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How/why does $\pi$ vary with different metrics in p-norms? Full question is below.

Background

Long ago I did an investigation on Taxicab Geometry using basic geometry. One think I recall is that a circle (as defined by all points equal distance from a centre point) 'looks' like a diamond. The 'circumference' of this circle is 8. As an extension I looked at other metrics of the form:

$$D_n\left((x_1,y_1),(x_2,y_2)\right)=(|x_2-x_1|^n+|y_2-y_2|^n)^\frac{1}{n}$$

(My limited reading of wikipedia suggests I should call this a p-norm.)

More recently using differing values of $n$ I calculated the 'circumference' of unit circles in these metrics. I took the definition of a unit circle to be all points a distance of one unit from the origin. This gave me a formula for a semi-circle:

$$y=\left(1-|x|^n\right)^\frac{1}{n}$$

I took the normal arc length formula of:

$$\int_a^b\sqrt{1+\left(\frac{dy}{dx}\right)^2}$$

and replaced all the powers of $2$ with powers of $n$ to get:

$$\int_a^b\left(1+\left|\frac{dy}{dx}\right|^n\right)^\frac{1}{n}$$

Combining the circle with the arc length formula (and take a quarter circle and times it by 4 gave) the following integral:

$$4\int_0^1\left(1+\left|\frac{d}{dx}\left(1-x^n\right)^\frac{1}{n}\right|^n\right)^\frac{1}{n}dx$$

Then $\pi(n)$ is found by dividing 'circumference' by two (twice the radius). Doing so led to this graph of $\pi(n)$ against $n$.

enter image description here

Interestingly $n=2$ is a minimum (both local and absolute) making our commonly thought of value of $\pi$ special.

Question

(EDIT) Math Question: Is my distance formula for a different metric correct? (Moishe Cohen's comment suggests it might not be).

Math Question: Assuming the math above is ok, is there a reason for $(2,\pi)$ to be a minimum?

Math/Philosophy Question: Assuming above ok, is this why we observe the metric $D_2$ in the real world?

Note

I have not formally studied metrics, tensors or vector spaces or related topics (but am happy to do some light reading if your answer requires it).

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    $\begingroup$ See math.stackexchange.com/questions/254620/… $\endgroup$ – levap Dec 5 '16 at 2:32
  • $\begingroup$ Thanks @levap. Tough reading (for me) but I'll see what I can make out of it. I also found this one: math.stackexchange.com/questions/8856/… $\endgroup$ – Ian Miller Dec 5 '16 at 4:59
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    $\begingroup$ Just our of curiosity: Why did you use this integral? It computes with respect to the (standard) $\ell_2$ metric the circumferences of the unit disk which is defined with respect to the $\ell_n$ metric. $\endgroup$ – Moishe Kohan Dec 9 '16 at 9:46
  • $\begingroup$ @MoisheCohen I've updated my question with my reasoning. Your comment seems to imply I have not computed the arc length correctly in the new metric. If so can you show me how it should have been done. Thanks. $\endgroup$ – Ian Miller Dec 9 '16 at 15:30
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    $\begingroup$ Thanks @Zach466920 Sorry for the slow response. I've read it (and related wikipedia pages) several times as a significant part of the terminology is beyond me at the moment. I appreciated your answer and gave you an upvote. I have also been exploring (in my own fashion) some of the ideas you expressed. I am currently resolving whether I need to seek minor clarification or actually ask a whole new related question. Again thank you for your answer and your patience, I will post one way or the other soon. $\endgroup$ – Ian Miller Dec 12 '16 at 23:29
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Math Question: Assuming the math above is ok, is there a reason for (2,π) (2,π) to be a minimum?

The $p=2$ is the only p-norm that has the $SO(3)$ Lie Group structure. In other words, it is rotation invariant. Try it, you can rotate the coordinate system without changing the length. You can't do that with the Taxicab metric, the length you get will change.

The deep answer to your question is that only $p=2$ has a continuous symmetry, rotation. All the other p-norms have a either a finite number of symmetries or symmetries. Here's the thing, a p-unit circle is something like every possible vector that can generated from the group glued together. Imagine starting with a small vector then copying and rotating it just a bit, attaching it to the original vector, then repeating with the copy. Eventually, you'll generate the circle. You generate any of the p-norm circles using a similar process using their symmetry, albeit a bit more opaque. So why is $p=2$ special? The circle is made with the shortest vectors, and only the shortest vectors available to construct it. Why? Because they're all the same length! This isn't true unless you're rotationally invariant and guess what? Only $p=2$ is!!

Math/Philosophy Question: Assuming above ok, is this why we observe the metric D2
D2 in the real world?

Quick run through physics. In physics, you have a theory described by a lagrangains that describes the amount of action particles do over a length of time. It's a scalar function of distance, speed, and time. The symmetries of the lagrangian correspond to conserved quantities according to Noether's Theorem. In standard classical mechanics, we assume three things,

(1) The action of a particle moving through some path along space is unchanged if you shift the start of the motion forward or backward in time. (Energy is Conserved)

(2) The action is unchanged if you shift the translate the start of the motion to another location. (Momentum is Conserved)

(3) The action is unchanged if you rotate the motion of the particle. (Angular Momentum is Conserved)

The last one isn't talked about to much, however, it's very important and highly relevant to our discussion. Since, our lagrangian, the theory for our physical theory, is invariant under rotation, there can never be a preferred direction in our theory. This means that means the metric we are using to measure distance and speed must also be rotation invariant. The $p=2$ norm is the only p-norm that'll work.

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    $\begingroup$ Thanks @Zach466920 I've decided my follow up thoughts deserved a new question. If you would love to help me further you can find it here. $\endgroup$ – Ian Miller Dec 13 '16 at 15:34
  • $\begingroup$ Really great answer. The physics discussion was very informative. $\endgroup$ – Aaron Hendrickson Dec 29 '18 at 21:49
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Having read (as best I can understand) the various related questions/answers linked by others and other links from those answers I have come to the following answer.

First of interesst There are two other integral forms of writing the integral for $pi_n$:

  • From Michael in this answer.

$$2\int_0^1\left(1+\left(\frac{x^n}{1-x^n}\right)^{n-1}\right)^\frac{1}{n}dx$$

$$4\int_0^1\frac{1}{n}\left(x^{1-n}+(1-x)^{1-n}\right)^\frac{1}{n}dx$$

Adler & Tanton also go on to prove that a minimum occurs when $2-\frac{1}{n}=1+\frac{1}{n}$ which leads to the value of $n=2$ being a minimum. For others curious about this question I recommend reading this paper. (You can create a free account to read it if you do not have JStor access.)

They conclude their paper with the unproven proposal that $\pi_n=\pi_m$ if $\frac{1}{n}+\frac{1}{m}=1$. I'll see what I can do on that problem.

Edit: Interesting thought: If $\pi_\frac{1}{n}=\pi_{1-\frac{1}{n}}$ then $\pi_n$ can be defined for negative values.

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The definition of the p-norm is, for a vector $\vec{x}$ in $n$ dimensional space, $$\|\vec{x}\|_p = \left( \sum_{i = 0}^n |x_i|^p \right)^{1/p}$$ where $x_i$ is the $ith$ entry of the vector $\vec{x}$. The metric that you called the 'p-norm' is indeed the metric induced by the p-norm, and is generalized to $n$ dimensional spaces as follows: $$d(\vec{x}, \vec{y}) = \|\vec{x} - \vec{y}\|_p = \left( \sum_{i = 0}^n |x_i - y_i|^p \right)^{1/p}.$$

I learned this from Fundamentals of Applied Mathematics, Volume I by Tyler Jarvis, Jeffrey Humphries, and Emily Evans. Unfortunately it is not published yet, but it will be within a few months.

Your graph of $\pi$ for the different values of is fascinating! I am likewise amazed that ($2,\pi$) is the minimum of the graph. The paper you talk about proves that this is the case, and therefore gives every mathematical reason 'why' this is true.

As far as your philosophy question goes, I don't know, but that is a cool idea!

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