# Implementation of Cauchy product on $\cos x\cdot \sin x$

I have a mistake that I can't find somewhere along the way. Please help me find the place things go wrong.

Find the product of $$\cos x$$ and $$\sin x$$ as defined: $$\cos(x) = \sum_{k=0}^{\infty} \frac{\alpha_k}{k!} x^k$$ $$\sin(x) = \sum_{k=0}^{\infty} \frac{-\alpha_{k+1}}{k!} x^k$$ with $$\alpha_k = \frac{i^k}{2} \left( 1 + (-1)^k \right)$$

My attempt, using the following tricks:

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

2) $$e^{z+w} = \sum_{n=0}^{\infty} \frac{z^n}{n!} \cdot \sum_{n=0}^{\infty} \frac{w^n}{n!} = \sum_{n=0}^{\infty} \sum_{k=0}^{n} \frac{z^k}{k!} \frac{w^{n-k}}{(n-k)!} = \sum_{n=0}^{\infty} \frac{1}{n!} \sum_{k=0}^{n} n! \cdot \frac{z^k}{k!} \frac{w^{n-k}}{(n-k)!} = \sum_{n=0}^{\infty} \frac{1}{n!} \sum_{k=0}^{n} \frac{n!}{k!(n-k)!} \cdot z^k \cdot w^{n-k} = \sum_{n=0}^{\infty} \frac{1}{n!} \sum_{k=0}^{n} {n\choose k} \cdot z^k \cdot w^{n-k} = \sum_{n=0}^{\infty} \frac{1}{n!} \sum_{k=0}^{n} (z+w)^n = e^{z+w}$$

Here is what I did:

$$\cos(x) \cdot \sin(x) = \sum_{k=0}^{\infty} \frac{\alpha_k}{k!} x^k \cdot \sum_{k=0}^{\infty} \frac{-\alpha_{k+1}}{k!} x^k = \sum_{n=0}^{\infty} \sum_{k=0}^{n} \frac{-\alpha_{k+1}}{k!} x^k \cdot \frac{\alpha_{n-k}}{(n-k)!} x^{n-k} = \\ \sum_{n=0}^{\infty} \frac{1}{n!} \sum_{k=0}^{n} \frac{n!}{k!(n-k)!} \left[ -\alpha_{k+1} \cdot x^k \cdot \alpha_{n-k} \cdot x^{n-k} \right] = \\ \sum_{n=0}^{\infty} \frac{1}{n!} \sum_{k=0}^{n} {n \choose k} \cdot \left[-1 \cdot \frac{i^{k+1}}{2} \left( 1 + (-1)^{k+1} \right) \cdot x^k \right] \cdot \left[ \frac{i^{n-k}}{2} \left( 1 + (-1)^{n-k} \right) \cdot x^{n-k} \right] = \\ \sum_{n=0}^{\infty} \frac{1}{n!} \sum_{k=0}^{n} {n \choose k} \cdot \left[-1 \cdot \frac{1}{2i} \cdot \left( i^k + (-i)^{k} \right) \cdot x^k \right] \cdot \left[ \frac{1}{2} \left( i^{n-k} + (-i)^{n-k} \right) \cdot x^{n-k} \right] = \\ -\frac{1}{2i} \cdot \frac{1}{2} \cdot \sum_{n=0}^{\infty} \frac{1}{n!} \sum_{k=0}^{n} {n \choose k} \cdot \left[ \left( i^k + (-i)^{k} \right) \cdot x^k \right] \cdot \left[ \left( i^{n-k} + (-i)^{n-k} \right) \cdot x^{n-k} \right] = \\ -\frac{1}{2i} \cdot \frac{1}{2} \cdot \sum_{n=0}^{\infty} \frac{1}{n!} \sum_{k=0}^{n} {n \choose k} \cdot \left[ (ix)^k + (-ix)^{k} \right] \cdot \left[ (ix)^{n-k} + (-ix)^{n-k} \right] = \\ -\frac{1}{2i} \cdot \frac{1}{2} \cdot \sum_{n=0}^{\infty} \frac{1}{n!} \sum_{k=0}^{n} {n \choose k} \cdot \left[ (ix)^k (ix)^{n-k} + (ix)^k (-ix)^{n-k} + (-ix)^{k} (ix)^{n-k} + (-ix)^{k} (-ix)^{n-k}\right] = \\ -\frac{1}{2i} \cdot \frac{1}{2} \cdot \left[ \sum_{n=0}^{\infty} \frac{1}{n!} \sum_{k=0}^{n} {n \choose k} \cdot \left[ (ix)^k (ix)^{n-k} \right] + \sum_{n=0}^{\infty} \frac{1}{n!} \sum_{k=0}^{n} {n \choose k} \cdot \left[ (ix)^k (-ix)^{n-k} \right] + \sum_{n=0}^{\infty} \frac{1}{n!} \sum_{k=0}^{n} {n \choose k} \cdot \left[ (-ix)^{k} (ix)^{n-k} \right] + \sum_{n=0}^{\infty} \frac{1}{n!} \sum_{k=0}^{n} {n \choose k} \cdot \left[ (-ix)^{k} (-ix)^{n-k} \right] \right] = \\ -\frac{1}{2i} \cdot \frac{1}{2} \cdot \left[ \sum_{n=0}^{\infty} \frac{1}{n!} (ix + ix)^n + \sum_{n=0}^{\infty} \frac{1}{n!} (ix-ix)^{n} + \sum_{n=0}^{\infty} \frac{1}{n!} (-ix+ix)^{n} + \sum_{n=0}^{\infty} \frac{1}{n!} (-ix-ix)^{n} \right] = \\$$

What do I do about $$\sum_{n=0}^{\infty} \frac{1}{n!} (ix-ix)^{n}$$ and $$\sum_{n=0}^{\infty} \frac{1}{n!} (-ix+ix)^{n}$$ Do they converge to 1 or diverge? How do I get the expected answer: $$\frac{1}{2} \sin(2x)$$

Thanks

Note that $$\alpha_{k}=0$$ if $$k$$ is odd and $$(-1)^{\frac{k}{2}}$$ otherwise. Applying Cauchy product yields
\begin{align*} \displaystyle \sum_{n=0}^{\infty} \displaystyle \sum_{k=0}^{n} \frac{(-1)^k x^{2k}}{(2k)!}\cdot \frac{(-1)^{n-k}x^{2n-2k+1}}{(2n-2k+1)!}=&\displaystyle \sum_{n=0}^{\infty} \displaystyle \sum_{k=0}^{n} \frac{(-1)^{n}x^{2n+1}}{(2n+1)!} \binom{2n+1}{2k} \\ =&\displaystyle \sum_{n=0}^{\infty} (-1)^{n} \frac{x^{2n+1}}{(2n+1)!} \displaystyle \sum_{k=0}^{n} \binom{2n+1}{2k} \\ =&\displaystyle \sum_{n=0}^{\infty} (-1)^{n} \frac{x^{2n+1}}{(2n+1)!} \cdot 4^{n} \\ =&\displaystyle \sum_{n=0}^{\infty} (-1)^{n} \frac{x^{2n+1}}{(2n+1)!}2^{2n} \\ =&\frac{1}{2} \cdot \displaystyle \sum_{n=0}^{\infty} (-1)^{n} \frac{x^{2n+1}}{(2n+1)!}2^{2n+1} \\ =&\frac{1}{2} \cdot \displaystyle \sum_{n=0}^{\infty} (-1)^{n} \frac{(2x)^{2n+1}}{(2n+1)!}\\ =&\frac{\operatorname{sin}(2x)}{2}. \end{align*}