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 $f(x)=\dfrac{\alpha\cos x-\alpha^2}{1-2\alpha\cos x+\alpha^2}$, $|\alpha|>1$.
Explanation of Method
Since $|\alpha|>1$, the geometric series in $\alpha$ used in Problems 346–347 does not converge; instead substitute $\beta=1/\alpha$ (so $|\beta|<1$) and rewrite $f$ in terms of $\beta$. Factoring $\alpha^2$ out of numerator and denominator reduces $f$ to $-(1-\beta\cos x)/(1-2\beta\cos x+\beta^2)$, which is evaluated directly by splitting $1-\beta\cos x$ into a symmetric half-sum of $(1-\beta e^{\pm ix})$ and applying the geometric series to each factor separately.
- Substitute $\beta=1/\alpha$ and rewrite $f$ as $-(1-\beta\cos x)/(1-2\beta\cos x+\beta^2)$.
- Split $1-\beta\cos x$ symmetrically and expand each geometric factor.
- Combine into a single cosine series and verify numerically.
Step 1 · Substitution
Let $\beta=1/\alpha$, so $|\beta|<1$. Dividing numerator and denominator of $f$ by $\alpha^2$: $$ f(x) = \frac{\beta\cos x-1}{\beta^2-2\beta\cos x+1} = -\frac{1-\beta\cos x}{1-2\beta\cos x+\beta^2}. $$
Step 2 · Symmetric splitting
Since $\beta\cos x=\dfrac\beta2(e^{ix}+e^{-ix})$: $$ 1-\beta\cos x = \frac12\bigl[(1-\beta e^{ix})+(1-\beta e^{-ix})\bigr]. $$ Dividing by $(1-\beta e^{ix})(1-\beta e^{-ix})=1-2\beta\cos x+\beta^2$: $$ \frac{1-\beta\cos x}{1-2\beta\cos x+\beta^2} = \frac12\left[\frac1{1-\beta e^{-ix}}+\frac1{1-\beta e^{ix}}\right] = \frac12\left[\sum_{n\ge0}\beta^ne^{-inx}+\sum_{n\ge0}\beta^ne^{inx}\right] = \sum_{n=0}^\infty\beta^n\cos(nx). $$
Step 3 · Fourier series
$$ f(x) = -\sum_{n=0}^\infty\beta^n\cos(nx) = -1-\sum_{n=1}^\infty\beta^n\cos(nx). $$ $$ \boxed{\frac{\alpha\cos x-\alpha^2}{1-2\alpha\cos x+\alpha^2} = -1-\sum_{n=1}^\infty\frac{\cos(nx)}{\alpha^n}} $$ (Verified numerically at $\alpha=2.5$: $f(0.5)\approx-1.41716$, $f(1.5)\approx-0.88064$, $f(2.5)\approx-0.73321$, matching this series exactly. An earlier attempt with an extra factor of $2$ in front of the sum — i.e. $-1-2\sum\cos(nx)/\alpha^n$ — was checked and found numerically incorrect; the corrected coefficient has no factor of $2$, since the symmetric split in Step 2 already accounts for both halves.)
Summary Table
| Property | Value |
|---|---|
| Fourier series | $f(x)=-1-\sum\dfrac{\cos(nx)}{\alpha^n}$ |
| Fourier coefficients | $a_0=-2$, $a_n=-1/\alpha^n$, $b_n=0$ |
Additional Notes
- The substitution $\beta=1/\alpha$ is the standard device for extending a series valid inside the unit circle to the region outside it, exactly analogous to a Laurent-series expansion in complex analysis.
- Comparing with Problem 347's kernel $1+2\sum\alpha^n\cos(nx)$ (valid for $|\alpha|<1$), this result for $|\alpha|>1$ has coefficients half the magnitude and an extra additive constant of $-1$ — reflecting that $f(x)$ here is a genuinely different function from the Poisson kernel, not merely its analytic continuation.
- Since $|\beta|<1$, the series converges absolutely and geometrically fast, just like the $|\alpha|<1$ series in Problems 346–347.