$\pi$ in arbitrary metric spaces

Question

Whoever finds a norm for which $\pi=42$ is crowned nerd of the day!

Can the principle of $\pi$ 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 $d(x,y):=\|x-y\|$. 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,

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

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 1

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 $\pi_p=42$.

Answer 2

I think your definition is reasonable and might produce some interesting results.

For example, if we put the taxicab metric on the plane, we should get $\pi_X$ = 4. Also, people say that if $\pi$ were 3, circles would be hexagons, which suggests that there's a metric on the plane that would make both of those true. I'd be curious to know what other $\pi_X$ values you could get.

You'll probably want some more conditions on your metric space. For example, probably the symmetries should act transitively on $X$ so that it doesn't matter where you put your disk. And maybe there should be scaling maps which fix a point but multiply all distances by a constant factor. In any case, to get a consistent $\pi_X$ you should be able to take a disk of any size.

Answer 3

One setting where this can be worked out is on 2-dimensional Riemannian manifolds (which have a metric in the metric space sense that's induced by the Riemannian metric). If you haven't heard about this before, Wiki has an introduction to what this means, with lots of pictures.

In that setting, you can pick a point $p \in M$, and there's a reasonable notion of a circle of radius $r$ centered at $p$, which I think agrees with yours. Then ask what is the circumference of a circle $L_r$ of radius $r$ centered at $p$. The corresponding value of $\pi$ would be $\pi_r = L_r/2\pi$. If you do a bit of work, you can work out that for small $r$, $\pi_r \approx \frac{L_r}{2 r} \approx \pi - \frac{\pi K}{6} r^2$, where $K$ is the sectional curvature of the surface at $p$. (This is exercise 5.7 in do Carmo's book on Riemannian Geometry)

So as $r$ gets small you get closer and closer to the usual value of $\pi$. Maybe the more interesting way to think about this is that the value of your $\pi$ tells you something about curvature: a space has positive curvature if small circles have circumference less $\pi r$, and negative curvature they have circumference more than $\pi r$.

Answer 4

When you finish doing these yourself, you can look up some literature by searching for "girth" of normed spaces.