For which $L^p$ is $\pi=3.2$?

Question

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''?

Answer 1

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 $=3.2001$

Answer 2

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 $y'' = 0.$

Answer 3

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...

Answer 4

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