Solving Polynomial Equations 8: Does the cubic polyseries yield approximate solns?
Wild Egg Maths
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We have a tentative power series of a general cubic equation. Could this actually be put to use to find solutions - necessarily approximate! of real life cubic equations? We will see that some algebraic techniques allow us to extend the usefulness of our proposed solution.
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hello everyone i'm norman wahlberger and
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thank you for joining me on this
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adventure
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in exploring research level mathematics
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we are looking at
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trying to solve a general polynomial
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equation
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and we've got some interesting formula
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and now we're
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checking out whether it actually kind of
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works in the cubic case
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so here's the polynomial that we're
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interested in c 0 plus c 1 x plus c
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2 x squared plus c 3 x cubed equals 0.
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and we saw we could generate some kind
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of solution
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kind of a power series we might call
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them poly series
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where we have a a sum of terms of form c
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m 2 m 3 and m 2 and m 3 are just some
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variables that
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are ranging through natural numbers
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starting with
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- and what is this cm2 m3 well it's
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this
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expression here involving a sine
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involving a quotient of
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factorials and involving
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this particular kind of expression which
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has powers of
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c0 c2 and c3 in the numerator
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and power of c1 in the denominator
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and these exponents depend on m2 and m3
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now to make this more explicit we can
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actually look at all these terms we can
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arrange them at least the initial ones
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in a kind of a matrix so here is a small
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initial matrix
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that captures some of the first terms
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i'm calling it m okay so this would be
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uh you go
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m going from uh 0 1 2 3
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4 5 6 i guess m goes from 0 to
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5 i suppose and n goes also from 0
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to 5. now we can actually
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write out this expression
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x at least up to the terms that we're
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looking at
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by multiplying this matrix m on the left
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by
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a row vector ones and on the right
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by a column vector also once and then we
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get this
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expression here so we just um basically
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expand this out basically just
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summing all these terms putting them
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together to form a
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a polynomial actually not really a
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polynomial because it has these
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terms also in the denominator but it's
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some kind of rational function
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expression
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in c0 c1 c2 and c3 and this is just
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the starting point of you know a series
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which
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which carries on we could we could keep
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going and and to take bigger and bigger
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snapshots of this thing okay
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now we want to ask the question today
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you know whether
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we can actually use this i mean
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practically speaking to actually
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solve some cubic equations okay so
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that's a very good question we've
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we've been doing abstract algebra
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now can we actually you know utilize
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this
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um go down to earth and solve some some
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cubic equations
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so let's have a look you know here is a
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rather complicated series i've kind of
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arranged things
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so that the the powers of c3 are
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increasing so here things don't involve
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c3
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corresponds to that first column the
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catalan column we might
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say and then these next terms involve c3
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they're all the terms coming from this
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second column
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and then the next one coming from the
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third column and so on
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to the last one that we have showing
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here all the things involving c3 to the
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fifth
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corresponding to this last column
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okay and if we look at the terms we see
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that you know there are these
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powers of c one in the in the
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denominators
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and the coefficients are generally
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getting big so if we're
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thinking about this generating some kind
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of approximate
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solution we want these terms to
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start getting small right that's the
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general kind of principle that analysts
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use when they're trying to get a
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convergent
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series that the terms themselves had
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better get small so that we're
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we're always adding some small amount
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and hopefully heading towards
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something or at least an approximate
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sort of
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area so this is going to
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have a best chance of working when this
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denominator here is
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you know bigger than the the numerators
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so that we have a chance of these
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coefficients getting small
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so let's start with that in mind trying
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a
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cubic equation that that
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maybe has a hope of getting somewhere so
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let's
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try to solve a cubic equation
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i've just i've already made this x so i
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won't use x i'll use maybe say z so
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maybe 1
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plus say 5 z minus z
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squared plus 3 z
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cubed equals zero
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okay so let's first of all have a go at
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just
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solving this numerically okay
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we get three solutions and uh
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two of them are complex involving eyes
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and this one here this
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w goes minus uh five
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what do we just what is w
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oh i've got lots of things i better i
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better
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uh kill my definitions definitions
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uh clear definitions because
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probably i had some previous definition
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involving w so we'll get rid of that
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now i have to re-um put this thing in
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here
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in order for me to do that i have to put
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this all on one line just bear with me
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one
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second i'm going to
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put this all on one line
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so that i can then
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make a scientific workplace understand
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that i want this entire thing to be a
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definition
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okay there's x and let's def make that
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definition and let's just check that we
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actually have
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something sensible so x is this big
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expression
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all the way up to 273 which is that last
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column okay that looks
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kind of reasonable i'll put some spaces
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in again so we can just
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admire it
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okay we've defined this function x which
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depends on c0 c1 c2 and c3
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okay here is our cubic that we're trying
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to solve
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let us ask our computer to
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solve it numerically now we get okay we
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get solutions in z which is good
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two complex solutions and one z which is
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um
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minus 0.188 and that carries on
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all right so now let's see what happens
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if we
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try this i mean after all this is
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something like a formula right
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so if we put in x and then we stick in
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the coefficients that we have
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so c0 is in this case 1 c1 is
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in this case 5 c2 is -1 and
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c 3 is 3 then we can ask
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the computer to plug these values in and
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see what we
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what we get so evaluate numerically
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we get minus one eight eight eight two
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eight eight nine three
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that compares quite nicely with this
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thing here we're correct
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up to uh six up to six decimal places
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cool uh let's try a variant of that so
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let's try to conf
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let's try to confirm our guess that it's
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really important for that
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coefficient of z to be bigger than the
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other ones
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let's make that 5 a 15 and
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solve this now okay we get
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two solutions again that are complex and
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one which is a
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small number minus 6.631 times 10 to the
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minus 2.
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okay now we're going to look at our
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formula see what our formula tells us
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but we have to change this 5 to a 15 and
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we're going to
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evaluate numerically we get minus
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6.63151598 times 10 minus 2.
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we get exact correspondence between the
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actual solution to the eight decimal
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places that i have my computer sort of
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giving me a resolution
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and we're getting agreement to that
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level so you
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it is sort of the case that you know if
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this
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coefficient of of uh zed
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is a lot bigger than the others then we
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have a good chance for
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there's some kind of convergence
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happening here and we can see this
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demonstrated right here so we are happy
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because
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this is demonstrating that we actually
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have a viable technology here
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okay maybe not in all cases but at least
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in these two cases we see that we can
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solve a cubic equation essentially with
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a formula that only involves high school
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algebra okay that's the only ingredients
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here are addition subtraction
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multiplication and some division
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otherwise so
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you could calculate this by hand if you
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had a lot of patience
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in particular you don't need some extra
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technology for computing
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cubed roots which you can only do
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approximately
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or square roots which you can only do
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approximately okay
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so admittedly we're not getting a final
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exact solution and that's because in
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general
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these equations don't have final exact
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solutions all right they don't
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they only you can only aspire to getting
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approximate solutions that's just
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the nature of the mathematical world
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okay we don't have to try to deny that
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we're working with it just like applied
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mathematicians do
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okay so let's try some variance to this
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you know
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what happens if we try something
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different so let's grab this thing again
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maybe the first one where was it
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and let's change it up a little bit okay
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so let's say um we'll
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drop this down maybe to three and maybe
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we'll bump this one up two to five
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see what effect that has let's just
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put a four in here just for fun okay so
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that was something different
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all right let's compute solve this
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numeric okay so there's a
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a real kind of solution when i say real
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i just mean non-complex
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the same way descartes started it's
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minus 0.23610
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etc okay now let's take our x
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function we should plug in one sorry one
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three minus four and five
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and see what it gives us compute
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evaluate numerically minus 33.980
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that is nowhere close to this thing here
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nowhere close and why is that happening
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well if we look at the terms here right
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c one is about three that's the the
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denominator this is the 3 in the
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denominator
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but the c 3
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is a is
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is 5. okay and the
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c 0 okay that's 1 and the c 2 is three
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so um let's have a look
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at a term which is likely um sort of
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problematic
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um you see if you have if you have
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something like like this
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um okay that's the that's a big
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c one uh
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the point is that some of these terms
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the the
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the fact that the denominator is
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is has a big power but it's only three
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to that big power
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so if we have um you know
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some terms that sort of compensate for
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that like a c2 to the fifth for example
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and c3 squared again with this fact this
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big
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constant in front that thing can end up
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bumping things up
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okay so we can kind of understand that
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if
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the c one is not a lot bigger than the
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other ones we
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can expect that things are not going to
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work out very well
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however i want to point out something
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that might be interesting like let's say
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you
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let's say let's try another variant okay
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and i suppose i could also change that
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first one doesn't have to be one so
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let's change that that's two
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um that's maybe three i'm gonna put a
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seven here and uh and since
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cubed okay so 2 plus 3 z minus
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7 z squared plus z cubed equals 0.
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there's a cubic
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compute solve numeric
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okay here we're getting three solutions
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which
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are visible so three real solutions by
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which i mean that they are not
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complex and
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okay so what what does our method give
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us if we take
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x of 2 3
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minus 7 and 1 what do we get
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okay that's the exact thing that we get
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and we evaluate numerically
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okay we're getting 12 000 which is
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nowhere near
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any of these okay so this is not even
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close
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all right but let's not despair let's
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let's think about it for a second okay
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so our method
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works well when the coefficient of z is
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is bigger now that opens up the
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possibility that we might be able to
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transform our
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equation okay but maybe
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we can transform our equation
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to get it in a better
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form i.e one more likely
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to produce
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a solution so i want to show you
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what that might look like so suppose
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that we take this
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this expression here and let's say let's
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let
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let's say u equal or might be z
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okay let's let z equal 1 over
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u we'll introduce the reciprocal of z as
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a um
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a new object and then this
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original equation here then this
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becomes that and if we factor that
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and we set it equal to zero this is
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equivalent
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to the new equation where we just take
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this
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numerator and set it equal to zero
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so solving the original equation is the
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same as solving this
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altered equation and maybe i'll put the
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one at the front so it's more in
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standard form
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solving this modified equation for u
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which happens to be 1 over z
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but now this is better
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why is it better well because the
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coefficient of the linear term u
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is now bigger than the other ones it's
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- so let's try x
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of x of
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1 minus 7
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3 and 2 and we'll evaluate that
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numerically let's see how that compares
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with solving this thing
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numerically
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one of the solutions is 1.54076
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we are getting 0.154074
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that means the the solution we are
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getting is actually very close to one of
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the solutions
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to this equation namely this one right
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here
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and that's telling us therefore so
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z which is one over u
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which is say one over what we just found
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let's evaluate that that should be a
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solution to the original one
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and or at least close and here is the
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original one it has an approximate
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solution 6.4902922
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and now we get 6.4903585
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so the lesson is that
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yes our current solution does not
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necessarily
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cover all equations but it might be
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that if we're a little bit flexible
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about transforming the equation
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that we can move it into a position
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where our formula will give us a
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solution
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just as we have here now there's a lot
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of questions that are raised here
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okay there's a lot of questions namely
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you know
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what are the conditions um on the
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coefficients that will ensure that we're
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getting
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close to a solution with our um
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with our poly series there's also the
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question of
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how many terms in the poly series do we
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have to go
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in order to try to you know get a
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certain kind of
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a rather small window of approximation
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so that we've
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we've squeezed our solution into a
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rather small interval
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so there's a lot of questions to be
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asked here um but
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the the main thing is that we are we are
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confident that we are
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in possession of a technology which
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actually solves
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cubic equations okay and that's
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really what we're after right that's
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what we're after
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we have found a methodology for solving
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cubic equations which is very different
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from the radical procedure which
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ultimately comes down
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to working with an expression like
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like this one here which ultimately
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comes down to
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computing a matrix of
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values coefficients and uh
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and terms and then creating this poly
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series from it
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so we understand the terms that are
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involved here
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and it's rather predictable but what's
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not so clear for us at this stage are
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these uh
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coefficients so now we really want to
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bear down
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and and start studying these
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coefficients we have something
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interesting
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and we know that it works so now it's a
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deserving of considerable study