Expand sinh x in a power series in x, using known
Expand $\sinh x$ in a power series in $x$, using known expansions, and indicate when it is valid.
Explanation of Method
Since $\sinh x=\frac{e^x-e^{-x}}2$, substituting the Maclaurin series for $e^x$ and $e^{-x}$ and subtracting cancels every even-power term, doubling the odd-power terms; dividing by $2$ leaves a clean series with only odd powers.
- Write the Maclaurin series for $e^x$ and $e^{-x}$.
- Subtract to cancel even powers and double odd 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 · Subtract
Even-power terms cancel; odd-power terms double: $$ e^x-e^{-x} = 2\sum_{n=0}^{\infty}\frac{x^{2n+1}}{(2n+1)!}. $$
Step 3 · Divide by $2$
$$ \sinh x = \frac{e^x-e^{-x}}2 = \sum_{n=0}^{\infty}\frac{x^{2n+1}}{(2n+1)!} = x+\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots $$ Since $e^x$ and $e^{-x}$ both have infinite radius of convergence, so does $\sinh x$: valid for all $x\in\mathbb R$.
Summary Table
| Property | Value |
|---|---|
| Series | $\sum x^{2n+1}/(2n+1)!$ |
| Region of validity | $\mathbb R$ |
Additional Notes
- Comparing term-by-term with $\sin x=\sum(-1)^nx^{2n+1}/(2n+1)!$: $\sinh x$ has exactly the same terms but with all signs made positive — the two are related by $\sinh(ix)=i\sin x$.
- All coefficients being positive (rather than alternating) means $\sinh x$ grows monotonically and without oscillation, unlike $\sin x$, matching its geometric interpretation as the odd half of exponential growth.
- This is the direct companion to Problem 253's $\cosh x$: together they split $e^x$ into its odd and even parts exactly as $\sinh x+\cosh x=e^x$.