By expanding in a power series in alpha, find the Fourier
By expanding in a power series in $\alpha$, find the Fourier series (in $x$) of $f(x)=\dfrac{1-\alpha^2}{1-2\alpha\cos x+\alpha^2}$, $|\alpha|<1$.
Explanation of Method
This is the classical Poisson kernel of harmonic analysis: the real part of $\dfrac{1+\alpha e^{ix}}{1-\alpha e^{ix}}$. Expanding via geometric series in $\alpha e^{ix}$ and $\alpha e^{-ix}$ separately and combining term by term recovers a pure cosine series with geometric coefficients.
- Expand $1/[(1-\alpha e^{ix})(1-\alpha e^{-ix})]$ as a double geometric series and collect by frequency.
- Multiply by $(1-\alpha^2)$ to obtain the Poisson kernel.
- Verify at the endpoints $x=0,\pi$ against the closed form.
Step 1 · Double geometric expansion
Using $\dfrac1{1-\alpha e^{\pm ix}}=\sum_{n\ge0}\alpha^ne^{\pm inx}$, and collecting terms by frequency $k=m-n$ (each nonzero $k$ receiving a factor $\alpha^{|k|}/(1-\alpha^2)$ from summing the geometric tail): $$ \frac1{1-2\alpha\cos x+\alpha^2} = \frac1{1-\alpha^2}\left[1+2\sum_{n=1}^\infty\alpha^n\cos(nx)\right]. $$
Step 2 · The Poisson kernel
Multiplying both sides by $(1-\alpha^2)$: $$ \boxed{\frac{1-\alpha^2}{1-2\alpha\cos x+\alpha^2} = 1+2\sum_{n=1}^\infty\alpha^n\cos(nx)} $$
Step 3 · Verification
At $x=0$: LHS $=\dfrac{1-\alpha^2}{(1-\alpha)^2}=\dfrac{1+\alpha}{1-\alpha}$; RHS $=1+2\dfrac\alpha{1-\alpha}=\dfrac{1+\alpha}{1-\alpha}$. ✓ At $x=\pi$: LHS $=\dfrac{1-\alpha}{1+\alpha}$; RHS $=1-2\dfrac\alpha{1+\alpha}=\dfrac{1-\alpha}{1+\alpha}$. ✓ (Also verified numerically at $\alpha=0.4,x=1.2$: both sides give $\approx0.96539$.)
Summary Table
| Property | Value |
|---|---|
| Fourier series | $f(x)=1+2\sum\alpha^n\cos(nx)$ |
| Fourier coefficients | $a_0=2$, $a_n=2\alpha^n$, $b_n=0$ |
Additional Notes
- This is precisely the real part of the complex generating function whose imaginary part produced Problem 346's sine kernel — the two problems are the cosine/sine halves of the same identity $\frac{1+\alpha e^{ix}}{1-\alpha e^{ix}}=1+2\sum\alpha^ne^{inx}$.
- As $\alpha\to1^-$, the kernel concentrates all its mass near $x=0$ (approximating a Dirac delta), the standard mechanism by which the Poisson kernel solves the Dirichlet problem for harmonic functions on the disk.
- Problem 348 treats the complementary regime $|\alpha|>1$: substituting $\beta=1/\alpha$ reduces that case back to this one, but the resulting series has a different overall structure (see that problem's notes).