34
CHAPTER 3. FOURIER ANALYSIS
The procedure follows the one above. First, write the partial sum of the first N terms in Eq. (94) as fN (x) =
N ³ 2(2m − 1)πx ´ 1 4A X sin . π m=1 2m − 1 L
(102)
Then define y≡
2(2m − 1)πx L
=⇒ dy =
4πx dm . L
(103)
Then multiply Eq. (102) by dm (which doesn’t affect anything, since dm = 1) and change variables from m to y. The result is (for large N ) 2A fN (x) ≈ π
Z
(4N −2)πx/L
0
sin y dy. y
(104)
As double check on this, if we let x = L/4N then the upper limit of integration is equal to π (in the large-N limit), so this result reduces to the one in Eq. (98). We were concerned with the N → ∞ limit in the above derivation of the overshoot, but now we’re concerned with R ∞just large (but not infinite) N . If we actually set N = ∞ in Eq. (104), we obtain (2A/π) 0 (sin y)/y · dy. This had better be equal to A, because that is the value of the step function at x. And indeed it is, because this definite integral equals π/2 (see the paragraph preceding Eq. (88); it boils down to the fact that (sin y)/y is the Fourier transform of a square wave). But the point is that for the present purposes, we want to keep N finite, so that we can actually see the x dependence of fN (x). (So technically the integral approximation to the sum isn’t perfect like it was in the N = ∞ case. But for large N it’s good enough.) To make the integral in Eq. (104) easier to deal with, let’s define n via x ≡ n(L/4N ). The parameter n need not be an integer, but if it is, then it labels the nth local maximum or minimum of fN (x), due to the reasoning we used in Fig. 31. The number n gives the number of half revolutions the vectors in Fig. 31 make as they wrap around in the plane. If n is an integer, then the sum of the horizontal projections (the cosines in Eq. (95)) is zero, and we therefore have a local maximum or minimum of fN (x). With this definition of n, the upper limit of the integral in Eq. (104) is essentially equal to nπ (assuming N is large), so we obtain fN (n) ≈
2A π
Z 0
nπ
sin y dy y
(where x ≡ n(L/4N )).
(105)
Again, n need not be an integer. What does this function of n look like? The integral has to be computed numerically, and we obtain the first plot shown in Fig. 33 (we have chosen A = 1). The second plot is a zoomed-in (vertically) version of the first one. Note that there is actually no N dependence in fN (n). It is simply a function of n. Therefore, the height of the nth bump is independent of N , assuming that N is reasonably large.8 The N (and L) dependence comes in when we convert from n back to x. 8 If N isn’t large, then we can’t make the approximation that the upper limit of the integral in Eq. (105) equals nπ. From Eq. (104), the limit actually equals nπ(1 − 1/2N ). But if N isn’t large, then the integral approximation to the sum isn’t so great anyway.