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.

  • $\begingroup$ TeXForm everything and put dollar signs around them. $\endgroup$ – UserX Sep 13 '14 at 18:57
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    $\begingroup$ Thanks Menni, I want the Fourier series and especially the cosine coefficient. I correct the problem statement to avoid misunderstanding. $\endgroup$ – Afreeka Sep 14 '14 at 8:21
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    $\begingroup$ Thanks Mehnni. Sorry for the typo in your name. $\endgroup$ – Afreeka Sep 14 '14 at 9:02
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    $\begingroup$ If I didn't make any mistake, we have $$\begin{align} &\frac{1}{2\pi}\int_{-\pi}^{\pi}\sqrt{1-k^2\sin^2t}\cos(2nt)dt\\ = & \frac{1 + \sqrt{1-k^2}}{2} \frac{(-\frac12)_n(-\lambda^2)^n}{n!} {}_2F_1\left(-\frac12, n-\frac12; n+1; \lambda^4\right) \end{align}$$ where $\displaystyle\;\lambda = \frac{k}{1 + \sqrt{1-k^2}},\;$ $(\gamma)_n = \gamma(\gamma+1)\cdots(\gamma+n-1)$ is the rising Pochhammer symbol and ${}_2F_1$ is the hypergeometric function. $\endgroup$ – achille hui Sep 16 '14 at 2:20
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    $\begingroup$ It seems for each $n$, the expression on RHS can be rewritten as linear combination of complete elliptic integrals $K(k)$ and $E(k)$ but I haven't figure out the correct coefficients. $\endgroup$ – achille hui Sep 16 '14 at 2:21

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.


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