COMPUTING FOURIER SERIES

Transcription

1 COMPUTING FOURIER SERIES Overview We have see i revious otes how we ca use the fact that si ad cos rereset comlete orthogoal fuctios over the iterval [-,] to allow us to determie the coefficiets of a Fourier series. Now, let's use this iformatio to evaluate some examles of Fourier series. We will, of course, show how usig Mathematica ca make our lives easier, esecially i the evaluatio of itegrals. Our First Examle Let' s begi by comutig the Fourier series of f (x) = x over the iterval [-,]. Let's thik about this for a momet; we are about to rereset a straight lie by a ifite sum of siusoidal curves. We will first do the calculatios outlied i revious otes, ad the show that these sums i fact give us the lie y=x over the iterval [-,]. First, let's write dow the geeral form of a Fourier series. We ca write ay fuctio that is eriodic as: where we have to evaluate the coefficiets : f HxL = a 0 + cos HxL + b si HxLD = a 0 = - f HxL dx () () a = - f HxL cos HxL dx (3) b = - f HxL si HxL dx (4) Evaluatig these three coefficiets will require that you be review itegratig trig fuctios, ad also review the value of trig fuctios at certai values (e.g., kowig the value of si ad cos at 0, ± ; ± ( /)). Let's roceed with our articular examle. Eqs. ()-(4) will rovide us with the coefficiets that we substitute ito () to fid the Fourier series for f(x)=x. We will ow comute these coefficiets, ad the roduce the fial Fourier series. ü Fidig the value of a 0 : We use eq. () to fid the value of a 0 that we will use i our fial Fourier series:

2 mfourierseries.b a 0 = - xdx = B x - x - F = 0 (5) Or, usig Mathematica to verify : H ê πl Itegrate@x, 8x, π, π<d 0 Some of you may have bee able to solve this usig symmetry cosideratios; more o the use of symmetry i aother classote. ü Fidig the values of a : We use eq. H3L alog with Mathematica to fid the values of a : a = f HxL cos HxLdx fl a = x cos HxLdx - - (6) Itegrate@x Cos@ xd, 8x, π, π<, Assumtios Elemet@, ItegersDD 0 Ad we fid that all over the coefficiets a are zero; agai, those of you roficiet with the use of symmetry argumets might have see this immediately. This result meas that the oly terms that will cotribute to the Fourier series of f(x)=x will cosist of si ( x) terms. We ow determie those coefficiets. ü Fidig the values of b : b = f HxL si HxL dx fl b = x si HxL dx - - (7) NOTE TO CLASS : The calculatio that aears below is icorrect. Immediately below I will correct this error ad roceed through the rest of the comutatio usig the correct values. Clear@D H ê πl Itegrate@x Si@ xd, 8x, 0, π<, Assumtios Elemet@, ItegersDD H L Ca you fid the error i the Mathematica statemet above? Look at the limits : I have itegrated betwee 0 ad istead of betwee - ad. The correct calculatio should be: I[]:= H ê πl Itegrate@x Si@ xd, 8x, π, π<, Assumtios Elemet@, ItegersDD Out[]= H L The correct aswer is double the error above; exactly what you would exect from symmetry argumets. (See the classote o

3 mfourierseries.b 3 "Itegratig Odd ad Eve Fuctios"). The correct versio of the itegral tells us that our coefficiets are : - H-L b = We will see may exressios like this whe we calculate Fourier coefficiets. Let' s write out the first few coefficiets exlicitly to see what they look like : b =-H-L = ; b = -H-L - H-L3 =-; b 3 = = 3 3 If we substitute these coefficiets ito eq. (), we ca write out exlicitly the first three terms of the Fourier series for f (x) = x : x = b si HxL + b si H xl + b 3 si H3 xl +... si H xl si H3 xl = Bsi HxL F 3 Equatio (0) reresets the first three terms of the Fourier series for f (x) = x over the iterval [-,]. You might woder how it is ossible that a straight lie ca be rereseted by a sum of sie curves. Let's lot these three terms out ad see if we ca recogize the straight lie y=x. Remember, these are oly the first three terms of a ifiite series, so we should ot exect a erfect correlatio with oly three terms...so, here we go: (8) (9) (0) I[]:= Plot@ HSi@xD Si@ xdê + Si@3 xdê3l, 8x, π, π<d 3 Out[]= Well, sort of. Let' s try a few more terms ad see what we get : -3

4 4 mfourierseries.b I[3]:= xdê + Si@3 xdê3 Si@4 xdê4 + Si@5 xdê5 Si@6 xdê6 Si@7 xdê7l, 8x, π, π<d 3 Out[3]= Closer. We could add more terms like this, but it would be laborious. Let' s lear how to write this series i closed form (e.g., equatio ()) ad work with it. I closed form, this series becomes : -H-L si HxL x = S =Ç () ad we ca ow lot it usig the Mathematica commad for summatio : I[6]:= Plot@Sum@ H L^ Si@ xdê, 8,, 00<D, 8x, π, π<, Eilog Poit@8, <DD 3 Out[6]= Now that' s a straight lie. (Quick Mathematica ote: the fial art of the Plot statemet i lie[6] itroduces us to the "Eilog" otio. This allows us to add somethig to a grah after it has bee lotted. I this case, I wated to lot the oit {,} to show that it does i fact lie o the lie y=x as exected. This hels verify that ot oly does our Fourier series roduce a straight lie, but the roer straight lie.) Let' s study iut lie[6] : it cotais ested fuctio calls. The "ier" call comutes the sum idicated i eq. (). The exressio i braces "{,,00}" meas to sum the fuctio from = to 00. The outer call lots those calculatios over the iterval [-,]. The ew (ad curret) versio of Mathematica has a very owerful ew feature, called "Maiulate". You should coy the followig commad exactly i your otebooks to see how it works. I this case, we will treat the uer idex of the summatio as a variable, i other words, we wat to see how this Fourier series coverges to f(x)=x as we vary the umber of terms we use i the summatio:

5 mfourierseries.b 5 I[5]:= Maiulate@Plot@Sum@ H L^ Si@ xdê, 8,, ul<d, 8x, π, π<d, 8ul,, 00<D ul Out[5]= Notice that i iut lie[5] I used "ul" as a variable to idicate the uer limit of the sum; the fial braces "{ul,,00}" will cause Mathematica to recomute the lot. Whe ul =, the lot will be based o oly term i the summatio; whe ul = 0, the lot is based o summig the first 0 terms of the series. This is a owerful ew fuctio that allows for more detailed studies of these sorts of series tha I have ever ecoutered. This is oe reaso why I wated to use Mathematica this term. A secod Examle For this examle, let' s work out the aswer to questio #7 o. 355 of Boas. This is a little differet from the first examle sice we have two regimes for our fuctio : 0, π < x < 0 f HxL = : x, 0 < x < π Because of this, we will eed to be careful i evaluatig our coefficiets eqs. () - (4). It should be obvious that the cotributios to all coefficiets will be zero from the iterval [-,0], so we eed to itegrate oly from 0 to : ü Fidig a 0 a 0 = 0 xdx = ÿ = fl a 0 = 4 () ü Fidig a : a = 0 x cos HxL dx (3)

6 6 mfourierseries.b H ê πl Cos@ xd, 8x, 0, π<, Assumtios ItegersDD + H L π The outut lie above tells us the values of all the coefficiets a. But this exressio bears further scrutiy. Notice that there is a term ivolvig H-L. This term will be - if is odd, ad + if is eve. Thus, the value of a will be zero for all eve values of, ad will equal -/( M for all odd values of. We will eed to be very careful to take all this ito accout i writig out our Fourier series. ü Fidig b : b = 0 x si HxL dx (4) Clear@D H ê πl Itegrate@x Si@ xd, 8x, 0, π<, Assumtios Elemet@, ItegersDD H L ü Puttig it together : We take our values for the coefficiets calculated above ad substitute ito equatio () to get our Fourier series. Writig out the first three (o - zero) terms i each art of the series, we get : f HxL = 4 - cos HxL B + cos H3 xl 3 + cos H5 xl si H xl si H3 xl F + Bsi HxL F 5 3 Let' s try to lot this sum for say the first 0 terms. We will have to be careful to take ito cosideratio the fact that the cosie art of the series ivolves oly odd itegers. The simlest Mathematica way to hadle this is : Plot@Hπ ê 4L H ê πl Sum@Cos@ xd ê ^, 8,, 0, <D + Sum@ H L^ Si@ xdê, 8,, 0<D, 8x, π, π<d (5) Cosider the Sum bracket above : the otatio "{,,0,}" meas to sum over, from to 0 idexig by two. This is

7 mfourierseries.b 7 equivalet to the statemet : 0 S Cos HxL =,3,5... (6)