How could one find $$\sum_{n=0}^{\infty}{\frac{\cos(nx)}{n!}}\,?$$
I tried to use Fourier series and integrals depending on a parameter to reduce the problem to a differential equation, but that didn't work.
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1$\cos(nx)$ is the real part of $e^{inx}$ which is the real part of $(e^{ix})^n$. The sum of $\frac{(e^{ix})^n}{n!}$ is easy to get so you just need to take the real part of that. – Ian May 19 '19 at 13:23
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Using Intuition behind euler's formula
$$\sum_{n=0}^{\infty}{\frac{\cos(nx)}{n!}}=$$
$$=\text{ real part of }\sum_{n=0}^{\infty}\dfrac{(e^{ix})^n}{n!}$$
$$\sum_{n=0}^{\infty}\dfrac{(e^{ix})^n}{n!}=e^{\cos x+i\sin x}=e^{\cos x}(\cos(\sin x)+i\sin(\sin x))$$
lab bhattacharjee
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Thank you for your answer, but does there exist a method, that does not attract complex analysis? – Sergey Grigoryants May 19 '19 at 13:31
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\begin{align*} \sum_{n=0}^{+\infty}\frac{\cos(nx)}{n!} =&\Re\left[\sum_{n=0}^{+\infty}\frac{(e^{ix})^n}{n!}\right]\\ =&\Re\left[e^{e^{ix}}\right]\\ =&\Re[e^{\cos x+i\sin x}]\\ =& e^{\cos x}\cdot\cos(\sin x) \end{align*}
Joe
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