FOURIER SERIES ON ANY INTERVAL

Transcription

1 FOURIER SERIES ON ANY INTERVAL Overview We have spent considerabe time earning how to compute Fourier series for functions that have a period of 2p on the interva (-p,p). We have aso seen how Fourier series generate some very interesting resuts, but you might have wondered how genera these resuts are. After a, there is a ot of physics that invoves functions that are not 2p periodic on (-p,p); does Fourier anaysis work for them? This cassnote is designed to iustrate some of the main principes deveoped in sections 8 and 9 of Boas (Ch. 7) showing how to generaize Fourier anaysis to any "reasonaby we behaved function." The ast section of this cassnote deas with the materia in Boas' section 9. I think this section (of this cassnote) is particuary important since I don't think Boas sufficienty stressed the significance of this materia. Fourier Series on Other Intervas So far we have considered functions ony on the interva (-p,p). There is no reason we coud not have chosen another interva; we just used (-p,p) as a starting point. As your text notes, though, it is important to understand both the function and its interva in anayzing it; for instance, if we consider the simpe function f HxL = x 2 on the two intervas (-p, p) and (0,2p), we wi get different Fourier series for them. First, et's pot severa cyces of the function on the interva (-p,p): Fig. 1 Pot of y = x 2 defined on H-p, pl As we woud expect, the function is even on this interva, and if we cacuate the Fourier series for this function, we find : f HxL = S H-1L n cos HnxL n=1 p2 If we transate this function by p, our function is now defined on the interva (0,2p). We can sti write a Fourier series for this function in famiiar terms: n 2 (1)

2 2 fourierintervas.nb f HxL = a S an cos HnxL + b n sin HnxL n=1 (2) The ony difference now is that our imits of integration for computing coefficients are aso shifted : a 0 = 1 p 0 2 p f HxL dx; a n = 1 2 p p f HxL cos HnxLdx; b n = 1 2 p 0 p f HxL sin HnxL dx 0 (3) Make sure you notice that we are now integrating between 0 and 2p. We can compute these coefficients easiy : In[238]:= Cear@fD f@x_d := x^2 Integrate@8f@xD, Cos@n xd f@xd, Sin@n xd f@xd<, 8x, 0, 2 π<, Assumptions Eement@n, IntegersDD Out[240]= : 8 π3 3, 4 π n 2, 4 π2 n > We observe that the b n are non zero as they were on the interva (-p,p). The reason for this becomes cear when we pot severa cyces of y = x 2 defined on (0,2p): Fig. 2 A graph of y = x 2 defined on H0, 2 pl Unike the Fourier series in equation (1) which invoves ony cos terms (i.e., even terms) because the function is even, the Fourier series defined on (0,2p) invoves both cos and sin terms since the function is neither even nor odd when defined and graphed on this interva. We can use the coefficients computed immediatey above and write the Fourier series for this interva as: f HxL = 4 p2 3 + S 4 p cos HnxL - 4 p2 sin HnxL n=1 n 2 n (4) This exercise shows that we can compute Fourier series for other intervas, but that we have to be carefu to recompute the coefficients

3 fourierintervas.nb 3 Fourier Series with different Periodicities Suppose we have a function that is periodic on the interva (-1, 1), or some other interva not invoving simpe mutipes of p. The extension of Fourier series to such instances is quite simpe. Suppose again we are deaing with our function of y = x 2 ; this time we wish to consider it its behavior if it is defined on the interva (-1,1). In this case, its period is 2. How can we define its Fourier series for this interva? For the more genera case of a function whose periodicity is 2, we can show that both sin( n p x ) and cos( n p x ) are 2 periodic. We know that a function is periodic with period p if f(x+p)=f(x). In this case, we can show that sin( n p x ) is 2 periodic because: sinb n p Hx + 2 LF = sin K n p x + 2 npo = sin K n p x O cos H2 npl + sin H2 n pl cos K n p x O We know that cos(2 n p) is equa to 1 for a integer vaues of n, and sin (2 n p) is zero for a integers, so that the expression in eq. (5) reduces to: sinb n p Hx + 2 LF = sin K n p x O (6) and we have shown that sin( n p x ) is in fact 2 periodic. A simiar anaysis wi show that cos( n p x ) is aso 2 periodic. It is a simpe enough matter to write the genera form of the Fourier series for any periodic interva L. The genera form of the series is: and our coefficients are defined as : f HxL = a S n=1 Ban cos K n p x L O + b n sin K n p x L OF (5) (7) L a 0 = 1 L f HxL dx -L (8) a n = 1 Lf L HxL cos K n p x -L L O dx b n = 1 Lf L HxL sin K n p x -L L O dx Notice that I am using upper case "L" to avoid confusion between "" and "1". More substantivey, notice that if the ength of the interva is 2 p, then L=p, and the definitions of the coefficients and of the Fourier series reduce to the famiiar forms we know for functions on the interva (-p,p). Let's try now to compute the Fourier series for y = x 2 on the interva (-1,1). Here, L=1, so we have: (9) (10)

4 4 fourierintervas.nb In[241]:= a 0 = 1 Integrate@x^2, 8x, 1, 1<D 1 Out[241]= 2 3 So the first term in the Fourier series wi be 1/2 of this, or 1/3. In[242]:= Out[242]= a n = Integrate@x^2 Cos@Hn π xlê1d, 8x, 1, 1<, Assumptions Eement@n, IntegersDD 4 H 1L n n 2 π 2 Where I do not expicity write the factor of "1" before the integra. In[243]:= b n = Integrate@x^2 Sin@Hn π xlê1d, 8x, 1, 1<, Assumptions Eement@n, IntegersDD Out[243]= 0 Not surprisingy, the b coefficients are zero. We coud have predicted this since our function is even on this interva, and we expect ony even terms, thus no odd (sin) terms. Verifying this Fourier series by potting three cyces of its sum over 100 terms: In[251]:= Pot@1 ê 3 + H4 ê π ^2L Sum@H 1L^n Cos@Hn π xl ê 1D ê n^2, 8n, 1, 100<D, 8x, 3, 3<, Epiog 8Red, PointSize@LargeD, Point@881 ê 2, H1 ê 2L^2<, ê 2, H1 ê 2L^2<, ê 3, H 2 ê 3L^2<<D<D Out[251]= The obvious red dots verify that each point ies on the curve defined by y = x 2 over the interva (-1,1) ü One more exampe As a fina exampe, et' s work out probem 15 a) on p. 363 of Boas. We are asked to find the Fourier series for the function defined by : f HxL = x, -1 x < 1 (11) Our interva is 2, so L = 1. Since our function is ceary odd over this interva, we know we need to compute ony the b n terms: In[247]:= b n = 1 Integrate@x Sin@Hn π xlê1d, 8x, 1, 1<, Assumptions Eement@n, IntegersDD 1 Out[247]= 2 H 1Ln n π

5 fourierintervas.nb 5 And our Fourier series is : f HxL =- 2 p S H-1L n sin Hn p xl n=1 Potting three cyces of this function for verification : n = -2 sin H2 p xl sin H3 p xl B-sin Hp xl F p 2 3 (12) In[249]:= Pot@ 2 ê π Sum@H 1L^n Sin@n π xdên, 8n, 1, 100<D, 8x, 3, 3<, Epiog 8Red, PointSize@LargeD, Point@881 ê 2, 1 ê 2<, ê 2, 1 ê 2<, ê 3, 2 ê 3<<D<D Out[249]= And the bright beacons of Mathematica te us we have chosen our coefficients we. Functions that are not periodic The previous sections have shown us how to compute the Fourier series for functions that are periodic on some interva other than (-p,p). What do we do if we want to use the Fourier series for a function that is not periodic on any interva? Suppose we need to find the Fourier series of a function that is defined on an interva (0,L)? (We wi encounter such situations in Ch. 13; some of you have aready encountered these in Ch. 3 of Griffiths.) This is the essentia question posed in Sect. 9 of Boas. So, suppose we need to find the Fourier series for the function f(x)=x defined on the interva (0,2). How can we do this since this function is not periodic? Simpe: we make it periodic. Consider one possibe way of doing this. The graph of our origina function is: Fig. 3. Graph of y = x on (0, 2) Now, we can make this periodic by extending the graph in the negative haf pane; for instance, we can make this an even

6 6 fourierintervas.nb function : Fig. 4. Graph of y = Abs[x] on (-2, 2) Or we can make this an odd function : Fig. 5. Graph of y = x on (-2, 2) You know how to find easiy the Fourier series for the functions defined in Figs. 4 and 5. For instance, for the function defined as in Fig.4, the Fourier series on the interva (-2, 2) is : f HxL = 1-8 p 2 B S cos I n p x n=1,3,5 M 2 n 2 Not surprisingy, the even extension of the function into the eft haf pane produces a Fourier series that consists of ony cos (even) terms. The graph of this series is: F (13) Fig. 6. Fourier series of y = Abs[x] on (-6, 6) We can just as easiy find the Fourier series for the odd function described by the graph in Fig. 5. Since this is an odd function, we expect to find ony sin terms, and the computed Fourier series for this function is :

7 fourierintervas.nb 7 f HxL = -4 p S H-1L n SinA n p x E 2 n=1 n (14) and yieds the graph : Fig. 7. Fourier series for y = x on (-6, 6) As we see from the preceding discussion, we can take any function and make it periodic in order to find its Fourier series. The key point is that we are interested ony in the Fourier series that defines our region of interest. So if we ony care about the behavior of the function on (0, 2), we can sti construct the Fourier series assuming the function is periodic on (-2, 2), and just use that portion of the Fourier series that appies to (0, 2). Notice that thefourier series, eqs. (13) and (14), have very different behaviors (one is even, one is odd; one varies as 1/n, one as 1 ë n 2 ), yet they give us exacty the same behavior on the interva (0,2). So how do we know whether to extend the function to make it an even or an odd function? Sometimes it doesn't matter; sometimes the symmetry of the probem wi make it cear that we must use either the even or odd extension. In fact, we have an infinite array of ways we can make a function periodic. For instance we coud have taken our initia function (y=x on (0,2)) and defined it on (-2,2) as: f HxL = x, 0 x < 2 0, -2 x < Fig. 8. Pot of function in (15) on (-2, 2) (I pot the function with red dashes so you can distinguish the portion of the graph on (-2,0) from the negative x axis). If we compute the Fourier series for this series, we get a much more compicated expression :

8 8 fourierintervas.nb f HxL = p 2 S cos I n p x 2 n=1,3,5 n 2-2 n S H-1L n sin I n p x M 2 n=1 n which we might expect since the function defined immediatey above is neither even nor odd, so our Fourier series incudes both sin and cos terms. Yet, this Fourier series produces exacty the same behavior on the interva (0, 2) : (15) -5 5 Fig. 9. Graph of Fourier series (15) on (-8, 8) So why not choose this formuation of y = x on (0, 2)? We coud, but why make our ives more difficut than necessary. Assuming even or odd symmetry makes our Fourier functions much simper (and the cacuation of those series much easier). Of course, as noted before, sometimes the constraints of the probem wi make it cear to us whether we shoud make use of the even or odd extension, but we can worry about those detais in Chapter 13.