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Classical to Quantum | Harmonic Analysis on a Circle: simplified! | Wild Egg Maths

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Let's now see how the complex basis for the polynumbers on the plane allows a much simplified view of harmonic analysis on the unit circle. We are able to recover some important foundational results about the decomposition of a general homogeneous polynomial into a harmonic part and another part which involves a polynomial of lower degree. Only elementary algebra is really involved here: in fact at this stage we don't even require any theory of integration on the circle. In fact we manage to solve the Dirichlet problem for the unit circle here in a very simple way. 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 ...*****************

0:00

Hello everyone, thanks for joining me on this adventure. We're learning physics, but actually,

0:05

we're taking a little detour now and reviewing some important harmonic analysis on a circle, with

0:12

a view to extending it to spheres, and in fact, extending it also to hyperbolas and hyperboloids

0:18

later on. So, our emphasis has been that we're going to shift gears, looking at the plane from a

0:24

complex number point of view. So, we've changed coordinates. So, instead of having sort of X

0:29

and Y, or correspondingly Alpha and Beta basis elements that generate the algebra of polynomial

0:36

numbers, we've changed gears. We've introduced U equal to Alpha plus I Beta and V equal to Alpha

0:42

minus I Beta, which is, of course, just the complex conjugate of U. And then, we're going

0:46

to use these guys here as the generating set for the polynomial numbers. Inverting the story,

0:55

we get Alpha equals U plus V over 2 and Beta equals U minus V over 2 I. So, we can go from

1:00

UV expressions to Alpha Beta expressions and vice versa, back and forth. And then, here is the story

1:06

that emerged, where we have a linear basis for our polynomial numbers starting with degree 0,

1:13

degree 1, degree 2, degree 3, etc., on any row, just decreasing powers of U, increasing powers of

1:20

V. And what makes this so appealing is that when we act on one of these typical elements, say U to

1:29

the m V to the n, we get 4 times m times n times U to the m minus 1 times V to the n minus 1. So,

1:37

both powers just come down by one, and we get this multiplicative factor here. Alternatively,

1:44

if we take the factorial basis point of view, just a slight renormalization, so we have U the m V

1:51

the n underscore, which means that there's really implicitly an M factorial N factorial there. Then,

1:57

in that case, the pass end just for times the one that you get by dropping the lines. Okay, and that

2:04

has the consequence then that the harmonic ones, the harmonic polynomial numbers in this case,

2:10

are just the top fringe of this triangle which extends, of course, further down consisting of

2:18

pure powers of U or pure powers of V or just one by itself. Okay, so this is a very attractive

2:24

setup. And now we're going to utilize this to show you some fundamental facts about Fourier

2:32

analysis on the circle, okay? And these facts are usually developed in a very different way,

2:38

usually with prior theories of integration and a prior sort of investigation of the properties of

2:43

harmonic functions and so on. But we're going to really cut to the chase and just see that just

2:49

elementary algebra and basically the sort of the geometry of this little triangle here allows us

2:55

to extract all kinds of really important things. And I'm going to express these things in a way

3:00

that's almost immediately generalizable to higher dimensional spheres, okay? So, we're going to see

3:06

the same patterns, the same phenomena appear in higher dimensions where, of course, the situation

3:11

is much more complicated. Okay, so let's get going on this. Now, it's really useful to identify an

3:16

important special polynomial, let's call it Q, which is U times V or Alpha squared plus Beta

3:24

squared, the sort of rotationally invariant polynomial which is quadratic. Okay, so here

3:31

is our first theorem. So, if we have P an element of CPN, which is complex polynomial numbers of

3:38

homogeneous degree n, so all the terms are degree n, then there are unique polynomial numbers H

3:46

which are harmonic, so H is in Hn, and R which is in CPN minus 2, so the space of polynomial numbers

3:56

two levels down in degree, such that P is equal to this harmonic one plus this Q, which is quadratic,

4:06

times this R. Okay, so this being homogeneous of degree n minus 2, this being homogeneous of

4:12

degree 2, means that the product here is still homogeneous of degree n. So, everything still is

4:17

of degree n in this equation, but we've isolated P and sort of extracted a harmonic part of it,

4:23

which is a rather special thing, and then the sort of the remainder, it's almost like a division kind

4:29

of thing, the remainder is Q times R, something of lower degree. Okay, let me illustrate this. Okay,

4:37

so you see why it's true. So, suppose that we have P in CP4 to be explicit, all right? So,

4:43

what does that mean? Well, it means it's a linear combination of... Here are the various degree four

4:49

polynomial numbers: U to the fourth, U cubed V, U squared V squared, U V cubed, and V to the fourth,

4:54

and we assuming we have some coefficients of that thing, right? So, there's a typical degree

4:59

four thing. Then, how are we going to write this thing as a harmonic plus something else? Well,

5:05

we know exactly what the harmonic ones look like, right? The harmonic ones are just linear

5:10

combinations of these two. So, these are the guys which are on the outside of that row, so that's

5:17

H. H is A times U to the fourth plus E times V to the fourth. And what's left over? Now, we see it

5:25

obviously has a factor of U times V. Okay, these guys here are the ones that have either powers

5:33

of U or powers of V. All the other ones have to have a power of U and a power of V. In particular,

5:38

they are divisible by this Q. Okay, so if we take the Q out of here because that's what we want to

5:44

do, then it's sort of obvious what we have. We just have B times U squared, we've taken UV out,

5:50

plus C times UV, we've taken UV out, plus D times V squared, we've taken UV out. So, it's automatic

5:58

why this is true once we look at it from this perspective, this triangle of coefficients

6:04

with respect to this complex basis, and our prior understanding of the simple nature of the harmonic

6:11

polynomial in this particular basis. And this is, of course, a very general argument. Alright,

6:17

so our next theorem says that if we take again this polynomial of degree n, homogeneous,

6:24

then there are unique harmonic polynomials H sub m such that we can write P sub n in the following

6:31

way as a sum of terms involving products of these harmonic ones with powers of Q. So, H sub n, which

6:42

is the same degree, plus Q times H sub n minus 2, plus Q squared times H sub n minus 4, and so on,

6:49

the powers of Q increase, the powers of H decrease by two each time until there's some final one,

6:55

say Q to the k times H to the n minus 2k, and this equality, k is less than or equal to n/2,

7:01

has to be valid. And it's sort of obvious why this is true, right? Because the previous theorem

7:05

asserted that we can write P sub n as H sub n plus Q times some other thing, something R, which is of

7:11

degree n minus 2. But then, we just apply the same reasoning to that R, and we write it as a harmonic

7:19

polynomial of its degree plus something else of degree 2 less times another power of Q. So, we

7:26

just sort of march down here and collect powers of Q, and we extract harmonic polynomial numbers. So,

7:35

let's see how this sort of works. Okay, so again, going back to that example that we just looked at,

7:41

here is some typical degree 4 polynomial number in our space. And we've already found what the

7:49

harmonic part of this is. The H4 is just this term plus this term. Okay, so let's imagine taking

7:54

those out. Now, the remainder. The remainder we want to write it as Q times something harmonic

8:01

plus Q squared times something else harmonic. Okay, so the remainder had these two things here,

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but once we've taken the UV's out here, then we have left over here A times B times U squared,

8:12

and over here, we have D times V squared. Okay, and so those are harmonic, so we can take those

8:20

things out and call that H2, right? That's going to be of degree two, where if the original one

8:27

was of degree four, and there's still something left over, and that's associated with this term.

8:33

And if we take Q out of this twice, so UV squared out, well, then what's left is just C. So, it's

8:41

really instructive to see what's going on here in this little triangle of coefficients, okay? It's

8:46

actually very simple, it's like a high school kind of stuff. What we're doing is we're starting with

8:51

a polynomial number which is a linear combination of these guys here along this row here, okay? And

8:59

the first thing that we're doing is we're sort of peeling off the harmonic part. And then,

9:03

the remaining things, what we're doing is we're extracting the UV term, right? And that

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essentially takes us two rows up in this triangle. So now, we're getting a combination of these guys

9:15

with these same coefficients, okay? So there's a B here and a D here, that's nothing there, but this

9:23

thing here can still be moved up. We can still go up by taking out another factor of UV, and then we

9:30

get just this thing here, with we'll have a factor of C. So, although what ends up happening is that

9:36

these coefficients kind of bubble up to the top of the column in which they're in, and then these

9:45

various pieces are just the sort of the harmonic pieces that we see on the boundary as we read this

9:52

thing going from the bottom up. All right, so let's take that result and just extract from it

10:00

an important object called the Pon-Harmonic or the Pon-Integral of a polynomial. So, if P sub n is,

10:10

as before, homogeneous of degree n, and here is that decomposition that we've seen in terms

10:17

of a sum of harmonics times powers of Q, Q is UV, which is Alpha squared plus Beta squared,

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then we're going to define the Pon-Harmonic or the Pon-Integral of P sub n, denoted by P with square

10:33

brackets P sub n, to be what we get when we just write down this same sum but without the Q's. So,

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we just take the Q's away. So, we just have H sub n plus H sub n minus 2, and so on,

10:47

up to H sub n minus 2K. This is then obviously a harmonic polynomial. That means it's killed by

10:54

the Laplacian because all of the ingredients are harmonic. Now, this thing here has an important

11:01

additional property with respect to its relation with P sub n, namely, it's the unique harmonic

11:08

polynomial that actually agrees with P sub n when evaluated on the unit circle S1. And

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that's an immediate consequence of the fact that this Q polynomial that we're looking at,

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this UV or Alpha squared plus Beta squared, if you restrict that to the unit circle, well,

11:25

obviously you're getting one because the unit circle is, you know, those X and Y such that

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X squared plus Y squared equals one. So, that polynomial is identically one on the unit circle.

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That means if you evaluate this thing at a point on the unit circle, and you evaluate this thing at

11:41

a point on the unit circle, you are going to get the same answer. So, if you're just on the circle,

11:46

these two things are not distinguishable. They're exactly the same polynomial when restricted to the

11:52

circle. But this one has the very special property that it's harmonic. So, what we've done here,

11:59

in fact, is solved a major problem in harmonic analysis on the plane, the Dirichlet problem. So,

12:06

what is the Dirichlet problem? Well, it's a famous problem that's usually stated as:

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given a continuous function in quotes, okay, say F on the circle, the unit circle, but there are

12:18

various extensions, maybe to higher dimensions or in fact to other curves, suitably defined curves

12:24

in the plane, but let's just stick to a circle, find a continuous function on the closed ball. So,

12:30

if this is the circle, then maybe B, think of that as the open ball, so those interior points A B,

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so that A S plus B S is less than one. So, what we ordinarily say is the interior of the circle,

12:45

then the closed ball just refers to the B, the interior together with the boundary S1. Okay,

12:53

so the problem is, given a function F on the boundary, represented by this sort of green wiggly

12:59

line representing values on the circle, we want to extend that function to a continuous function on

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the closed ball, which is first of all, harmonic inside, harmonic so you know it's killed by

13:12

the Laplacian, and agrees with F on S1. So, it really is an extension of the given function on

13:18

the boundary. So, that's the Dirichlet problem. And if you take a course in harmonic analysis,

13:23

you might come across this solution to the Dirichlet problem, which is in terms of something

13:29

called a Poisson kernel or Poisson integral. So, the Poisson kernel, in this situation,

13:34

there are lots of different Poisson kernels, but for this situation with this circle, it has

13:39

a reasonably pleasant form. So, it's a function of two points, X and Z, X is in the interior and

13:47

Z is on the boundary. Okay, then the definition is that it's one minus the quadrants of X, so the

13:53

coordinates of X, just the ordinary quadrants in the plane, so sum of squares of coordinates. So,

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we take one minus that in the numerator, and down here in the denominator, it's the quadrants of X

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minus Z, so we look at this vector between this point X, which is interior, and Z, which is on the

14:09

boundary. It's a vector, and that has a quadrants also. So, if X is sort of, you know, close to the

14:16

circle, well, then this quadrants will be very small, and so the whole thing will be rather big.

14:23

It's giving a rather large contribution. So, what this represents is roughly, um, we're integrating

14:29

over the circle, so Z is running around the circle. Okay, D mu represents a so-called

14:35

invariant measure, okay, on the circle with total measure 2 pi. That's the usual story. Okay,

14:43

and then what we're doing is we're integrating the values of F on the circle times this kernel

14:51

function. Okay, so what you end up getting, if you evaluate this for a given X, is a certain value U

14:58

of X. This is the usual story, and it requires the usual integration theory of the circle. And

15:04

I'm going to explain to you why that's actually a very seriously problematic issue, okay? Very

15:09

seriously problematic issue. So, physics students flag this, all right? Things are not as pleasant

15:16

and logical as you may hope that they are here in the world of pure mathematics. Okay, but anyway,

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so that's how this U of X is defined. For X in B in the interior, and then on the circle,

15:28

it's just defined to be F of X. So, this is the classical solution to the Dirichlet problem. Now,

15:36

what I'm showing you is a completely different approach, which doesn't involve any integrations

15:41

or kernel functions, but it's restricting ourselves to the world of polynomials. Okay,

15:49

an analyst would say, "Yeah, but I'm working with continuous functions." Okay, and I would object

15:55

very strongly to that because I don't believe that the notion of continuous function, as

15:59

is usually promoted by analysts, makes any shred of sense. It's just not a well-defined notion

16:05

mathematically, but we'll talk about that at some other point. Okay, but anyway, I'm just pointing

16:09

out that there's an elaborate established theory, and what I'm providing you here is an alternative,

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much simplified, completely algebraic, which admittedly just works in this nice, proper realm

16:21

of, you know, polynomials or polynomial numbers in the plane. Okay, so let's go through an example,

16:27

so we see exactly how this works. So, we're going to take an explicit polynomial of degree 4,

16:33

this one here. 4 Alpha to the 4th plus 3 Alpha cubed Beta minus 16 Alpha Beta cubed. That's a

16:39

polynomial on the plane, and we're interested in applying the Dirichlet problem story to it,

16:46

finding a harmonic polynomial which agrees with this polynomial on the unit circle but

16:53

is harmonic in the interior. So, our business first of all involves changing coordinates to

17:01

the complex basis. So, remember, we know exactly how to replace the alphas and the

17:06

betas with U plus V over 2 or U plus V over 2i. So, we do that. So, a little bit of algebra,

17:13

your computer does it immediately, and we get this degree 4 expression in powers of U and V. Okay,

17:24

1 over 16 minus i times U to the 4th plus 1 plus 2i times U cubed V plus 15 over 8 times U

17:27

squared V squared plus 1 minus 2i times U V cubed plus 1 over 16 plus i times V to the 4th. Okay,

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and now we're going to extract those various harmonic parts. Okay, so the first part is

17:39

the H4, the degree 4 stuff. We just take these pure powers, okay? There they are right there.

17:47

But we'd like to sort of reconfigure that to bring it back to the alpha-beta coordinates. So,

17:54

we do an inverse transformation. We replace the U with Alpha plus i Beta, and the V with Alpha

18:00

minus i Beta, and expand that out, and we get this polynomial in Alpha and Beta. 1/8 Alpha to the 4th

18:10

plus 8 Alpha Beta minus 3 Alpha cubed Beta minus 8 Alpha Beta cubed plus 1/8 Beta to the 4th. And

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you can check that this is a harmonic polynomial in the sense that it's killed by the Laplacian.

18:24

That's H4. So, once we've extracted that, well, what's left is this here. And then,

18:29

remember what we do is we sort of peel the next layer out. So, these ones here, we take a UV out,

18:35

and we're left with 1 plus 2i U squared plus 1 minus 2i V squared. We again reinterpret that

18:42

in terms of alphas and betas, and we get this quadratic expression, 2 Alpha squared minus

18:45

8 Alpha Beta minus 2 Beta squared. You can check that that's harmonic also. And finally,

18:51

there's one thing left, this thing here. Okay, we pull out the UV squared, and we just have 15 over

18:57

8. That's also harmonic. So, now, what is the Pon integral of P sub 4, or maybe we'd like to

19:08

call it the Pon harmonic of P sub 4? Well, it's H4 plus H2 plus H0. So, we just have to take the

19:15

sum of this one and this one and this one. There it is there. This is a polynomial of degree 4,

19:21

but it's no longer homogeneous. Okay, it's got different degrees. It doesn't look very close to

19:28

P4. So, you might have a hard time guessing that it's this thing here that solves the Dirichlet

19:33

problem for that polynomial. But this has two important properties. First of all, it's harmonic,

19:39

and secondly, that it has the same values on the circle as does P4. And we can check that P of 3/5,

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4/5 is this number here, which also is the same value that you get when you replace

23:38

alpha and beta here with 3/5 and 4/5 and a different one, say, 5/13 minus 12/13. That's

23:44

another Pythagorean triple. If we evaluate both of these expressions at this point,

23:49

we get the same fraction there it is there. Okay, so, we start with polynomial P. It's

23:57

defined on the entire plane, but we really are only interested in it as a function on the circle,

24:02

because we're interested in functions on the circle. And now, we have a kind of canonical

24:06

way of writing it, because there's lots of different polynomials that have the same

24:10

value on the circle. What we're doing is we're extracting maybe the best one, okay? The one

24:17

which has the additional property that it is harmonic. That's killed by the Laplacian. So,

24:25

that's a very nice story. We see it's purely algebraic and turns out that it extends very

24:31

naturally to spheres, in fact to hyperboloids too, in some fashion, as we eventually see.