Minimum of function given by a Fourier Series

We have $\begin{aligned} 1+f(x)&=\sum_{n=0}^\infty \frac{\cos nx}{n!} \\ &= \text{Re}\left[\sum_{n=0}^\infty \frac{e^{inx}}{n!}\right] \\ &= \text{Re}\left[ e^{e^{ix}}\right] \\ &= \text{Re}\left[ e^{\cos x + i\sin x }\right] \\ &= \text{Re}\left[ e^{\cos x} e^{i \sin x}\right] \\ &=e^{\cos x} \cos(\sin x) \end{aligned}$

So $f(x)=e^{\cos x} \cos(\sin x)-1$ .

Differentiating, $\begin{aligned} f'(x) &=e^{\cos x} \left[ -\sin x \cdot \cos(\sin x)-\cos x \cdot \sin(\sin x)\right] \\ &=e^{\cos x} \sin(x+\sin x) \end{aligned}$

Setting this equal to zero, we need $x+\sin x=k\pi$ for some integer $k$ .

Normally, this type of equation (transcendental) is difficult to solve; however, here, note that $x=k\pi$ is trivially a solution, and the function $x+\sin x$ is increasing, so for each $k$ there is at most one solution.

So, the turning points occur at $x=k\pi$ .

When $k$ is even, we have $f(k\pi)=e-1$ ; when $k$ is odd, we have $f(k\pi)=e^{-1}-1$ ; hence $c=\boxed{-1}$ .

Let us start doing the following calculations

$e^{ \cos{x} } \cos{(\sin x)}+ie^{ \cos{x} } \sin{(\sin x)}=e^{e^{ix}}=\sum_{n=0}^{\infty}\frac{ e^{inx}}{n!}=\sum_{n=0}^{\infty}\frac{ \cos{nx}+i\sin{nx}}{n!}=\sum_{n=0}^{\infty}\frac{ \cos{nx}}{n!}+i\sum_{n=0}^{\infty}\frac{ \sin{nx}}{n!}$

Making the real parts equal, we obtain that $\sum_{n=0}^{\infty}\frac{ \cos{nx}}{n!}=e^{ \cos{x} } \cos{(\sin x)}$

Therefore, $f(x)=e^{ \cos{x} } \cos{(\sin x)}-1.$ Using the derivative of the function, we obtain that the minimum value of the function is $e^{-1}-1.$ Therefore, the value of $c=\boxed{-1}.$