Expand cosh x in a power series in x, using known
Expand $\cosh x$ in a power series in $x$, using known expansions, and indicate when it is valid.
Explanation of Method
Since $\cosh x=\frac{e^x+e^{-x}}2$, adding the Maclaurin series for $e^x$ and $e^{-x}$ cancels every odd-power term and doubles the even-power terms; dividing by $2$ leaves a series with only even powers.
- Write the Maclaurin series for $e^x$ and $e^{-x}$.
- Add to cancel odd powers and double even powers.
- Divide by $2$ and note the region of convergence.
Step 1 · Series for $e^x$ and $e^{-x}$
$$ e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!}, \qquad e^{-x} = \sum_{n=0}^{\infty}\frac{(-x)^n}{n!}. $$
Step 2 · Add
Odd-power terms cancel; even-power terms double: $$ e^x+e^{-x} = 2\sum_{n=0}^{\infty}\frac{x^{2n}}{(2n)!}. $$
Step 3 · Divide by $2$
$$ \cosh x = \frac{e^x+e^{-x}}2 = \sum_{n=0}^{\infty}\frac{x^{2n}}{(2n)!} = 1+\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots $$ Since $e^x$ and $e^{-x}$ both have infinite radius of convergence, so does $\cosh x$: valid for all $x\in\mathbb R$.
Summary Table
| Property | Value |
|---|---|
| Series | $\sum x^{2n}/(2n)!$ |
| $\cosh0$ | $1$ |
| Region of validity | $\mathbb R$ |
Additional Notes
- Comparing with $\cos x=\sum(-1)^nx^{2n}/(2n)!$: $\cosh x$ has the same terms with all signs made positive — the two are related by $\cosh(ix)=\cos x$.
- Unlike $\sinh0=0$ (Problem 252), $\cosh0=1$ is the nonzero starting value, since $\cosh x$ is the even part of $e^x$ and picks up the constant term directly.
- Together, $\sinh x+\cosh x=e^x$ exactly reconstructs the exponential series, confirming that these two expansions are precisely the odd and even halves of Problem 249's Maclaurin series for $e^x$.