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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.2011-11-23
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    This is the geometric mean of the radii *squared*. Is this what you want?2011-11-23
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    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: http://mathoverflow.net/questions/29534/the-product-of-n-radii-in-an-ellipse It turns out that this expression is related to Fermat's Last Theorem. I was curious what happens when you take the limit.2011-11-24

2 Answers 2

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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.2011-11-23
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    Thanks. It turns out that the value I quoted for the integral in the previous revision was wrong. Hopefully that's fixed now. :)2011-11-23
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    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!2011-11-23
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    I believe your answer is now correct. I have posted [a verification](http://math.stackexchange.com/q/85078/).2011-11-23
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    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.2012-12-26
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    @user54331 Math SE is no mere forum ;o) We can oblige with a citation in either BibTeX or amsrefs style, whichever your preference may be! Look up, at the end of each answer, and note on the left, where it says "share edit flag". Click "share". There will be a pop-up, with a shortened URL, inset in a rectangular blue-gray colored field. On the lower left corner of that blue-gray field, it says "cite". Click on that, and choose your preferred citation style. This feature is available for every Math SE answer.2013-01-04
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    @user54331 Not exactly sure what you want, but my answer, together with robjohn's, provides an essentially complete solution. Of course, I am assuming a little background in integrating standard functions, and the theory of Riemann integration. These are typically covered in an intro course in analysis; consult your favourite text book.2013-01-11
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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 $[2\pi,4\pi]$.2011-11-23