# Limit of geometric mean of N radii of an ellipse

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?

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This basically is the average square radius of an ellipse with parameters $a,b$, picked uniformly with respect to angle from the center. It wouldn't make sense for this to scale linearly with $a$ and $b$ as the expression $ab$ does. – anon Nov 23 '11 at 19:31
This is the geometric mean of the radii squared. Is this what you want? – wnoise Nov 23 '11 at 19:55
Yes, it's the geometric mean of the radii squared that I want. The reason I asked it is because of another question I asked a while back here: mathoverflow.net/questions/29534/… It turns out that this expression is related to Fermat's Last Theorem. I was curious what happens when you take the limit. – Craig Feinstein Nov 24 '11 at 1:59

The answer has been substantially revised. In particular, the previous revision quoted a wrong value for the integral $I(a,b)$, which resulted in an overall incorrect answer. I hope that this revision is correct. :-)

The limit is $\Big(\frac{a+b}{2} \Big)^2$ and is not $ab$ (as claimed). I break down the solution into multiple steps for ease of understanding.

(1) A Riemann sum and integral. We first convert the product to a sum by taking logs: $$\frac{1}{N} \sum_{n=1}^N \ln \left( a^2 \cos^2 (2\pi n/N) + b^2 \sin^2 (2\pi n/N) \right).$$ This is the Riemann sum of the function $$f: [0,1] \to \mathbb R : x \mapsto \ln \left( a^2 \cos^2 (2\pi x) + b^2 \sin^2 (2\pi x) \right) ,$$ corresponding to the uniform partition of $[0,1]$ into $N$ parts. Since $f$ is integrable (being a continuous function over a compact interval), as $N \to \infty$, this sum tends to

$$\int_0^1 \ln \left( a^2 \cos^2 (2\pi x) + b^2 \sin^2 (2\pi x) \right) \ dx = \frac{1}{2 \pi} \int_0^{2\pi} \ln \left( a^2 \cos^2 y + b^2 \sin^2 y \right) \ dy .$$ Call this integral $I(a,b)$.

(2) Evaluating $I(a,b)$. We use the idea of differentiating under the integral sign. \begin{align*} \frac{\partial }{\partial a} I(a,b) &= \frac{1}{2 \pi} \int_0^{2\pi} \frac{\partial}{\partial a} \ln \left( a^2 \cos^2 y + b^2 \sin^2 y \right) \ dy \\ &= \frac{1}{2 \pi} \int_0^{2\pi} \frac{2a \cos^2y}{a^2 \cos^2 y + b^2 \sin^2 y } \ dy . \\ &= \frac{a}{\pi} \int_0^{2\pi} g(y) \ dy , \end{align*} where $g(y) = \frac{\cos^2y}{a^2 \cos^2 y + b^2 \sin^2 y }$. Since $g(y) = g(\pi + y)$, we can write this integral as $\frac{2a}{\pi} \int_0^{\pi} g(y) \ dy$. Similarly, since $g(y) = g(\pi - y)$, we can further simplify it to $\frac{4a}{\pi} \int_0^{\pi/2} g(y) \ dy$.

(3) Evaluating $\int_0^{\pi/2} g(y)dy$. Using Wolfram|Alpha, we can find the indefinite integral $$\int g(y) \ dy = \int \frac{\cos^2 y}{ a^2 \cos^2 y + b^2 \sin^2 y} \ dx = \frac{ay - b \ \arctan \Bigl( \frac{b \tan y}{a} \Bigr)}{a(a^2 - b^2)} \color{\Grey}{+ \mathrm{const}}.$$ By plugging in the limits $0$ and $\pi/2$, we get $$\int_0^{\pi/2} g(y) \ dy = \frac{ \left. ay - b \; \arctan \Bigl( \frac{b \tan y}{a} \Bigr) \right|_{0}^{\pi/2}}{a(a^2 - b^2)} = \frac{(a-b) \frac{\pi}{2}}{a(a^2 - b^2)} = \frac{\pi}{2a(a+b)}.$$ Therefore, $\frac{4a}{\pi} \int_0^{\pi/2} g(y) \ dy = \frac{2}{a+b}$.

(4) Back to $I(a,b)$. We have $\frac{\partial}{\partial a} I(a,b) = \frac{2}{a+b}$. Similarly, $\frac{\partial}{\partial b} I(a,b) = \frac{2}{a+b}$. Combining these two observations, we conclude $$I(a,b) = 2 \ln (a+b) + \mathrm{const}.$$ When $a=b=1$, the integral is $0$, which gives the constant to be $- 2\ln 2$. Therefore

$$I(a,b) = 2\ln \Bigl( \frac{a+b}{2} \Bigr).$$

(5) Final answer. To get the final answer, we only need to exponentiating this answer. That is, the limit mentioned in the question is equal to $\Bigl(\frac{a+b}{2} \Bigr)^2$.

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Thanks. (i) Right, that's $n$ in the index. (ii) As you say, the final answer is NO. I have now added that to the beginning of the answer. – Srivatsan Nov 23 '11 at 19:21
Thanks. It turns out that the value I quoted for the integral in the previous revision was wrong. Hopefully that's fixed now. :) – Srivatsan Nov 23 '11 at 20:20
Very interesting! So the geometric mean of the radii of an ellipse when the angles are equally spaced approaches the arithmetic mean of the minimum and maximum radii! – Craig Feinstein Nov 23 '11 at 22:47
I believe your answer is now correct. I have posted a verification. – robjohn Nov 23 '11 at 23:47
Both the answers of @Srivatsan and robjohn are very helpful and interesting, thanks a lot! Could you please provide any reference (book, journal article, paper, letter, ..) about the proofs you made? I would be interested to use this result to a paper, but a link to a forum wouldn't be acceptable.. I really hope there is something out there in the literature.. Sorry if this is not the correct place to ask such a question, but I'm new to this forum. – user54331 Dec 26 '12 at 22:25

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 $$\frac{1}{2\pi}\int_0^{2\pi}\log\left(a^2\cos^2(y)+b^2\sin^2(y)\right)\mathrm{d}y=2\log\left(\frac{a+b}{2}\right)\tag{6}$$

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I should point out that I left out a bit of explanation in getting to $(3)$. When substituting $y\mapsto y/2$, the integral gets a factor of $1/2$ and the range of integration becomes $[0,4\pi]$. Since the integral is the same on $[0,2\pi]$ and $[2\pi,4\pi]$, the factor of $1/2$ was used to remove the integral on $[2\pi,4\pi]$. – robjohn Nov 23 '11 at 23:29