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 $\dfrac{x\cos\alpha-x^2}{1-2x\cos\alpha+x^2}$ in a power series in $x$ and indicate when the expansion is valid.
Explanation of Method
The denominator factors over $\mathbb C$ as $(1-xe^{i\alpha})(1-xe^{-i\alpha})$, so this whole family of Poisson-kernel-type rational functions is best handled by writing a single complex geometric series and reading off its real and imaginary parts.
- Form the complex geometric series $g(x)=xe^{i\alpha}/(1-xe^{i\alpha})$.
- Rationalize $g(x)$ over the real denominator $1-2x\cos\alpha+x^2$ and identify its real part.
- Match the real part to the given numerator and determine the region of convergence.
Step 1 · A complex generating series
For $|x|<1$, the geometric series starting at $n=1$ gives $$ g(x) = \frac{xe^{i\alpha}}{1-xe^{i\alpha}} = \sum_{n=1}^{\infty}x^ne^{in\alpha} = \sum_{n=1}^{\infty}x^n\cos(n\alpha) + i\sum_{n=1}^{\infty}x^n\sin(n\alpha). $$
Step 2 · Rationalize the denominator
Multiplying numerator and denominator of $g(x)$ by $1-xe^{-i\alpha}$: $$ g(x) = \frac{xe^{i\alpha}\bigl(1-xe^{-i\alpha}\bigr)}{(1-xe^{i\alpha})(1-xe^{-i\alpha})} = \frac{xe^{i\alpha}-x^2}{1-2x\cos\alpha+x^2}. $$ The numerator expands as $$ xe^{i\alpha}-x^2 = x(\cos\alpha+i\sin\alpha)-x^2 = (x\cos\alpha-x^2)+i\,(x\sin\alpha). $$
Step 3 · Identify the real part
Comparing real parts of the two expressions for $g(x)$: $$ \frac{x\cos\alpha-x^2}{1-2x\cos\alpha+x^2} = \operatorname{Re}\bigl[g(x)\bigr] = \sum_{n=1}^{\infty}\cos(n\alpha)\,x^n. $$ (This is the numerical check: at $\alpha=60^\circ$, $x=0.1$, direct substitution gives $f(0.1)\approx0.04396$, matching $\sum\cos(n\cdot60^\circ)(0.1)^n\approx0.04396$ — confirming the cosine form, not a sine form.)
Summary Table
| Property | Value |
|---|---|
| Series | $\sum_{n=1}^{\infty}\cos(n\alpha)x^n$ |
| Singularities | $x=e^{\pm i\alpha}$ (on unit circle) |
| Region of validity | $|x|<1$ |
Additional Notes
- This is the companion Poisson-kernel formula to Problem 263: that problem's numerator $x\sin\alpha$ picks out $\operatorname{Im}[g(x)]=\sum\sin(n\alpha)x^n$, while this problem's numerator $x\cos\alpha-x^2$ picks out $\operatorname{Re}[g(x)]=\sum\cos(n\alpha)x^n$ — both come from the same complex generating function $g(x)=xe^{i\alpha}/(1-xe^{i\alpha})$.
- Unlike the related identity $\frac{1-x\cos\alpha}{1-2x\cos\alpha+x^2}=\sum_{n=0}^\infty\cos(n\alpha)x^n$ (which includes the $n=0$ term, $\cos0=1$), this problem's series starts at $n=1$ because the numerator here vanishes to first order in $x$ at the origin.
- The singularities $x=e^{\pm i\alpha}$ lie exactly on the unit circle for every real $\alpha$, so the radius of convergence is always exactly $1$ regardless of the parameter's value.