dimensions, you will choose the bsis tht is most convenient for the problem you re working with. If you do simple mechnics problem with mss moving on n incline, you cn choose bsis ˆx nd ŷ tht re rrnged horizontlly nd verticlly. OR, you cn plce them t n ngle so tht they point down the incline nd perpendiculr to it. The ltter is often simpler choice in tht type of problem. The sme pplies to Fourier series. The intervl on which you re working is not necessrily from zero to, nd even on the intervl < x < you cn choose mny sets of function for bsis: sin nπx/ cos nπx/ sin(n + / 2 )πx/ e 2πinx/ (n =, 2,...) s in equtions (5.) nd (5.2), or you cn choose bsis (n =,, 2,...) s in Eq. (5.), or you cn choose bsis (n =,, 2,...), or you cn choose bsis (n =, ±, ±2,...), or n infinite number of other possibilities. In order to use ny of these you need reltion such s Eq. (5.5) for ech seprte cse. Tht s lot of integrtion. You need to do it for ny intervl tht you my need nd tht s even more integrtion. Fortuntely there s wy out: Fundmentl Theorem If you wnt to show tht ech of these respective choices provides n orthogonl set of functions you cn integrte every specil cse s in Eq. (5.6), or you cn do ll the cses t once by deriving n importnt theorem. This theorem strts from the fct tht ll of these sines nd cosines nd complex exponentils stisfy the sme differentil eqution, u = λu, where λ is some constnt, different in ech cse. If you studied section 4.5, you sw how to derive properties of trigonometric functions simply by exmining the differentil eqution tht they stisfy. If you didn t, now might be good time to look t it, becuse this is more of the sme. (I ll wit.) You hve two functions u nd u 2 tht stisfy u = λ u nd u 2 = λ 2 u 2 Mke no ssumption bout whether the λ s re positive or negtive or even rel. The u s cn lso be complex. Multiply the first eqution by u * 2 nd the second by u*, then tke the complex conjugte of the second product. u * 2 u = λ u * 2 u nd u u * 2 = λ * 2 u u * 2 Subtrct the equtions. Integrte from to b u * 2 u u u * 2 = (λ λ * 2 )u* 2 u b dx ( u * 2 u u u * ) b 2 = (λ λ * 2 ) dx u * 2 u (5.4)

7 Now do two prtil integrtions. Work on the second term on the left: b dx u u * 2 = u u * 2 b b dx u u * 2 = u u * 2 b Put this bck into the Eq. (5.4) nd the integrl terms cncel, leving u u * 2 u u * 2 b 5 Fourier Series 7 = (λ λ * 2 ) u u * 2 b b + dx u u * 2 b dx u * 2 u (5.5) This is the centrl identity from which ll the orthogonlity reltions in Fourier series derive. It s even more importnt thn tht becuse it tells you wht types of boundry conditions you cn use in order to get the desired orthogonlity reltions. (It tells you even more thn tht, s it tells you how to compute the djoint of the second derivtive opertor. But not now sve tht for lter.) The expression on the left side of the eqution hs nme: biliner concomitnt. You cn see how this is relted to the work with the functions sin(nπx/). They stisfy the differentil eqution u = λu with λ = n 2 π 2 / 2. The intervl in tht cse ws < x < for < x < b. There re generliztions of this theorem tht you will see in plces such s problems 6.6 nd 6.7 nd.2. In those extensions these sme ides will llow you to hndle egendre polynomils nd Bessel functions nd Ultrsphericl polynomils nd mny other functions in just the sme wy tht you hndle sines nd cosines. Tht development comes under the generl nme Sturm-iouville theory. The key to using this identity will be to figure out wht sort of boundry conditions will cuse the left-hnd side to be zero. For exmple if u() = nd u(b) = then the left side vnishes. These re not the only possible boundry conditions tht mke this work; there re severl other common cses soon to pper. The first consequence of Eq. (5.5) comes by tking specil cse, the one in which the two functions u nd u 2 re in fct the sme function. If the boundry conditions mke the left side zero then b = (λ λ * ) dx u * (x)u (x) The λ s re necessrily the sme becuse the u s re. The only wy the product of two numbers cn be zero is if one of them is zero. The integrnd, u * (x)u (x) is lwys non-negtive nd is continuous, so the integrl cn t be zero unless the function u is identiclly zero. As tht would be trivil cse, ssume it s not so. This then implies tht the other fctor, (λ λ * ) must be zero, nd this sys tht the constnt λ is rel. Yes, n 2 π 2 / 2 is rel. [To use nother lnguge tht will become more fmilir lter, λ is n eigenvlue nd d 2 /dx 2 with these boundry conditions is n opertor. This clcultion gurntees tht the eigenvlue is rel.] Now go bck to the more generl cse of two different functions, nd drop the complex conjugtion on the λ s. b = (λ λ 2 ) dx u * 2 (x)u (x) This sys tht if the boundry conditions on u mke the left side zero, then for two solutions with different eigenvlues (λ s) the orthogonlity integrl is zero. Eq. (5.5) is specil cse of the following eqution. b If λ λ 2, then u2, u = dx u * 2 (x)u (x) = (5.6)

8 5 Fourier Series 8 Apply the Theorem As n exmple, crry out full nlysis of the cse for which = nd b =, nd for the boundry conditions u() = nd u() =. The prmeter λ is positive, zero, or negtive. If λ >, then set λ = k 2 nd u(x) = A sinh kx + B cosh kx, then u() = B = No solutions there, so try λ = nd so u() = A sinh k = A = u(x) = A + Bx, then u() = A = nd so u() = B = B = No solutions here either. Try λ <, setting λ = k 2. u(x) = A sin kx + B cos kx, then u() = = B, so u() = A sin k = Now there re mny solutions becuse sin nπ = llows k = nπ/ with n ny integer. But, sin( x) = sin(x) so negtive integers just reproduce the sme functions s do the positive integers; they re redundnt nd you cn eliminte them. The complete set of solutions to the eqution u = λu with these boundry conditions hs λ n = n 2 π 2 / 2 nd reproduces the result of the explicit integrtion s in Eq. (5.6). ( nπx ) u n (x) = sin n =, 2, 3,... un, u m = dx sin ( nπx ) sin nd ( mπx ) = if n m (5.7) There re other choices of boundry condition tht will mke the biliner concomitnt vnish. (Verify these!) For exmple u() =, u () = gives u n (x) = sin ( n + / 2 ) πx/ n =,, 2, 3,... nd without further integrtion you hve the orthogonlity integrl for non-negtive integers n nd m ( ) ( ) (n + / 2 )πx (m + / 2 )πx un, u m = dx sin sin = if n m (5.8) A very common choice of boundry conditions is u() = u(b), u () = u (b) (periodic boundry conditions) (5.9) It is often more convenient to use complex exponentils in this cse (though of course not necessry). On < x < u(x) = e ikx, where k 2 = λ nd u() = = u() = e ik The periodic behvior of the exponentil implies tht k = 2nπ. The condition tht the derivtives mtch t the boundries mkes no further constrint, so the bsis functions re u n (x) = e 2πinx/, (n =, ±, ±2,...) (5.2)

9 5 Fourier Series 9 Notice tht in this cse the index n runs over ll positive nd negtive numbers nd zero, not just the positive integers. The functions e 2πinx/ nd e 2πinx/ re independent, unlike the cse of the sines discussed bove. Without including both of them you don t hve bsis nd cn t do Fourier series. If the intervl is symmetric bout the origin s it often is, < x < +, the conditions re This sys tht 2k = 2nπ, so u( ) = e ik = u(+) = e +ik, or e 2ik = (5.2) u n (x) = e nπix/, (n =, ±, ±2,...) nd f(x) = The orthogonlity properties determine the coefficients: um, f = u m, c n u n = cm um, u m dx e mπix/ f(x) = c m um, u m c n u n (x) = c m dx e mπix/ e +mπix/ = c m dx = 2c m In this cse, sometimes the rel form of this bsis is more convenient nd you cn use the combintion of the two sets u n nd v n, where u n (x) = cos(nπx/), (n =,, 2,...) v n (x) = sin(nπx/), (n =, 2,...) (5.22) un, u m = (n m), vn, v m = (n m), un, v m = (ll n, m) nd the Fourier series is sum such s f(x) = nu n + b nv n. There re n infinite number of other choices, few of which re even useful, e.g. u () = = u (b) (5.23) Tke the sme function s in Eq. (5.7) nd try different bsis. Choose the bsis for which the boundry conditions re u() = nd u () =. This gives the orthogonlity conditions of Eq. (5.8). The generl structure is lwys the sme. f(x) = n u n (x), nd use um, u n = (n m) Tke the sclr product of this eqution with u m to get um, f = u m, n u n = m um, u m (5.24) This is exctly s before in Eq. (5.3), but with different bsis. To evlute it you still hve to do the integrls. um, f ( ) (m + / 2 )πx ( ) (m + / = dx sin = m dx sin 2 2 )πx = m um, u m [ ( cos (m + / 2 )π )] = (m + / 2 )π 2 m 4 m = (2m + )π Then the series is [ 4 sin πx π 2 + 3πx sin ] 5πx sin (5.25)

10 5 Fourier Series 5.4 Musicl Notes Different musicl instruments sound different even when plying the sme note. You won t confuse the sound of pino with the sound of guitr, nd the reson is tied to Fourier series. The note middle C hs frequency tht is 26.6 Hz on the stndrd equl tempered scle. The ngulr frequency is then 2π times this, or rdins/sec. Cll it ω = 644. When you ply this note on ny musicl instrument, the result is lwys combintion of mny frequencies, this one nd mny multiples of it. A pure frequency hs just ω, but rel musicl sound hs mny hrmonics: ω, 2ω, 3ω, etc. Insted of e iω t n instrument produces? n e niω t n= (5.26) A pure frequency is the sort of sound tht you her from n electronic udio oscilltor, nd it s not very interesting. Any rel musicl instrument will hve t lest few nd usully mny frequencies combined to mke wht you her. Why write this s complex exponentil? A sound wve is rel function of position nd time, the pressure wve, but it s esier to mnipulte complex exponentils thn sines nd cosines, so when I write this, I relly men to tke the rel prt for the physicl vrible, the pressure vrition. The imginry prt is crried long to be discrded lter. Once you re used to this convention you don t bother writing the rel prt understood nywhere it s understood.?? p(t) = R n e niωt = n cos ( ) nω t + φ n n= n= where n = n e iφn (5.27) I wrote this using the periodic boundry conditions of Eq. (5.9). The period is the period of the lowest frequency, T = 2π/ω. A flute produces combintion of frequencies tht is mostly concentrted in smll number of hrmonics, while violin or reed instrument produces fr more complex combintion of frequencies. The size of the coefficients n in Eq. (5.26) determines the qulity of the note tht you her, though oddly enough its phse, φ n, doesn t hve n effect on your perception of the sound. These represent couple of cycles of the sound of clrinet. The left grph is bout wht the wve output of the instrument looks like, nd the right grph is wht the grph would look like if I dd rndom phse, φ n, to ech of the Fourier components of the sound s in Eq. (5.27). They my look very different, but to the humn er they sound like. You cn her exmples of the sound of Fourier series online vi the web site: courses.ee.sun.c.z/stelsels en Seine 35/wordpress/wp-content/uplods/jhu-signls/ nd isten to Fourier Series You cn her the (lck of) effect of phse on sound. You cn lso synthesize your own series nd her wht they sound like under such links s Fourier synthese nd Hrmonics pplet found on this pge. You cn bck up from this link to lrger topics by using the links shown in the left column of the web pge.

11 5 Fourier Series Rel musicl sound is of course more thn just these Fourier series. At the lest, the Fourier coefficients, n, re themselves functions of time. The time scle on which they vry is however much longer thn the bsic period of oscilltion of the wve. Tht mens tht it mkes sense to tret them s (lmost) constnt when you re trying to describe the hrmonic structure of the sound. Even the lowest pitch notes you cn her re t lest 2 Hz, nd few home sound systems cn produce frequencies nerly tht low. Musicl notes chnge on time scles much greter thn /2 or / of second, nd this llows you to tret the notes by Fourier series even though the Fourier coefficients re themselves time-dependent. The ttck nd the decy of the note gretly ffects our perception of it, nd tht is described by the time-vrying nture of these coefficients.* Prsevl s Identity et u n be the set of orthogonl functions tht follow from your choice of boundry conditions. f(x) = n n u n (x) Evlute the integrl of the bsolute squre of f over the domin. b [ ] b * [ ] dx f(x) 2 = dx m u m (x) n u n (x) m n = b * m n dx u m (x) * u n (x) = m n n b n 2 dx u n (x) 2 In the more compct nottion this is f, f = m u m, m n n u n = * m n um, u n = n 2 u n, u n m,n n (5.28) The first eqution is nothing more thn substituting the series for f. The second moved the integrl under the summtion. The third eqution uses the fct tht ll these integrls re zero except for the ones with m = n. Tht reduces the double sum to single sum. If you hve chosen to normlize ll of the functions u n so tht the integrls of u n (x) 2 re one, then this reltion tkes on simpler ppernce. This is sometimes convenient. Wht does this sy if you pply it to series I ve just computed? Tke Eq. (5.) nd see wht it implies. f, f = dx = = k 2 u n, u n = k= ( 4 ) 2 π(2k + ) k= ( ) (2k + )πx dx sin 2 = ( k= 4 π(2k + ) ) 2 2 Rerrnge this to get k= (2k + ) 2 = π2 8 (5.29) * For n enlightening web pge, including complete nd impressively thorough text on mthemtics nd music, look up the book by Dvid Benson. It is vilble both in print from Cmbridge Press nd online. mth92/ (University of Aberdeen)

12 5 Fourier Series 2 A bonus. You hve the sum of this infinite series, result tht would be quite perplexing if you see it without knowing where it comes from. While you hve it in front of you, wht do you get if you simply evlute the infinite series of Eq. (5.) t /2. The nswer is, but wht is the other side? = 4 π k= (2k + )π(/2) sin = 4 2k + π k= 2k + ( )k or = π 4 But does it Work? If you re in the properly skepticl frme of mind, you my hve noticed serious omission on my prt. I ve done ll this work showing how to get orthogonl functions nd to mnipulte them to derive Fourier series for generl function, but when did I show tht this ctully works? Never. How do I know tht generl function, even well-behved generl function, cn be written s such series? I ve proved tht the set of functions sin(nπx/) re orthogonl on < x <, but tht s not good enough. Mybe clever mthemticin will invent new function tht I hven t thought of nd tht will be orthogonl to ll of these sines nd cosines tht I m trying to use for bsis, just s ˆk is orthogonl to î nd ĵ. It won t hppen. There re proper theorems tht specify the conditions under which ll of this Fourier mnipultion works. Dirichlet worked out the key results, which re found in mny dvnced clculus texts. For exmple if the function is continuous with continuous derivtive on the intervl x then the Fourier series will exist, will converge, nd will converge to the specified function (except mybe t the endpoints). If you llow it to hve finite number of finite discontinuities but with continuous derivtive in between, then the Fourier series will converge nd will (except mybe t the discontinuities) converge to the specified function. At these discontinuities it will converge to the verge vlue tken from the left nd from the right. There re vriety of other sufficient conditions tht you cn use to insure tht ll of this stuff works, but I ll leve tht to the dvnced clculus books. 5.5 Periodiclly Forced ODE s If you hve hrmonic oscilltor with n dded externl force, such s Eq. (4.2), there re systemtic wys to solve it, such s those found in section 4.2. One prt of the problem is to find solution to the inhomogeneous eqution, nd if the externl force is simple enough you cn do this esily. Suppose though tht the externl force is complicted but periodic, s for exmple when you re pushing child on swing. m d2 x = kx bdx dt2 dt + F ext(t) Tht the force is periodic mens F ext (t) = F ext (t + T ) for ll times t. The period is T. Pure Frequency Forcing Before ttcking the generl problem, look t simple specil cse. Tke the externl forcing function to be F cos ω e t where this frequency is ω e = 2π/T. This eqution is now m d2 x + kx + bdx dt2 dt = F cos ω e t = F [ e iω et + e iωet] (5.3) 2 Find solution corresponding to ech term seprtely nd dd the results. To get n exponentil out, put n exponentil in. for m d2 x + kx + bdx dt2 dt = eiωet ssume x inh (t) = Ae iωet

13 Substitute the ssumed form nd it will determine A. [ m( ω 2 e ) + b(iω e ) + k ] Ae iωet = e iωet 5 Fourier Series 3 This tells you the vlue of A is A = mω 2 e + b iω e + k (5.3) The other term in Eq. (5.3) simply chnges the sign in front of i everywhere. The totl solution for Eq. (5.3) is then x inh (t) = F [ ] 2 mωe 2 + b iω e + k eiωet + mωe 2 b iω e + k e iωet (5.32) This is the sum of number nd its complex conjugte, so it s rel. You cn rerrnge it so tht it looks lot simpler, but there s no need to do tht right now. Insted I ll look t wht it implies for certin vlues of the prmeters. Suppose tht the viscous friction is smll (b is smll). If the forcing frequency, ω e is such tht mωe 2 + k =, or is even close to zero, the denomintors of the two terms become very smll. This in turn implies tht the response of x to the oscillting force is huge. Resonnce. See problem In contrsting cse, look t ω e very lrge. Now the response of the mss is very smll; it brely moves. Generl Periodic Force Now I ll go bck to the more generl cse of periodic forcing function, but not one tht is simply cosine. If function is periodic I cn use Fourier series to represent it on the whole xis. The bsis to use will of course be the one with periodic boundry conditions (wht else?). Use complex exponentils, then u(t) = e iωt where e iω(t+t ) = e iωt This is just like Eq. (5.2) but with t insted of x, so et ω e = 2π/T, nd this is u n (t) = e 2πint/T, (n =, ±,...) (5.33) u n (t) = e inωet The externl force cn now be represented by the Fourier series e inωet, k= F ext (t) = k= k e ikωet, where k e ikωet = n T = e inωet, F ext (t) T = dt e inωet F ext (t) (Don t forget the implied complex conjugtion in the definition of the sclr product,,, Eq. (5.)) Becuse the force is periodic, ny other time intervl of durtion T is just s good, perhps T/2 to +T/2 if tht s more convenient. How does this solve the differentil eqution? Plug in. m d2 x dt 2 + bdx dt + kx = n= n e inωet (5.34)

14 5 Fourier Series 4 All there is to do now is to solve for n inhomogeneous solution one term t time nd then to dd the results. Tke one term lone on the right: m d2 x dt 2 + bdx + kx = einωet dt This is wht I just finished solving few lines go, Eq. (5.3), but with nω e insted of simply ω e. The inhomogeneous solution is the sum of the solutions from ech term. x inh (t) = n= Suppose for exmple tht the forcing function is simple squre wve. n m(nω e ) 2 + binω e + k eniωet (5.35) F ext (t) = { F ( < t < T/2) F (T/2 < t < T ) nd F ext (t + T ) = F ext (t) (5.36) The Fourier series for this function is one tht you cn do in problem 5.2. The result is F ext (t) = F 2 πi The solution corresponding to Eq. (5.35) is now x inh (t) = F 2πi n odd n odd n eniωet (5.37) ( m(nωe ) 2 + ibnω e + k ) n eniωet (5.38) A rel force ought to give rel result; does this? Yes. For every positive n in the sum, there is corresponding negtive one nd the sum of those two is rel. You cn see this becuse every n tht ppers is either squred or is multiplied by n i. When you dd the n = +5 term to the n = 5 term it s dding number to its own complex conjugte, nd tht s rel. Wht peculir fetures does this result imply? With the simply cosine force the phenomenon of resonnce occurred, in which the response to smll force t frequency tht mtched the intrinsic frequency k/m produced disproportiontely lrge response. Wht other things hppen here? The nturl frequency of the system is (for smll dmping) still k/m. ook to see where denomintor in Eq. (5.38) cn become very smll. This time it is when m(nω e ) 2 + k =. This is not only when the externl frequency ω e mtches the nturl frequency; it s when nω e mtches it. If the nturl frequency is k/m = rdins/sec you get big response if the forcing frequency is rdins/sec or 33 rdins/sec or 2 rdins/sec or 4 rdins/sec etc. Wht does this men? The squre wve in Eq. (5.36) contins mny frequencies. It contins more thn just the min frequency 2π/T, it contins 3 times this nd 5 times it nd mny higher frequencies. When ny one of these hrmonics mtches the nturl frequency you will hve the lrge resonnt response. Not only do you get lrge response, look t the wy the mss oscilltes. If the force hs squre wve frequency 2 rdins/sec, the mss responds* with lrge sinusoidl oscilltion t frequency 5 times higher rdins/sec. * The next time you hve ccess to pino, gently depress key without mking sound, then strike the key one octve lower. Relese the lower key nd listen to the sound of the upper note. Then try it with n intervl of n octve plus fifth.

15 5 Fourier Series Return to Prsevl When you hve periodic wve such s musicl note, you cn Fourier nlyze it. The boundry conditions to use re nturlly the periodic ones, Eq. (5.2) or (5.33), so tht f(t) = n e inω t If this represents the sound of flute, the mplitudes of the higher frequency components (the n ) drop off rpidly with n. If you re hering n oboe or violin the strength of the higher components is greter. If this function represents the sound wve s received by your er, the power tht you receive is proportionl to the squre of f. If f represent specificlly the pressure disturbnce in the ir, the intensity (power per re) crried by the wve is f(t) 2 v/b where v is the speed of the wve nd B is the bulk modulus of the ir. The key property of this is tht it is proportionl to the squre of the wve s mplitude. Tht s the sme reltion tht occurs for light or ny other wve. Up to known fctor then, the power received by the er is proportionl to f(t) 2. This time verge of the power is (up to tht constnt fctor tht I m ignoring) f 2 = lim T 2T +T dt f(t) 2 T Now put the Fourier series representtion of the sound into the integrl to get [ +T ] 2 lim dt T 2T n e inω t T The sound f(t) is rel, so by problem 5., n = * n. Also, using the result of problem 5.8 the time verge of e iωt is zero unless ω = ; then it s one. f 2 [ ] [ ] +T = lim dt T 2T n e inω t m e imω t T n m +T = lim dt T 2T n e inωt m e imω t T n m = +T n m lim dt e i(n+m)ω t T 2T n m T = n = n n n n 2 (5.39) Put this into words nd it sys tht the time-verge power received is the sum of mny terms, ech one of which cn be interpreted s the mount of power coming in t tht frequency nω. The Fourier coefficients squred (bsolute-squred relly) re then proportionl to the prt of the power t prticulr frequency. The power spectrum. Other Applictions In section.2 Fourier series will be used to solve prtil differentil equtions, leding to equtions such s Eq. (.5).

16 5 Fourier Series 6 In quntum mechnics, Fourier series nd its generliztions will mnifest themselves in displying the discrete energy levels of bound tomic nd nucler systems. Music synthesizers re ll bout Fourier series nd its generliztions. 5.7 Gibbs Phenomenon There s picture of the Gibbs phenomenon with Eq. (5.). When function hs discontinuity, its Fourier series representtion will not hndle it in uniform wy, nd the series overshoots its gol t the discontinuity. The detiled clcultion of this result is quite pretty, nd it s n excuse to pull together severl of the methods from the chpters on series nd on complex lgebr. 4 π k= (2k + )πx sin =, ( < x < ) 2k + highest hrmonic: 5 highest hrmonic: 2 highest hrmonic: The nlysis sounds stright-forwrd. Find the position of the first mximum. Evlute the series there. It relly is lmost tht cler. First however, you hve to strt with the finite sum nd find the first mximum of tht. Stop the sum t k = N. 4 π N k= For mximum, set the derivtive to zero. (2k + )πx sin = f 2k + N (x) (5.4) f N (x) = 4 N cos (2k + )πx Write this s the rel prt of complex exponentil nd use Eq. (2.3). N e i(2k+)πx/ = N z 2k+ = z N z 2k = z z2n+2 z 2 Fctor these complex exponentils in order to put this into nicer form. = e iπx/ e iπx(n+)/ e iπx(n+)/ e iπx/ e iπx/ e iπx(n+)/ e iπx/ = sin(n + )πx/ sin πx/ e iπx(n+)/ The rel prt of this chnges the lst exponentil into cosine. Now you hve the product of the sine nd cosine of (N + )πx/, nd tht lets you use the trigonometric double ngle formul. f N (x) = 4 sin 2(N + )πx/ 2 sin πx/ (5.4) This is zero t the mximum. The first mximum fter x = is t 2(N + )πx/ = π, or x = /2(N + ).

17 Now for the vlue of f N t this point, f N ( /2(N + ) ) = 4 π N k= 5 Fourier Series 7 (2k + )π/2(n + ) sin = 4 2k + π N k= (2k + )π sin 2k + 2(N + ) The finl step is to tke the limit s N. As k vries over the set to N, the rgument of the sine vries from little more thn zero to little less thn π. As N grows you hve the sum over lot of terms, ech of which is pproching zero. It s n integrl. et t k = k/n then t k = /N. This sum is pproximtely 4 π N k= 2Nt k sin t k π = 2 π N t k sin t t k π 2 dt k π t sin πt In this limit 2k + nd 2k re the sme, nd N + is the sme s N. Finlly, put this into stndrd form by chnging vribles to πt = x. 2 π dx sin x π x = 2 Si(π) =.7898 π x dt sin t t = Si(x) (5.42) The function Si is clled the sine integrl. It s just nother tbulted function, long with erf, Γ, nd others. This eqution sys tht s you tke the limit of the series, the first prt of the grph pproches verticl line strting from the origin, but it overshoots its trget by 8%. Exercises A vector is given to be A = 5 ˆx + 3ŷ. et new bsis be ê = (ˆx + ŷ)/ 2, nd ê 2 = (ˆx ŷ)/ 2. Use sclr products to find the components of A in the new bsis: A = A ê + A 2 ê 2. 2 For the sme vector s the preceding problem, nd nother bsis f = 3 ˆx + 4ŷ nd f 2 = 8 ˆx + 6ŷ, express A in the new bsis. Are these bsis vectors orthogonl? 3 On the intervl < x <, sketch three grphs: the first term lone, then the second term lone, then the third. Try to get the scle of the grphs resonble ccurte. Now dd the first two nd grph. Then dd the third lso nd grph. Do ll this by hnd, no grphing clcultors, though if you wnt to use clcultor to clculte few points, tht s ok. sin ( πx/ ) 9 sin ( 3πx/ ) + 25 sin ( 5πx/ ) 4 For wht vlues of α re the vectors A = α ˆx 2ŷ + ẑ nd B = 2α ˆx + αŷ 4ẑ orthogonl? 5 On the intervl < x < with sclr product defined s f, g = dx f(x)* g(x), show tht these re zero, mking the functions orthogonl: x nd 3 2x, sin πx/ nd cos πx/, sin 3πx/ nd 2x

18 6 Sme s the preceding, show tht these functions re orthogonl: 5 Fourier Series 8 e 2iπx/ nd e 2iπx/, 4 (7 + i)x nd ix 7 With the sme sclr product the lst two exercises, for wht vlues of α re the functions f (x) = αx ( α)( 3 2 x) nd f 2(x) = 2αx + ( + α)( 3 2x) orthogonl? Wht is the interprettion of the two roots? 8 Repet the preceding exercise but use the sclr product f, g = 2 dx f(x)* g(x). 9 Use the sclr product f, g = dx f(x)* g(x), nd show tht the egendre polynomils P, P, P 2, P 3 of Eq. (4.6) re mutully orthogonl. Chnge the sclr product in the preceding exercise to f, g = dx f(x)* g(x) nd determine if the sme polynomils re still orthogonl. Verify the exmples stted on pge 5.

19 5 Fourier Series 9 Problems 5. Get the results in Eq. (5.8) by explicitly clculting the integrls. 5.2 () The functions with periodic boundry conditions, Eq. (5.2), re supposed to be orthogonl on < x <. Tht is, u n, u m = for n m. Verify this by explicit integrtion. Wht is the result if n = m or n = m? The nottion is defined in Eq. (5.). (b) Sme clcultion for the rel version, u n, u m, vn, v m, nd un, v m, Eq. (5.22) 5.3 Find the Fourier series for the function f(x) = s in Eq. (5.), but use s bsis the set of functions u n on < x < tht stisfy the differentil eqution u = λu with boundry conditions u () = nd u () =. (Eq. (5.23)) Necessrily the first step will be to exmine ll the solutions to the differentil eqution nd to find the cses for which the biliner concomitnt vnishes. (b) Grph the resulting Fourier series on 2 < x < 2. (c) Grph the Fourier series Eq. (5.) on 2 < x < () Compute the Fourier series for the function x 2 on the intervl < x <, using s bsis the functions with boundry conditions u () = nd u () =. (b) Sketch the prtil sums of the series for, 2, 3 terms. Also sketch this sum outside the originl domin nd see wht this series produces for n extension of the originl function. Ans: Eq. (5.) 5.5 () Compute the Fourier series for the function x on the intervl < x <, using s bsis the functions with boundry conditions u() = = u(). How does the coefficient of the n th term decrese s function of n? (b) Also sketch this sum within nd outside the originl domin to see wht this series produces for n extension of the originl function. Ans: 2 π ( ) n+ n sin(nπx/) 5.6 () In the preceding problem the sine functions tht you used don t mtch the qulittive behvior of the function x on this intervl becuse the sine is zero t x = nd x isn t. The qulittive behvior is different from the bsis you re using for the expnsion. You should be ble to get better convergence for the series if you choose functions tht more closely mtch the function tht you re expnding, so try repeting the clcultion using bsis functions tht stisfy u() = nd u () =. How does the coefficient of the n th term decrese s function of n? (b) As in the preceding problem, sketch some prtil sums of the series nd its extension outside the originl domin. Ans: 8 ( π 2 ( ) n /(2n + ) 2) sin ( (n + / 2 )πx/ ) 5.7 The function sin 2 x is periodic with period π. Wht is its Fourier series representtion using s bsis functions tht hve this period? Eqs. (5.2) or (5.22). 5.8 In the two problems 5.5 nd 5.6 you improved the convergence by choosing boundry conditions tht better mtched the function tht you wnt. Cn you do better? The function x vnishes t the origin, but its derivtive isn t zero t, so try boundry conditions u() = nd u() = u (). These conditions mtch those of x so this ought to give even better convergence, but first you hve to verify tht these conditions gurntee the orthogonlity of the bsis functions. You hve to verify tht the left side of Eq. (5.5) is in fct zero. When you set up the bsis, you will exmine functions of the form sin kx, but you will not be ble to solve explicitly for the vlues of k. Don t worry bout it. When you use Eq. (5.24) to get the coefficients ll tht you need to do is to use the eqution tht k stisfies to do the integrls. You do not need to hve solved it. If you do ll the lgebr correctly you will probbly hve surprise.

20 5 Fourier Series () Use the periodic boundry conditions on < x < + nd bsis e πinx/ to write x 2 s Fourier series. Sketch the sums up to few terms. (b) Evlute your result t x = where you know the nswer to be 2 nd deduce from this the vlue of ζ(2). 5. On the intervl π < x < π, the function f(x) = cos x. Expnd this in Fourier series defined by u = λu nd u( π) = = u(π). If you use your result for the series outside of this intervl you define n extension of the originl function. Grph this extension nd compre it to wht you normlly think of s the grph of cos x. As lwys, go bck to the differentil eqution to get ll the bsis functions. Ans: 4 π k= 2k+ (2k+3)(2k ) sin ( (2k + )(x + π)/2 ) 5. Represent function f on the intervl < x < by Fourier series using periodic boundry conditions f(x) = n e nπix/ () If the function f is odd, prove tht for ll n, n = n (b) If the function f is even, prove tht ll n = n. (c) If the function f is rel, prove tht ll n = * n. (d) If the function is both rel nd even, chrcterize n. (e) If the function is imginry nd odd, chrcterize n. 5.2 Derive the series Eq. (5.37). 5.3 For the function e αt on < t < T, express it s Fourier series using periodic boundry conditions [u() = u(t ) nd u () = u (T )]. Exmine for plusibility the cses of lrge nd smll α. The bsis functions for periodic boundry conditions cn be expressed either s cosines nd sines or s complex exponentils. Unless you cn nlyze the problem hed of time nd determine tht it hs some specil symmetry tht mtches tht of the trig functions, you re usully better off with the exponentils. Ans: [( e αt )/ αt ][ + 2 [α2 cos nωt + αnω sin nωt] / [α 2 + n 2 ω 2 ] ] 5.4 () On the intervl < x <, write x( x) s Fourier series, using boundry conditions tht the expnsion functions vnish t the endpoints. Next, evlute the series t x = /2 to see if it gives n interesting result. (b) Finlly, wht does Prsevl s identity tell you? Ans: 4 2 n 3 π 3 [ ( ) n ) ] sin(nπx/) 5.5 A full-wve rectifier tkes s n input sine wve, sin ωt nd cretes the output f(t) = sin ωt. The period of the originl wve is 2π/ω, so write the Fourier series for the output in terms of functions periodic with this period. Grph the function f first nd use the grph to nticipte which terms in the Fourier series will be present. When you re done, use the result to evlute the infinite series ( )k+ /(4k 2 ) Ans: π/4 /2 5.6 A hlf-wve rectifier tkes s n input sine wve, sin ωt nd cretes the output sin ωt if sin ωt > nd if sin ωt The period of the originl wve is 2π/ω, so write the Fourier series for the output in terms of functions periodic with this period. Grph the function first. Check tht the result gives the correct vlue t t =, mnipulting it into telescoping series. Sketch few terms of the whole series to see if it s

21 heding in the right direction. Ans: 4 /π + / 2 sin ωt 8 /π n even > cos(nωt)/(n2 ) 5 Fourier Series For the undmped hrmonic oscilltor, pply n oscillting force (cosine). This is simpler version of Eq. (5.3). Solve this problem nd dd the generl solution to the homogeneous eqution. Solve this subject to the initil conditions tht x() = nd v x () = v. 5.8 The verge (rithmetic men) vlue of function is f = lim T 2T +T dt f(t) T or f = lim T T T dt f(t) s pproprite for the problem. Wht is sin ωt? Wht is sin 2 ωt? Wht is e t2? Wht is sin ω t sin ω 2 t? Wht is e iωt? 5.9 In the clcultion leding to Eq. (5.39) I ssumed tht f(t) is rel nd then used the properties of n tht followed from tht fct. Insted, mke no ssumption bout the relity of f(t) nd compute f(t) 2 = f(t) * f(t) Show tht it leds to the sme result s before, n The series n cos nθ ( < ) n= represents function. Sum this series nd determine wht the function is. While you re bout it, sum the similr series tht hs sine insted of cosine. Don t try to do these seprtely; combine them nd do them s one problem. And check some limiting cses of course. And grph the functions. Ans: sin θ/ ( cos θ ) 5.2 Apply Prsevl s theorem to the result of problem 5.9 nd see wht you cn deduce If you tke ll the elements u n of bsis nd multiply ech of them by 2, wht hppens to the result for the Fourier series for given function? 5.23 In the section 5.3 severl bses re mentioned. Sketch few terms of ech bsis A function is specified on the intervl < t < T to be f(t) = { ( < t < t ) (t < t < T ) < t < T On this intervl, choose boundry conditions such tht the left side of the bsic identity (5.5) is zero. Use the corresponding choice of bsis functions to write f s Fourier series on this intervl Show tht the boundry conditions u() = nd αu() + βu () = mke the biliner concomitnt in Eq. (5.5) vnish. Are there ny restrictions on α nd β? Do not utomticlly ssume tht these numbers re rel.

22 5.26 Derive Fourier series for the function { Ax ( < x < /2) f(x) = A( x) (/2 < x < ) 5 Fourier Series 22 Choose the Fourier bsis tht you prefer. Evlute the resulting series t x = /2 to check the result. Sketch the sum of couple of terms. Comment on the convergence properties of the result. Are they wht you expect? Wht does Prsevl s identity sy? Ans: (2A/π 2 ) k odd ( )(k )/2 sin(kπx/) / k Rerrnge the solution Eq. (5.32) into more esily understood form. () Write the first denomintor s mω 2 e + b iω e + k = Re iφ Wht re R nd φ? The second term does not require you to repet this clcultion, just use its results, now combine everything nd write the nswer s n mplitude times phse-shifted cosine. (b) Assume tht b is not too big nd plot both R nd φ versus the forcing frequency ω e. Also, nd perhps more illuminting, plot /R Find the form of Prsevl s identity pproprite for power series. Assume sclr product f, g = f(x)* g(x)dx for the series f(x) = nx n, nd g(x) = b nx n, expressing the result in terms of mtrices. Next, test your result on simple, low-order polynomil. Ans: (...)M(b b...) where M = 2, M 2 = 2 / 3 M 4 = 2 / 5,... nd is trnspose () In the Gibbs phenomenon, fter the first mximum there is first minimum. Where is it? how big is the function there? Wht is the limit of this point? Tht is, repet the nlysis of section 5.7 for this minimum point. (b) While you re bout it, wht will you get for the limit of the sine integrl, Si( )? The lst result cn lso be derived by complex vrible techniques of chpter 4, Eq. (4.6). Ans: (2/π) Si(2π) = Mke blown-up copy of the grph preceding Eq. (5.4) nd mesure the size of the overshoot. Compre this experimentl vlue to the theoreticl limit. Sme for the first minimum. 5.3 Find the power series representtion bout the origin for the sine integrl, Si, tht ppered in Eq. (5.42). Wht is its domin of convergence? Ans: π 2 ( )n( x 2n+/ (2n + )(2n + )! ) 5.32 An input potentil in circuit is given to be squre wve ±V t frequency ω. Wht is the voltge between the points nd b? In prticulr, ssume tht the resistnce is smll, nd show tht you cn pick vlues of the cpcitnce nd the inductnce so tht the output potentil is lmost exctly sine wve t frequency 3ω. A filter circuit. Recll section 4.8. b 5.33 For the function sin(πx/) on ( < x < 2), expnd it in Fourier series using s bsis the trigonometric functions with the boundry conditions u () = = u (2), the cosines. Grph the resulting series s extended outside the originl domin For the function cos(πx/) on ( < x < 2), expnd it in Fourier series using s bsis the trigonometric functions with the boundry conditions u() = = u(2), the sines. Grph the resulting series s extended outside the originl domin.

23 5 Fourier Series () For the function f(x) = x 4, evlute the Fourier series on the intervl < x < using periodic boundry conditions ( u( ) = u() nd u ( ) = u () ). (b) Evlute the series t the point x = to derive the zet function vlue ζ(4) = π 4 /9. Evlute it t x = to get relted series. Ans: [ ( )n 8 n 2 π 48 ] 2 n 4 π cos nπx/ Fourier series depends on the fct tht the sines nd cosines re orthogonl when integrted over suitble intervl. There re other functions tht llow this too, nd you ve seen one such set. The egendre polynomils tht ppered in section 4. in the chpter on differentil equtions stisfied the equtions (4.62). One of these is dx P n (x)p m (x) = 2 2n + δ nm This is n orthogonlity reltion, P n, P m = 2δnm /(2n+), much like tht for trigonometric functions. Write function f(x) = n P n (x) nd deduce n expression for evluting the coefficients n. Apply this to the function f(x) = x For the stndrd differentil eqution u = λu, use the boundry conditions u() = nd 2u() = u (). This is specil cse of problem 5.25, so the biliner concomitnt vnishes. If you hven t done tht problem, t lest do this specil cse. Find ll the solutions tht stisfy these conditions nd grph few of them. You will not be ble to find n explicit solution for the λs, but you cn estimte few of them to sketch grphs. Did you get them ll? 5.38 Exmine the function on < x < given by ( < x < /2) nd (/2 < x < ) f(x) = ( < x < /2) ( /2 < x < ) Drw it first. Now find Fourier series representtion for it. You my choose to do this by doing lots of integrls, OR you my prefer to strt with some previous results in this chpter, chnge periods, dd or subtrct, nd do no integrls t ll In Eq. (5.3) I wrote cos ω e t s the sum of two exponentils, e iωet + e iωet. Insted, write the cosine s e iωet with the understnding tht t the end you tke the rel prt of the result, Show tht the result is the sme. 5.4 From Eq. (5.4) write n pproximte closed form expression for the prtil sum f N (x) for the region x but not necessrily x N, though tht extr-specil cse is worth doing too. 5.4 Evlute the integrl dx x2 using the series Eq. (5.) nd using the series (5.2) The Fourier series in problem 5.5 uses the sme bsis s the series Eq. (5.). Wht is the result of evluting the sclr products, nd, x with these series? 5.43 If you evluted the n = m cse of Eq. (5.6) by using different trig identity, you cn do it by n lterntive method: sy tht n nd m in this eqution ren t necessrily integers. Then tke the limit s n m.