Folks, We Have the Best Π

I had the same thoughts when studying physics (I have a PhD). Math was some kind of a toolbox for my problems - I used it without too many thoughts and a deeper understanding. Some of the "tools" were wonderful and I was amazed that it worked; some were voodoo (like the general idea of renormalisation, which was used as a "Deus ex machina" to save the day when infinities started to crawl up).

Math is very cool but I think it requires a special (brilliant) mind to go through, and a lot of patience at the beginning, where things seem to go at a glacial pace with no clear goal.

> How to calculate π for n-metrics numerically. The general idea of "divide the circle into segments and calculate the length by the metric" is explained, but the exact algorithm or formulas are not shown.

I feel like that would have been a bit in the weeds for the general pacing of this post, but you just convert each angle to a slope, then solve for y/x = that slope, and the metric from (0,0) to (x,y) equal to 1, right? Now you have a bunch of points and you just add up the distances.

> The article shows this graphically, but there is no formal proof (although the Adler & Tanton paper is mentioned).

Well, if that interested you, you could have downloaded the paper and read it. To me your comment sounds a shade entitled, as if the blog author is under an obligation to do all the work. Sometimes one has to do the work themselves.

There's a stackexchange thread that touches on the first two questions. It's got the integral form of the circumference calculation, but I doubt there's a closed-form solution in general: https://math.stackexchange.com/questions/2044223/measuring-p...

For point one the reason is of course that π(p) has a [global] minimum at 2. Actually showing that is not that easy because the involved integrals have no closed form solution but in principle it is not too hard. The circumference of the circles is 2 π(p) and equals four times the length of the quarter circle in the first quadrant which has all x and y positive and allows dropping the absolute values. The quarter circle is y(x) = (1 - x^p)^(1/p) with x in [0, 1]. Its length is the integral of ds over dx from 0 to 1 where ds is the arc length in the Lp norm ds = (dx^p + dy^p)^(1/p) which yields ds = (1 + (dy/dx)^p)^(1/p) dx. For dy/dx you insert the derivative of the quarter circle dy/dx = -x^(p - 1)(1 - x^p)^(1/p - 1) and finally you have to compute the derivative of the integral with respect to p and find the zeros to figure out where the extrema are. Well, technically you have to also look at the second and third derivative to confirm that it is a minimum and check the limiting behavior. The referenced paper works around the integral by modifying the function in a way such that it still agrees with the original function in some relevant points but yields a solvable integral and shows that using the modified function does not alter the result.

I think lcantuf has looked at the first two and decided that the answer is too complex for a post like this. He linked to the article.

The third one we can reason about: For all cases where x and y aren't 0, |x|^n goes to 1 as n goes to 0, so (|x|^n + |y|^n) goes to 2 , and 1/n goes to infinity, so lin n->0 (|x|^n + |y|^n)^(1/n) goes to infinity. If x and y are 0 it's 0, if x xor y are 0 it's 1.

To phrase this in a mathematically imprecise way, if all distances are either 0, 1, or infinite the concept of distance no longer represents how close things are together.

The font stack is just "Verdana, Geneva, sans-serif;" so either one of those or a system font is the culprit.

Yeah I noticed that too, and had to comment (now deleted) to test it was the website here, and not the original copied and pasted text.

Once you understand the math, continuous functions are much simpler than discrete maths (the CS-y things like how many strings with such and such a property are there, or how many ways can you program a Turing machine to take a long time but eventually finish work).

Especially if they are complex differentiable functions - then they are wholly determined by their values in any tiny (complex) neighborhood around 1 (complex) value. Basically just equivalent to power series at that point.

While even finding the number of ways to give change is extremely challenging.

Well, it's pi parameterised by the distance metric, Pi(d)

You can parameterise it by other concerns if you wish, and other things follow. But as a matter of fact, this is how pi depends on the distance metric.

I can't edit my comment anymore so let me elaborate a bit here.

What is this sneaky connection between squared Euclidean and Cartesian coordinates that I mentioned ? Why are they such a compatible pair ?

The answer is the Pythagorean theorem.

The squared Euclidean distances decomposes nicely along orthogonal (perpendicular) directions.

    d^2 = x^2 + y^2.

The Cartesian coordinates decomposes a point along orthogonal (perpendicular) axes as well, which we know is special for squared Euclidean distances.

The other metrics considered in the blog post decompose as, for lack of a better name, Fermat's last theorem decomposition.

    d^n = x^n + y^n

Now if we were to use a coordinate system that decomposes points like that, that would be interesting to explore. I don't know of coordinate systems that do that.

This much is true, forget about integral triples (lattice points) for integral n > 2.

One of these ways (from which Cauchy-Schwarz and the other Hilbert Space results follow) is that d_2 is the only metric that satisfies the Parallelogram Law [1]:

2 d_2(x) + 2 d_2(y) = 2 d_2(x + y) + 2 d_2(x - y)

[1] https://en.wikipedia.org/wiki/Parallelogram_law

Most of the special properties can be traced back to its special relationship with the inner product. And inner products have somewhat more elementary properties, so in that sense it explains the special position of the euclidean norm.

This has nothing to do with the coordinates by the way. If you want a different norm you'll first have to figure out an alternative to the bilinearity that gives the inner product its special properties.

Though bilinearity is pretty special itself, given the link between the tensor space and the linear algebra equivalent of currying.

The special properties extend into statistics, where you have the Gaussian distribution which feels both magical and universal, and is precisely the exponential of a (squared) Euclidean distance, i.e. exp(-(x - x0)^2).

I have the same feeling, that Cartesian coordinates and Euclidian distances are inherently connected as a natural pairing that is uniquely suited for producing the familiar reality that we inhabit and experience.

In my opinion it holds the same place in mathematics that water holds in biology and chemistry.

Very nice. Thank you.

Is 2 a number?

Is 4 a number?

Is 4/2 a number?

Is 3 a number?

Is 3/2 a number?

etc...

All of these symbols represent precise points on the numberline. Pi also represents a precise point on the numberline, so is it not a number?

If you believe that real numbers are numbers, then, yes, pi is a number. Indeed because pi is computable, it’s actually "more" real than almost all real numbers because there is only a countable infinity of computable reals.

Anyway, in modern math what a real number is, is defined as the limit of a "process", namely a Cauchy sequence. Of course, for the rational subset of reals the limit is trivial.

To answer your question you need to define what a number is to you. There are many different kinds of numbers, naturals, integers, rationals, irrationals, computable reals, reals, infinitesimals... Not even getting into complex, quaternions, octonions etc.

Is sqrt(2) a number to you ?

If you accept computable reals as numbers then \pi is definitely a number. So is the golden ratio.

There are mathematical definitions of the terms "real number", "rational number", etc., but there is no mathematical definition of the word "number".

> we can approximate it with better and better precision

In one of the three common formal definitions of the real numbers, that's what a real number is: a Cauchy sequence of rational numbers, which approximate that real number with increasing precision. (Well, a real is an equivalence class of such sequences.)

(The other two common definitions are the Dedekind reals and the reals as the unique complete ordered field.)

It's an infinite family of metrics - you provide an n (a positive integer) and get back a metric.

So if you pick n=1 you get d(x, y) = |x| + |y|, which is the taxicab metric. You can apply this metric to a Euclidean space of whatever dimension you like, just substituting the appropriate definition of |x| and |y|. For 1-dimensional space you would use |x| = abs(x[0]), for 2-dimensional space you would use |x| = sqrt(x[0]**2 + x[1]**2), etcetera. Hope that helps.

For anyone else curious, I found this [0]. Not a full answer but a starting point.

0: https://en.wikipedia.org/wiki/Lp_space#The_p-norm_in_finite_...

The definition of "circle" they are using is the set of points at an equal distance (the radius) from a given point (the centre). This definition works in any setting in which some sort of "distance" (metric) is defined. They are also using an implicit definition of "circumference" that works for the cases being considered here: split the "circle" into sections, measure the sum of their lengths according to the metric and take the limit as you use more and more, smaller and smaller sections. There are details that aren't covered in the article, but it works.

Having defined what a "circle" is and what its "circumference" and "radius" are, "pi" is defined: it's half the ratio of the circumference to the radius.

(I don't think it was very nice of whoever downvoted you, presumably because you're wrong, given you explicitly allowed that you might not be getting it.)

All of these metrics are just functions of a vector. Pop in an (x, y) pair and you get a number out. d₁, or as it's usually called, ℓ¹, |x| + |y|, gives you 7 when you pop in (3, 4). ℓ² = √[|x|² + |y|²] gives you √[3² + 4²] = 5. ℓ³ gives you ∛[|3|³ + |4|³] ≈ 4.498. None of them are moving or otherwise changing with time. They don't pertain to different numbers of dimensions; all of them are defined (in this post) as functions of two dimensions. They just assign a distance metric to every point in a two-dimensional plane.

I think where you're coming from is that the ℓ¹ metric tells you how far a taxicab would have to move, changing in one dimension at a time, while the ℓ² metric tells you how far you have to go if you go in a straight line. But the ℓ³ metric doesn't correspond to anything similar, not even in three or four or five dimensions, and neither does, for example, ℓ¹·⁵. To get to them you have to go through the formulas above.

The curves drawn are "level sets", which connect points that have the same distance metric.

The point of the post is not that "we can approximate [π] best" with a particular distance metric. Rather, it says that every metric (of this family of metrics—you can invent an infinite number of other metrics) has its own ratio of the circumference of a ball to its diameter, which we could jokingly call its "π", and that ratio is lowest for ℓ².

The ratio itself is a notion that really only makes sense in two dimensions; in three dimensions, for example, a ball has a surface rather than a circumference, and dividing the surface by the diameter gives you a length, not a number.

I am completely mystified by your remark that something "has everything to do with the nature of pi, not with the math". What is the nature of π if not math?

The usual distance in 3 dimensions is still d₂ = √(x²+y²+z²)

This is a nice analog but unfortunately I think it breaks down in a way the "π" calculation does not.

In the article 2π(d) = the ratio of the circumference to the radius. This is dimensionless, in the sense that the circumference and the radius are both lengths (measured in meters, or whatever), so 2π(d) is really just a number.

But the (hyper)volumes you're talking about depend on dimension, which is exactly why you say "hyper". In 2 dimensions the volume is the area, πr^2, which has dimensions L^2 [measured in m^2 or whatever]. But in 3 dimensions the volume is 4/3 πr^3, which has dimensions L^3. The 5 dimensional (hyper)volume has dimensions L^5, and so on.

So, "comparing" these to find out which is bigger and which smaller is not really meaningful---just like you shouldn't ask which is the bigger mass: a meter or a second? Neither is, they aren't masses.

I hate this result because it's not actually saying what it's advertized to be saying

sure, you already got the complaint about comparing values of different units - but observe HOW this question is actually sidestepped! We divide hyper-volume of n-sphere by hyper-volume of n-cube!

Now this raises the question: WHAT n-cube are we taking?

If you take hyper-cube with side-length of sphere's diameter, you'll have nice relation between cube and its inscribed sphere - and, predictably, as n goes up, number of cube's "corners" also goes up. So this ratio consistently goes down

But what about your numbers? Well that result happens when you take cube with side-length of sphere's RADIUS. That way you arbitrarily add a scaling factor 2^n - and there's nothing geometric about this behaviour

The post is using MathJax. If you long press (presumably right click on desktop?) you get a menu.

If weird fonts bother you, consider disabling custom fonts in your browser settings. I do it occasionally, but it also breaks fontawesome (and similar), so it's not a clear win.

Think of irrationality as a form of infinite precision, such that no matter how many numbers you write down it’s always an approximation because there is always a more precise form. Now think of the number of sides as increasing your precision. Meaning there is an n such that every approximation after will read 3.14… and another for 3.141….

There is a local metric. The value of length/2/r depends on how big the circle is:

Imagine the Earth is a sphere. You make circles centered in the north pole:

* If the circle is tiny, the Earth is almost flat and you get almost pi.

* If the circle is the equator, you have to walk 1/4 of length the circle from the pole to the equator, so the result is 4/2=2

* If the circle is so big that you walked almost to the south pole, the result is almost 0.

Yes. The angle subtended by the two points at the center of the sphere. It's the angular displacement made to go from one point to the other along the great circle joining the two points on the surface.

The great circle is the one that passes through those points and has the center of the sphere as it's center.