Solving Polynomial Equations15
Wild Egg Maths
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How do we extend this power series approach to solving polynomial equations to the quartic case? Well a natural step is to consider a three-dimensional extension of the BiTri array, now including polygonal subdivisions involving not just triangles and quadrilaterals, but pentagons as well.
At the beginning of the video we also report on a new development --one of our Members has found a form of the BiTri array in the Online Encyclopedia of Integer Sequences. And an intriguing paper by Jian Zhou seems to be involved in the story too --a connection with mathematical physics!
00:00
hello everyone i'm norman wahlberger and
00:02
thank you for joining me on this
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adventure
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towards solving general polynomial
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equations
00:08
in this lecture i want to talk about the
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quartic equation i want to extend our by
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try array
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to a bigger story which we're going to
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call the by
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tri quad numbers kind of
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moving up a ladder from triangles to
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quadrilaterals
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and now pentagons but before
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i get into that there is some very
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interesting breaking news that's just
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come to my attention that i want to
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share with you
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this is from one of our members elias
00:36
ammana titus
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many thanks to elias he has discovered
00:40
the following very
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interesting fact that the by try array
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that we've been talking about is
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actually on the
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online encyclopedia of integer sequences
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and it's found at
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a 104978
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where we see an entry that says triangle
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where
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gf that stands for generating function
01:02
satisfies
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and then this cubic equation in a and x
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y
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and then read by rows and then here's a
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string of numbers
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so what does this string of numbers have
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to do with our buy try
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array well it's just a way of reading
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this array
01:18
starting from the top left corner and
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kind of going up the
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cross diagonals so we start with this
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one and then we read up like
01:26
it's one one and then two five three
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and follow along here two five three and
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then five twenty one
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twenty eight twelve and then fourteen
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eighty four
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one eighty um 165 55 etc
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so this one sequence captures in a nice
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way the entire array and that's
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how the online encyclopedia uh deals
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with
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capturing arrays as well okay it's very
01:51
interesting there's not a lot of
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information but
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importantly there is a reference to a
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paper
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by john zao in
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the archive published there 2018 and the
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paper is called
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fat and thin emergent geometries of
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hermitian
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one matrix models it's actually in uh
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in mathematical physics so it's really
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coming from string theory
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and involves feynman diagrams and
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various curves and
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and some you know rather sophisticated
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things but having a look through this
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i found that uh dr zo really does
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uh find these numbers in in formulas
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that he sets up
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okay so um his notation is just a little
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bit different but without understanding
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the paper i think it's really intriguing
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but there's definitely some connections
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here at least with
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some of the things happening in
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mathematical physics which all
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is good for us we like to have
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connections with as many things so we
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can
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and it motivates us to try to you know
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go further and that's what we're going
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to do today
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so we're going to look at the quartic
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equation so let me remind you
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that that is obtained by the equation c
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zero plus c one x plus c two x squared
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plus c three x cubed plus c four x to
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the fourth equals zero
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we've already solved this at least if at
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least found one simple
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solution we looked at this power series
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say a0 plus a1 t up to a8
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to the t to the eight where t is equal
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to
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c zero and then we came up with this
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uh solution right here x is this uh sum
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involving these various um
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kind of terms but now involving all the
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coefficients c0 up to c4 in a somewhat
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predictable way
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and these these numbers here are
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hopefully some kind of variance of the
03:47
by tri
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numbers we want to look at this a little
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bit more carefully
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today and and see that in fact the by
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try situation
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does extend very naturally to this more
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complicated thing
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so the general term here
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is of the form okay it's a it's a
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fraction
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it has some power of c zero let's
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say that that power is m sub
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zero and then we have some similar terms
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one for c2 one for c3 one for c4
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and one for c1 in the denominator
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okay so let's see 0 and then we have to
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change the
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sub 2 and then
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sub 3
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and then sub 4.
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sub 4.
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so that's the the general form of the
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terms that are appearing here
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where we have the two conditions
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and the two conditions were first of all
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that the total degree
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is equal to well for example like here
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the total degree is eight plus four plus
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one
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minus thirteen that's zero so m sub
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zero plus
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m sub two plus and this should be
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sub three here
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plus m sub 3
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plus m sub 4 minus m sub 1
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that should all be zero and the other
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slightly more
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complicated thing is that the the index
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degree should be minus
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one so 0
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times m sub 0
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plus 1 times m sub 1.
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uh
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oh one time
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it's actually the actual exponent is
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negative uh m sub one so which put a
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minus sign there uh plus two uh
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m sub two
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plus three m sub three
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plus four m sub four that's the total
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index degree
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and that should be equal to minus one
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okay so it's a it's not a quartic
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extension
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of the kinds of things we were talking
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about already in the cubic case
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and now let's solve this just as we did
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before we're going to solve
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for um
06:32
[Music]
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there's two equations so we can solve
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for
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for two things um for m0
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and m1 okay so
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put the cursor in there and we'll solve
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exact for m sub 0
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and m comma m sub 1.
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okay and we get these two
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things we get m0 equals this let's
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make that equation i might put the one
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there
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first
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and the second equation is this one here
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the m1 equals say 1 plus
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2 m2 plus 3 m3 plus 4m4
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so hopefully this is very familiar right
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this is a
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very natural extension of the cubic
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thing in fact if you had to guess what
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it would be you would guess exactly this
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right if we drop off the m4 terms then
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we get the
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cubic story and we're going up one by
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one here so it's natural that there's
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three m4 here and a four m4 here and so
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now
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so that general term
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is given by
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okay the m0 is this thing here
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that was m0 and the m1
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is this thing here
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okay and what was the coefficient out
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front
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well there was a minus one to the
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uh i think it was a coefficient of m 3 m
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sub 3
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plus 1. and then there was a fraction
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and that fraction had some factorials
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the top
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factorial was the
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exponent of c1 factorial
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but one removed
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and the bottom we just have the product
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of all these
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factorials that are appearing here in
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the numerator
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so that one and then m sub 2
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factorial m
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sub 3
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factorial at m sub 4 factorial
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okay and that is
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what we think is the general expression
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that's appearing in this
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solution to a general quartic equation
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okay let's make a definition out of this
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so we can work with it maybe say
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f of m sub 0
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m sub 1 m sub 2
09:34
m sub 3
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m sub 4. it's actually five
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inputs but actually m0 and m1 are all
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determined right so there's now actually
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only three inputs
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that are three m2 m3 at
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m4
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now so let me uh
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let's make this as a definition first of
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all clear the definitions
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then i will make this a definition
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okay and let me just check
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check a few more a few of them let's put
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go up here
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so if we put in these are the
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coefficients of
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c2 c3 and c4 so if we choose this term
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here for example which has coefficients
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of 1 1
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- we get 7
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2 c the c c y c
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um
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oh it's because i didn't put a c 0 here
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sorry ok c
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sub zero should be c sub zero so let me
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rewrite that
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okay i'm just going to augment the
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definition
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okay now let's try this again
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equals 72 c 0 to the 7 yes
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c1 to the 10th and the bottom c2 c3 c4
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good
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one more
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cup let's say this one right here where
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we have a
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three um actually a zero
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a three and a a one
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zero three and one whoops
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um oh sorry
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that was that was wrong it was three c2s
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and then zero c3s and one c4 let's try
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that instead
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minus 120
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c 0 to the 7 c 1 11 on the bottom c 2
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cubed and c to the 4. that looks right
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okay that looks um
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kind of reasonable okay so
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when we were dealing with the cubic we
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were rather you instead of using m2 m3
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and m4 we were using m and
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n you may recall so let's be a little
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bit consistent let's just
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sort of change notation a little bit
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instead of
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m 2 m 3 m 4 let's write m
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and n and say p
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okay so that's the coefficient
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um there should be a factorial here oops
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a pardon me
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i did that wrong also uh let's correct
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this again
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okay and now let's try this again
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minus one to the n plus one
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power of c0
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[Music]
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c2 c3 sc4
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move them up there this thing over there
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also
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and then c1 on the bottom okay
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and then we can move this factorial over
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there
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i might write this as one plus instead
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it's a little bit easier for me
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how's that look okay
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so um great
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and that's that's the general expression
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and
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so the maybe i should put the
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coefficient in front
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there the coefficients
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are what we now might
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call by try
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quad numbers
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let's introduce that little terminology
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i will have to explain why bi tri and
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quad but uh
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so it turns out that has to do more with
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a
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graph theoretic or tree theoretic
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approach to uh
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these catalan kinds of objects which
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we're going to talk about
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but um there's binary trees and ternary
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trees and uh sort of cortex kind of
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trees and
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so on so that's that's kind of the
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reason for the terminology
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okay and what what are these things so
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let's let's define um
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say maybe c
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m n p equals and we'll just pick out
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these coefficients
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and which are sub
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multinomial
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numbers
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and there's a multinomial because
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well the sum of the terms in the bottom
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is one plus two m plus three n plus four
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p which is one
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more than the factorial that's appearing
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in
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the numerator all right
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so let's make that now definition
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c
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and let's pause okay so what we now have
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what i'm proposing now is that this
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array that we've been looking at
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the by triarray which we were starting
15:37
our lecture
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with namely this array here what we're
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doing is we're extending this into now a
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three-dimensional
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kind of array okay there's now indices
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going in three directions
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m the n and the p and
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our natural conjecture would be that
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since the m
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and the n correspond to subdivisions of
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a polygon
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involving triangles and quadrilaterals
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that this p
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very likely or at least possibly
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has an interpretation of counting
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numbers of pentagons which are also
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included so we're thinking about
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having a polygon and subdividing it into
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triangles
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quadrilaterals and pentagons and this is
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somehow counting the number of such
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subdivisions when you have m triangles
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and n
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quadrilaterals and p pentagons
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at least that's our naive conjecture
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that's sort of just
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following our nose and guessing that
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what's happening here is as close as
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it's possible to be
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to what we've already established okay
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so we don't know whether this is true or
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not this is just our first initial guess
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but in any case these numbers are
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going to be interesting especially if or
16:54
if you know they do appear to be
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natural numbers okay they're certainly
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natural numbers in the uh
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as we see the coefficients of this thing
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but you know more generally how can we
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tell
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so let's have a look at this and i want
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to i want to sort of
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show you some of these numbers but now
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we're talking about a three-dimensional
17:13
array so how am i going to show
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that to you okay so i'm
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let's say we are going to inspect this
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3d array sort of one
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level at a time and
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so to get the levels i might um
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let's vary
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uh well yeah i could uh
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could vary p okay so let's say very p so
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if p equals zero let's make that
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definition
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and then i'm going to make an array so
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let's say m
17:58
of m and n
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equals
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this thing here
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but with an m minus 1 and an n minus 1
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so that my array starts from 0. so i'm
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going to make that definition
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and this is the case have i defined p
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equals 0 yet let me just check
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yes p equals 0 is the case so now we've
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defined this and now that means we're
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just
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sort of reading the p equals zero so if
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you think of p
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is like the third dimension like the the
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z coordinate we're now
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reading this array at the bottom level
18:39
okay
18:40
and let's see what that is
18:44
compute matrices fill matrix
18:48
eight by eight is fine defined by
18:51
function w
18:52
oops not w
18:56
m pardon me
18:59
compute matrices fill matrix
19:04
defined by function m okay
19:08
okay we get the same array that we've
19:10
been talking about which is which is
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fine um that's the base level so
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the base level is essentially the cubic
19:18
okay and now we're building
19:19
up from the cubic in the same way that
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the cubic was built from
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starting with the quadratic case here
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essentially then we were
19:27
extending it and now we're extending the
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cubic case to get to the quartic
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i hope you appreciate how beautiful this
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is or at least potentially beautiful if
19:35
it does really work
19:36
it means that all of these successive
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equations are somehow going to be nested
19:41
inside each other
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so the solution of aquatic is is
19:45
intrinsically going to have the solution
19:47
of the cubic in a very nice way inside
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and somehow builds up from it
19:51
okay um so if we go to the the quintic
19:55
equation which of course we're
19:57
very keen on doing as soon-ish then
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we're going to have to extend this
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three-dimensional array into a sort of a
20:02
fourth dimension
20:03
okay but still they're all building up
20:05
on top of each other and there's some
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kind of consistency
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if it's all true okay so that's
20:12
that's very familiar now let's go to
20:14
something less familiar let's let's look
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at p equals one
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what does that mean that means we're now
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considering
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uh how to decompose a polygon into
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triangles
20:25
quadrilaterals and pentagons where we're
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having exactly one pentagon
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okay so p equals one we're going to make
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that definition
20:33
and now i'm still going to use the same
20:35
formula to
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show now what the values of these
20:40
numbers are
20:43
fill matrix m
20:55
that's kind of cool let's grab one of
20:57
these things okay um
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how about this this first column
21:06
let's take the transpose
21:09
and let's copy this and let's go to neil
21:12
sloane's online encyclopedia and see if
21:14
we can find
21:15
this sequence
21:18
so here we are at the online
21:20
encyclopedia and i've
21:22
input essentially the first column
21:25
of this array that i've showed you 1 6
21:27
28
21:28
120 495 and so on and we
21:31
find that well there's uh it's a
21:34
binomial coefficient
21:36
a double zero two six
21:39
nine four and let's see what
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kinds of things it says
21:46
okay so there's a few things about it um
21:48
and again it's sort of dyke paths
21:50
convex polynomials binary trees
21:54
number of dissections of a convex n plus
21:57
three gone
21:58
by non-crossing diagonals into several
22:00
regions exactly
22:01
n minus two of which are triangular
22:12
okay that that uh is probably very close
22:15
to
22:15
you know our in in intention um where
22:18
we've got
22:19
um triangles and and a pentagon
22:22
basically the other
22:24
block is a pentagon great
22:27
okay so let's go back and that was a
22:29
double zero
22:30
two six nine four
22:35
this is a double zero
22:38
two 6 9 4.
22:42
okay let's grab the first
22:47
row as well and come back
22:51
to the encyclopedia and see if
22:56
we can find this one here
23:00
that's a binomial and
23:04
what does it say
23:08
it doesn't say anything about
23:09
subdivisions of any
23:11
um polygons
23:16
so i think you know we're probably
23:19
touching base with with things that
23:21
people have looked at in some at some
23:24
level
23:24
but maybe we have a bigger
23:26
interpretation of this particular
23:28
kind of story here and especially this
23:30
strong connection with this very simple
23:32
problem extending euler's
23:34
subdivision kind of thing to considering
23:37
subdivisions
23:38
into more than just triangles to
23:40
triangles quadrilaterals pentagons and
23:42
and perhaps higher ones as well okay and
23:46
this was all
23:47
two three six one nine four two three
23:50
six
23:50
one nine four
23:59
two three six oops two three six
24:02
194 the first
24:08
column the first row rather is this one
24:14
okay great now let's go on to p equals
24:16
two and see what happens if we
24:18
require our polygonal subdivision to
24:20
have two pentagons
24:22
okay so we'll make that definition and
24:25
we'll just create a matrix
24:28
compute matrix fill matrix
24:34
so now m again okay
24:40
so we grab one of these see what happens
24:42
in the first top
24:43
top row 4 55 5 546.
24:57
did i grab that right hang on so let's
25:01
try to get copy
25:07
paste okay there we go search
25:11
okay so you see we've gotten already to
25:14
the point now where it seems as if we've
25:17
lost contact with the online
25:20
encyclopedia
25:21
we're at it's not a very stratospheric
25:24
level but it's just a
25:25
different kind of point of view than
25:27
probably what people have been looking
25:28
at
25:29
so this should be a very nice uh
25:31
sequence right if
25:32
if what we're saying um you know has any
25:35
truth to it
25:36
um what we're conjecturing is well i've
25:39
actually stated the conjecture let me
25:42
show it to you here so whoop
25:45
back that's a conjecture
25:54
and give let's give this a name
25:58
let's call this the buy try quad
26:01
conjecture
26:03
that this buy try quad number
26:07
quad number c mnp counts the number of
26:11
diagonal subdivisions of a okay a
26:14
certain polygon
26:15
and the polygon is going to have 2 plus
26:17
m plus 2n plus 3p
26:19
sides into m triangles and n
26:22
quadrilaterals and p pentagons so
26:24
remember the way we counted the
26:26
the number of sides of the polygon we
26:28
start with two and then we add one for
26:30
every triangle that we're creating
26:32
okay if we have one triangle we have
26:33
three sides we have two triangles we
26:35
have four sides and so on
26:36
so the the triangles add one each the
26:39
quadrilaterals add
26:40
two new
26:43
vertices or edges and the pentagons will
26:46
add three new vertices or edges
26:48
so that makes perfect sense and that's
26:51
what we're
26:52
conjecturing without a lot of evidence
26:55
so far
26:56
um but there's a kind of a sense of you
26:58
know
26:59
optimism here and and maybe a sense of
27:02
aesthetic value to it i mean
27:03
it seems reasonable and in particular if
27:05
this is true then this shows that
27:07
cmnp is actually an integer in general
27:09
we can see that it is at least for a
27:11
certain range but in general it will
27:13
still be an integer if this is true
27:15
okay so now already there's lots for you
27:19
to do
27:20
along the lines of what i was showing
27:21
you you know before you take a
27:23
polygon of a certain size and you start
27:26
asking
27:26
you know how many ways can you subdivide
27:28
it into say
27:30
one pentagon and uh you know and
27:34
and two triangles okay
27:38
maybe there's 495 of those
27:41
so i you could try to
27:45
verify some of these numbers and see if
27:47
we can gather some support
27:49
for this enticing conjecture which
27:52
is really exciting because now i think
27:54
we're kind of on top if this is true
27:57
then we're probably on top of a very
27:59
general situation because it's now quite
28:01
clear
28:02
how we can extend this to higher degree
28:06
equations
28:06
right the form of our of our numbers are
28:10
so
28:10
regular 2m plus 3n plus 4p if we add go
28:14
to the
28:14
fifth level we should add a five q and
28:17
over here we should add a
28:18
four q we should add a q factorial okay
28:22
so there's lots to do to check whether
28:24
any of these
28:26
ideas might actually be connected with
28:29
computational explicit reality