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Solution (continued)

On May 4th, 2008 JSteinhardt says:

We see that the eigenvalue of the eigenfunction $f_{(r,s)}$ is given by

$\lambda_{(r,s)} = \int_{-\pi}^{\pi} \sum_{k=0}^{\infty} a^{-k} e^{i(2k+1)(r\cos(t)+s\sin(t))} dt = \int_{-\pi}^{\pi} \sum_{k=0}^{\infty} a^{-k} e^{i(2k+1)\sqrt{r^2+s^2}\cos(t+\phi)} dt$

for an appropriately chosen $\phi$ . Thus we need only actually consider $\lambda_{(r,0)}$ , which we from now on denote $\lambda(r)$ . Then some calculation gives us that the integral is

$\int_{-\pi}^{\pi} \frac{a(a-1)\cos(r\cos(t))}{(a-1)^2+4a\sin^2(r\cos(t))} dt$

We can show that (and this will suffice) that when $a$ is near $1$ ,

$\int_{0}^{\frac{\pi}{2}} \frac{(a-1)\cos(r\cos(t))}{(a-1)^2+4a\sin^2(r\cos(t))} dt \geq -4(a-1)^{-\frac{3}{4}}-\frac{\pi}{2}$

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Solution (continued)

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