3
$\begingroup$

I have that the Taylor series of $\sin x$ and $\cos x$ are \begin{equation*} \sin x = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}x^{2n+1} \\ \cos x = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}x^{2n} \end{equation*} which I understand yields the product series $\sum_{n=0}^{\infty} c_{n}x^n$ where \begin{equation*} c_n = \begin{cases} \sum\limits_{k=0}^{m} \frac{(-1)^{k}(-1)^{m-k}}{(2k+1)!(2m-2k)!} & n = 2m + 1 \\ \hspace{32 pt} 0 & n = 2m \end{cases} \end{equation*} $\forall \hspace{3 pt} m \in \mathbb{N}$. I then know by simple substitution that \begin{equation*} \frac{1}{2} \sin 2x = \sum_{n=0}^{\infty} \frac{(-1)^{n}2^{2n}}{(2n+1)!}x^{2n+1} \end{equation*} I know I need to show that the odd $c_n$'s and the terms of the above series equal (the even ones are irrelevant as they are all $0$ and so have no bearing on sum) but I am having trouble doing so. Can someone please show how to reduce said equality? Thanks in advance.

$\endgroup$
3
$\begingroup$

Note that $$ \begin{align} c_{2n+1}&=\sum_{k=0}^{n} \frac{(-1)^{k}(-1)^{n-k}}{(2k+1)!(2n-2k)!}\\ &=\sum_{k=0}^{n} \frac{(-1)^{n}(2n+1)!}{(2n+1)!(2k+1)!(2n-2k)!}\\ &=\sum_{k=0}^{n} \frac{(-1)^{n}}{(2n+1)!}{2n+1 \choose 2k+1} \end{align} $$ and that $$ \begin{align} \sum_{k=0}^n{2n+1 \choose 2k+1}&=\sum_{k=0}^n({2n \choose 2k}+{2n \choose 2k+1})\\ &=\sum_{k=0}^{2n}{2n \choose k}\\ &=2^{2n} \end{align} $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.