By considering $f(z)=exp(z-\frac{1}{z})$ show that

$$ \frac{1}{2\pi}\int_{0}^{2\pi}cos(n\theta-2sin\theta)d\theta=\sum_0^{\infty}\frac{(-1)^k}{k!(n+k)!}\ \forall n\ge1$$

f is holomorphic in $\gamma (0,R)$\{$0$} $\forall R$ so I tried the Laurent series formula about $0$:

$$c_n=\frac{1}{2i\pi}\int \frac{f(w)}{w^{n+1}}dw$$

now setting $w = exp(i\theta)$ and $f(z)=exp(z - \frac{1}{z})$ as per the hint

$$c_n=\frac{1}{2i\pi}\int_{\gamma(0,1)} \frac{e^{e^{i\theta}-e^{-i\theta}}}{e^{n{i\theta}}}id\theta=\frac{1}{2\pi} \int e^{i(2sin\theta-n\theta)} d\theta$$

and then I guess

$$\sum_{-\infty}^{\infty} c_n z^n = f(z) = exp (z - 1/z) = \sum_{0}^{\infty} \frac {(z-1/z)^n}{n!}=\sum_{0}^{\infty}\frac{\sum_{k=0}^n{C_n^k z^k (-1/z)^{n-k}}}{n!}$$

i then tried to collect $c_n$ and $c_{-n}$ but really I am no longer sure :(

  • $\begingroup$ Hint $\Re[exp(i((n\phi)-2\sin(n \phi))]=\cos((n\phi)-2\sin(n \phi))$ the rest seems to be correct $\endgroup$ – tired Jun 22 '15 at 14:25
  • $\begingroup$ yes but I thought I would leave as is to allow the collection of c_{n} and c_{-n} , since the Laurent series goes to +/_ infinity , unlike the power series ? $\endgroup$ – user3203476 Jun 22 '15 at 14:29
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    $\begingroup$ You have to compute a single coefficient of the whole Laurent series, i.e. the coefficient of $z^n$. Things went bad just because you used $n$ as summation variable in too many places. Try expanding $f(z)$ as $$\sum_{m\geq 0}\frac{\left(z-\frac{1}{z}\right)^m}{m!}$$ and now ask yourself what is the coefficient of $z^n$ in the last sum. $\endgroup$ – Jack D'Aurizio Jun 22 '15 at 14:38
  • $\begingroup$ But for a given n there will be contributions from multiple m's ? $\endgroup$ – user3203476 Jun 22 '15 at 15:32

Following the suggestion of user Jack d'Aurizio (thank you), I am trying to identify the coefficient of a particular $z^n$ in:

$$ \sum_{m-0}^{\infty} \frac {(z-1/z)^m}{m!}=\sum_{m=0}^{\infty}\frac{\sum_{k=0}^m{C_m^k z^k (-1/z)^{m-k}}}{m!} = \sum_{m=0}^{\infty}\frac{\sum_{k=0}^m{C_m^k z^{2k-m} (-1)^{m-k}}}{m!}$$

now for a particular $z^n$ there will be one contribution each from each $m$ : the contribution from the k in the inner sum such that $2k-m=n$

So the coefficient of a $z^n$ is, summing over all the $m$:

$$ \sum_{m=0}^{\infty}\frac{C_{m}^{(m+n) \over 2} (-1)^{ {m-n} \over 2}}{m!} $$

for the cases where $ {m+n} \over 2$ is even but that is still not the neat answer that i was asked to reach ...


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