# 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 $$\sinh^2 x$$?

• There are well known identities for $\cosh^2(x) + \sinh^2(x) = \cosh(2x)$ and $\cosh^2(x) - \sinh^2(x) = 1$.
– PC1
Nov 3, 2021 at 1:27
• Yes, but the question required the use of only Taylor series expansions Nov 3, 2021 at 1:30
• You can prove using the expansions that $\cosh(x)-\sinh(x)=e^{-x}$ $$\implies \cosh^2(x)-\sinh^2(x)=(\cosh(x)-\sinh(x))(\cosh(x)+\sinh(x))=1$$ Nov 3, 2021 at 1:40
• I see! But then how do we handle the problem of $cosh^2 + sinh^2$? Nov 3, 2021 at 1:43

In order to multiply two power series, say \begin{align*} \def\bl#1{\color{blue}{#1}} \def\gr#1{\color{green}{#1}} \bl{A}(x) &= \bl{a_0} + \bl{a_1}x + \bl{a_2}x^2 + \cdots \\ \gr{B}(x) &= \gr{b_0} + \gr{b_1}x + \gr{b_2}x^2 + \cdots, \end{align*} we have to imagine opening parentheses: \begin{align*} \bl{A}(x) \, \gr{B}(x) &= \bigl( \bl{a_0} + \bl{a_1}x + \bl{a_2}x^2 + \cdots \bigr) \, \bigl( \gr{b_0} + \gr{b_1}x + \gr{b_2}x^2 + \cdots \bigr) \\ &= \bl{a_0} \, \bigl( \gr{b_0} + \gr{b_1}x + \gr{b_2}x^2 + \cdots \bigr) \\ &\quad {}+ \bl{a_1}x \, \bigl( \gr{b_0} + \gr{b_1}x + \gr{b_2}x^2 + \cdots \bigr) \\ &\qquad {}+ \bl{a_2}x^2 \, \bigl( \gr{b_0} + \gr{b_1}x + \gr{b_2}x^2 + \cdots \bigr) \\ &\qquad\quad {}+\cdots \\ &= \bigl( \bl{a_0}\gr{b_0} + \bl{a_0}\gr{b_1}x + \bl{a_0}\gr{b_2}x^2 + \cdots \bigr) \\ &\quad {}+ \bigl( \bl{a_1}\gr{b_0}x + \bl{a_1}\gr{b_1}x^2 + \bl{a_1}\gr{b_2}x^3 + \cdots \bigr) \\ &\qquad {}+ \bigl( \bl{a_2}\gr{b_0}x^2 + \bl{a_2}\gr{b_1}x^3 + \bl{a_2}\gr{b_2}x^4 + \cdots \bigr) \\ &\qquad\quad {}+\cdots \end{align*} Now, we collect like terms: \begin{align*} \bl{A}(x) \, \gr{B}(x) &= \bigl( \bl{a_0}\gr{b_0} \bigr) \\ &\quad {}+ \bigl( \bl{a_0}\gr{b_1} + \bl{a_1}\gr{b_0} \bigr) x \\ &\qquad {}+ \bigl( \bl{a_0}\gr{b_2} + \bl{a_1}\gr{b_1} + \bl{a_2}\gr{b_0} \bigr) x^2 \\ &\qquad\quad {}+\cdots \end{align*} In general, the $$x^n$$ coefficient in $$\bl{A}(x) \gr{B}(x)$$ is $$\bl{a_0}\gr{b_n} + \bl{a_1}\gr{b_{n-1}} + \cdots + \bl{a_{n-1}}\gr{b_1} + \bl{a_n}\gr{b_0},$$ i.e. a sum of terms $$\bl{a_i}\gr{b_j}$$, where $$\bl{i} + \gr{j} = n$$.
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 $$\cosh^2(x)-\sinh^2(x)=\frac 1 2 (\cosh{2x}+1) - \frac 1 2 (\cosh{2x} - 1)=1$$