Classical to Quantum
46 views May 3, 2024
The classical approach to integration on a circle involves transcendental functions, such as the "Gamma function", and irrational numbers, such as "pi". An alternate formulation, developed by Dr Kevin Limanta and myself, accomplishes the same job only in the context of finite algebraic operations involving rational numbers, and avoiding entirely any "completed infinite processes". In this video we compare these two approaches, and show that they are roughly equivalent (aside from the obvious logical superiority of the algebraic approach) and develop basic facts about the important Circle Integration Numbers, which appear when we integrate standard bi polynumbers over the unit circle. This allows us to compute recursively, by hand even, a table of values of these important numbers. 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, classical to quantum. We're having
0:04
a look at harmonic analysis on the circle with the intention of going to spheres and
0:09
applying this all to quantum mechanics. We've outlined a new approach to Fourier
0:17
series harmonic analysis on a circle from a purely rational point of view, avoiding all
0:22
notions of transcendental functions and irrational numbers, and basically anything that requires the
0:30
supposition that we're supposed to do an infinite amount of work in order to calculate stuff.
0:35
So, I want to compare today the classical versus the rational circle integration approaches. The
0:43
classical results are encapsulated by this formula going back to Fejér and Baker, and probably other
0:51
people, for the integral of a general monomial 𝑥2𝑘 𝑦2𝑙 over the unit circle with respect to
1:00
the invariant measure, which is normalized so that the total mass is 2π. It's given in terms of the
1:08
gamma function: 2 * gamma of 2k + 1/2 * gamma of 2l + 1/2 divided by gamma of 2k + 2l + 1. Now,
1:21
we want to contrast that with this new alternative formulation due to Kevin Lamata and myself, where
1:30
we introduce a linear functional on the space of polynomials satisfying just three conditions: a
1:35
normalization condition, a localization condition, and a rotation invariance condition. Then we claim
1:41
that these conditions are enough to specify I uniquely, and it actually does exist. And here is
1:48
this integration value on the polynomial Alpha2k Beta2l: it's some number which we're going to
1:55
call Omega KL because it's non-zero only if these exponents here are even. It's given by 2 * k
2:03
factorial * 2 * l factorial divided by (k + l)! * k factorial * l factorial * (k + l)! We're calling
2:11
these things Circle Integration Numbers. There's a whole table of them; k and l are natural numbers,
2:17
0, 1, 2, 3, etc. So, we have a whole table, and our claim is that these are really at the heart
2:23
of harmonic analysis on a circle, and they're really at the center of integration on a circle.
2:30
And it's a purely rational picture because the functional I has been normalized so that the
2:38
weight or the mass of the circle is represented by the integral of the constant function one,
2:45
and that's equal to one. So, there's no pies around, there's no transcendental functions
2:52
that need infinite processes in order to define. The story is entirely rational.
2:58
Today, what I want to do is I want to talk more about these important Circle Integration Numbers
3:02
and connect these two formulas so that you see that they are essentially equivalent. Okay,
3:08
they're possessing the same mathematical content except this one's in a rational context and this
3:13
is in a transcendental context. There's a big difference between those two things. So, let's
3:19
see how we go from the classical approach, how we can morph that into this more rational approach.
3:24
The gamma function is given by this integral, which is a transcendental thing. It involves an
3:29
exponential function, involves an integration. So, I put in quotes, this is not in my view an
3:35
exact function. Okay, it's only approximate because you can only calculate this thing in
3:41
general approximately. Nevertheless, it has some traditional properties. So, Gamma of k when k is
3:49
a natural number is (k - 1) factorial, and when k is a natural number, Gamma of k + 1/2 is something
3:56
like a factorial. It's (k - 2)(k - 3)(k - 5) going down by one each step until you get to a half,
4:06
that product times the square root of Pi. So, some more irrationalities there. And here then is that
4:14
formula of Fejér and Baker and other people. Okay, and we want to rewrite this thing. Okay,
4:21
so let's have a look at this term here: Gamma of 2k + 1/2. Well, the twos can cancel, and we
4:28
can think of this as k + 1/2. So, we're exactly in this position here. So, this Gamma of 2k + 1/2 can
4:34
be replaced with this product (k - 1/2)(k - 3/2) down to 1/2 times the square root of Pi. Okay,
4:43
and we do exactly the same thing with this term here: Gamma 2l + 1/2. That's equally
4:48
replaced with (l - 1/2)(l - 3/2) down to 1/2 with a square root of Pi. There's also a two out here,
4:55
and in the bottom, we have... Well, this is k + l + 1, Gamma of that, and that can be simplified
5:02
using this first relation so that's just (k + l) factorial. Okay, so there's two square roots of
5:11
Pi's together. They make a Pi, and we can bring that out here, join this two so we have 2 * Pi.
5:18
And now what we're going to do is we're going to sort of bring the common factors of two out
5:24
of these all these factors here. Okay, so we're going to take 2k out. There are k terms in this
5:31
product. Okay, when we do that, we're going to get 2k - 1, then 2k - 3 all the way down to 1,
5:38
and similarly, we're going to take a two out of each one of these factors. There's l of
5:43
them all together. So altogether, we have 2(k+l) in the denominator here, and this becomes 2l - 1,
5:50
2l - 3 all the way down to 1, and we still have the (k + l)! in the bottom. And now we
5:57
do essentially the same trick that I showed you in my last video. Okay, we convert these things
6:02
into factorials by sort of interjecting some more double factorials. So, we multiply this
6:09
thing here by 2k double factorial, in other words, 2k * (2k - 2) * (2k - 4) etc. down to two. Okay,
6:18
then when we multiply by that, we're going to get a 2k factorial, and on the bottom,
6:23
what we introduce to compensate is another 2k * k factorial, that product of even numbers,
6:32
you can take the twos out, and you get a k factorial times 2k. And we do exactly
6:37
the same thing with this product here. It becomes 2l factorial, we have to introduce
6:42
an l factorial and another 2l. So altogether, we have this 2(k+l) and another term just like it,
6:49
2(k+l), so together they make 4(k+l), and left over is the (k + l) factorial. There, great.
6:58
So what do we recognize? Well,
7:00
we recognize this is exactly... well, there's a 2 * Pi, okay, and otherwise, it's just Omega of k,
7:09
l. So aside from the renormalization of the integral giving the circle a mass of 2 * Pi
7:19
versus giving it a mass of one, our formula is in agreement with the classical formula,
7:29
and either of them can sort of be used to derive the other one. Okay, so, that's good. So that
7:37
means that our formula is consistent with the known facts, but again emphasize that this formula
7:44
here, without the 2 * Pi, is purely a rational thing, something you get just by finite algebra.
7:50
Alright, so this summarizes what we've just found. This is the classical integral with the d mu
7:55
measure that has weight to 2 * Pi. We found that that integral is the same as 2 * Pi * Omega(k, l),
8:01
and that is just the value of our integration functional I that's normalized, localized,
8:09
and rotationally invariant on the polynomial Alpha to the 2k Beta to the 2l. So this is
8:16
another way of thinking about how you integrate something like this on the circle, and here is
8:23
explicitly just to remind you again, the formula for Omega(k, l), these are the Circle Integration
8:30
Numbers, and there's a whole table of them. k and l in general are natural numbers starting
8:36
with zero. There are two important special cases that we need to look at.
8:42
The first is when one of these is zero, let's say l is zero. It's symmetric, so it's the same the
8:47
other way. If l is zero, well this 2l factorial is just one, this l factorial is just one,
8:55
and this k + l factorial is the same as k factorial, so we end up getting 2k factorial
9:02
over k factorial k factorial, and there's now 4k in the denominator so we get 1 over 4k
9:09
times this Central binomial coefficient 2k choose k. So if you have a Pascal triangle like this,
9:17
you know coming down like this, one, one, one, one, two, one, one, three, three, one, if you go
9:21
down right down the middle, okay, those are the central binomial coefficients. They're arguably
9:27
the most important binomial coefficients. They're right in the middle of the story. Okay, they have
9:31
all manner of combinatorial interpretations, they're very important combinatorial numbers.
9:38
The other thing we want to look at is the special case when the two indices are the same,
9:44
so Omega(k, k). Okay, so now there's going to be a 2k factorial here, another 2k factorial,
9:50
but there's also going to be a 2k factorial here which will cancel with one of those, and these
9:56
two are then going to be the same k factorial l factorial just becomes k factorial k factorial,
10:02
and here we're going to get 42k, so we're getting that same Central binomial coefficient because we
10:10
have 2k factorial over k factorial k factorial that's that Central binomial coefficient but
10:15
now the factor in front is 1 over 4(2k), so these things are closely connected but
10:23
there's a power of four that separates them and precisely Omega(k, k) is this one here Omega(k,
10:31
0) divided by 4k. So in this table, we can go from say the First Column the zeroth column to
10:43
the diagonal reasonably simply by multiplying or dividing by an appropriate power of four
10:52
and of course the First Column and the first row are the same because of the symmetry of
10:57
the table. So this is an important relation in the this world of circle integration numbers.
11:06
Let me just review the other properties that we're already familiar with: the reverse Pascal
11:11
property. I gave that to you sort of as an exercise. Let me just clarify why that is the
11:15
case. So if you have I of Alpha to the m Beta to the n and M and N are arbitrary natural numbers
11:22
here you can rewrite that by multiplying by Alpha square + Beta square because on the circle that's
11:29
the same as one it means that the I of this polom is going to be equal to the I of this
11:34
polom because they agree on the circle. Okay, and then by linearity you expand out you get I
11:39
of Alpha to the m + 2 * Beta the N + Alpha to I of Alpha to the m Time Beta to the n + 2 okay so
11:50
since Omega KL is by definition I of alpha to the 2K beta to the 2 L we get that Omega KL when you
11:59
replace M and N here with 2 K and 2 L okay then what you're going to get here is Omega k + 1 L and
12:08
here Omega k l + 1 so this is the Pascal property that says that if you have a table the Omega value
12:16
at any point is the sum of the thing just below it and to the right and the other two important
12:23
properties the Symmetry property you didn't change K andl it doesn't make any difference
12:27
and the normalization property that when K and L are both zero we're getting one okay now I can
12:33
show you a pleasant result that shows that the table of circle integration numbers is in fact
12:39
fully determined by these four properties that we've just talked about namely the normalization
12:48
property the Symmetry property the reverse Pascal property and this diagonal relation that Omega k k
12:56
is 1 4 the K * Omega k0 so that means that if you were on a desert island and you were trying to do
13:05
some integration on a circle and you remember there's some formula involves some factorials
13:10
and some powers of four but what exactly was it you might be able to remember these properties
13:16
because they're not so hard to remember this is maybe the trickiest one but it's not so bad and
13:23
the claim is that you can recover the table from just this information all right so let me convince
13:28
you why that theorem is correct and also show you how you can just explicitly recursively construct
13:34
this table starting up here in the top left corner okay so here is the normalization condition that
13:41
Omega of 0 0 is equal to 1 okay so that's the starting point now we want to continue sort
13:48
of in this direction going down like this along these cross diagonals recursively or successively
13:54
creating more and more of these entries okay let's look at this next cross diagonal this one
13:59
here what do we know about these two numbers well they first of all have to be equal because of the
14:03
symmetry of the table and secondly the reverse Pascal property says that these two guys have to
14:08
add up to one so obviously they have to be both a half so those are good okay maybe I'll Circle
14:18
them instead how's that okay so we we know these guys and we know this one here okay good okay now
14:23
we're at the next cross diagonal what do we know about these guys well it's symmetric so this one
14:29
has to equal this and we know that these two have to add up to this and these two have to
14:32
add up to this but that's not enough to determine these three things we need another condition and
14:38
that's the diagonal relation you see if you look at this column okay and you look at this
14:45
diagonal okay here is the entry uh Omega 1 0 here is the entry Omega 1 one what do we know about the
14:55
relation between these things we know that this one is one qu times this one 1 over 4 to the 1
15:02
one because we're in that row so we just multiply this one by a half which we've already determined
15:08
and we get this value so we know this value okay but now we know this plus this equals this so that
15:15
determines that and this is the same as that or this these two add up to that either one and we
15:20
get 38 okay let's look at the next one uh in this case we don't need to use that diagonal proper
15:29
property the diagonal relation because if we go to the middle of it okay there's not an entry exactly
15:36
in the middle there're these two that are sort of straddling the middle symmetrically and they
15:41
have to add up to 1/8 so because they're equal by symmetry we already can immediately deduce that
15:48
they're just half of this one and once we have these two well these two have to add up to this
15:52
one so that's determined and these two have to add up to this one so that one's determined okay what
15:57
about the next one that's the last one we do okay well again these two have to add up to this these
16:04
two have to add up to this these two have to add up to this and these two have to add up to this
16:07
that's not quite enough information to determine things we need one more piece of information and
16:13
that's this one here how do we get this one here well this one here is related to this one here
16:20
this one is what you get when you take this one and you multiply by 1 over 4 sared 4 SAR is 16
16:27
so 16 * 8 is um 128 you multiply this by 1 over 16 you get this one here so now that we have this one
16:35
well these two add up to this so we get this one these two add up to this so we get this one and
16:40
by symmetry or the same argument we go like this and now I hope it's completely clear we just keep
16:45
going so at the next level uh these two things have to be equal and they have to add up to that
16:50
so these are just going to be half of half of this one okay so just carry on exactly like that
16:59
so um that's cool I mean so really we're in possession now of if you like you can think
17:05
about it like this that the classical uh first year or second year course in calculus you want
17:10
to integrate powers of cosine and S over the circle well just the question of finding the
17:16
appropriate entry on this table and multiplying by two Pi presuming you have two Pi under your
17:23
control which you don't but let's just pretend okay so you find the corresponding value here
17:29
how well you can do it recursively or you can just use the formula it's all rational arithmetic
English
AllWatched