Value of $p(y) = \sum_{n=0}^{\infty} \dfrac{2^n \cos(ny)}{n!}$

Question

The problem is prove that, for $y \in \mathbb{R}$,$ p(y)$ is absolutly and uniformly convergent (this part is cleared with M-test) and show that

$$p(y) = \cos(2\sin(y))e^{2\cos(y)}$$

This part is my problem =/

Thanks for yours hint/answers.

Answer 1

Hint $$e^z=\sum_{n=0}^\infty \frac{z^n}{n!} $$

Set $z=\cos(2z)+i\sin(2z)$ and take the real part.

Answer 2

Recall that when $z = a + ib \in \mathbb{C}$ we have

$$ e^{z} = e^a (\cos(b) + i \sin(b)) = \sum_{n=0}^{\infty} \frac{z^n}{n!}. $$

Thus,

$$ \cos(2 \sin y)e^{2 \cos y} = \Re \left( e^{2 \left( \cos(y) + i \sin(y) \right) }\right) = \Re \left( e^{2 e^{iy}} \right) = \Re \left( \sum_{n=0}^{\infty} \frac{(2e^{iy})^n}{n!} \right) \\ = \Re \left( \sum_{n=0}^{\infty} \frac{2^n e^{iny}}{n!} \right) = \sum_{n=0}^{\infty} \Re \left( \frac{2^n e^{iny}}{n!} \right) = \sum_{n=0}^{\infty} \frac{2^n\cos(ny)}{n!}.$$