Advice Maths| The conjugation connection between the exp polyseries and Hermite on-maxels

In studying Harriot Pascal maxels in Algebraic Calculus One, we realized that a suitable conjugation simplifies their study significantly, and sheds light on the close connection between the exponential on-series and the Binomial theorem and associated coefficients. Using the same reasoning, we can unravel some of the mystery with the Hermite polynomial maxel, where we focus on the probabilists version of these polynomials.

Again we see the power of this two-dimensional linear algebraic approach to the rich number theory contained in many families of classical orthogonal and related polynomials (or as we prefer to say -- polynumbers!)

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0:01

Hello everyone i'm Norman Wildberger thanks for joining me on this adventure

0:05

in my last video i showed you a lovely connection between the exponential poly series and the Pascal

0:10

Harriet Maxel we want to take that same understanding now and apply it to the

0:17

Hermete polynomial maxel situation we think we can do this because there's this fundamental

0:24

beautiful relation that seems to occur with the Hermete Maxel which also occurred with

0:30

the Pascal Harriet matrix namely that if you take one of these top left blocks okay

0:37

this matrix here for example and and you take its inverse then you get basically the same

0:44

matrix back again except with some checkerboard pattern of minus ones so what we're going to do is

0:50

we're going to investigate the same kind of thing do a little bit of exploring okay and see what we

0:55

get so we're going to start with a bigger example so instead of that one there i'm going to just

0:59

choose this slightly bigger one let's call it h all right so then my scientific workplace

1:05

program can deal with it nicely so first thing i'm going to do is that let's clear

1:10

our definitions and i'm going to make this h here okay definition so h is there and i'm going to

1:19

try to see if we can conjugate this okay by something suitable to get something simpler that's

1:27

exactly what happened in the Pascal Harriet story and okay you can conjugate by you know any matrix

1:34

in principle but in the Pascal Harriet story we restricted ourselves to conjugating bi-diagonal

1:40

matrices which have particularly simple effects sort of multiplying rows or columns by just

1:47

constants depending on whether you're multiplying on the left or right so let's introduce a generic

1:53

diagonal matrix of the same size as this let's call it d okay so there's d let's define that

2:00

and let me just uh let's just check maybe here check that indeed this h matrix does have this

2:07

remarkable property that we're really interested in understanding so h inverse there it is there's

2:14

h inverse it's exactly the same thing as this thing here without these distinctive

2:20

minus signs uh along sort of the odd things but we also know that you know

2:24

that this thing here to start with has already a checkerboard pattern actually actually the odd uh

2:29

entries down here in this lower triangle thing are already zero so it's it's really the check a

2:34

checkerboard pattern of science where we're going to this uh first uh non-zero sub diagonal so it's

2:42

really the second sub diagonal here that that were that we're talking about and then the fourth

2:46

sub-diagonal and then the sixth and so on okay so there's h inverse you know it's the same pattern

2:51

as before so we don't really need this okay so what are we going to do now well we're going to um

2:59

multiply h by d and its inverse and i'm going to multiply by d inverse and then h and then d sort

3:07

of motivated by what happened in our last video we we found that it was maybe easier to have the

3:14

variable entries on the right hand side so let's compute d inverse hd

3:22

or that's not our computer computed so there it is there okay so we note that the diagonal entries

3:29

are one and that's kind of obvious because you're multiplying a a column by say alpha

3:37

to alpha sub 3 okay and you're multiplying by the the corresponding row by the inverse

3:43

of that namely alpha sub 3 inverse and so this enter here gets multiplied by alpha

3:49

sub 3 and by alpha sub 3 inverse and so it doesn't change but all the other entries

3:56

do change here and so we're getting these entries here so what we're going to planning on

4:02

doing is is to try to make uh this first non-zero sub-diagonal it's actually the second sub-diagonal

4:10

um constant okay let's see if we can arrange that all right so let's try

4:17

let's try to make the second sub-diagonal

4:26

constant now there's all these uh negative signs here so actually i think maybe

4:33

uh maybe we'll choose minus one um instead of um plus one that seems to be more in accord

4:38

with um you know what the mathematics is trying to suggest to us well we get

4:46

various equations such as say the first one if we set this equal to minus 1 then

4:52

we really getting a sub 0 equals a sub 2. okay and then this next one if we set this equal to

5:01

1 up to -1 then let's see what that means that means that a3 will be equal to 3a

5:09

so let me write three a one equals a three and then the next one okay if this is going to be

5:17

equal to minus one that means um six a two six a 2 equals a sub 4. and then the next one 10

5:30

a sub 3 equals a sub 5 and then 15 a sub 4 equals a sub 6 sub 6. okay and i think we're seeing the

5:45

pattern here right what's happening is that each one of these equations involves two of the these

5:53

unknown numbers a one a two a a zero a one a two eight center uh they'll draw on the diagonal of d

6:00

and there's a relation between like a two and a four between a three and a five is separated

6:04

by two and the uh the triangular numbers are appearing simply as coefficients here okay um so

6:16

a solution might be more than one solution in fact there probably is going to be solutions

6:19

right this one here is telling us that if we know a0 then we know a2 and then if we know

6:23

a2 we know a4 and then if we know a4 we know a6 etc but also this this uh sort of these odd

6:31

equations with the odd terms that's sort of a separate thing so they're also sort of separate so

6:37

this is sort of a second system of equations there but um so one solution might be obtained by by

6:43

choosing uh a zero equals and a 1 both equal to 1. a solution may be obtained let's say by setting

6:54

a sub 0 equals a sub 1 equals 1. in which case we get what do these things become

7:04

uh well we get uh a sub 2 will be equal to uh to 1 and then a sub 3 equals 3 and then

7:21

a sub 4 what was that a sub 4 is going to be 6 a 2 a 2 is 1 so a 4 is going to be 6 and then a5

7:33

so we get 30 and then a6

7:39

a6 is 15 times a4 a4 was 6 so this thing is going to be 90.

7:45

okay maybe we do want to go a little bit further here to see what the pattern is

7:50

all right so a7 is equal to 21a5 but a5 was 30. so this is 21 times 30. i'll just write that down to

8:04

and that's what is that 630

8:11

and then a sub 8 equals from here 28 times the the previous uh even one which is 90.

8:23

uh can i do that in my head um 72 18 maybe 252520

8:32

is that correct let me check compute check equality that's true okay good i did it right

8:42

okay so that's maybe enough okay so that is the sequence of diagonal elements that

8:48

we're proposing to stick down this matrix d okay so let's do that so then the matrix

8:59

d equals so we get the sequence one three six thirty ninety six thirty which

9:06

continues two five two zero and then 22 680 etc this is a090932

9:15

in the online encyclopedia and uh what is it well it's a

9:22

number of permutations of the nth row of pascal's triangle

9:27

can also be seen as the multiplicative equivalent to the generalized pentagonal numbers

9:33

okay so the number of permutations of n in which all ascents start at an even position

9:41

formulas for it here's a formula a sub n equals binomial n minus 1

9:48

2 times oh this is a recursive definition a sub n equals n minus 1 choose 2 times a

9:56

n minus 2. actually that's probably exactly the thing that we we have been using right

10:00

because we we've developed this recursively and we chose one that was two previous and

10:06

we multiplied it by a triangular number which is just a binomial coefficient of this

10:10

kind so that's that's exactly you know we're touching base with um with this formula here

10:15

that's how we generated this this thing here and there's more interesting information there's an

10:21

exponential generating function and uh okay some some more information okay so this thing

10:28

is around um but there's no obvious you know connection at least that i can see uh

10:37

immediately with the the hermete polynomials so maybe we are learning something uh

10:42

interesting here okay so what was this last one let's complete the story that was 2 2 6 8

10:51

0 okay and now we're going to redefine b there we go and now we can look at the

10:58

moment of truth so then we get what actually happens when we now take this new d inverse h d

11:10

there it is so we have indeed successfully managed to conjugate so that things are much simpler

11:17

and we arrange that this the second sub-diagonal here i'll consist just of minus ones but we now

11:25

see very very pleasantly that the other sort of even sub-diagonals also also

11:32

are are constant that's gets constant one-half that's constant minus 1 6 that's constant 1 over

11:39

24. so we we we can see the pattern here very clearly the pattern here is clear

11:50

what's the pattern well um the even sub diagonals

11:58

have constant values which are just reciprocals

12:06

of factorials

12:13

and in terms of j you may recall okay here is the j matrix of the corresponding size so it's

12:20

just this maxel that has a single sub-diagonal of ones just the first sub-diagonal of course

12:27

it carries on so let's define j okay then what we have up here what what this

12:34

this matrix really is um then d to the minus one h d actually is equal to what is it it's

12:48

uh x of not j but um actually minus j squared see if this thing is j then j squared will have

13:00

a string of ones in the in the second sub diagonal in other words exactly along here

13:05

and minus j squared will just exactly consist of of this this sub-diagonal of minus ones

13:11

and then the various powers of this minus j squared are going to give us these further even

13:18

sub-diagonals so i'm just going to check this by evaluating this and it should be the same and

13:24

it looks like it is okay so this d inverse hd is x minus j squared that's true and so

13:34

we can actually sort of come full circle and um and state that h what we're really interested in

13:42

is obtained by conjugating we can undo the conjugation by multiplying by d on the on the left

13:50

taking the x of minus j squared and then multiplying by d on the d inverse on the right

13:58

and i'm just going to check this if i take this thing here and i evaluate it

14:03

does that look like the original h yes it does so this is great people so you see how

14:10

how powerful and useful this this technology is so if we wanted to uh make a much bigger

14:18

version of h okay this is a really good formula to do that

14:22

with okay because the j just carries on in the obvious nice way we can make a you know

14:27

a j matrix of any size very easily and then we've got this x minus j squared so we're connecting

14:33

with the exponential series with exponential poly series again but now sort of you know with

14:38

the minus j squared in there and then we're having this this conjugation by d which just

14:44

sort of introduces factorials you know in just the right places and so this is um this also explains

14:52

explains why um you know why why we get what we get when we do h inverse let me explain that so

15:00

if we look at h inverse well then that's going to be if you take the inverse of something like this

15:07

you take the inverse of something like this then you take the product of the inverses

15:12

in the opposite order so you this d inverse comes to the front with an inverse becomes a

15:16

d this d comes to the back with an inverse and then this x above minus j squared is inverted

15:23

and that just becomes x of plus j squared which is that thing there.