# Pi - No Patterns, because Pi is the pattern

Discussion in 'Pseudoscience' started by Quantum Quack, Jul 23, 2013.

1. ### Quantum QuackLife's a tease...Valued Senior Member

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They often say that the ratio of Pi has never revealed a pattern with in its prodigious list of digits as it seeks to resolve itself.
In this sense pi offers a rather amazing insight IMO
and I disagree with the assessment that Pi offers no pattern, as Pi, it self IS the pattern.

The issue though of this OP is to propose the notion that whilst Pi's digital sequence may not have the ability to conform with statistical probabilities of randomness Pi is in itself a pattern.
By this I mean that regardless of how many times you generate Pi the digits are always the same. That the digital sequence of Pi is the pattern regardless of how many digits you may allow the sequence to run to.

Care to discuss?

3. ### CheezleHab SoSlI' Quch!Registered Senior Member

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745
Your idea of pi as a pattern in itself, seems reasonable to me. One interesting development in calculating digits of pi is the spigot algorithm. It can generate any digit of pi that you wish without needing the previous digits. See http://www.cs.ox.ac.uk/jeremy.gibbons/publications/spigot.pdf which even has some programming examples. To me the idea of the spigot algorithm bolsters the your idea. But I can't say why I feel that way and have no justification for the statement. You might also enjoy this video of the history of pi, examples of how to calculate it, and the problematic nature of the beast.

5. ### Beer w/StrawTranscendental Ignorance!Valued Senior Member

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Most of the time I think this forum sucks, but I'm really intrigued by those algorithms.

Thank you very much.

7. ### someguy1Registered Senior Member

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I've noticed that every time I generate the number 3, the digits are always exactly the same. Is that amazing, or is it an obvious triviality following from the fact that 3 is a particular real number, just as pi is?

It may (or may not; the question is open) be the case that pi is "normal" in the sense that its digits are statistically random. But pi itself is not random. It's a particular real number, characterized by a deterministic algorithm. It would be truly shocking if it came out differently each time. It doesn't, because it's a specific, deterministic real number.

Pi is in fact a computable number. It's the output of a finitely-describable algorithm.

I can't post links yet but please look up "computable number" to understand this important point. Pi is not random. It's the result of a deterministic algorithm.

Of much more interest to those who care about true randomness are the non-computable numbers, of which there are plenty. "Almost all" real numbers are non-computable, meaning that the probability of picking a random real number and getting a computable one is zero.

The non-computable numbers are the numbers that are truly random. Their digits can not be compressed and generated by any algorithm. The shortest description of a non-computable number is the number itself.

Pi is a specific, computable real number that is the output of a deterministic algorithm. It's no more surprising that the digits of pi are the same every time you compute them; than it is that the digits of 3 are the same every time you compute them.

8. ### Quantum QuackLife's a tease...Valued Senior Member

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True. except... that it is a fundamental of describing a fundamental geometry of space time. A natural outcome of 3 dimensional space is the sphere thus Pi whilst a computational number has direct and "natural relevance" to the structure of space time.
Also can I ask as you are obviously conversant with algorithms.
Do you know of any other computational number that can derive such an array of digits that demonstrate no pattern forming?
If so I would dearly like to read of it...

The fact that calculating Pi produces a repetitive solution, is indeed, trivial...
However Pi in Nature is not calculated as it is already existent. [in any curved surface I would imagine]
and it emulates something that is already existent and that makes it a very remarkable number. [concerning the fundamentals of this universe] IMO

Last edited: Jul 26, 2013
9. ### CheezleHab SoSlI' Quch!Registered Senior Member

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745
The problem with Pi is that it is incommensurate. It is a ratio that is irrational. Or better stated the diameter of a circle is incommensurate with the circumference.

In the video I linked to, professor Wildberger suggests that pi is what he calls a meta number. It isn't exactly a number at all. He points out that all irrational numbers are very problematic in mathematics. I remember from a math class I once took where this was explained in this way. A line that crosses a circle does not necessarily intersect the circle. It passes between the points on the circle, and the circle passes between the points on a line. For mot lines there is not a point on the line that is also on the circle. The line is a rational object while the circle is composed on mostly irrational points (the points [1,0], [0,1], [-1,0] and [0,-1] being exceptions). But those circle points were all the points found by the pythagorean theorem. Only a small subset (I think) of these irrational points are transcendental. The circle's as defined by angle are transcendental. So if you define a circle by (rational or irrational) angles, it is a different circle than the one found through squareroots. I don't remember the whole thing exactly. Wildberger's point of view is considered by most to be unconventional but his view is that his view is more rigorous and that rigor is needed in mathematics. (See his videos on rational trigonometry.) Anyway, I would think that Wildberger would say that Pi is not a pattern at all, and it isn't even a number. Only rational numbers are patterns. He seems pretty open to questions from the public. You might even ask him directly what he thinks about the subject.

10. ### someguy1Registered Senior Member

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Or for that matter sqrt(3), sqrt(5), sqrt6), etc. etc. etc. There are many many irrational numbers, far more of them than rational numbers. You can pick any computable number, meaning any number generated by an algorithm, and chances are its digits are random. How about e, the base of the natural logarithm? Or phi, the golden ratio.

Wildberger is a crank, though I agree with you that he's sometimes an interesting one.

That's just insane. The lines y = x and y = -x intersect at the origin. Do you really think they "hop over" the origin instead?

To be fair, and to forestall your obvious objection, you did say that Wildberger believes this is only true for "some" intersections. Presumably he believes in rational points on the circle but not irrational points. I get that he believes that. It's still insane.

I am sorry but I can't get into a debunking of all the math cranks out there. But the terrific math blogger Mark Chu-Carroll took the trouble to thoroughly debunk Wildberger on his blog Good Math, Bad Math. Regrettably I can't post links yet but if you google "Wildberger Good Math Bad Math" you'll get the link to Chu-Carroll's excellent debunking of Wildberger.

11. ### CheezleHab SoSlI' Quch!Registered Senior Member

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I watched a few of Wildberger's Math History videos and thought they were interesting. I was not really that interested in his rational trigonometry though it looked like it might make computer calculations more exact. He mentioned this about the line not intersection the circle (not the interior but the circle itself). I did not question it because I heard this in a unversity math class I took in about 1975. The professor mentioned that there were problems with irrationality in the same way that Wildberger said. Maybe my professor was a crank too and since I think the example was in the text the author must have been a crank also. I think it was probably worded differently than I said. Perhaps the line was between 2 rational points. So the function of the line would map a rational x to a rational y. But the circle could map that same rational x to an irrational y. So there would be no intersection. Like I said, long time ago, don't remember the specifics. Sounded reasonable at the time. And it was mentioned in the class as a warning that sometimes lack or rigor would lead to some errors. And that irrational numbers were somewhat problematic even if they were general accepted.

So if you can find an x value that a rational line gives a rational number for y (obvious), but plugging that rational value of x into the circle gives an irrational value for y, then the line crosses the circle but does not intersect it for that value of x. It is probably easy to find said non-intersections. Where the rational line goes between the irrational points on the circle. Does not matter how close the points are to each other. But like I said, my memory is fuzzy on the subject after 30+ years.

Last edited: Jul 26, 2013
12. ### someguy1Registered Senior Member

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The problem with this point of view is that a continuous line would then be full of holes.

To put this in a larger perspective, think of the original Euclidean line, intended to model the idea of a continuum. The line is made of individual points; but they are not lined up "next to each other" one after the other. Rather, between any two points on the line there is an entire continuum of points. The idea is to model the path of a point moving through classical Euclidean space. We are doing math and not physics so we are not concerned with quantum theory or the Planck length or any other kind of physical consideration. This is purely a mental game ... we are imagining the path of a point moving continually through space, the way people thought of it before quantum physics.

If you don't believe in irrational numbers, then your line looks like swiss cheese except that it's got more holes than points. If there are no irrationals then two lines could pass through each other without intersecting. But that violates the idea of the continuum. The intermediate value theorem in calculus would be false because a continuous function could be negative at one point and positive at another; yet never attain the value zero. This would be bad.

In standard modern math we model the classical number line with the set of real numbers, irrationals at all. It does do a very good job of modeling the idea of continuity; at the expense of having to acknowledge the existence of irrational numbers. This is the prevailing modern view.

There are some mathematicians who are finitists or even ultra-finitists. They deny the use of infinitary processes and they deny noncomputable numbers ... that is, they deny the existence of numbers that can not be generated by an algorithm. Even an ultra-finitist believes in sqrt(2) and pi, because those numbers can be generated by algorithms to any desired degree of accuracy. However an ultra-finitist would not believe in the non-computable numbers. Those are real numbers that take up a point on the number line; yet can not be generated by any algorithm.

That's a respectable (yet definitely minority) viewpoint in math. However, Wildberger is still a crank. He's not a respectable finitist. I'm not familiar enough with his work to comment in detail, which is why I directed your attention to Mark Chu-Carroll's detailed debunking of Wildberger.

That's pretty much what I know about this. In standard modern math we accept the reality of the entire set of real numbers, which are admittedly very strange. Each real number is a point on the number line; and the real number line does do a good job of representing the classical idea of the continuum.

If one wanted to study ultra-finitism that would be a perfectly respectable thing do to. But studying the work of cranks would not be productive if you're trying to understand how contemporary mathematicians think about the real numbers.

Don't know if any of this helped ... but if you want to understand math, then study math; not cranks.

I can't comment on what your prof and/or textbook said. But if they didn't believe in irrational numbers and/or the continuous real number line, they would be far outside the mainstream of math.

13. ### CheezleHab SoSlI' Quch!Registered Senior Member

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I never said I didn't believe in irrational numbers. The example I gave just says that the rational line may not intersect the circle. And that the prof warned us to be wary that this was true. If you don't realize this, then it might lead to some mistakes. And that it was an example that was quite problematic. He did seem to agree with Wildberger and this was long before anyone had heard of Wildberger. I do remember the prof gave a couple examples where this was a problem and it came up in the homework problems. But I can't remember it all exactly. Do you disagree with the premise? I am sure that there are a few math types here that can verify it as fact. It sounds strange until you actually consider it. I consider it an interesting fact. Nothing more than that. Pure math was never my cup of tea.

14. ### someguy1Registered Senior Member

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Yes I do disagree. A line must intersect a circle or else miss it completely. It can't "hop over" or "drive through" holes in the circle for the reason that there are no holes in the circle. This is expressed technically as the completeness of the real numbers. There are no holes in the real number line.

That's why I went to great (and probably too wordy) lengths to explain that the real numbers embody our modern idea of the continuum ... the path of a point moving through classical Euclidean space. There are no holes or gaps. That's essential.

So if you take the unit circle and any line emanating from the origin, that line must necessarily intersect the circle at a point ( or two points, one in each direction).

You can also verify this algebraically by simultaneously solving the equation of the line and the formula for the circle. There's a point of intersection. The fact that the coordinates of the point of intersection may well be irrational is simply not relevant.

Note that all I'm doing is explaining the standard accepted mathematical orthodoxy. I'm not claiming that it's "true" in any generalized sense of the word, other than that it follows from our basic axioms of mathematics as they are understood and accepted by mathematicians today. I haven't refuted any alternate crank theories; all I've done is state the orthodoxy.

But I am trying to point you at the right mental picture. A line (or a circle) is a continuum, a continuous set of points with no holes in it. If your mental picture of the circle has holes in it, or is composed of a discrete set of points jammed "shoulder to shoulder," that's the wrong picture.

Is that our point of confusion here? The mental picture of the continuum?

15. ### CheezleHab SoSlI' Quch!Registered Senior Member

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745
Hmmm. Well, I am not sure what to tell you. But I will try one more time. Imagine a rational line y=mx where m is rational. Any rational x you put into the formula will yield a rational y. That is pretty easy to see. Now suppose that this rational line crosses the circle ${x^2} + {y^2} = 1$, plug a rational x into the circle formula $y = \sqrt {1 - {x^2}}$. Do you see that for all rational values of 0<x<1 not all values of y will be rational. So the rational coordinate is [x,mx] where y has to be rational, and the irrational coordinate of the circle [x.$y = \sqrt {1 - {x^2}}$] is not the same. So the question is, on the circle does every point with a rational x pair with a rational y of the form $y = \sqrt {1 - {x^2}}$. If there is an interesection then if the x values are the same, then the y values are also the same. y can't be irrational and rational at the same time.

That is probably not right, but I am not gonna try any harder to remember. Maybe someone will comment and confirm deny.

16. ### Fednis48Registered Senior Member

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All $m$ I'm about to refer to are rational by assumption. The problem with your argument is that while all lines of the form $y=mx$ will intersect the unit circle, it does not follow that every point on the circle will be intersected by some line of the form $y=mx$. As an easy example, the point $\langle 1/2, \sqrt{3}/2\rangle$ has rational $x$ and irrational $y$. From this, you can deduce that no line of the form $y=mx$ intersects the unit circle at $x=1/2$.

In fact, given how rare rational coordinates on the unit circle are, I feel confident in saying most lines of the form $y=mx$ intersect the unit circle at irrational $x$. I can actually only think of four points on the unit circle with rational $x$ and $y$: the top, bottom, left, and right extremities. Can anyone think of any others? Or is there a proof that those four are all there are?

17. ### CheezleHab SoSlI' Quch!Registered Senior Member

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Yeah I realized as I was writing that last reply that what I was saying was wrong for just the reason you said. Like I said the math class was 30+ years ago, but I remember the fact that some line crossed some circle but had no points of intersection. It was a remarkable thing and that is why I remember it. But old memories are sometimes wrong. Like Layman's memory of an old navy radar manual that convinced him that electromagnetic waves are beams of electrons. So maybe it is one of those situations. One other thing I remember is that the prof said that this proof was well known in the past but had just fallen through the cracks because it cast doubt on other areas of math. Hmm. Maybe he was a crank, or as I say, maybe I just misremember. It is just that the memory seems so clear, I can even see his face and him drawing the circle and line. Oh well, probably some malfunctioning brain cells.

18. ### Quantum QuackLife's a tease...Valued Senior Member

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Yes... I have heard this in the past as well...but I think it was more about entertaining the audience with a sort of fudgy application of the two concepts more than good math... a bit like squaring the circle type challenges.

19. ### CheezleHab SoSlI' Quch!Registered Senior Member

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745
I suppose it could have been a deception or a joke. But it was not presented that way. It was a 3rd or 4th year math analysis class that involved a lot of writing proofs in number theory and other areas like topology. The professor was kind of a jerk so maybe it was a deception to see if any body caught it. I was not the best student in the class so maybe the joke went over my head. Hmm. Does not sound very professional.

20. ### someguy1Registered Senior Member

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Even a line with rational slope contains plenty of irrational points. For example the line y = 5x contains the point (pi, 5pi). Because of the square root involved in the formula for a circle, most of the time even a line with rational slope will intersect the unit circle at a point that has at least one irrational coordinate.

21. ### CheezleHab SoSlI' Quch!Registered Senior Member

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745
Well, for a rational line all rational x map to rational y, and irrational x map to irrational y. But for a circle we know that some rational x map to irrational y and some irrational x map to rational y. But I am not sure, gotta get going. People to see, things to do, Friday night.

22. ### Quantum QuackLife's a tease...Valued Senior Member

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@someguy1,
my question:

"For randomness to be truly random, pattern formation must be possible" in other words there should be no constraint either way. Whether patterns form or not. However if a number sequence runs into millions of digits then pattern formation is a statistical necessity.

This is why I asked the question that I did.
Pi appears to have an ability when derived to in excess of a million digits no ability to form patterns with in that sequence.
This defies normal statistical requirements for the term usage of "randomness". [I believe]

Pi is not an accident, as it appears to be specifically devoted to prevent pattern formation.

So therefore one could conclude that the formation of Pi which is a mathematical "emulation" of intrinsic natural phenomena is indicative of some fundamental of universal structure. And is a pattern in itself, but only when taken holistically.

Example:

Challenge:

To formulate an algorithm that produces and "infinite" sequence of digits that MUST NOT allow the formation of any patterns what so ever.
My bet is that the only way to do this is to use Pi as no other method would be sufficient.

Last edited: Jul 27, 2013
23. ### Beer w/StrawTranscendental Ignorance!Valued Senior Member

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Euler's number?