Classical to Quantum

47 views May 2, 2024

47 views • May 2, 2024

This video explores some of the remarkable consequences of taking a purely rational approach to the problem of integration on a circle. It represents joint work with Dr Kevin Limanta, who is currently at UNSW Sydney. I hope to get him to explain further aspects of this remarkable situation, especially how this rational theory brings together the integration theories for the rational numbers and for finite (prime power) fields. The crucial orientation is to let go of the usual "limit of Riemann type sums" that dominates the current approach to harmonic analysis and replace it with the more functorial approach coming from the Algebraic Calculus (the first course is online and available at Open Learning, and will transform your understanding of integration). In this video we show how some elementary computations strongly suggest that a rational approach is possible, by essentially evaluating some simple integrals, the rational versions of the integrals of cos^2(x) and cos⁴ (x) and cos² (x) sin² (x), around the unit circle, completely without any need for "completed infinite processes". (Let's give up religion in mathematics, friends!) The crucial numbers for the general integrals turn out to be what we call the Circle Integration Numbers. These are absolutely central objects in mathematics, and all students of the subject, and physicists too, should become familiar with them. We show how the array of these numbers again connects naturally with the Discrete Calculus (also laid out in the Algebraic Calculus One course) and a surprising symmetry relating to the Gregory Newton Interpolation formula. My research papers can be found at my Research Gate page, at https://www.researchgate.net/profile/... My blog is at http://njwildberger.com/, where I will discuss lots of foundational issues, along with other things. Online courses will be developed at openlearning.com. The first one, already underway is Algebraic Calculus One at https://www.openlearning.com/courses/... If you would like to support these new initiatives for mathematics education and research, you could also consider become a Patreon supporter. Thank you. Here are the Wild Egg Maths Playlists (some available only to Members!) • Algebraic Calculus One • Classical to Quantum (Members only)[ • Algebraic Calculus Two](https://www.youtube.com/playlist?list=PLzdiPTrEWyz5VLVr-0LPPgm4T1mtU_DG-) • Advice for mathematics exploration (M... • Solving Polynomial Equations (Members...[ • The Hexagrammum Mysticum (Members only)](https://www.youtube.com/playlist?list=PLzdiPTrEWyz5j1BJdXBw1MFst_nQAqzZ_) • Algebraic Calculus and dCB curves (Me... • Dynamics on Graphs (Members only) • Playlist • Triangle Geometry • Explorations with q-series (Members o... • Six: An elementary course in Pure Mat... • Maxel Inverses and Orthogonal Polynom...Here are the Insights into Mathematics Playlists: • MathHistory: A course in the History ... • WildTrig: Intro to Rational Trigonometry • Math Foundations • Wild Linear Algebra • Famous Math Problems • Box Arithmetic • Elementary Mathematics (K-6) Explained • Year9Maths • Ancient Mathematics • Wild West Banking • Sociology and Pure Mathematics • Sociology and Pure Physics • Old Babylonian Mathematics • Probability and Statistics: an introd... • Boole's Logic and Circuit Analysis • Universal Hyperbolic Geometry • Differential Geometry • Algebraic Topology • Diffusion Symmetry: A bridge between ...*****************

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Hello everyone, thanks for joining me on this adventure. So we're going from classical to Quantum with a view of understanding physics from a mathematical

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point of view. An important topic here that we're now developing is some basic harmonic analysis on circles and spheres, but in a very novel way where we emphasize

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rational approaches that allow us to stay within purely the algebraic framework,

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avoiding transcendental functions as much as possible, and also rational numbers.

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So today I'm going to talk about the Circle Integration numbers. These are a very surprising array of numbers that emerge from this rational point of view towards the

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harmonic analysis on a circle. They have remarkable properties; they're connected in interesting ways with classical formulas both from analysis, but it turns out also in

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combinatorics. And this is all work with my former student Kevin Lamont, and part of it

1:00

is in his master's thesis which he did with me and also in his PhD thesis which he also did with me.

1:06

Okay, so the main theorem that I told you about last time is this alternate version

1:12

of how we integrate on a circle. It's all having to do with a particular linear

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function along the space P of bipolar on the plane. So a bipolar is something in Alpha and

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beta polynomial in Alpha and beta if you like. And so we're looking at whether we can find a

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linear map from this space P to the rational numbers which has these three properties:

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First of all, it's normalized so that the function one gets sent to the number one. Secondly, it's localized so that if a polynomial happens to be zero when you restricted to the circle

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then I of that will be zero. And finally, it's rotation invariant that if you have

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some polynomial in the plane and you rotate it in the obvious way by composing with rotation,

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the I functional will not change. And the claim is that such a functional exists and it is unique,

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and that functional then is how we integrate polynomial on the circle. And specifically,

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we have this very precise formula for I of this monomial Alpha to the 2k beta to

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the 2l. We have even exponents here because if one of these exponents is odd then the I

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is easily seen to be zero. Okay, so this is the only case of interest really and

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the formula is that what we get is in terms of double factorials: (2k+2l)!!/(2k−1)!!(2l−1)!!

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Double factorial means that if you have an even number inside here like 10 then you

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have to multiply 10 * 8 * 6 * 4 * 2 going down by twos and until you get to two, and if it's an odd

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number like 11 then you have to multiply 11 * 9 * 7 * 5 * 3 * 1 down by twos until you get to one.

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Okay, so it's going to be pretty interesting for us to look at these numbers and sort of

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try to understand where this is coming from and to connect it also with traditional integration,

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and then I think some really surprising things are going to emerge here. So a good place to start with an investigation of this is just to work through some simple examples

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and see if we can get a feel for what this magical linear functional I looks like, what it actually

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does to some simple polynomial numbers. So the simplest one is just the number one. We know

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by the normalization condition that I of one is one. Okay, what about linear ones like Alpha and

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beta? Well, these guys are, you know, have odd powers so the I of them is automatically zero,

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and the reason for that is if you take something like Alpha and you rotate it by a half turn,

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okay essentially sending every point to its reflection in the origin, then what happens to

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such a polynomial is it basically is negated in value. Now if I of that is going to be a

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constant then we have that the value has to have the property that it's negative equals itself,

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and the only thing that has that property is zero, so it's pretty easy to see that that's the case. And also if we have some combination like Alpha times beta quadratic thing,

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since the alphas and betas are doing that, then they're still going to be zero. So for example,

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you could think about just reflecting in the x or the y-axis to determine that.

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Okay, so then what about these other quadratic ones, say Alpha squared and beta squared? What

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should I of them be? Well, pretty clearly, whatever I of Alpha squared is, it should

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be equal to what I of beta squared is because Alpha squared and beta squared are rotations of each other. So, but what else we know? Well, we know that Alpha squared plus beta squared

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equals 1 when we restrict to the circle, and then the localization property implies that

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therefore I of this has to equal I of that because the difference is zero on the circle,

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and so I of the difference will be zero, and this thing is linear so I of this has to equal I of one which is one. So we have these two quantities here by linearity

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we can deduce that I of Alpha squared and I of beta squared both have to be equal to a half.

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Okay, so we've been able to establish that pretty easily. Cubic cases, again, it's similar. It's not

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really with any interest because they get killed. So the quartic case is the next,

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so let's have a look at the quartic case. So can we find I of something like this Alpha squared?

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Well, because of linearity, this is I of Alpha to the fourth plus I of Alpha squared beta squared,

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but also because of this locality property, the Alpha squared plus beta squared equals

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1 on the circle so I of this has to equal I of just Alpha squared which we've just determined

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up here is equal to 1. So we've deduced that whatever I of Alpha to the fourth is,

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whatever I of Alpha squared beta squared is, those two rational numbers have to add up to a half.

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Now another thing that we might try is to say well why don't we just take Alpha squared plus beta squared and square it? We're going to get something that's of degree four. Of course,

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this thing here is equal to one. If we square it we still get something one when we restrict to the circle, so we know that I of that should equal one by the normalization condition. If we

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expand this thing out and use linearity, we're going to get I of Alpha to the fourth plus 2

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I of Alpha squared beta squared plus I of beta to the fourth, and then when we stare at this

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and realize well Alpha to the fourth and beta to the fourth are just rotations of each other so they're going to be equal. So we really have two of these plus two of these equals 1. Well,

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that's really the same condition that we had there. In other words, that I of Alpha to the fourth plus I of Alpha squared beta squared is going to equal a half.

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So that may have looked like it was a promising direction but it turns out that it doesn't give us any more information than just this did, so we still are not in possession

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of these individual values even though we know they're some. Well, let's try to utilize that

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rotation property of this functional. So let's take a point on the unit circle,

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say RS. And because we know how complex number multiplication works, we know that if we multiply

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by a point on the unit circle it's essentially a rotation. Okay, so the effect of rotation

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by this point RS on a general point XY is really just the same as complex number multiplication

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of this complex number with this complex number, so that's R * x- s * y and R * y + S * x. Okay,

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so if we say start with the polynomial Alpha to the fourth one of our quartic ones and we

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compose with this rotation, okay so it's going to send point XY to, well here's the x coordinate

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of that rotation, then when we take Alpha to the fourth we're taking that to the fourth power so in

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terms of just a polynomial what we've determined is that the composition of the original polynomial

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Alpha to the fourth with its rotation is the new polynomial R Alpha minus s beta to the fourth. And

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the rotation invariance of the functional I says that I of this had better equal I of that. Okay,

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so I of Alpha to the fourth is going to be I of this. Okay, so what is going to be I of this? Well, this we can expand this using the binomial theorem of course and then use linearity so we're

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going to get a term here R to the fourth times I of Alpha to the fourth then we're going to get a

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minus 4 R cubed s times I of Alpha cubed beta plus 6 R squared s squared times I of Alpha squared

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beta squared minus 4 R s cubed times I of Alpha beta cubed plus finally s to the fourth times I

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of beta to the fourth. Now some of these are going to be zero. This I of Alpha cubed beta is going to

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be zero because like the beta actually both Alpha and beta are appearing with odd exponents so I of

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that is going to be zero and similarly this one here is going to be zero. Okay, so there's really only three nonzero terms. There's this one here R to the fourth I of Alpha to the fourth, there's

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s to the fourth I of beta to the fourth, but Alpha to the fourth and beta to the fourth are going to have the same I value so we just group those together R to the fourth plus S

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to the fourth times I of Alpha to the fourth and then the other one that's remaining here is this one here 6 R squared S squared times I of Alpha squared beta squared but remember in

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the previous page we determined that there was a relationship between this one and this one, they add up to a half so we can rewrite this one here as 12 minus I of alpha to the fourth.

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Okay, so when we take all of this and sort of put the I of Alpha to the fourth on one side, okay,

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we have 1 - R to the fourth - s to the fourth plus 6 R squared s that'll all happen on the

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left-hand side and on the right-hand side, all that's left is this 6 R squared s squared. Okay,

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so this is going to be a relation that's true whenever R and S are on the unit circle in

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other words when R squared plus S squared equals 1. Okay, so maybe we can simplify

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this to evaluate I of Alpha to the fourth. Okay, so what we've just determined is that I of Alpha

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to the fourth times 1 - R to the fourth - S to the fourth plus 6 R squared S squared equals 3 R squared S squared, and we said that R squared plus S squared equals 1 so if we square it we

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get 1 equals R to the fourth plus S to the fourth plus 2 R squared S squared.

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Okay, so in this equation so we can rewrite it as I of Alpha to the fourth time well this

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1 - R to the fourth - s to the fourth by this equation here is just equal to 2 R

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squared S squared so we can replace this with 2 R squared S squared there's still a plus 6 R

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squared S squared and on the right-hand side 3 R squared S squared and then happily we get 8

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R squared s squared cancel with those r s s squ and we find that I of alpha to the fourth equal

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38 and so the other quadratic expression that we don't know I of Alpha squared Beta

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squared had better be equal to 1/8 because we know that the sum of these two equals a half.

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Well, that's pretty cool so we've been able to calculate these quartic values just by some

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algebraic manipulations and sort of judicious use of these three basic properties that we've

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assumed this I has, namely the normalization, the locality, and the rotation invariance. And

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having gone this far, it's reasonably clear that one could probably with some more cleverness

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extend this argument, actually calculate higher values for these this I functional.

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Okay, so we don't actually have to rely on some prior classical result we could determine this

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functional by first principles just with algebraic manipulations. Now I'd like to point out that

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the things that we've established already are just rational versions of reasonably familiar

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classical integrals that you probably have learned or dealt with in an undergraduate education. Okay,

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so this one here that we've just determined is really the rational version of this

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thing here you see because Alpha to the fourth what's alpha alpha is just the x coordinates like X okay and when you convert if you want to convert everything to radians

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angles going around the circle of course I don't recommend this there's all kinds of logical problems with that but if you did that then the alpha or the X would be replaced with cosine

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Theta and so this Alpha 4 becomes cosine to the 4th Theta we have to divide by 2 pi because this

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integral the standard integral is normalized so that the weight the total mass of the circle is 2 pi and we want the mass to be one so we have to divide by 2 pi so the content of what we've

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found is that this integral is 38 if you didn't have the 2 pi here well then the integral from 0

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to 2 pi of cosine to the 4th Theta D Theta is 38 times 2 pi or 3/4 pi and similarly this one

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here relates to the classical formula that the integral from 0 to pi of cosine squared theta

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sine 2 Theta D Theta is 1/8 times 2 pi and these are quartic versions of the simple

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thing that we already established that I of alpha squared equals a half that was an analog of this

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very familiar integral that the integral from 0 to 2 pi of cosine 2 Theta D Theta equals pi.

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Okay, but we prefer to write it this way here because we're renormalizing this thing the total mass of the circle as far as we're concerned is more naturally one. Okay, so you can see that

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what we're doing here is non-trivial that we're somehow able to get these integrals without any of the machinery of classical calculus. Well, there's no re and sums there's

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no subdivisions of the circle into little pieces there's no taking limits of some kind. Okay,

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we've avoided all of that but we're still getting the correct answers in fact rational

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numbers because we've oriented ourselves to have a purely rational point of view. See there's some

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advantages in thinking rationally in looking for a rational way of approaching classical subjects.

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Traditional mathematicians are maybe not inclined to look this way because they're happy with the transcendental irrational approach you that's what they're used to they don't see anything

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wrong with that but I see that as seriously wrong. I need to find a better way and if you

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need to find a better way well then you're more likely to look for it and find it you can join me.

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Okay, so it's time to officially introduce these rational numbers that are going to be playing this

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central role for us here in harmonic analysis on Circle done rationally we might call them

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Circle Integration numbers CI numbers for short. Ω_kl=(2k+2l)!/(2k−1)!(2l−1)! And our convention

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just to remind you is that 0 double factorial is one and in case we put a zero in here we get

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a - one double factorial that's one also that ends up being consistent so this is a rational

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number not an integer it's a rational number like the 3 over 8 that we found or 1 over 8 or 1/2.

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However, there's an alternate form for this which actually is preferable for us

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and connects more naturally with sort of more familiar notation involving ordinary

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factorials not these double factorials. So to go from here to here is a nice

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and you'll be able to do that once you look carefully at this example so let me do this example for you which illustrates exactly what we're going to do so let's look at Omega 5, 8.

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Okay, so that's the relevant thing. If you wanted to integrate Alpha to the 10 times Beta to the 16,

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okay, so what is this thing here? Well, according to this, it's 2K - 1, so 9 double factorial,

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2L - 1, so that's 15 double factorial over 2K + 2L, that's 26 double factorial. And what is

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9 double factorial? Well, it's 9 * 7 * 5 * 3 * 1. And what is 15 double factorial? Well,

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it's 15 * 13 * 11 all the way down to 1. And what is 26 double factorial? Well, it's 26 * 24 * 22

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all the way down to 2. And now I'm judiciously just going to rewrite this a little bit but

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introducing some other factors in the numerators and denominators which will allow us to convert

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these double factorials into ordinary factorials. So first of all, let's have a look at this. Let

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me actually look at this denominator first. Okay, so this denominator 26 * 24 all the way down to 2,

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so they're all even numbers, right? So we could take a multiple of two out of each

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and every one of them. In fact, how many are there? Well, that's 2 * 1, and that's 2 * 13,

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so we're having 13 numbers in this product. If we take a factor of two out of each one of them,

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we've taken all together 2 to the 13th out, and what will be left is 1 * 2 * 3 all the way up

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to 13, in other words, 13 factorial. Okay, so we've rewritten the denominator. Let's look at

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the numerator now. Here's the 9 * 7 * 5 * 3 * 1. I'm going to introduce this thing here,

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which is actually 10 double factorial, 10 * 8 * 6 * 4 * 2. But to compensate, I'm going to write

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the same thing in the denominator, but I'm going to do the same thing that I did to the 26 double factorial. I'm going to take a multiple of two out of each of these five factors, so I get 2

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to the 5th, and then what's left over is 5 * 4 * 3 * 2 * 1, in other words, five factorial. So

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this thing here exactly balances this thing here, which I've inserted, but I just prefer to have it in this form here and have prefer to have this in this form in the numerator. Okay, over here is 15,

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13 is the 15 double factorial as before, and now I'm going to insert in the numerator 16

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double factorial, 16 * 14 * 12 all the way down to 2, and correspondingly down here is the same

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16 double factorial, but now written as 2 to the 8th * 8 factorial. There's 8 terms here.

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If I take a two out of each of them, I get 1 * 2 all the way up to 8, so that's that thing there.

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And now what I see is that this thing here, together with this thing here, is just a 10 factorial. Here are the odd ones, here are the even ones, together they make 10

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factorial, and here and here, these two multiplied together is 16 factorial, that's the odd ones,

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those are the even ones. So we have 10 factorial * 16 factorial in the numerator,

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altogether in the denominator, well, we have 2 to the 13 here, we also have a 2 to the 13th here,

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so that's altogether 4 to the 13, and we still have a 5 factorial and an 8 factorial and a 13

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factorial. So what are we getting here in terms of the original things here? Well, we're getting a U,

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a 4 to the sum of these things, 13, we're getting a 5 factorial, an 8 factorial, and a sum,

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the sum 13 factorial. And in the numerator, we're getting a 2 * 5 factorial and a 2 * 8 factorial.

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So that's the general rule, the Omega KL is 2K factorial * 2L factorial over 4 to the K + L times

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K factorial times L factorial and K + L factorial also in the denominator. So an excellent exercise

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to write out a proof of that, following exactly this example. Okay, so here is the start of a

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table of the circle integration numbers, these Omega KL's, so K down here, L across there,

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starting with 0, 1, 2, 3, 4, 0, 1, 2, 3, 4. Okay, so here's Omega 00, here's Omega 01,

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Omega 02, etc. So this one here and this one here, we've already seen that that's what you

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need when you're integrating Alpha squared or Beta squared. These ones here, we've also seen what you

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need when you integrate Alpha to the 4th or Beta to the 4th or Alpha squared Beta squared. Well,

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here are some more, allowing you to integrate quickly higher monomials on the unit circle. Now,

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so here's the formula again, just to remind you, it's 2K factorial * 2L factorial. In the

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numerator, in the denominator, 4 to the K + L * K factorial * L factorial and K + L factorial.

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Now, there's much to say about this. So in my view, this is a central table for the subject

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of harmonic analysis, and I think, in fact, that electrical engineers who are being taught Fourier

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series should see this table and appreciate it. This table allows you to integrate on the circle,

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okay, integrate polynomials on the circle or in a classical setting, it allows you to integrate

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products of cosines and sines. In other words, polynomials and cosine theta and sine theta on

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the unit circle. So, for example, if we wanted to integrate, say, cos to the 4th times sine

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to the 6th, okay, they're both even, so we're going to get something nonzero. 4 and 6 are the exponents, half of those are 2 and 3, so we'd look into 2 and 3. If we integrate in our sense using

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the I functional, we get 3 over 256, that's our integral of cos to the 4th sine to the 6th over

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the unit circle. But the standard formula would be that you'd want to multiply that by 2 pi to

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correlate with a measure of 2 pi, so it would be 6 pi over 256, or, you know, divide a factor

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of two. Okay, so there's much to be said about this, and I will have to get Kevin to say some

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things about this, such a beautiful thing. Okay, but first of all, there's some obvious things to

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observe about this table. Um, the most obvious, I suppose, immediately is that it's symmetric. And that's kind of obvious. The role of the K and L are completely interchangeable, so this

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is symmetric. Okay, that row there is exactly the same as that column, etc. Something that's more

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subtle but something that you would observe pretty quickly if you stared at this for a few minutes is that this table has a remarkable property which is somewhat reminiscent of the Pascal triangle.

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So let me remind you. This was a Pascal table with one here and one, one and one, two, one, and one,

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three, three, one, and one, four, six, four, etc. Then this six here would be the sum of this three

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and this three. Okay, there's a three here, there's a one here. There'd be a four here. So any entry is the sum of the entry above it and to the left. That's with an ordinary Pascal triangle.

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This thing is related to the Pascal triangle, but it has a kind of a reverse Pascal property in that

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an entry like this is actually the sum of the guy below it and to the right. This 1/8 is a sum of

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1 over 16 and 1 over 16. The half is a sum of 1 over 8 and 3 over 8. This one is a sum of these

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two. This one is the sum of these two. This one's the sum of these two, etc. And it's a very nice

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exercise to see why that's true. It's not hard to see at all. In fact, we sort of already use that

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in some sense in one of the arguments in today's video. So that's a really great exercise to think

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for a little while and convince yourself there's a very natural and easy reason why that's true,

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but it's a very powerful fact. And one of the interesting consequences is that we

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are immediately connecting with the algebraic calculus, which makes us very excited. In the

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algebraic calculus, we spent quite a bit of time on the discrete calculus where we're considering

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sequences and we have difference operators and so on, right? So we like to think of, say, a sequence

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given as a vertical thing like this, and then we take differences, you know, this minus this, and we write it there or there, and we're doing something similar here, right? The difference

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between this and this is this because this is the numbers that when you add, added to this,

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you get this, right? So this minus this is this, this minus this is this, this minus this is this,

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this minus this is this. That's just another way of saying what this reverse Pascal property is, but that's telling us that these various columns are more or less just successive differences of

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the previous columns, except that it's a backward difference instead of a forward difference. So

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ordinarily in the discrete calculus, we take this one minus this one and this one minus this one, etc. Now, you may remember if you've done the algebraic calculus course that there's a sort of

27:32

a Gregory Newton formula which is really beautiful which starts with a sequence like this and does

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the sort of successive differences, and then these coefficients that we end up with are a kind of

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very powerful summary of some of the information required to generate this sequence. Okay, that

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in certain situations we could find polynomial equations for this sequence, if such a thing

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exists, by sort of looking at the appropriate ones over here and hopefully eventually they get zero at some point, may remember that, okay? So notice that in this case here what we're getting

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is that this sort of top row here generated by these successive difference operations has the

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remarkable property that it's actually the same as the original sequence column that we started with.

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It's another consequence of this reverse Pascal property, but it's also combined

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with the evidence symmetry of the table to begin with. So that's actually a very strong constraint,

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those two things, the fact that it's symmetric and that it has its Pascal property really is

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something very special about this table. But it turns out that there's something even more

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remarkable about this table, that in fact it connects in a wonderful and surprising

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way with a largely sort of unsolved mystery in combinatorics with interesting number theoretical

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consequences. So that was one of the things that Kevin investigated, and it turns out that,

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you know, speaking of number theory, that this entire story of harmonic analysis on the circle that I'm presenting you here it turns out miraculously to have a direct analog

29:28

over finite fields. Okay, and it turns out as Kevin has shown that essentially the same

29:37

formulas, more or less the same formulas that are applying here also work in the prime field

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case of a finite field given by some prime or even some prime power. So there's a remarkable

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kind of bringing together of the discrete and the continuous by this table here. This table somehow

30:00

is looking at the continuous world and also the discrete world and somehow incorporating both

30:06

of them into a big harmonic analysis. So lots to think about here and if you're a physicist,

30:14

this should be pretty important. Okay, this should be of considerable interest.

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