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Continuing the post here let us see the series:

$$\mathcal{S}(x)=\sum_{n=0}^{\infty} \frac{x^{4n}}{(4n)!}$$

By using Discrete Fourier Transform (used in the http://www.mathimatikoi.org/forum/viewt ... f=27&t=686 )

\(\mathcal{S}(x)=\sum_{n=0}^{\infty} \frac{x^{4n}}{(4n)!}\)

\(= \frac{1}{2}\left ( \cosh x +\cos x \right )\)

Correct. I am filling in the details. Let $\zeta$ be a fourth root of unity, that is $\zeta^4=1$. Then we note that:

$$\sum_{\zeta} \zeta^n =4_{n \equiv 0 \pmod 4}$$

and $0$ otherwise. Also,

$$\sum_{\zeta} e^{\zeta x} = \sum_{n=0}^{\infty}\frac{x^n }{n!} \sum_{\zeta} \zeta^n = 4 \sum_{n=0}^{\infty} \frac{x^{4n}}{(4n)!}$$

This, however, means that

$$ \sum_{n=0}^{\infty} \frac{x^{4n}}{(4n)!} = \frac{e^x+e^{-x} +e^{-ix}+e^{ix}}{4}= \frac{1}{2} \left ( \cosh x + \cos x \right )$$

Just add one remark. This problem is a special case of the Simpson's dissection method: If $f(x) = \sum_{n=0}^\infty\,a_nx^n$, then
$$\sum_{n=0}^\infty\,a_{kn+m}x^{kn+m} = \frac{1}{k}\,\sum_{j=0}^{k-1}\,\omega^{-jm}f(\omega^jx),$$
where $\omega = e^{2\pi i/k}$ is a primitive $k$th root of unity. Appealing to $f(x) = e^x, k = 4, m= 0$ solves the proposed problem. By the way, Hamza's problem is the special case of $f(x) = e^x, k = 3, m= 0$.