Mathematical Methods for Physicists: A concise introduction - Site Map

FOURIER SERIES AND INTEGRALS
Solution:
The Fourier coecients are readily calculated:
a n ˆ 1 Z 0
Z
…k† cos nx dx ‡ k cos nx dx
0
"
ˆ 1
sin nx
0
k sin nx
n ‡ k n ˆ 0
0
b n ˆ 1 Z 0
Z
…k† sin nx dx ‡ k sin nx dx
0
"
ˆ 1
cos nx
0
k cos nx
n k n ˆ 2k …1 cos n†
n
0
Now cos n ˆ1 for odd n, and cos n ˆ 1 for even n. Thus
b 1 ˆ 4k=; b 2 ˆ 0; b 3 ˆ 4k=3; b 4 ˆ 0; b 5 ˆ 4k=5; ...
and the corresponding. Fourier series is
4k
sin x ‡ 1 3 sin 3x ‡ 1
sin 5x ‡ :
5
For the special case k ˆ =2, the Fourier series becomes
2 sin x ‡ 2 3 sin 3x ‡ 2 sin 5x ‡:
5
The ®rst two terms are shown in Fig. 4.4, the solid curve is their sum. We will see
that as more and more terms in the Fourier series expansion are included, the sum
more and more nearly approaches the shape of f …x†. This will be further demonstrated
by next example.
Example 4.3
Find the Fourier series that represents the function de®ned by
0;