� � � 75 16. More on Fourier series The Mathlet Fourier Coefficients displays many of the effects described in this section. 16.1. Symmetry and Fourier series. A function g(t) is even if g(t) = g(−t), and odd if g(t) = −g(−t). Fact: Any function f(t) is a sum of an even function and an odd function, and this can be done in only one way. The even part of f(t) is f+(t) = f(t) + f(−t) . 2 and the odd part is f−(t) = f(t) − f(−t) . 2 It’s easy to check that f+(t) is even, f−(t) is odd, and that f(t) = f+(t) + f−(t). We can apply this to a periodic function. We know that any periodic function f(t), with period 2ν, say, has a Fourier expansion of the form ↑ a0 + (an cos(nt) + bn sin(nt)). 2 n=1 If f(t) is even then all the bn’s vanish and the Fourier series is simply ↑ a0 + an cos(nt). 2 n=1 If f(t) is odd then all the an’s vanish and the Fourier series is ↑ bn sin(nt). n=1 Most of the time one is faced with a function which is either even or odd. If f(t) is neither even nor odd, we can still compute its Fourier series by computing the Fourier series for f+(t) and f−(t) separately and adding the results.� � � � 76 16.2. Symmetry about other points. More general symmetries are often present and useful. A function may exhibit symmetry about any fixed value of t, say t = a. We say that f(t) is even about a if f(a + t) = f(a − t) for all t. It is odd about a if f(a + t) = −f(a − t). f(t) is even about a if it behaves the same as you move away from a whether to the left or the right; f(t) is odd about a if its values to the right of a are the negatives of its values to the left. The usual notions of even and odd refer to a = 0. Suppose f(t) is periodic of period 2ν, and is even (about 0). f(t) is then entirely determined by its values for t between 0 and ν. When we focus attention on this range of values, f(t) may have some further symmetry with respect to the midpoint ν/2: it may be even about ν/2 or odd about ν/2, or it may be neither. For example, cos(nt) is even about ν/2 exactly when n is even, and odd about ν/2 exactly when n is odd. It follows that if f (t) is even and even about ν/2 then its Fourier series involves only even cosines: a0f(t) = + an cos(nt). 2 n even If f(t) is even about 0 but odd about ν/2 then its Fourier series involves only odd cosines: f(t) = an cos(nt). n odd Similarly, the odd function sin(nt) is even about ν/2 exactly when n is odd, and odd about ν/2 exactly when n is even. Thus if f(t) is odd about 0 but even about ν/2, its Fourier series involves only odd sines: f(t) = bn sin(nt). n odd If it is odd about both 0 and ν/2, its Fourier series involves only even sines: f(t) = an sin(nt). n even 16.3. The Gibbs effect. The Fourier series for the odd function of period 2ν with F (x) = ν − x for 0 < x < ν 2 is ↑ � sin(kx)F (x) = . k k=177 In Figure 10 we show the partial sum n � sin(kx)Fn(x) = k k=1 with n = 20 and in Figure 11 we show it with n = 100. The horizontal lines of height ±ν/2 are also drawn. −8 −6 −4 −2 0 2 4 6 8 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 Figure 10. Fourier sum through sin(20x) −8 −6 −4 −2 0 2 4 6 8 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 Figure 11. Fourier sum through sin(100x) Notice the “overshoot” near the discontinuities. If you graph Fn(t) for n = 1000 or n = 106, you will get a similar picture. The spike near x = 0 will move in closer to x = 0, but won’t get any shorter. This is the “Gibbs phenomenon.” We have F (0+) = ν/2, but it seems that for any n the partial sum Fn overshoots this value by a factor of 18% or so. A little experimentation with Matlab shows that the spike in Fn(x) occurs at around x = x0/n for some value of x0 independent of n. It� � � 78 turns out that we can compute the limiting value of Fn(x0/n) for any x0: Claim. For any x0, � � � x0x0 sin t lim Fn = dt. n�↑ n 0 t To see this, rewrite the sum as � � n x0 sin(kx0/n) x0Fn = . n kx0/n · n k=1 Using the notation sin t f(t) = t this is n � ⎨ � x0 � kx0 x0Fn = f n n · n k=1 You will recognize the right hand side as a Riemann sum for the func-tion f(t), between t = 0 and t = x0. In the limit we get the integral, and this proves the claim. To find the largest overshoot, we should look for the maximal value x0 sin t sin t of dt. Figure 12 shows a graph of : t t0 −20 −15 −10 −5 0 5 10 15 20 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 sin t Figure 12. t� 79 The integral hits a maximum when x0 = ν, and the later humps are smaller so it never regains this size again. We now know that � � αν sin t lim Fn = dt. n�↑ n 0 t The actual value of this definite integral can be estimated in various ways. For example, the power series for sin t is t3 t5 sin t = t − 3! + 5! − ··· . Dividing by t and integrating term by term, � x0 3sin t x x5 0 t dt = x0 − 3 0 3! + 5 0 5! − ··· . · · Take x0 = ν. Pull out a factor of ν/2, to compare with F (0+) = ν/2: � α sin t ν dt = G, t 2 · 0 where � ⎨ ν2 ν4 G = 2 1 − 3 3!+5 5! − ··· . · · The sum converges quickly and gives G = 1.17897974447216727 . . . . We have found, on the graphs of the Fourier partial sums, a sequence of points which converges to the observed overshoot: �ν �ν �� � ν � n, Fn n � 0, (1.1789 . . .) · 2 , that is, about 18% too large. As a proportion of the gap between F (0−) = −ν/2 and F (0+) = +ν/2, this is (G − 1)/2 = 0.0894 . . . or about 9%. It can be shown that this is the highest overshoot. The Gibbs overshoot occurs at every …
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