Mathematical Tools for Physics

5—Fourier Series 119

1

3

5

highest harmonic: 5

1

3

5

highest harmonic: 5

The same function can be written in terms of sines with another series:

x^2 =

2 L^2

π

∑∞

1

[

(−1)n+1

n

−

2

π^2 n^3

(

1 −(−1)n)

)

]

sin

nπx

L

(2)

and again you can see how the series behaves by taking one to several terms of the series. (right graph) The
graphs show the parabolay=x^2 and partial sums of the two series with terms up ton= 1 , 3 , 5.
The second form doesn’t seem to work as smoothly as the first example, and there’s a reason for that.
The sine functions all go to zero atx=Landx^2 doesn’t, making it hard for the sum of sines to approximate
the desired function. They can do it, but it takes a lot more terms in the series to get a satisfactory result. The
series Eq. ( 1 ) has terms that go to zero as 1 /n^2 , while the terms in the series Eq. ( 2 ) go to zero only as 1 /n.

5.2 Computing Fourier Series
How do you determine the details of these series starting from the original function? For the Taylor series, the trick
was to assume a series to be an infinitely long polynomial and then to evaluate it (and its successive derivatives)
at a point. You require that all of these values match those of the desired function at that one point. That
method won’t work here. (Actually it can work here too, but only after a ridiculous amount of labor.)
The idea of the procedure that works here is like one that you can use to determine the components of a
vector in three dimensions. You write such a vector as

A~=Axˆx+Ayyˆ+Azˆz