Solving Polynomial Equations 3: Wrapping the Quadratic Case, and on to the Cubic
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We continue our new approach to solving polynomial equations via power series which finesses the classical dependence on radicals. We look at how to obtain two solutions to the general quadratic, and then move to the much more challenging case of the general cubic equation.
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hello everyone i'm norman wahlberger and
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thanks for joining me
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on this exploration of research level
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mathematics
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we are looking at solving polynomial
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equations very classical subject
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in a potentially new way so last time we
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were looking at the quadratic equation
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and we wrote it as t plus c one x
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plus c two x squared equals zero and we
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got these two solutions x1 and x2
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through a method of a kind of a power
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series method proceeding step by step
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solving some equations so we have a
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power series in
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t here with coefficients which uh
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little functions of c1 and c2 and then
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there's also these numbers which are the
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catalan numbers
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c n is 1 over n plus 1 times the
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binomial coefficient 2
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n choose n and we had two solutions
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the second one a little bit more
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complicated than the first one
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due to this additional sort of constant
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term
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but otherwise actually they're very
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similar this wasn't just sort of the
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negative of
01:02
of this one here okay so what we want to
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do today is we want to
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wrap up this quadratic case and move
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to the cubic case which is a lot more
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challenging
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okay so i want to rewrite this x1
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as well in terms of uh
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more sort of standard mathematical
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notation involving summation
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so let's rewrite
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there it is as the summation
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okay x1 equals so there's going to be
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well there's a minus sign let's pull
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that out and then there's a summation
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sign
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and maybe we'll use the dummy variable k
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so k is going to go say from one one
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will be the
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exponent of t so k goes starting from
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one in this case going up to
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a nine okay and what
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are we actually doing well there's a
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power of t
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so t to the k and then in front of that
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there's some coefficients uh there's
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first of all
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a catalan number and
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so we decided that 42 was the fifth one
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uh fifth catalan number and so
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that's corresponding to the t to the
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sixth term so probably we should put
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c k minus 1 here all right
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and then there's also this quotient of
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of c's so there's some power of c
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sub 2 on the top and some power of c
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sub 1 on the bottom let's figure out
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what these powers are
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so in case uh say here when k is two
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then the power of c2 is one one less
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and that's carrying on the power of c2
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is always one less than the power of t
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so
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we should put up here the exponent k
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minus 1. and what about the denominator
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this power of c1 they are all odd powers
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so the power here should be 2 to the
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something
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minus 1 or plus 1. say 2 to the k
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in case k is 3 we're going to get
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the power of 5. so we should choose 2
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times k
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minus 1 it looks like
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all right and let's just check that
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let's see what we
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have in the way of um definition so we
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have
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cn right and these ones we don't care
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about
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so let's just evaluate this thing
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and just compare minus t over c one
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minus t squared c two c one cubed so
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scientific workplace arranges the terms
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here a little bit differently from what
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we've got but it's the same
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and the last one 14 thirty t to the nine
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c
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two to the eighth divided by c one to
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the seventeen that looks like a good
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formula
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okay what about our x two
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[Music]
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well that was much the same so it was
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really the negative of what we got up
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here so we should just put that without
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a negative sign this time
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and then we should throw in this
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additional constant term minus
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c1 over c2 we'll put that out front with
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a plus sign
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and we'll check that okay what's that
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evaluate
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do we get the same thing oh here's the
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minus c1 c2 okay
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you can move it over there if we prefer
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and then
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otherwise it looks uh much the same as
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this
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okay so that's for um
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going up to sort of level nine or degree
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nine and
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in general we like to write well we want
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to write some more general formula
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so we'll write x one
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equals this thing here and instead of
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having a 9 i'll just put
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a plus sign which is my way of
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indicating that we have
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on series okay series which are ongoing
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but we are definitely not pretending
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that we are summing all of these things
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to infinity absolutely not okay that's
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that's a an arithmetic that's not
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allowed in proper
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thinking despite the fact that it's
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commonplace in modern pure mathematics
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a very sad story okay so we're not
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pretending that
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this is this is an expression okay which
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is
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something like a polynomial or a poly
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number
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but it has an additional aspect and
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ultimately we'll have to describe
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you know how the arithmetic of such
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things work but crucially that
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arithmetic is going to be
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finitely specifiable and finitely
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checkable okay that's the crucial
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distinction here
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okay and there's also the of course the
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other one
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that's x2
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there it is there and we should go up
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to nine we'll just keep on going
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okay and just a note so you know
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eventually
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we will interpret
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these once we have
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proven this formula for these formulas
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in some kind of in some kind of
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truncated algebra
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of on series okay but we're just
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exploring right now so we don't have to
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nail everything down we just want to get
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some formulas
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and uh and then stare at them and you
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know and contemplate them
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but right now we have some really nice
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formulas for the quadratic case
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and i do want to emphasize that we have
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not yet proven these so
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what we've done is we've made some
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calculations going up to degree five
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up to degree nine and certainly those
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calculations suggest some kind of
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general formula like this but
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it's still a question to be proven okay
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and those of you with some mathematical
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uh experience
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will be able to prove them okay so you
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may want to sort of delve into just
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proving these things
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without any reference to the usual
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classical formula
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the quadratic formula just starting with
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algebra and of course
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employing properties of the catalan
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numbers because they're central objects
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in this story
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fabulous okay now it's time to move on
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to the cubic case
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okay the cubic case so
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here is our general cubic equation
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t plus c one x plus c two x squared plus
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c
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three x cubed equals zero where again
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we're
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using t instead of c 0 because we're
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going to follow
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the rather successful row that we travel
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down with our
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quadratic case where we're going to
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expand things in powers of t
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and we're going to solve this for x
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hopefully
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and maybe you will try the same kind of
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story we'll just
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write x as some unknown power series in
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t
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and we'll choose say going up to degree
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7
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a little bit more than five but less
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than nine so it's a compromise
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and we'll see what happens when we plug
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this thing into this equation
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and see if we can find out what the a
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i's are in terms of the
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c's okay so there's a condition here so
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subject
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to the condition that
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t to the eight equals zero
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we want to solve for
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a for it let's say the
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a sub i's
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in terms of c1
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c 2 and c
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- okay
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are we ready to do that okay let's have
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a look so here we'll take
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t plus c one x plus c two
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x squared plus c three x cubed okay
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and we'll write it again and now we're
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going to replace the x here
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with this expression
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okay and we'll just copy this thing and
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i'll put it
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here and i'll also
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put it here okay
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let's expand this whole thing so it's
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got cubic terms so there'll be lots of
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terms i just want to show you what that
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actually looks like
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so let's expand this out
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to give you a sense of the power of
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these kinds of programs okay it's
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absolutely fabulous what they can do
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with polynomial arithmetic and that's
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absolutely crucial if you're
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an amateur mathematician wanting to do
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experimentation exploration in
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mathematics you
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you have to deal with polynomials and
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matrices those are the
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the main kinds of arithmetic that go
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beyond our
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usual arithmetic of numbers okay so
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polynomial arithmetic matrix arithmetic
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those are your two
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fundamental arithmetical tools
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okay and that's pretty well it people
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you know it doesn't
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it really doesn't get much more than
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that
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i know there's all these fancy things
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but ultimately they come down to those
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two things
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okay so that's all too much isn't it
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right so
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we are going to uh shortly remove this
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but i'm going to
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truncate this right because we've agreed
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that t to the eight is zero so a lot of
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these terms here like this one here with
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t12 so i'm not going to go through
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individually and kill them i'm going to
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just use our power
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series little tool okay so we're going
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to expand this
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number of terms let's say
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8 powers of t
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okay what do we get
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okay there there's the equal sign we
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look for the equal sign okay
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this is the stuff that we had before and
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i'm now about to um
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to uh selectively
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delete
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that
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delete it and now let's have a look at
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what we've got
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here in front of us the ota that
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means that there's terms of degree eight
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above that
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okay so what we see is a constant term
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a multiple of t a multiple of t squared
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all the way down to a multiple of t
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cubed t to the seven now
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what do we have to do well we have to
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set all of this equal to zero
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so open up here
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so we get
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zero
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zero equals all of this stuff
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that's a polynomial in t of degree seven
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all the terms all the coefficients had
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better be
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zero
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so this gives us
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a number of equations
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okay and we'll write them down just like
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we did before so this first one
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is the constant term and while we're at
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it we're going to factor that thing
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that's the same thing that we did in the
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quadratic case
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and
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that's going to be a zero in fact i'll
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write just write it like that okay
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and then
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we'll write down the other ones and then
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we'll stare at it okay and then there's
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t coefficient of that
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is zero so this first equation allows us
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potentially at least to solve for a zero
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okay and you see what's involved in
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doing that there's two possibilities
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a zero is going to be zero from this
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factor or a 0 is going to be a
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solution to this quadratic equation
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it's a quadratic equation it's written a
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bit funny let me
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just write it a little bit more smoothly
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put the linear term
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there it's c1 plus a0
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c2 plus a0 squared c3
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so it's a quadratic equation with
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constant terms c1
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then c2 and then c3 so do we know how to
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solve quadratic equations
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well yes we do because we've just done
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that okay so in principle we could write
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down a solution for
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a zero from this factor but it'd be a
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lot simpler if we wrote down
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the other solution which is that a0
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equals zero so again there's going to be
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this separation into these two cases the
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simple cases when a
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0 is 0 and the more complicated case is
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when this
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factor is 0.
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okay what are the other ones um okay
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anyway oh yes and so
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by the time we've sort of decided what a
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0 is
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then this next equation that we've just
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written down let's have a look at it
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as an equation in the a's there's an a1
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here there's an a0 there's an a1 here
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and a0 and a1
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so this is a linear equation in a1
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that has an a0 involved so if we've
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already
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got a value for a0 we can simply solve
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this equation because it's a linear
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equation
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in a1 okay and then the next one
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is this thing here
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equals zero and then the cubic one
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and then the degree four one
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these are getting a bit longer aren't
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they the degree five one
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[Music]
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degree six one
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okay we're just almost at the limits of
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visibility here
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and let's say this one here
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and just to make it temporarily visible
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i put a line break in there so that's
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that one there
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okay what are we going to do we want to
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solve
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this for a 0 a 1 a 2
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up to a 7. so i'm actually going to go
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back up to my earlier
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work where at some point i wrote down a
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list of the a i's i'm just going to
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borrow that again so let's just copy
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that
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come back down okay
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let's get rid of this
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and i'll put these a's here
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okay so let's solve this i'm going to
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first copy this
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and put it
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and also there so just another copy
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so this first one we're going to do the
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simplest thing okay the simplest thing
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is to agree that a0
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is actually equal to zero a sub zero
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equals zero okay so let's start
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let's start with the possibility
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for the case the case
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a sub zero equals 0. all right
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so that's included there so this a0
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okay that this one is then redundant but
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it doesn't matter and then we'll be able
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to or the computer will be able to
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successively solve for a1 and then this
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next one has an a2 in it which is
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appearing only linearly
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and then the other things are a a1 and
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a0
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and then the next one has an a3 which is
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appearing linearly
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and so on so it should be a snap for the
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computer to
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solve each one of these sequentially and
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to
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make sure that it doesn't mess up i'm
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going to put this back
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as a one-line thing even though you
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can't see it but it doesn't matter
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because we're just going to ask the
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computer
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to solve this compute solve
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oh i have to before i do that i should
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copy this
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we want going to solve a0 up to a7
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because the last term is a7
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okay now i have to put the cursor back
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in here somewhere
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anywhere it doesn't matter and now i'm
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going to ask
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our brilliant friend to
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solve for a0 up to a7
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okay you can do it
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okay let's see what happens here we get
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okay we get something here and
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i'll get rid of the brackets
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and maybe i'll get rid of the solution
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and so we can see this maybe i will i
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will put it um
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a little bit more organized way so we
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can
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see them sequentially a3
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we can see this is actually quite a lot
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more
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involved slash interesting than the
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quadratic
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case remember in the quadratic cases all
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these a's turned out to be
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uh so some power of c2 divided by some
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power of c1 and it was rather
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predictable
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okay it's suddenly not so predictable at
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all
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these a i's are now polynomials
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in the c1 c2 c3 with
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also denominators the denominators are
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powers of c one actually that is very
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familiar
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okay um but there's then these other
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terms
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and now if we actually stare at this for
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a second you say but wait a minute
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actually we see something that we
19:21
recognize
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already have a look see if you can see
19:25
what we've
19:26
already seen before
19:30
[Music]
19:32
you see that the coefficients if you
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just look down this first
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column here where
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the terms involved are just pure powers
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of c2
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we're getting 2 5 well 1 1 1 1
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2 5 14 42 132 we recognize the catalan
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numbers
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so this is just what we've already
19:52
gotten before
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and then there are some additional terms
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that also involve
19:58
c3 so we note that
20:01
[Music]
20:02
that this extends the
20:05
quadratic solution
20:09
in that if
20:10
[Music]
20:12
if we set c sub 3
20:15
equal to zero in other words if we
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reduce the equation from being a cubic
20:21
to a quadratic that all these other
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terms will disappear and we'll just get
20:26
what we
20:27
got before we get
20:30
what we got before
20:39
okay so great now we can write down some
20:42
kind of solution
20:43
in the same kind of way that we did in
20:45
the quadratic case so let's go back up
20:47
to our expression for x
20:48
here it is here
20:54
and
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we get a solution
21:06
x one say oh let's write it like this
21:09
x sub one equals okay the a
21:12
zero we said is equal to zero so we'll
21:14
eliminate that one
21:16
and then we'll copy these various ones
21:18
so there's a1
21:20
put that in there for that and a2
21:28
instead of this and a3
21:35
that one and a4 minus sign
21:41
oops sorry that was t cubed going this
21:45
way
21:48
and a5
21:55
and a6
22:04
and finally a7 maybe off the page a
22:08
little bit
22:12
okay and i'm going to define x1 to be
22:16
that
22:18
and just so we can see it let me wrap
22:20
this around so there is a solution
22:22
of our cubic which sort of has
22:26
one of the solutions to the quadratic
22:29
inside it if c3
22:31
is zero then we sort of reduce to the
22:34
previous quadratic equation um solution
22:36
with just the catalan numbers which are
22:38
the first
22:39
elements in each of these brackets but
22:41
for the cubic there's more
22:43
terms now but we're only going up to
22:44
degree seven so
22:46
i want to check whether this actually
22:48
works okay so we want to
22:50
constantly use opportunities to check
22:53
that what we've done is correct because
22:55
there's lots of ways of making mistakes
22:57
we can make logical mistakes we can make
22:59
algebraic mistakes we can make cut and
23:01
paste mistakes and so on
23:02
all right so let's go back up to the
23:04
equation that we are trying to solve
23:06
which is our pleasant little cubic there
23:09
it is there
23:17
and let's check
23:25
that there it is there this is going to
23:28
be equal to
23:29
i'm just going to okay i'm going to just
23:31
maybe over here just replace the x with
23:34
x1 that's the easy way of doing it
23:36
i suppose
23:39
then we don't need this
23:43
okay and so let me just copy that so
23:48
it's just plugged this expression the
23:50
expression for x1 let me just check
23:52
what that looks like so that's this
23:55
thing
23:55
carries on here i don't think you can
23:57
see all of it but anyway
23:59
and um we plugged it in for x1 here and
24:02
x1 here and x1 here we got these terms
24:03
here and now we want to kind of
24:05
expand this and you know and simplify it
24:09
but to make life easier here i'm going
24:11
to
24:13
just restrict to all our terms of degree
24:16
7 or less in t so that we don't have to
24:19
worry about all those higher powers of t
24:21
which would make this blow out
24:24
actually maybe let me illustrate that
24:25
actually so let's actually just for fun
24:27
let me make a little space here that's
24:30
fine let's just
24:31
expand this out okay
24:34
compute expand
24:40
it's a big polynomial in t
24:47
all right and
24:50
okay we want to understand this so let's
24:52
have a look
24:54
at it as a polynomial in
24:57
t just polynomial collect
25:00
terms organize in terms of t
25:04
what do we get well there are these all
25:08
these higher powers of t
25:10
but notice the lowest power of t is t to
25:12
the eight
25:14
what's that tell us that tells us that
25:19
this is equal to zero up to
25:23
degree seven
25:27
in other words you make this big
25:28
computation all the terms involving t
25:31
t squared t cubed up to t to the seven
25:33
including the constant term also are all
25:35
zero
25:37
okay so how should we uh we don't need
25:38
all this stuff here so let's uh
25:40
summarize that let's check that this
25:42
thing here
25:44
okay it does actually
25:49
it's equal to zero up to degree seven
25:53
so we have indeed
25:56
right here let's put this
26:07
let's put this in a
26:14
display it
26:18
there we go that is a solution
26:21
oh t seven better not forget that one
26:25
that is a solution to our cubic equation
26:29
which is valid up to degree
26:32
7 expressing it as a
26:36
an on series in powers of t whose
26:39
coefficients
26:40
are
26:44
rational functions of the ci c1 c2 c3
26:48
and we see the catalan numbers are
26:49
involved but there are other constants
26:51
too
26:52
so a very big question is how to make
26:55
sense
26:57
of all these new
27:01
constants and
27:04
terms but in any case it really does
27:08
appear as if we have now a
27:10
new kind of solution to a cubic equation
27:14
quite different from that obtained by
27:17
del ferro and tartaglia
27:20
etc and purely in the algebraic
27:24
world without any square roots cube
27:26
roots or any kind of other radicals that
27:29
you can't actually calculate that this
27:31
is very exciting
27:32
but obviously we still have a lot more
27:35
work to do to
27:36
to understand fully this uh this cubic
27:39
situation
27:42
i'm norman wahlberger thanks for
27:45
listening