Fourier series of $\sqrt{1 - k^2 \sin^2{t}}$

Question

I'm struggling with a Fourier series. I need to find the Fourier series of the following function.

That's the function under study: $f(t)=\left[\sqrt{1-k^2\sin^2t}\,\right]$.

The function is even and $\pi$-periodic.

The Fourier series should be in this form: $f(t)=\frac{a_0}2+\sum\limits_{i=1}^\infty a_n\cos[2nt]$.

In $t\to0$, the Taylor series is: $$f(t)=\left[\frac{2E[k^2]}\pi+\sum_{i=0}^\infty\frac1{2^{2i-1}}\pmatrix{1/2\\ i}(k)^{2i}\sum_{j=0}^{i-1}(-1)^j\pmatrix{2i\\j}\cos(2(i-j)t) \right].$$

It's pretty close to the final Fourier serie but I cannot find the coefficient $a_n$ by identification. Can someone give a help on this? PS: $k\ll 1$ is real and $E(k^2)$ is the complete elliptic integral of the second kind.

Answer 1

Here is an approach. We expand the function $f(x)$ using its Taylor series as

$$ f(x)= \sum_{m=0}^{\infty}{ -\frac{1}{2} \choose m }(-1)^m k^{2m}\sin^{2m}(t) . $$

We need to find $a_n$ which are given by

$$ a_n=\frac{1}{\pi} \int_{-\pi}^{\pi} f(x)\cos(nx)dx =\sum_{m=0}^{\infty}{ -\frac{1}{2} \choose m }(-1)^m k^{2m}\frac{1}{\pi} \int_{-\pi}^{\pi} \sin^{2m}(t)\cos(nx)dx . $$

Now your job to evaluate the last integral and try to finish the problem.