$\sin x \cos x$ defined as a Cauchy Product of their Taylor Series.

Question

The goal of this project is to show that the Cauchy Product of two Taylor Series, $\sin x$ and $\cos x$, is equal to the Maclaurin Series of $\sin x\cos x$ . I am having trouble simplifying the definite sum component of my current result.

Start with: $$\sin x\cos x = \left(\sum_{i=0}^{\infty}\frac{(-1)^ix^{2i+1}}{(2i+1)!}\right)\left(\sum_{j=0}^{\infty}\frac{(-1)^jx^{2j}}{(2j)!}\right)$$

The Cauchy Product in this case is equal to the product of the two sums:

$$\left(\sum_{i=0}^{\infty}a_i\right)\left(\sum_{j=0}^{\infty}b_j\right)=\sum_{k=0}^{\infty}\left(\sum_{n=0}^{k}a_nb_{k-n}\right)$$

So, we have: $$\sin x \cos x = \sum_{k=0}^{\infty}\left(\sum_{n=0}^{k}\left(\left(\frac{(-1)^nx^{2n+1}}{(2n+1)!}\right)\left(\frac{(-1)^{k-n}x^{2k-2n}}{(2k-2n)!}\right)\right)\right)$$ $$=\sum_{k=0}^{\infty}\left(\sum_{n=0}^{k}\frac{(-1)^kx^{2k+1}}{(2n+1)!(2k-2n)!}\right)$$ $$=\sum_{k=0}^{\infty}\left((-1)^kx^{2k+1}\sum_{n=0}^{k}\left(\frac{1}{\Gamma(2n+2)\Gamma(2k-2n+1)}\right)\right)$$

I would like to simplify the definite sum component of the above: $\sum_{n=0}^{k}\left(\frac{1}{\Gamma(2n+2)\Gamma(2k-2n+1)}\right)$ into a closed form which should let the whole equation simplify to the McLaurin Series of $\sin x \cos x$: $\sum_{k=0}^{\infty}\left((-4)^k\frac{x^{2k+1}}{(2k+1)!}\right)$.

I have tried applying discrete calculus techniques to simplify the sum but have had no success. I am only a Calc 3 student so I am not familiar with many techniques for simplifying definite sums. Any help is very appreciated.

 

Answer 1

Notice that $\displaystyle\frac{1}{(2n+1)!(2k-2n)!}=\frac 1 {(2k+1)!}C_{2k+1}^{2n+1}$.

So the sum $\displaystyle\sum_{n=0}^k \frac{1}{(2n+1)!(2k-2n)!} = \frac 1 {(2k+1)!} \sum_{n=0}^k C_{2k+1}^{2n+1}$.

To solve $\displaystyle \sum_{n=0}^k C_{2k+1}^{2n+1}$, we will use the following expansions: $\displaystyle (1+x)^{2n+1}=\sum_{n=0}^{2k+1} C_{2k+1}^{n}x^n$, substituting for x=1 and -1 respectively yields:

$2^{2n+1} = \displaystyle (1+1)^{2n+1}=\sum_{n=0}^{2k+1} C_{2k+1}^{n}=\sum_{n=0}^k C_{2k+1}^{2n} + \sum_{n=0}^k C_{2k+1}^{2n+1}$ and $0^{2n+1} = \displaystyle (1-1)^{2n+1}=\sum_{n=0}^{2k+1} C_{2k+1}^{n}(-1)^n = \sum_{n=0}^k C_{2k+1}^{2n} - \sum_{n=0}^k C_{2k+1}^{2n+1}$.

Therefore $2^{2n+1}-0 = 2\sum_{n=0}^k C_{2k+1}^{2n+1} \implies \frac 1 {(2k+1)!} \sum_{n=0}^k C_{2k+1}^{2n+1} = \frac 1 {(2k+1)!} 2^{2n}= \frac 1 {(2k+1)!} 4^n$. And that proves the Cauchy sum.

Answer 2

We use the power series expansion of $sin$ and $cos$ to show \begin{align*} \color{blue}{\sin(x)\cos(x)=\frac{1}{2}\sin(2x)}\tag{1} \end{align*}

We start with the left-hand side of (1) and obtain \begin{align*} \color{blue}{\sin(x)\cos(x)} &=\left(\sum_{k=0}^{\infty}(-1)^k\frac{x^{2k+1}}{(2k+1)!}\right) \left(\sum_{l=0}^{\infty}(-1)^l\frac{x^{2l}}{(2l)!}\right)\\ &=\sum_{n=0}^{\infty}\left(\sum_{{2k+1+2l=n}\atop{k,l\geq 0}}\frac{(-1)^{k+l}}{(2k+1)!(2l)!}\right)x^n\\ &=\sum_{n=0}^{\infty}\left(\sum_{{2k+1+2l=2n+1}\atop{k,l\geq 0}}\frac{(-1)^{k+l}}{(2k+1)!(2l)!}\right)x^{2n+1}\tag{2}\\ &=\sum_{n=0}^{\infty}(-1)^n\left(\sum_{{k+l=n}\atop{k,l\geq 0}}\frac{1}{(2k+1)!(2l)!}\right)x^{2n+1}\tag{3}\\ &=\sum_{n=0}^{\infty}(-1)^n\left(\sum_{k=0}^n\frac{1}{(2k+1)!(2n-2k)!}\right)x^{2n+1}\tag{4}\\ &=\sum_{n=0}^{\infty}(-1)^n\left(\sum_{k=0}^n\binom{2n+1}{2k+1}\right)\frac{x^{2n+1}}{(2n+1)!}\tag{5}\\ &=\frac{1}{2}\sum_{n=0}^{\infty}(-1)^n2^{2n+1}\frac{x^{2n+1}}{(2n+1)!}\tag{6}\\ &\,\,\color{blue}{=\frac{1}{2}\sin(2x)} \end{align*} and the claim (1) follows.

Comment:

• In (2) we note that the left-hand side $2k+1+2l$ of the index of the inner sum is odd for each integer $k,l\geq 0$. This implies that even $n$ do not contribute. We skip even $n$ and consider odd $n$ in the form $2n+1$ only.

• In (3) we simplify the index region and factor out $(-1)^{k+l}=(-1)^n$, since $k+l=n$.

• In (4) we eliminate the index variable $l$ by using $l=n-k$.

• in (5) we introduce the binomial coefficient $\binom{2n+1}{2k+1}$ and divide by $(2n+1)!$ as compensation.

• In (6) we use the binomial identity $\sum_{k=0}^{n}\binom{2n+1}{2k+1}=2^{2n}$, since \begin{align*} \color{blue}{\sum_{k=0}^{n}\binom{2n+1}{2k+1}} &=\frac{1}{2}\sum_{k=0}^{n}\binom{2n+1}{2k+1}+\frac{1}{2}\sum_{k=0}^{n}\binom{2n+1}{2n-2k}\tag{7}\\ &=\frac{1}{2}\sum_{k=0}^{n}\binom{2n+1}{2k+1}+\frac{1}{2}\sum_{k=0}^{n}\binom{2n+1}{2k}\tag{8}\\ &=\frac{1}{2}\sum_{k=0}^{2n+1}\binom{2n+1}{k}\tag{9}\\ &\,\,\color{blue}{=2^{2n}} \end{align*}

• In (7) we split the sum and use for the right-most sum the binomial identity $\binom{p}{q}=\binom{p}{p-q}$.

• In (8) we change the order of summation of the right-most sum $k\to n-k$.

• In (9) we merge odd and even summands into one sum and apply the binomial theorem.