Expand e^xsin x in a power series in x, using known
Expand $e^x\sin x$ in a power series in $x$, using known expansions, and indicate when it is valid.
Explanation of Method
Rather than laboriously multiplying the two Maclaurin series term by term, writing $\sin x=\mathrm{Im}(e^{ix})$ turns the product into $\mathrm{Im}\bigl(e^{x(1+i)}\bigr)$, a single exponential series whose coefficients can be extracted in closed form using the polar form of $1+i$.
- Rewrite $e^x\sin x=\mathrm{Im}\bigl(e^{x(1+i)}\bigr)$.
- Expand the complex exponential and take the imaginary part using the polar form $1+i=\sqrt2\,e^{i\pi/4}$.
- Verify the first few coefficients directly.
Step 1 · Complex exponential rewriting
Since $\sin x=\mathrm{Im}(e^{ix})$: $$ e^x\sin x = \mathrm{Im}\bigl(e^x\cdot e^{ix}\bigr) = \mathrm{Im}\bigl(e^{x(1+i)}\bigr). $$
Step 2 · Expand and extract the imaginary part
$$ e^{x(1+i)} = \sum_{n=0}^{\infty}\frac{(1+i)^nx^n}{n!}. $$ Using $1+i=\sqrt2\,e^{i\pi/4}$, so $(1+i)^n=2^{n/2}e^{in\pi/4}$, and $\mathrm{Im}\bigl[(1+i)^n\bigr]=2^{n/2}\sin\!\left(\dfrac{n\pi}4\right)$: $$ e^x\sin x = \sum_{n=0}^{\infty}\frac{2^{n/2}\sin(n\pi/4)}{n!}\,x^n. $$
Step 3 · Verify the first few terms
$n=1$: $\sqrt2\sin(\pi/4)/1!=\sqrt2\cdot\frac{\sqrt2}2=1$. $n=2$: $2\sin(\pi/2)/2!=2/2=1$. $n=3$: $2\sqrt2\sin(3\pi/4)/3!=2\sqrt2\cdot\frac{\sqrt2}2/6=2/6=\frac13$. $n=4$: $4\sin(\pi)/4!=0$. $n=5$: $4\sqrt2\sin(5\pi/4)/5!=4\sqrt2\cdot(-\frac{\sqrt2}2)/120=-\frac4{120}=-\frac1{30}$. So $$ e^x\sin x = x+x^2+\frac{x^3}3-\frac{x^5}{30}-\frac{x^6}{90}-\cdots $$ Since $e^x$ is entire, this series converges for all $x\in\mathbb R$.
Summary Table
| Property | Value |
|---|---|
| Closed-form coefficient of $x^n$ | $2^{n/2}\sin(n\pi/4)/n!$ |
| First few terms | $x+x^2+\frac{x^3}3-\frac{x^5}{30}-\cdots$ |
| Region of validity | $\mathbb R$ |
Additional Notes
- The complex-exponential shortcut used here is far less error-prone than a direct Cauchy product of the two Maclaurin series: instead of tracking a growing number of cross terms, a single polar-form computation yields every coefficient in closed form at once.
- The coefficients repeat with period $8$ in sign pattern (since $\sin(n\pi/4)$ has period $8$ in $n$) — the sequence $\sin(n\pi/4)$ for $n=0,1,2,\ldots$ cycles through $0,\frac{\sqrt2}2,1,\frac{\sqrt2}2,0,-\frac{\sqrt2}2,-1,-\frac{\sqrt2}2$ before repeating.
- This same polar-form technique ($1\pm i=\sqrt2\,e^{\pm i\pi/4}$) would work equally well for $e^x\cos x=\mathrm{Re}(e^{x(1+i)})$, giving the companion series $\sum2^{n/2}\cos(n\pi/4)x^n/n!$.