Prove the identity $\cosh(2x)=\cosh^2(x)+\sinh^2(x)$ using the Cauchy product. Prove the identity 
$$\cosh(2x)=\cosh^2(x)+\sinh^2(x)$$ 
using the Cauchy product and the Taylor series expansions of $\cosh(x)$ and $\sinh(x)$. The relations involving the exponential function are not to be used.
 A: Given two power series
$$f(x)=\sum_{n=0}^\infty{a_nx^n}, $$
$$g(x)=\sum_{n=0}^\infty{b_nx^n}$$
The Cauchy product is just their product
$$f(x)g(x)=\sum_{n=0}^\infty{a_nx^n}\sum_{n=0}^\infty{b_nx^n}=\sum_{n=0}^\infty\left(\sum_{k=0}^n{a_kb_{n-k}}\right)x^n$$
So consider $f(x)=\cosh{x}=\sum_{n=0}^\infty{\frac{x^{2n}}{(2n)!}}, g(x)=\sinh{x}=\sum_{n=0}^\infty{\frac{x^{2n+1}}{(2n+1)!}}$.  Now $f(x)^2=\cosh^2{x}$ is
$$\sum_{n=0}^\infty{\frac{x^{2n}}{(2n)!}}\sum_{n=0}^\infty{\frac{x^{2n}}{(2n)!}}=\sum_{n=0}^\infty\sum_{k=0}^n\frac{1}{(2k)!}\frac{1}{(2n-2k)!}x^{2n}=\sum_{n=0}^\infty\sum_{k=0}^n\binom{2n}{2k}\frac{x^{2n}}{(2n)!}$$
For $g(x)^2=\sinh^2{x},$ rewrite the sum as 
$$\sum_{n=0}^\infty{\frac{x^{2n+1}}{(2n+1)!}}=x\sum_{n=0}^\infty{\frac{x^{2n}}{(2n+1)!}}$$
It is easier to take the $x$ out and put it back, as it will stay in line with the Cauchy Product definition (won't change the power of $x$ in the expansion).  Thus
$$x\sum_{n=0}^\infty{\frac{x^{2n}}{(2n+1)!}}x\sum_{n=0}^\infty{\frac{x^{2n}}{(2n+1)!}}=x^2\sum_{n=0}^\infty\sum_{k=0}^n\frac{1}{(2k+1)!}\frac{1}{(2n-2k+1)!}x^{2n}$$
$$=\sum_{n=0}^\infty\sum_{k=0}^n\frac{1}{(2k+1)!}\frac{1}{(2n-2k+1)!}x^{2n+2}=\sum_{n=1}^\infty\sum_{k=0}^{n-1}\frac{1}{(2k+1)!}\frac{1}{(2n-2-2k+1)!}x^{2n}$$
$$=\sum_{n=1}^\infty\sum_{k=0}^{n-1}\frac{1}{(2k+1)!}\frac{1}{(2n-2k-1)!}x^{2n}=\sum_{n=1}^\infty\sum_{k=0}^{n-1}\binom{2n}{2k+1}\frac{x^{2n}}{(2n)!}$$
Now you can add the two sums together, and notice, you are getting a sum of binomial coefficients up to $2n$!  So
$$\sum_{k=0}^n\binom{2n}{2k}+\sum_{k=0}^{n-1}\binom{2n}{2k+1}=\sum_{k=0}^{2n}\binom{2n}{k}=2^{2n}$$
So $$\cosh^2{x}+\sinh^2{x}=\sum_{n=0}^\infty{2^{2n}\frac{x^{2n}}{(2n)!}}=\sum_{n=0}^\infty{\frac{(2x)^{2n}}{(2n)!}}=\cosh{2x}$$
