<s>[INST] Simplifying $\cosh x + \sinh x$, $\cosh^2 x + \sinh^2 x$, $\cosh^2 x - \sinh^2 x$ using only the Taylor Series of $\cosh,\sinh$ I was trying to solve the following question:
\begin{align*}
\sinh x &= x + \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots \\
\cosh x &= 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \cdots
\end{align*}
Using only this information, calculate
$\cosh x + \sinh x$, $\cosh^2 x + \sinh^2 x$, and $\cosh^2 x - \sinh^2 x$.
Calculating $\cosh x + \sinh x$ was easy because it's just the Taylor Series of $e^x$, but dealing with squaring is where it gets difficult because the Taylor Series are infinite. How can I circumvent the infinite portion to get $\cosh^2 x$ and $\sinh^2 x$?
 [/INST] We have
$$
\begin{align}
\cosh^2(x)&=\left(\sum_{k=0}^\infty \frac{x^{2k}}{(2k)!}\right)\left(\sum_{k=0}^\infty \frac{x^{2k}}{(2k)!}\right)
\\&= \sum_{k=0}^\infty \sum_{l=0}^k \frac{x^{2l}x^{2(k-l)}}{(2l)!(2(k-l))!}
\\&= \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!} \color{red}{\sum_{l=0}^k \binom{2k}{2l}}
\\&= 1 + \sum_{k=1}^\infty \frac{\color{red}{2^{2k-1}}x^{2k}}{(2k)!}
\\&= \frac 1 2 +\frac 1 2 \sum_{k=0}^\infty \frac{(2x)^{2k}}{(2k)!}
\\&= \frac 1 2 (\cosh{2x}+1)
\end{align}
$$
where the red is a well-known identity (when $2k \in \mathbb{Z^+}$). We similarly have
$$
\begin{align}
\sinh^2(x)&=\left(\sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!}\right)\left(\sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!}\right)
\\&= \sum_{k=0}^\infty \sum_{l=0}^k \frac{x^{2l+1}x^{2(k-l)+1}}{(2l+1)!(2(k-l)+1)!}
\\&= \sum_{k=0}^\infty \frac{x^{2(k+1)}}{(2(k+1))!} \color{red}{\sum_{l=0}^k \binom{2(k+1)}{2l+1}}
\\&= \sum_{k=0}^\infty \frac{\color{red}{2^{2k+1}}x^{2(k+1)}}{(2(k+1))!}
\\&= \frac 1 2 \sum_{k=0}^\infty \frac{(2x)^{2(k+1)}}{(2(k+1))!}
\\&= \frac 1 2 \sum_{k=1}^\infty \frac{(2x)^{2k}}{(2k)!}
\\&= \frac 1 2 (\cosh{2x} - 1)
\end{align}
$$
So
$$
\cosh^2(x)+\sinh^2(x)=\frac 1 2 (\cosh{2x}+1) + \frac 1 2 (\cosh{2x} - 1)=\cosh(2x)
$$
and
$$
\cosh^2(x)-\sinh^2(x)=\frac 1 2 (\cosh{2x}+1) - \frac 1 2 (\cosh{2x} - 1)=1
$$
</s>