More on Fourier Series

Transcription

1 More on Fourier Series R. C. Trinity University Partial Differential Equations Lecture 6.1

2 New Fourier series from old Recall: Given a function f (x, we can dilate/translate its graph via multiplication/addition, as follows. Geometric operation Dilate along the x-axis by a factor of a Dilate along the y-axis by a factor of b Translate (right along the x-axis by c units Translate (up along the y-axis by d units Mathematical implementation f (x/a bf (x f (x c f (x + d

3 One has the following general principles. Theorem If the graph of f (x is obtained from g(x by dilations and/or translations, then the same operations can be used to obtain the Fourier series of f from that of g. Theorem If f (x is a linear combination of g 1 (x, g (x,..., g n (x, then the Fourier series of f is the same linear combination of the Fourier series of g 1, g,..., g n. Remarks: These are both easily derived from Euler s formulas for the Fourier coefficients. These tell us that we can construct Fourier series of new functions from existing series.

4 Example Use an existing series to find the Fourier series of the π-periodic function given by f (x = x for 0 x < π. The graph of f (x: This function can be obtained from the earlier sawtooth wave by translating both up and to the right by π units.

5 The old sawtooth wave has Fourier series ( 1 n+1 sin(nx, n so the function f has Fourier series ( 1 n+1 sin(n(x π π + n ( 1 n+1 = π + (sin(nx cos(nπ sin(nπ cos(nx n ( 1 n+1 ( 1 n = π + sin(nx n sin(nx = π n

6 Example Use an existing series to find the Fourier series of the 4-periodic function satisfying { x if 1 x < 1 f (x =. x if 1 x < 3 The graph of f (x: We can obtain f from the graph of an earlier π-periodic triangular wave.

7 Earlier wave: g(x Dilation of /π along both axes: Translation by 1 along both axes: π g ( πx ( -1 + π g π(x 1 We already know that the Fourier series for g is π 4 π cos((k + 1x (k + 1. We simply transform it as above, and simplify.

8 This yields 1 + π ( π 4 π cos((k + 1π(x 1/ (k + 1 The cosine term inside the sum is ( ( ( (k + 1πx (k + 1π (k + 1πx (k + 1π cos = cos cos ( ( (k + 1πx (k + 1π + sin sin ( (k + 1πx = ( 1 k sin. So the series simplifies to 8 π ( 1 k (k + 1 sin ( (k + 1πx.

9 Example Use existing series to find the Fourier series of the π-periodic function satisfying { 0 if π x < 0, f (x = x if 0 x < π. The graph of f (x (left is the average of the sawtooth and triangular waves shown. = 1 + 1

10 So, the Fourier series of f is the average of our two previous series: ( 1 ( 1 n+1 sin(nx + π n 4 cos((k + 1x π (k + 1 = π 4 cos((k + 1x ( 1 n+1 π (k sin(nx. n We could combine these into one series, but it s easier to just leave the cosine and sine series separate.

11 Differentiating Fourier series Term-by-term differentiation of a series can be a useful operation, when it is valid. The following result tells us when this is the case with Fourier series. Theorem Suppose f is π-periodic and piecewise smooth. If f is also piecewise smooth, and f is continuous everywhere, then the Fourier series for f can be obtained from that of f using term-by-term differentiation. Remark: This can be proven by using integration by parts in the Euler formulas for the Fourier coefficients of f.

12 Example Use an existing series to find the Fourier series of the π-periodic function satisfying { 1 if π x < 0, f (x = 1 if 0 x < π. The graph of f (x (a square wave shows that it is the derivative of the triangular wave.

13 Since the triangular wave is continuous everywhere, we can differentiate its Fourier series term-by-term to get the series for the square wave. d dx ( π 4 π cos((k + 1x (k + 1 = 4 π = 4 π (k + 1 sin((k + 1x (k + 1 sin((k + 1x. (k + 1 Warning: The hypothesis that f is continuous is extremely important. For example, if we term-wise differentiate the Fourier series for the discontinuous square wave (above, we get 4 π cos((k + 1x which converges (almost nowhere!

14 Half-range expansions Goal: Given a function f (x defined for 0 x p, write f (x as a linear combination of sines and cosines. Idea: Extend f to have period p, and find the Fourier series of the resulting function.

15 Sine and cosine series We set f o = odd p-periodic extension of f, f e = even p-periodic extension of f. If we expand f o as a Fourier series, it will involve only sines: ( nπx b n sin. p This is the sine series expansion of f. According to Euler s formula the Fourier coefficients are given by b n = 1 p ( nπx f o (x sin dx = p ( nπx f (x sin dx. p p p p 0 p }{{} even

16 If we expand f e as a Fourier series, it will involve only cosines: a 0 + ( nπx a n cos p. This is the cosine series expansion of f. This time Euler s formulas give a 0 = 1 p a n = 1 p p p p p p f e (x dx = 1 }{{} p 0 even ( nπx f e (x cos p }{{} even f (x dx, dx = p p 0 ( nπx f (x cos dx. p If f is piecewise smooth, both the sine and cosine series converge f (x+ + f (x to the function (on the interval [0, p].

17 Example Find the sine and cosine series expansions of f (x = 3 x on the interval 0 x 3. Taking p = 3 in our work above, the coefficients of the sine series are given by b n = 3 ( nπx (3 x sin dx ( = 3(3 x ( nπx cos 9 ( nπx 3 nπ 3 n π sin 3 3 = 3 9 nπ cos(0 = 6 nπ. So, the sine series is 6 1 ( nπx π n sin. 3 0

18 The cosine series coefficients are a 0 = 1 ( 3 3 x dx = 1 3x x = 3 and for n a n = ( nπx (3 x cos dx ( = 3(3 x ( nπx sin 9 ( nπx 3 nπ 3 n π cos = ( 9 3 n π cos(nπ + 9 n π if n is odd, n π = 0 if n is even. 0

19 Since we can omit the terms with even n, we write n = k + 1 (k 0 and obtain the cosine series a 0 + ( nπx a n cos = π 1 (k + 1 cos ( (k + 1πx 3. Here are the graphs of f, f o and f e (over one period: Consequently, the sine series equals f (x for 0 < x 3, and the cosine series equals f (x for 0 x 3.