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This is a silly question that came to mind after watching numberphile video on How Pi was nearly changed to 3.2. For which $p\in(1,+\infty)$ is the ratio of the perimeter of the $L^p$ disc in $\Bbb R^2$ over its diameter equal to $3.2$?

EDIT 1 As suggested by @Maesumi's comment, for which value of $p$ is the area over the radius squared equal to $3.2$?

EDIT 2 As suggested by @DanielFischer's comment, what is the perimeter of the $L^p$ ball in the $L^q$ metric?

On a more serious note (?), are there any values of $p$ other than $1,2,\infty$ for which the perimeter of the $L^p$ ball in the plane is known in terms of other ``well-known mathematical constants''?

Olivier Bégassat
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  • I guess it just becomes a numerical problem. And it is not clear why should one use perimeter over diameter and not area over square of radius. Unless there is a "natural" definition of $\pi$ with respect to a metric. – Maesumi Nov 21 '13 at 19:49
  • @Maesumi Yes, it is a numerical problem! this whole question is somewhat tongue in cheek. What I would like to know though is wether there are $L^p$ balls of which we can calculate say their area, or their perimeter. I'll actually integrate your remark into my question. – Olivier Bégassat Nov 21 '13 at 19:53
  • By calculate, do you mean analytically as opposed to numerically? – RussH Nov 21 '13 at 20:01
  • @RussH With respect to the are and perimeter, yes, analytical expressions, if such things exist (Will Jagy tells us the area has an analytical expression.) – Olivier Bégassat Nov 21 '13 at 20:12
  • Just a note. To be consistent, the length of the perimeter should be determined using $$\sup \left\lbrace\sum_{k=1}^n \lVert x(t_k) - x(t_{k-1})\rVert_p : a = t_0 < t_1 < \dotsb < t_n = b \right\rbrace$$ for a parametrisation $x \colon [a,b] \to \mathbb{R}^2$ of the curve. No fair using the Euclidean metric for that. – Daniel Fischer Nov 21 '13 at 20:41
  • @DanielFischer That's an interesting twist :D – Olivier Bégassat Nov 21 '13 at 20:42
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    Then we have $\pi_1 = \pi_\infty = 4$. I would be interested if we always have $\pi_p = \pi_q$ for conjugate exponents $p,q$. – Daniel Fischer Nov 21 '13 at 20:44
  • @DanielFischer are you sure about $\pi_1$? Methinks it's $4\sqrt{2}$... But there might nonetheless be a rapport between the two perimeters. – Olivier Bégassat Nov 21 '13 at 20:47
  • $\lVert (1,0) - (0,1)\rVert_1 = 2$, so we have the perimeter $8$ computed in the $\ell^1$ norm. Divide by the diameter $2$, you get $\pi_1 = 4$. – Daniel Fischer Nov 21 '13 at 20:49
  • Oh, you were using the ''internal'' length. OK. – Olivier Bégassat Nov 21 '13 at 20:51
  • I am 100% sure that a very close question was asked before. EDIT: Found it. http://math.stackexchange.com/questions/254620/pi-in-arbitrary-metric-spaces – Asaf Karagila Nov 21 '13 at 22:47
  • @AsafKaragila you are correct and I just saw your comment right after posting my answer below. – Fixed Point Nov 22 '13 at 07:00
  • @DanielFischer Yes, looking at my answer below and then following the link, the $\pi_p=\pi_q$ if $p$ and $q$ are conjugates. In addition for $\pi=\pi_2=3.141592...$ is also the unique minimum. – Fixed Point Nov 22 '13 at 07:01

4 Answers4

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The area is known, $$ \frac{4 \Gamma \left( 1 + \frac{1}{p} \right)^2}{\Gamma \left( 1 + \frac{2}{p} \right)} $$

The method is due to Dirichlet (1839) but they appear to have blocked out the relevant pages in that article. Alright, found it ELSEWHERE, page 389. A little hard to read; that is the sort of thing that happens if you scan something and use the "despeckle" option, because some aspects of letters and symbols are of similar size to the speckles. My understanding is that this is also in Whittaker and Watson.

EEDDDDIITTT: I prefer the curves related to $$ \color{magenta}{ x^4 + x^2 y^2 + y^4 \leq 1}, $$ as $$ { x^4 + A x^2 y^2 + y^4 \leq 1} $$ with real $A.$ Mostly it is because these are real analytic. With $A=2$ we have the circle. With $A=0$ we have the $L^4$ ball. For $A>0$ we have nonzero curvature at $(0,1),$ implicit differentiation twice gives $y''(0) = -A/2.$ By the time we reach $A=14$ the curve is no longer convex. I think there is a way to rotate by $45^\circ$ and scale that will show the $A$ corresponding to $A=0,$ which is the convexity boundary in the other direction.

EEddIItteeDDiiTT: yes, that worked, the revised $A$ for the rotation is $$ \frac{12-2A}{2+A} $$ so the boundary case for large positive $A$ is $A=6.$ And, if we consider $$ x^4 + 6 x^2 y^2 + y^4 \leq 8 $$ at the point $(1,1)$ we do get $y'' = 0.$

Will Jagy
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Using Will Jagy's formula and a numerical solver, $p \approx 2.10134909469$ does the trick to within the accuracy of my calculator.

Incidentally, $p \approx 2.00208615381$ gives $\pi = 22/7$. And $p \approx 1.79147384986$ gives $\pi = 3$, compliant with 1 Kings 7:23...

Nate Eldredge
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Using matlab (to compute $4\int_0^1\sqrt[p]{1 - x^p}\,dx$), I got that the area of the $p$-ball is, for

$p = 2.1030$, area $=3.1995$

and

$p = 2.1040$, area $=3.2001$

BaronVT
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  • I suggest doing this in polar coordinates, less influence of the infinite derivative at $x=1.$ – Will Jagy Nov 21 '13 at 23:16
  • Yeah, this became especially apparent when I tried to compute perimeter using the usual arclength formula (and then someone else cleverly noted that's not even the right arclength for this metric!) – BaronVT Nov 22 '13 at 16:09
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    An easier alternative is to identify, for given $p,$ the value $x_1$ where $2 x_1^p = 1,$ which is where $x=y$ in the curve $x^p + y^p = 1$ with both positive. Then do $$ 8 \int_0^{x_1} \sqrt[p]{1-x^p} - x ; ; dx $$ – Will Jagy Nov 22 '13 at 20:36
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I have already posted the answer here where somebody wanted to know for which $p$ is $\pi=42$. Using

$$\pi_p=\frac{2}{p}\int_0^1 [u^{1-p}+(1-u)^{1-p}]^{1/p}du$$

we get $p=2.60513$ and $p=1.623$ where $\pi_p=3.2$ and we also know that those are the only two answers.

Reference: http://www.jstor.org/stable/2687579

Community
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Fixed Point
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  • Notice that $1/p_1+1/p_2=1$ – Empy2 Nov 22 '13 at 07:03
  • @Michael Yep, they are both conjugates of each other and since they are both bigger than one (because the required value of $\pi$ is less than four), we have two norms where $\pi=3.2$. – Fixed Point Nov 22 '13 at 07:15