Page 464 - Advanced engineering mathematics

444 CHAPTER 13 Fourier Series

The Fourier expansion of h on [−L, L] has only sine terms because h is odd on [−L, L].But
h(x) = f (x) on [0, L], so this gives a sine expansion of f on [0, L]. This suggests the following
definitions.

The Fourier sine coefficients of f on [0, L] are
2 L
b n = f (x)sin(nπx/L)dx. (13.12)
L 0
for n = 1,2,···. With these coefficients, the series
∞

b n sin(nπx/L) (13.13)
n=1
is the Fourier sine series for f on [0, L].

Again, we have a convergence theorem for sine series directly from the convergence theorem
for Fourier series.

THEOREM 13.3 Convergence of Fourier Sine Series

Let f be piecewise smooth on [0, L]. Then
1. If 0 < x < L, the Fourier sine series for f on [0, L] converges to
1
( f (x+) + f (x−)).
2
2. At x = 0 and x = L, this sine series converges to 0.

Condition (2) is obvious because each sine term in the series vanishes at x = 0 and at x = L,
regardless of the values of the function there.

EXAMPLE 13.10
2x
Let f (x)= e for 0≤ x ≤ 1. We will write the Fourier sine series of f on [0,1]. The coefficients
are
1
2x
b n = 2 e sin(nπx)dx
0
1
−2nπe cos(nπx) + 4e sin(nπx)
2x 2x
=
2
4 + n π 2 0
n 2
nπ(1 − (−1) e )
= 2 .
2
4 + n π 2
2x
The sine expansion of e on [0,1] is
∞ n 2
nπ(1 − (−1) e )
2 sin(nπx).
2
4 + n π 2
n=1
2x
This series converges to e for 0 < x < 1and to0at x = 0 and x = 1. Figure 13.16 shows
the function and the fortieth partial sum of this sine expansion.
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October 14, 2010 14:57 THM/NEIL Page-444 27410_13_ch13_p425-464