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Let $E$ be an ellipsoid in $\mathbb{R}^d$ defined by

$$\sum_{i=1}^d \frac{x_i^2}{a_i^2}=1$$

Is there a formula to express the mean width (or an approximation of the mean width) of $E$ in term of the lengths $a_i$ of the (semi-principal) axis?

The width in the direction of the principal axis are $2a_1,\ldots,2a_d$. So I am expecting the mean width to be approximately $2(a_1, \ldots, a_d)/d$.

Rodrigo de Azevedo
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Gilles Bonnet
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1 Answers1

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for the special case $n=2$ the answer is comparatively easy, but not elementary: The mean width of a convex body $K$ in $\mathbb R^2$ is equal to $\frac{\mathrm{perimeter}(K)}{\pi}$.

The perimeter of an ellipse with semi-axes $a_1>a_2$ can be calculated via an elliptic integral of the first kind, precisely $4a_1 E(e)$, where the excentricity $e=\sqrt{1-\frac{{a_2}^2}{{a_1}^2}}$ and

$$E(e)=\int_0^{\frac{\pi}{2}} \sqrt{1-e^2\sin^2 \theta}\ d\theta \; .$$

Elliptic integrals can be approximated by series or approximations be looked up in tables; furthermore there exist some good approximations by 'ordinary' elementary functions.

The answer for $n>2$ ought to be at least as complicated as this.

Yours, Stefan

Stefan Born
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  • It is written in wikipedia that $E(k) = \frac{\pi}{2}\left(1-\left(\frac12\right)^2 \frac{k^2}{1}-\left(\frac{1\cdot 3}{2\cdot 4}\right)^2 \frac{k^4}{3}-\cdots-\left(\frac{(2n-1)!!}{(2n)!!}\right)^2 \frac{k^{2n}}{2n-1}-\cdots\right)$, which means that $$ E(k) \leq \frac{\pi}{2}\left(1-k^2/4\right) $$ But then we have $$ width\leq \frac{4a_1}{\pi}\frac{\pi}{2}\left(1-k^2/4\right)=2a_1(1-(1-\frac{a_2^2}{a_1^2})/4) = 2a_1(\frac{3}{4}+\frac{a_2^2}{4a_1^2})=\frac{3}{2}a_1+\frac{a_2^2}{2a_1} $$ – qwerty43 Jun 05 '23 at 09:39