How would I find the closed form of

$$ f(\alpha) = \sum_{n=0}^{\infty} \frac{\cos(\alpha \, n)}{n!}$$

My old pal Wolfram tells me that

$$ f(\alpha) = \mathrm{e}^{\cos(\alpha)}\cos(\sin(\alpha))$$

I've attempted writing out the first couple terms for different values of $\alpha$, but I haven't quite figured out how to arrive at the result.

  • $\begingroup$ Use De Moivre's Theorem $ \exp(i \alpha) =\cos \theta + i\sin \theta$ and the look at the real part. $\endgroup$ Commented Feb 9, 2017 at 0:17

2 Answers 2


Consider \begin{eqnarray*} \sum_{n=0}^{\infty} \frac{\cos (n\alpha)}{n!}+i \sum_{n=0}^{\infty} \frac{\sin (n\alpha)}{n!}. \end{eqnarray*} Now by De Moivre's theorem this equals \begin{eqnarray*} \sum_{n=0}^{\infty} \frac{e^{(ni\alpha)}}{n!}= e^{e^{i\alpha}}=e^{\cos \alpha+i\sin \alpha} =e^{\cos \alpha} e^{+i\sin \alpha} =e^{\cos \alpha} (\cos (\sin \alpha)+i\sin (\sin\alpha)) \end{eqnarray*} Now equate the real parts & we have \begin{eqnarray*} \sum_{n=0}^{\infty} \frac{\cos (n\alpha)}{n!}=e^{\cos \alpha} \cos (\sin \alpha). \end{eqnarray*}



$$\cos(\alpha n)=\frac{1}{2}\left(\left(e^{i\alpha}\right)^n+\left(e^{-i\alpha}\right)^n\right)$$

Alternatively, this can be written as (when $\alpha$ is real) as

$\cos(\alpha n)$ is the real part of $\left(e^{i\alpha}\right)^n$.


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