TITLE: Limit of geometric mean of N radii of an ellipse
QUESTION [6 upvotes]: Is this equation correct?
$$\lim_{N \to \infty} \prod_{n=1}^N (a^2\cos^2 (2\pi n/N)+b^2\sin^2(2\pi n/N))^{1/N}=ab$$
If so, why?

REPLY [6 votes]: This is not an answer, but simply a verification of the integral in Srivatsan's answer.
Using $\cos^2(y)=\frac{1+\cos(2y)}{2}$ and $\sin^2(y)=\frac{1-\cos(2y)}{2}$,
$$
\frac{1}{2\pi}\int_0^{2\pi}\log\left(a^2\cos^2(y)+b^2\sin^2(y)\right)\mathrm{d}y\tag{1}
$$
becomes
$$
\log\left(\frac{a^2+b^2}{2}\right)+\frac{1}{2\pi}\int_0^{2\pi}\log\left(1+\frac{a^2-b^2}{a^2+b^2}\cos(2y)\right)\;\mathrm{d}y\tag{2}
$$
Letting $x=\frac{a^2-b^2}{a^2+b^2}$ and substituting $y\mapsto y/2$, the integral in $(2)$ becomes
$$
\frac{1}{2\pi}\int_0^{2\pi}\log\left(1+x\cos(y)\right)\;\mathrm{d}y\tag{3}
$$
Taking the derivative of $(3)$ with respect to $x$ yields
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}x}\frac{1}{2\pi}\int_0^{2\pi}\log\left(1+x\cos(y)\right)\;\mathrm{d}y
&=\frac{1}{2\pi}\int_0^{2\pi}\frac{\cos(y)}{1+x\cos(y)}\mathrm{d}y\\
&=\frac{1}{2\pi}\int_{-\infty}^\infty\frac{\frac{1-z^2}{1+z^2}}{1+x\frac{1-z^2}{1+z^2}}\frac{2\mathrm{d}z}{1+z^2}\\
&=\frac{1}{\pi(1+x)}\int_{-\infty}^\infty\frac{(1-z^2)\;\mathrm{d}z}{\left(1+\frac{1-x}{1+x}z^2\right)(1+z^2)}\\
&=\frac{1}{\pi}\int_{-\infty}^\infty\left(\frac{\frac{1}{x}}{1+z^2}-\frac{\frac{1}{x(1+x)}}{1+\frac{1-x}{1+x}z^2}\right)\mathrm{d}z\\
&=\frac{1}{x}-\frac{1}{x(1+x)}\sqrt{\frac{1+x}{1-x}}\\
&=\frac{1}{x}-\frac{1}{x\sqrt{1-x^2}}\tag{4}
\end{align}
$$
where $z=\tan(y/2)$.
Integrating $(4)$ gives
$$
\frac{1}{2\pi}\int_0^{2\pi}\log\left(1+x\cos(y)\right)\;\mathrm{d}y=\log\left(\frac{1+\sqrt{1-x^2}}{2}\right)\tag{5}
$$
Substituting $x=\frac{a^2-b^2}{a^2+b^2}$ into $(5)$ and $(5)$ into $(2)$ yields
$$
\frac{1}{2\pi}\int_0^{2\pi}\log\left(a^2\cos^2(y)+b^2\sin^2(y)\right)\mathrm{d}y=2\log\left(\frac{a+b}{2}\right)\tag{6}
$$