Using the power series expansions of e^x, sin x, cos x
Using the power series expansions of $e^x$, $\sin x$, $\cos x$, $\ln(1+x)$, $(1+x)^a$, expand $\dfrac{x\sin\alpha}{1-2x\cos\alpha+x^2}$ in a power series in $x$ and indicate when the expansion is valid.
Explanation of Method
This is the imaginary-part companion to Problem 262: the same complex generating function $g(x)=xe^{i\alpha}/(1-xe^{i\alpha})$, rationalized over the real denominator, splits into a real part (Problem 262's cosine series) and this imaginary part (a sine series).
- Form the complex geometric series $g(x)=xe^{i\alpha}/(1-xe^{i\alpha})$.
- Rationalize and identify the imaginary part.
- Verify with a special case and determine the region of convergence.
Step 1 · Complex generating series
For $|x|<1$: $$ g(x) = \frac{xe^{i\alpha}}{1-xe^{i\alpha}} = \sum_{n=1}^{\infty}x^ne^{in\alpha}. $$
Step 2 · Rationalize and take the imaginary part
As in Problem 262, rationalizing gives $$ g(x) = \frac{(x\cos\alpha-x^2)+i\,x\sin\alpha}{1-2x\cos\alpha+x^2}. $$ Comparing imaginary parts: $$ \frac{x\sin\alpha}{1-2x\cos\alpha+x^2} = \operatorname{Im}[g(x)] = \sum_{n=1}^{\infty}\sin(n\alpha)\,x^n. $$
Step 3 · Verify with a special case
At $\alpha=\frac\pi2$: $\sin\left(n\cdot\frac\pi2\right)$ cycles through $1,0,-1,0,\ldots$, and the left side becomes $\dfrac x{1+x^2}$. The series gives $$ \frac{x}{1+x^2} = x-x^3+x^5-x^7+\cdots, $$ which is exactly the geometric series $x\sum(-x^2)^n$ — a direct, elementary confirmation of the formula.
Summary Table
| Property | Value |
|---|---|
| Series | $\sum_{n=1}^{\infty}\sin(n\alpha)x^n$ |
| Singularities | $x=e^{\pm i\alpha}$ |
| Region of validity | $|x|<1$ |
Additional Notes
- Classical generating function for $\sin(n\alpha)$, fundamental to Fourier analysis and the theory of the Poisson kernel for the unit disk.
- Paired with Problem 262's cosine series, the two together reconstruct the full complex series: $\dfrac{1-xe^{i\alpha}\cdot(-1)}{\ldots}$, i.e. $g(x)=\sum\cos(n\alpha)x^n+i\sum\sin(n\alpha)x^n$.
- The special case $\alpha=\pi/2$ collapses the formula to the elementary geometric series for $x/(1+x^2)$, providing a quick sanity check independent of the complex-exponential derivation.