Math 414 - Spring 2004

Fourier Series - Orthogonality Relations and Coefficients

In the previous lecture, we discussed Fourier's solution to the problem of heat flow in a bar of length π, with a temperature of 0° at each end and an initial temperature u(x,0)=f(x). The solution to this problem is the temperature u(x,t) at any time t and any point x along the bar:

The idea is match the inital conditions to the series above with t=0,
u(x,0)=f(x)=∑_n b_n sin(nx).
The result is a series for f(x) involving only sines. Had we solved a similar heat flow problem for a ring of radius 1 instead of a bar, matching the initial condition u(x,0)=f(x) (x is the angle in radians) to the corresponding series yield a full Fourier series for f(x), a series having both sines and cosines:

In either case, we are left with the problem of determining the coefficients from the function.

Fourier's first approach was messy. He expanded both sides in Taylor series about x=0, collected terms, and then solved an infinte set of equations. In the end, he obtained the following expressions for the coefficients:

It occurred to Fourier that he might give a simpler proof for these formulas and derive them under more general conditions if he did two things. First assume that f(x) has such an expansion and second plug the series into the expressions he derived by the messy power series method. With luck, the coefficients should easily appear with much less work. That is exactly what happens.

Before we start our derivation, we will need to have availble a set of integrals that are called orthogonality relations for the trigonometric functions. These are in the text, but we'll list them here, too.

In the chain of equations below, we use what we expect to be the formula for b_n. In the first line we place f(x) by the corresponding series. We then interchange the sum and the integral. All of the integrals in the series on the second line can be evaluated using the orthogonality relations above. In the last, only πb_n remains. Dividing by π then gives the expression for b_n given earlier. A similar derivation applies for the a_n.

The point is that this is a fairly simple derivation and holds true under mild assumptions. It also applies in many more situations than Fourier's original power series method would have.

We will close this lecture with an example of finding a Fourier series. Let f(x)=x on the interval [-π,π]. The function itself has a symmtery property; namely, it's odd. The cosines are all even functions, and since the product of an even and odd function is odd, x cos(nx) is odd. Now, the integral of an odd function over a symmetric integral is odd. It follows that a_n=0 for all n. To compute b_n, we again employ the symmetries involved. Since sin(nx) is odd and x is odd, x sin(nx) is odd, too. Thus, we see that
b_n=(2/π)∫₀^πx sin(nx)dx.
We can evaluate this integral via integration by parts. The result is
∫ x sin(nx)dx = - x cos(nx)/n + sin(nx)/n² .
Evaluating this between x=0 and x=π and inserting this in the expression for b_n gives
b_n=(2/π)(-π cos(n π)/n)=2(-1)ⁿ⁺¹/n.
We arrive at x~∑_n2(-1)ⁿ⁺¹sin(nx)/n = 2sin(x)-2sin(2x)/2+2sin(3x)/3 ...