Fourier analysis – the real story III

We’re trying to get inside Fourier’s head, so to speak, and explore the origins of his methods. To do this, we’re going to look at his derivation of the identity

\frac{\pi}{4} = \cos x - \frac{1}{3} \cos 3x + \frac{1}{5} \cos 5x - \frac{1}{7} \cos 7x + \cdots.

(Again, this is from Paul J. Nahin’s wonderful “Hot Molecules, Cold Electrons”.)

First, we need the indefinite integral

\int \frac{1}{1+x^2} dx = \tan^{-1}x + C.

(Deriving this integral is an easy exercise, but worth knowing.) Now suppose you have a right-angled triangle with the two unspecified angles being \theta and, by necessity, \pi/2 - \theta. Letting the hypotenuse be 1, the side adjacent to \theta be x and the remaining side (necessarily) being \sqrt{1-x^2}, we have that

\frac{\pi}{2} = \tan^{-1} \left( \frac{\sqrt{1-x^2}}{x}\right)+ \tan^{-1} \left( \frac{x}{\sqrt{1-x^2}}\right).

By using the appropriate substitution, we get

\frac{\pi}{2} = \tan^{-1} u + \tan^{-1} \frac{1}{u}.

Now, using the “fact” that

\frac{1}{1+u^2} = 1 - u^2 + u^4 - u^6 + \cdots

(which you can get from, e.g., long division), we integrate to get the indefinite integral

\int \frac{du}{1+u^2} = \tan^{-1} u + C = u - \frac{1}{3}u^3 + \frac{1}{5}u^5 - \frac{1}{7}u^7 + \cdots.

Comparing the two expressions on the right, we see that C = 0. By replacing u with 1/u, we get

\tan^{-1} \left( \frac{1}{u} \right) = \frac{1}{u} - \frac{1}{3} \left( \frac{1}{u} \right)^3 + \frac{1}{5} \left( \frac{1}{u} \right)^5 - \frac{1}{7} \left( \frac{1}{u} \right)^7 + \cdots.

Combining our previous expressions yields

\frac{\pi}{2} = \left( u + \frac{1}{u} \right) - \frac{1}{3} \left( u + \frac{1}{u} \right)^3 + \frac{1}{5}\left( u + \frac{1}{u} \right)^5 - \cdots.

Replacing u by e^{ix}, we can immediately get Fourier’s identity:

\frac{\pi}{4} = \cos x -\frac{1}{3}\cos 3x +\frac{1}{5}\cos 5x - \frac{1}{7} \cos 7x + \cdots.

This is a remarkable identity, but is it true? It is very instructive to graph the right-hand side to see what happens. As it turns out, the identity is only kind-of true. It can be improved, and I suggest finding the mistakes in the derivation as a first step to doing so.

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