expansion of $\text{ cosh}(z+1/z)$

Question

It is given that if $|z|>0$, $$\text{cosh}(z+1/z) = c_0 + c_1 (z + 1/z) + c_2(z^2 + 1/z^2) + \dots $$ where $$c_n = \frac{1}{2\pi}\int_0^{2\pi} \cos (n \phi) \text{cosh}(2\cos \phi ) d\phi$$

I only know to expand $\text{cosh(z + 1/z)}$ by putting $w = z + 1/z$ and expand it as $\frac 12 (e^{w} + e^{-w})$. How do I introduce $c_n$, any hints?

ADDED:

Here's what I got $$c_n = \frac{1}{2 \pi i} \oint_{|z| = 1} \frac{\text{cosh}(z+1/z)}{z^{n+1}}dz = \frac{1}{2\pi} \int_0^{2\pi} \text{cosh}(2\cos \phi)e^{-in \phi}d\phi$$

It seems that $c_n = c_{-n}$ by the symmetricity of $z+1/z$ and of $\cos(-z) = \cos(z)$ but how to get rid of the imaginary part above?

And what is wrong or terrible with writing down the expansion of $,\cosh z,$ by using the very well known and nice expansion of the exponential function? — DonAntonio, May 04 '13 at 10:33
@DonAntonio i don't know, the exercise say so, maybe I should look into $c_n$ from Laurent theorem. — Monkey D. Luffy, May 04 '13 at 10:34

score 1 · Accepted Answer · answered May 04 '13 at 10:36

1

If you write $z=e^{i\phi}$, then $\cosh(z+1/z)=\cosh(2\cos\phi)=f(\phi)$ is an even $2\pi$-periodic function of $\phi$ which can therefore be expanded into Fourier cosine series: $$ f(\phi)=\frac{a_0}{2}+\sum_{n=1}^{\infty}a_n\cos n\phi,\tag{1}$$ with $$ a_n=\frac{1}{\pi}\int_0^{2\pi}f(\phi)\cos n\phi\,d\phi.$$ These $a_n$'s are related to your $c_n$'s by $a_n=2c_n$ (note that $z^n+z^{-n}=2\cos n\phi$).

Addendum: Fourier series can be seen as the restriction of Laurent series to the unit circle. Note, however, that once you know coefficients of the former, this also gives the coefficients of the latter.

answered May 04 '13 at 10:36

Start wearing purple

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could you elaborate the last?? what is the relation between Fourier series and Laurent series? – Monkey D. Luffy May 04 '13 at 10:39
I've added a small comment. Is it ok now? – Start wearing purple May 04 '13 at 10:44
thank you ... it is very interesting which i did not know. but before that could you check some of my thoughts? I have added in question. – Monkey D. Luffy May 04 '13 at 10:53
Concerning imaginary part: the function you integrate is periodic and you integrate over one period. Therefore instead of going from $0$ to $2\pi$ you could go from $-\pi$ to $\pi$. But the imaginary part is an odd function of $\phi$, so the corresponding integral is equal to $0$. – Start wearing purple May 04 '13 at 11:02
Hmm ... i see, thank you very much!! – Monkey D. Luffy May 04 '13 at 11:03

expansion of $\text{ cosh}(z+1/z)$

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