What numerical method can approximately compute the $(n-1)$-dimensional surface area of the $\ell_p$ ball $\{x\in\mathbb R^n: \sum_{i=1}^n |x_i|^p=1\}$, for $p\in[1,\infty)$? Ideally the method should handle $n$ and $p$ both in the range of 5 to 10.

One approach begins with the definition of surface area as $$ \lim_{\varepsilon\to 0^+} \frac{\mu_n(B_p + \varepsilon B_2)-\mu_n(B_p)}{\varepsilon}, $$ where $B_p$ and $B_2$ are unit $\ell_p$ and $\ell_2$ balls, using Monte Carlo to estimate the volume of both bodies. This method fails numerically, because when $\varepsilon\to 0^+$ the two volumes are very close to each other.

Another approach uses Cauchy's integration formula, which states that the volume of $\partial B_p$ is equal to $$ \frac{1}{\mu_{n-1}(B_2)}\int_{S_2}\mu_{n-1}(B _p|u) du, $$ where $\mu_{n-1}(B_p|u)$ is the volume of $B_p$ projecting onto the orthogonal complement of $u$. However, this projection seems difficult to numerically approximate.

What approaches would provide a better approximation?

  • 1
    $\begingroup$ Are you aware of the formula for the volume of a p-ball: ergodicity.net/2010/07/02/… Can this help? $\endgroup$ – Beni Bogosel Feb 27 '18 at 17:43
  • $\begingroup$ Hello Beni! Thanks for your suggestions. Yes I'm aware of the volume formula for p-ball but unfortunately it provides little help into the volume of a p-surface. A more extensive discussion can be found in this question mathoverflow.net/questions/234314/… $\endgroup$ – Yining Wang Feb 27 '18 at 18:27
  • $\begingroup$ What error in the estimation would you find acceptable? $\endgroup$ – Matt F. Mar 27 '18 at 11:44
  • $\begingroup$ Even in the 2d case I don’t see an especially good way to approximate the arclength, although the problem may be clarified by writing it as $$\frac{4}{p} \int_0^1 \sqrt{\frac{1}{u^{2-2/p}}+\frac{1}{(1-u)^{2-2/p}}}du.$$ $\endgroup$ – Matt F. Mar 27 '18 at 15:57
  • $\begingroup$ Maybe this could be of help? arxiv.org/abs/1511.02084 $\endgroup$ – Tomek Kania Mar 31 '18 at 11:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.