Mathematical Tools for Physics

5—Fourier Series 121

Now all you have to do is to evaluate the integral on the left.

∫L

0

dxsin

(mπx

L

)

1 =

L

mπ

[

−cos

mπx

L

]L

0

=

L

mπ

[

1 −(−1)m

]

This is zero for evenm, and when you equate it to ( 6 ) you get

am=

4

mπ

formodd

Relabel the indices so that the sum shows only odd integers and the Fourier series is

4

π

∑∞

k=0

1

2 k+ 1

sin

(2k+ 1)πx

L

= 1, (0< x < L) (7)

highest harmonic: 5 highest harmonic: 19 highest harmonic: 99
The graphs show the sum of the series up to 2 k+ 1 = 5, 19 , 99 respectively. It is not a very rapidly
converging series, but it’s a start. You can see from the graphs that near the end of the interval, where the
function is discontinuous, the series has a hard time handling the jump. The resulting overshoot is called the
Gibbs phenomenon. See section5.6.
There is a compact and powerful notation that makes this look much simpler. Further exploit the analogy
with the dot product in three dimensions. In that case I can evaluate a component of the given vector by using
the scalar product:
ˆx.A~=ˆx.

(

Axˆx+Ayyˆ+Azˆz

)

=Ax. OR

〈

x,ˆA~

〉

=Ax

The last, with the angle brackets, is an alternate notation for the scalar product.
The orthogonality integral for the sines is another scalar product written in the same compact notation:

〈

f,g

〉

=

∫L

0

dxf(x)*g(x) (8)