π\pi in arbitrary metric spaces

Whoever finds a norm for which π=42 is crowned nerd of the day!

Can the principle of π in euclidean space be generalized to 2-dimensional metric/normed spaces in a reasonable way?

For Example, let (X,||.||) be a 2-dimensional normed vector space with a induced metric d(x,y):=. Define the unit circle as
\mathbb{S}^1 := \{x\in X|\;\|x\|=1\}
And define the outer diameter of a set A\in X as
d(A):=\sup_{x,y\in A}\{d(x,y)\}=\sup_{x,y\in A}\{\|x-y\|\}
Now choose a continuous Path \gamma:[0,1]\rightarrow X for which the image \gamma([0,1])=\mathbb{S}^1.
Using the standard definition of the length of a continuous (not necessarily rectificable) path given by
L(\gamma):=\sup\bigg\{\sum_{i=1}^nd(\gamma(t_i),\gamma(t_{i+1}))|n\in\mathbb{N},0\le t_0\lt t_1\lt … \lt t_n\le 1\bigg\}
we can finally define \pi in (X;\|.\|) by
\pi_{(X,\|.\|)}:=\frac{L(\gamma)}{d(\mathbb{S}^1)}
(This is way more well-defined than the old definition (check the rollbacks))

Examples:

  • For the euclidean \mathbb{R}^2, \pi_{\mathbb{R}^2}=3.141592…
  • For taxicab/infinity norms, \pi_{(\mathbb{R}^2,\|.\|_1)}=\pi_{(\mathbb{R}^2,\|.\|_\infty)}=4
  • For a norm that has a n-gon as a unit circle, we have \pi_{(\mathbb{R}^2,\|.\|)}=?? (TODO: calculate)

While trying to calculate values for \pi for interesting unit circles, I have defined a norm induced by a unit circle: let \emptyset\neq D\subset X be star-shaped around 0\in D. Define \lambda D:=\{\lambda d|d\in D\}. Now the (quasi-)norm in X is defined as \|x\|:=\min\{\lambda\mid x\in \lambda D\}.

In other words: the scaling factor required to make x a part of the border of D.
This allows us to easily find norms for most geometric shapes, that have exactly that Geometric shape as a unit circle and have the property, that the choice of the radius for the definition of \pi is insignificant (for example \pi=3 is calculated for regular triangles with (0,0) in the centroid).

This allows for the following identity: let x\in\mathbb{R}^2, and S\in\partial D be the Intersection of the Line \overline{x0} with the border of D. Then we have
\|x\|_D = \frac{\|x\|_p}{\|S\|_p} where \|.\|p is any (quasi)-norm in \mathbb{R}^2 (This follows out of the positive scalability of norms)

EDIT: I have been thinking about this a bit more. However I found out that my induced norm defined earlier is not even a norm… It violates the subadditivity axiom: Let the unit circle be a equilateral triangle where the centroid marks the point (0,0). Here we find d(\mathbb{S}^1)=3 which violates the triangle equation, as d is measured in between 2 points of the unit sphere; Therefore, we have \|a\|=1 and \|b\|=1, but \|a+b\|=3\nleq1+1=\|a\|+\|b\|. Instead, we have a Quasinorm.

This indeed allows for \pi=42, if the unit circle is, for example, graph of a function \varphi\mapsto(r(\varphi),\varphi) in Polar coordinates where r(\varphi)=a*\sin(2\pi k).

Questions
Any other interesting norms?
Is this definition reasonable, and is there any practical use to this?
Feel free to share your thoughts. Mind me if I made some formal mistakes.
And especially, how do I define a norm with \pi=42?

About the \pi=42

Prince Alis Answer below shows that there can be no such p-norm, for which \pi_p=42, In fact this holds true for every norm. However, you can easily define a quasi-norm that has any arbitrary \pi_{\|.\|}=\kappa\gt\pi. For example,

$1 + \frac{1}{a} \sin(b \theta)$

r(\theta)=1 + \frac{1}{a} \sin(b \theta)

Defines a quasinorm with \pi_{\|.\|_{a,b}}=\kappa for every \kappa>\pi. \kappa can be increased by increasing a and b. The only thing left to do is find a and b for which we finally have \pi_{\|.\|_{a,b}}=42.

According to Mathematica, the Length of the boundary curve is 33.4581 (Took ~10 minutes to calculate) and the diameter is 4.5071, resulting in \pi=7.42342 for the norm given above (a=b=10). I doubt I will be able to easily find a solution for \pi=42 using this method… (Testing manually, I got exemplary values a=9.95, b=175 with \pi=42.0649 which comes very close…

On top of that, Prince Ali found a p-norm with p<1 for which \pi=42. Thank you very much!

Answer

I believe I do have a partial answer to your question but I do not claim credit for it. http://www.jstor.org/stable/2687579 is the paper I am referring to. Undoubtedly Miha Habič is referring to the same thing. And here I will summarize the relevant info from the paper.

Only working with the p-norms defined in \mathbb{R}^2 as
d_p((x_1,y_1),(x_2,y_2))=(|x_2-x_1|^p+|y_2-y_1|^p)^{1/p}
we already know that this is a norm if and only if p\geq 1 and the usual norms mentioned here like the taxicab, euclidean, and the max norm (by "setting" p=\infty) are all special cases so we only look at d_p for p\in [1,\infty).

The authors then derive the expression
\pi_p=\frac{2}{p}\int_0^1 [u^{1-p}+(1-u)^{1-p}]^{1/p}du
for \pi in any p-norm. Then they just numerically integrate and estimate \pi for different p and get

\begin{array}{ll}
p & \pi_p \\
1 & 4 \\
1.1 & 3.757... \\
1.2 & 3.572... \\
1.5 & 3.259... \\
1.8 & 3.155... \\
2 & 3.141...=\pi \\
2.25 & 3.155... \\
3 & 3.259... \\
6 & 3.572... \\
11 & 3.757... \\
\infty & 4
\end{array}

Then the authors prove that the (global) minimum value of \pi_p indeed occurs when p=2. And numerics seem to suggest that \pi_p is always between [\pi,4] so the answer to your question seems to be that there is no p-norm in which \pi_p=42.

Attribution
Source : Link , Question Author : Community , Answer Author : Fixed Point

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