Simplifying $\cosh x + \sinh x$, $\cosh^2 x + \sinh^2 x$, $\cosh^2 x - \sinh^2 x$ using only the Taylor Series of $\cosh,\sinh$

Question

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 $\sinh^2 x$?

Answer 1

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 $\bl{i} + \gr{j} = n$.

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Can you see how to use these observations to square the series for the hyperbolic trig. functions? There are some rather straightforward patterns that you should recognize before having to compute too many coefficients. Try it!

Answer 2

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 $$ \cosh^2(x)-\sinh^2(x)=\frac 1 2 (\cosh{2x}+1) - \frac 1 2 (\cosh{2x} - 1)=1 $$