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))


  • 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!

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    What if $X$ is a metric space so small that there are no circles of radius $1$? – Zhen Lin Dec 09 '12 at 16:02
  • Apparently my assumption is wrong. In cases where $\mathbb{D}=X$, $\partial\mathbb{D}$ is empty. I will have to require that this is not the case. – CBenni Dec 09 '12 at 16:09
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    I'm afraid that your question is written in too sloppy a way to be understood. First, you introduce a set $A$: what is the use of this? Then (before you define it), you speak about $\partial \mathbb{D}$ being a $1$-dimensional manifold. What about this, are you assuming it or what? Then, you define a length for a parameterized curve, but who said that $\partial \mathbb{D}$ is a parameterized curve? And even if it were, who guarantees that $L(\partial \mathbb{D})$ is finite? Who guarantees that it is independent of the parameterization? – Giuseppe Negro Dec 09 '12 at 23:58
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    If you clarify those points I will surely upvote. It is an interesting idea but I think that your metric space setting is too general. You could have more chances if you worked on a Riemannian manifold, maybe. – Giuseppe Negro Dec 10 '12 at 00:06
  • I agree with your doubts, this is not well-defined in any metric space. Iff 1 is a regular value of d (or rather, of the norm $||.||$ in normed vector spaces), $\partial \mathbb{D}$ is a 1-dimensional manifold. I was thinking about using the open unit circle instead of the closed, therefore, we would have a value of $\pi_X=0$ if we are using the discrete metric. – CBenni Dec 10 '12 at 18:26
  • Oh and, @Guiseppe Negro, the thing about my set A, yes that was indeed bad. Shall I rewrite my OP? – CBenni Dec 10 '12 at 18:31
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    Now it is much better. +1. Interesting question. – Giuseppe Negro Dec 10 '12 at 21:35
  • what is metric induced by the unit circle as a regular triangle which centroid is at (0,0)? –  Dec 20 '12 at 11:13
  • @seda read my description on norms that get induced by 1-dimensional submanifolds – CBenni Dec 20 '12 at 16:37
  • @CBenni I'm new here. I coundn't find your definiton Benni. Can you send the link , please. –  Dec 23 '12 at 19:13
  • @seda, It is in my questions text! Given a star-shaped set $S$ , $0\in S$, you can find a quasi-norm for any $x\in\mathbb{R}^2$ by calculating the minimal scaling factor $\|x\|:=\lambda$ that is required to make $x\in \lambda S$, where $\lambda S := \{\lambda s\mid s\in S\}$. – CBenni Dec 23 '12 at 21:44
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    I upvoted so that that the question now has 42 upvotes... – JP McCarthy Aug 14 '13 at 14:19
  • @JpMcCarthy : It is 1st September and it still has 42 upvotes! Any theories about $\pi=42$ and 42 votes? – jimjim Sep 01 '13 at 12:47
  • @Arjang aside from that this is the first question ever on Math.SE that has $\pi$ points? I guess not. – CBenni Sep 01 '13 at 20:58
  • Wanted to upvote this question, but it appeared to have 42 votes. Decided to not break the beauty :) – Ruslan Sep 06 '13 at 18:43
  • To define a norm a star-shaped region is not enough. The region should be convex, symmetric with respect to origin, and contain a neighborhood of the origin. – Maesumi Nov 22 '13 at 00:48

7 Answers7


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 $\pi_p=42$.

Fixed Point
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    Just noticed that if $\frac{1}{p}+\frac{1}{q}=1$ then $\pi_p=\pi_q$ so the minimum at $p=2$ is unique meaning for sure there is no $p$-norm for which $\pi_p=42$. – Fixed Point Dec 23 '12 at 23:39
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    Also for $p\in(0,1)$, we don't have a norm (because the triangle inequality is violated) but the value of $\pi_p$ increases without bound as $p\rightarrow 0$ so in this range there is a (unique) solution to $\pi_p=42$ and it is around $p=0.0804670...$ so here is a seminorm which gives $\pi=42$. – Fixed Point Dec 23 '12 at 23:52
  • Indeed, there is no p-norm with $\pi =42$, however, this doesnt hold for quasinorms. Thank you very much for your answer tho ;) – CBenni Dec 24 '12 at 11:05
  • Out of interest, how did you find that solution numerically? Mathematica is calculating since ~10 minutes without success? Awesome find however :) – CBenni Dec 24 '12 at 11:40
  • Define f[p_]:=(2/p)*NIntegrate[...]. And then I just used simple bisection method to approximate $p$. Mathematica evaluates f[p] rather fast so you can even plot it fairly fast. But using any built-in solver to solve f[p]=42 either fails, is too slow, or throws an error because of the setdelayed definition. – Fixed Point Dec 24 '12 at 21:14
  • With help by Mathematica.SE I was able to make Mathematica calculate it to any given precision, albeit it converges tiringly slow, I calculated it up to 1000 digits of precision and still had $\pi=41.99999999999999995...$ – CBenni Dec 25 '12 at 12:25
  • I can't see the article, can someone share the PDF? – Basj Dec 21 '16 at 08:56

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 $\pi r$.

some guy
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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.

Hew Wolff
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    Actually, pi in taxicab/maximum distance would be: $\frac{4\sqrt{2}}{2}=2\sqrt{2}\approx 2.828$ if I am not mistaken; Thank you for your answer! For hexagons, indeed $\pi=3$; For any regular n-gon the formula is $\pi_n=n*sin(\pi/n)$ – CBenni Dec 10 '12 at 18:21
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    @CBenni: It seems unprincipled to define the _shape_ of the circle by the taxicab metric but measure the _length_ of the perimeter by the Euclidean metric. In taxicab metric the "circle" is a diamond, but the length of each of the edges of the diamond is 2 according to the taxicab metric itself, so the entire perimeter is 8, which is 4 times the diameter. – hmakholm left over Monica Dec 10 '12 at 20:45
  • true. true. I wasnt thinking there. What value does pi take for a hexagonal unit circle? – CBenni Dec 10 '12 at 20:52
  • Still 3? That is what I am getting (drew a hexagon with 1cm edgelength, all edges are 1 cm long and for all corners, you can place a similar hexagon to that corner and it will intersect the first on in the adjacent points, so the curve length is 6. The radius is 2. So basically, we have $\pi=3$ – CBenni Dec 10 '12 at 20:59
  • New result: $\pi$ is 6 in a space induced by the unit circle as a regular triangle whichs centroid is at (0,0). I will add a definition for induced metric in my question. – CBenni Dec 10 '12 at 22:12
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    @CBenni I once calculated $\pi_p$ in the different $p$-norms during a boring talk. If I remember correctly, $\pi_1=\pi_\infty=4$, these are the maximal values and the minimum occurs at $p=2$. This seemed interesting at the time. – Miha Habič Dec 11 '12 at 02:36
  • @MihaHabič Did you find a formula in dependancy of p? – CBenni Dec 11 '12 at 08:46
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    @CBenni I did. It came to a fairly horrible integral. I'll see if I can rederive it. – Miha Habič Dec 11 '12 at 13:13
  • Posted below guys! – Fixed Point Dec 23 '12 at 23:35

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

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  • http://alvarezpaiva.wordpress.com/tag/geometry-of-normed-spaces/ Thanks for the tip :) BUT does Golabs work ($3\le\pi\le 4$ in normed planes) mean that ANY norm has a value of pi between 3 and 4? Or is that only the p-norms $\|(x,y)\|_p:=\sqrt[p]{x^p+y^p}$? I am VERY sad if this is true ;; I thought it were possible to define a "wavy" unit sphere that has a very high girth-to-diameter ratio (speak: $\pi$) – CBenni Dec 11 '12 at 23:43

A similar question was asked before on StackExchange Developing the unit circle in geometries with different metrics: beyond taxi cabs They came up with the ``elevator metric" and "bus "metric".

Any normed space should have notions of length, area and volume. So you can compute pi as

  • the ratio of circumference to diameter
  • the area of a unit circle
  • half the circumference of unit circle
  • 3/4 the volume of the unit circle (or whatever the appropriate fractions are in your normed space!)

Here is a paper called Volumes on Normed and Finsler spaces which looks helpful. There is also something called the isoperimetric problem, which describes a "circle" as maximizing the area given a length.

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  • I wanted to post a 50pt bounty on [your helpful answer to my question about the Hilbert space-filling curve](http://math.stackexchange.com/a/586221/25554), but SE has a restriction that bounty amounts must be monotonically increasing. So I am going to post it here instead. – MJD Dec 03 '13 at 15:53

The integral I found for the circumference of the $p$-norm 'circle' was
$$C(p)=4\int_0^1\left(1+\left[\frac{x^p}{1-x^p}\right]^{p-1}\right)^{1/p}dx$$ The asymptotics of this for large $p$ is $8-8p^{-1}\ln 2+O(p^{-2})$

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In the group of geometries of isocurves (ie euclidean, hyperbolic, spheric), one might measure distances in a variety of rulers.

The most common is to use a ruler of the same curvature as space, which is one of minimal curvature. That is, the ruler is 'straight'. But since a great circle in spherical geometry is a limiting case of a general circle, the variation consists of $f(r)$ with $f(R)$, where $r$ and $R$ are radii of small and large circles.

A different approach is to use a ruler of zero curvature, such as might be implemented by the circumference of circles is $4\pi$ of the diameter. This kind of ruler derives in a geometry of considering every circle the intersection of a euclidean and a non-euclidean geometry, and then measuring distances in euclidean measures.

This roughly equates to supposing a spherical tiling is a polyhedron, and as long as you stick to the vertices, all pentagons look the same, and straight lines are then invoke $R$ separately.

Curvature then becomes an intrinsic measure of comparing $D/R$, both measured in the euclidean manner. For a sphere, $D<2R$, which means that the circumference of a circle is less than $C < 2\pi R$. You can find the radius of the sphere by finding the diameter (ie circumference / $\pi$ of the circle that passes through the centre, against the diameter of the larger circle. This ratio can be used to find the diameter of the circle through these points, presumably straight.

Done in hyperbolic space, we have $D>2R$ , or $C = \pi D > 2 \pi R$. It's not a function of the underlying space, but a function of the equilavant of angle. That is, one can construct a circle where $D/R = 2/\phi$ where $R= 72$ degrees, or $D/R=2/sqrt{3}$ where $RR is 60 degrees etc. So is it with hyperbolic geometry.

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  • This is actually used quite heavily in the search for uniform hyperbolic polytopes. Just because the maths is different, it's not wrong. – wendy.krieger Aug 14 '13 at 12:42