Taylor expansion of $\sum_{k=0}^\infty \cos^k(x)$

Question

I want to derive the Taylor expansion of $\sum_{k=0}^\infty \cos(x)$. By using binomial theorem and Taylor expansion of $e^x=\sum_{i=0}^\infty \frac{x^i}{i!}$, we get $$\begin{align}&\sum_{k=0}^\infty \cos^k(x)\\ =&\sum_{k=0}^\infty (\frac{e^{ix}+e^{-ix}}{2})^k \\ =&\sum_{k=0}^\infty \sum_{j=0}^k2^{k}\binom{k}{j}e^{i(k-2j)x}\\ =&\sum_{k=0}^\infty \sum_{j=0}^k2^{k}\binom{k}{j}\sum_{l=0}^\infty \frac{(i(k-2j)x)^l}{l!}\\ \end{align}$$ However, I don't understand why odd terms appear because $\cos(x)$ have only even degree terms. What mistake did I made?

Answer 1

You can reason on a partial sum, so that the Taylor expansion is possible.

$e^{ix}+e^{-ix}$ is an even function and remains so at the $k^{th}$ power. In the binomial expansion of $(e^{ix}+e^{-ix})^k$, you indeed have terms like $\displaystyle\binom kje^{i(k-j-j)}$. But these come in pairs with $j$ and $k-j$, giving

$$\binom kje^{i(k-2j)}+\binom k{k-j}e^{i(2j-k)}=\binom kj\left(e^{i(k-2j)}+e^{-i(k-2j)}\right).$$

Then if you perform the Taylor expansion, all odd terms will cancel out. (In fact, you just linearized $\cos^k(x)$ as a sum of cosines.)