Solving Polynomial Equations 1
Wild Egg Maths
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How do you solve a polynomial equation? Let's play around with this famous problem and see if we can tinker our way to some new insights. Along the way we strengthen our understanding of polynomial or polynumber arithmetic, and its extension to on-polynumbers or power series, and get some experience with the power of Scientific Workplace for writing and working on mathematics.
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hello everyone i'm roman wilder
00:04
and thanks for joining me on this
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adventure where we're going to be
00:08
looking at
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polynomial equations so i'm going to
00:12
take you through
00:13
some ideas that we might have a look at
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this is a very classical problem the
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problem is
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how do you solve polynomial equations
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going back of course
00:22
uh you know to antiquity to the
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babylonians who are interested in
00:25
quadratic equations
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and then in the 15th 1600s
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a lot of work on the cubic and quartic
00:35
equations
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and then some sort of difficulties as
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people started to realize that
00:40
degree five and higher equations were
00:42
somehow different and then that
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the standard procedures that the
00:47
classical
00:48
mathematicians had developed weren't
00:49
working that well so
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right now i would say that we're not in
00:53
a very satisfactory situation
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and i'm going to look at this problem
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and hopefully with some fresh eyes
01:00
so let's start by just doing some simple
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things which will also allow me to
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generate a little bit of experience for
01:07
you working with
01:08
scientific workplace okay let's say uh
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the variable is x
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and let's have a very simple equation
01:14
say a plus
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b x equals zero okay there is a
01:18
polynomial equation of degree one
01:20
i've made it very simple let's see what
01:22
our computer
01:23
tells us is the solution so we're going
01:25
to ask it to solve
01:26
exact and we're going to solve for x
01:30
okay we find that x is equal to
01:34
right here x equals minus a over b
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provided that b is not equal to zero
01:40
fine
01:44
okay let's be a little bit more
01:46
adventurous let's uh
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look at a quadratic equation
01:52
but let's actually do a specific one
01:53
just to be a little bit more
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concrete so let's look at two minus
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three x
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plus um say five x squared
02:02
okay there's a quadratic equation let us
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try to solve it
02:07
solve it exact okay we get
02:10
two solutions and we see that they're
02:12
complex here why are they complex
02:14
there's some i's here it's because the
02:16
discriminant of this equation
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is 3 squared or minus 3 squared
02:21
minus four times two times five so we
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get
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complex solutions
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since the discriminant
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which is uh minus three
02:36
squared minus four times a
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times five and that that's equal to
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minus 31
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that is is negative
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okay so let's try a variant on that
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let's make the discriminant positive so
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we could do that
02:55
say for example by bumping this three up
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let's make it
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say i don't know nine
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and then what will happen is
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solve exact okay now we're getting um
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solutions in the standard form involving
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some sirs or some square roots so
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this involves thirds
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which for us are not
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exact numbers okay so we're going to
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draw upon
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our wealth of experience here you all
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know that i don't believe in square
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roots
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that these are actual numbers these are
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algorithms that can be run but they
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don't actually finish so this is a
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kind of a red herring and we can see
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that
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more clearly and directly by
03:47
by asking for a numerical solution
03:52
solve numeric okay now we get the two
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solutions but they're of course only
03:57
approximate so here are two solutions
03:59
to five decimal places or so okay
04:03
so we see that even the quadratic
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equation
04:07
has has come complications we're not
04:10
really able to solve it in some way
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okay let's be now more adventurous and
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let's go to a cubic equation
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say three plus x minus a two
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x squared okay plus okay uh
04:22
let's say four x cubed
04:26
equals zero all right what's going to
04:28
happen if we
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try to solve it let's first solve it
04:32
numerically
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okay so this thing here has three
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solutions but two of them are complex
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and one of them is
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real okay and and all of them are
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approximate
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so this the graph of this thing which
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seems to cross the
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x-axis in only one point around x equals
04:53
minus 0.69 okay so but
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that's still not really a solution i
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mean those are approximate solutions but
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they're not really exact solutions
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so let's have a look at trying to solve
05:05
this thing exactly
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solve exact okay and now we
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run into some complicated formulas that
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are really going back to cardano
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and uh antar taglia and uh and
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del faro and uh and the italians where
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they realize that the formulas for the
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cubic equations were rather complicated
05:26
they involve cube roots you can see some
05:28
cube roots here
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and also some square roots inside
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and of course you know this is very
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unwieldy i mean to actually use this
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is is almost impossible really at least
05:39
before the age of computers
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and so this is really a major stumbling
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block i mean people got
05:44
these cubic equations and they could
05:46
write down some formulas but
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they are really not very useful so um
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so it's a problem and and
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if you go higher higher degrees well
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let's say you go to four
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four degrees let's see that okay let's
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modify this a little bit
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five and let's take a quick a quarter
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term in here okay so now we have a
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quartic equation
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and let's solve it first
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numerically uh we've got two
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real solutions and two complex solutions
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okay and now i'm going to just
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solve it exact
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oh in this case this is actually
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relatively simple so maybe i've just
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chosen some
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really particularly simple ones here
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okay you see
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this solution is quite simple that's not
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interesting
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in that it only involves square roots so
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that's a happy accident
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let's modify it a little bit if i change
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this x to the
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fourth here to 7x to the fourth and i
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don't think that's going to happen
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again so now let's calculate the
06:55
solution of this
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exactly okay and
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you can see that it's somewhat analogous
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to the cubic thing in fact the cubic and
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the quartic equations are
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are closely related um and again we have
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some
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rather complicated um you know formulas
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involving
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nested uh radicals okay
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so this is somehow the the classical
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story i mean it's a very
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long and laborious story but um the the
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basic idea
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or direction is the classical idea
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is to try to
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reduce such equations
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to simpler equations involving just
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radicals
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involving just radicals such as
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i.e x
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to the n equals a
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okay so if you have an equation like
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that then you express the solution
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with an nth root and
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then you feel confident that you've
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gotten something even though
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we would not accept that as a really
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legitimate answer but anyway that was
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the orientation
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of the classical people
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and so we want to do something different
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okay we want to look at this entire
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story from a new point of view
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all right so let's try to be more
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algebraic
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okay so we're going to
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we'll dispense with the degree one
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equation and we're going to go to a
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degree
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two equation and see if we can kind of
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think about you know how to solve a
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quadratic equation maybe in a slightly
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different way
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so let's look at a general equation so
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instead of looking at particular numbers
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let's look at a general equation
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a general quadratic what does a general
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quadratic equation look like
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well it's got a linear term and i'm
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going to
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express that'll be term in mathematics c
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sub zero
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let's say that's the constant term and
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then plus say c
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sub 1 times x plus c sub 2
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x squared equals zero okay there's the
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our kind of generic quadratic equation
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where
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we're not specifying particular values
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of c0 c1 and c2
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now note that
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we really ought to assume
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assume that uh the c2
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first of all c even c zero that c zero
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does not equal
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does not equal zero because if it c zero
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equals zero then
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you can factor out an x and then it
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becomes a linear equation so it's so
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somehow already we've dealt with that
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and we can also assume that uh c2
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does not equal zero because if c2 equals
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zero then we hit it back to a linear
09:54
equation okay so
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we'll make these two assumptions but
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otherwise we'll keep the nature of c0 c1
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and c2 somewhat
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open in particular we might even
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eventually
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think about c0 and c1 and c2 as
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you know variables
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now you will know those of you who have
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watched my videos that i don't like to
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think about variables i rather think
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about
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poly numbers okay and so we would want
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to
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think about that those variables as
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being developed in a polynumber
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framework
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but we're not going to go in that
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direction that's on the side that maybe
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we'll talk about
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at some other point in in this at this
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stage here we're just going to sort of
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think about this as a fundamental little
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equation okay and we want to solve this
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okay so we want to solve for x we want
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to solve for x
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well in terms of c0
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c1 and c2
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so what what kind of
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thing are we looking for well we're not
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not going to allow radicals we've just
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said that because radicals are not
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something that we could actually
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calculate
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so what are the kinds of things that we
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can calculate
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right we're just interested in
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restricting ourselves to
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to basic algebraic operations which are
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addition
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subtraction multiplication and division
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so really what we're doing is we're
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at least you know one approach would be
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well what would happen if you
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just tried to express x as a polynomial
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in c zero c one and c two
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well it's not too hard to see that uh
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that's not going to work
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so if x is a polynomial or let's say a
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poly number if you allow me to do that
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in c0 c1 and
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c2 then
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this is not going to work well
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because there's going to be top degrees
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and then when you plug in the top
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degrees you can see that the top term is
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not going to be 0.
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but in a big but when
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poly numbers or polynomials don't work
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sometimes
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poly numbers that keep on going on poly
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numbers
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polynomials that keep on going formal
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power series
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they sometimes have a better chance of
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working
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okay that's an important kind of
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understanding that
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if you can't make polynomials
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fit what you're doing you should maybe
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think in terms of power series
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even though maybe what exactly a power
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series is is problematic but we're just
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playing around so let's let's see what
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we can do
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but um
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how about how about a on
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poly number or maybe
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or maybe usually called a power series
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okay um so let's try a power series
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and uh so for that we need a kind of a
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you know a variable and
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where you're maybe going to choose one
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of these
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things to be variable why don't we say
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try c0
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let's set
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c 0 equal to say
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t and think of this as a a variable for
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a power series
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and and we're looking for and we look
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for
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a solution which is a
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on poly number or a power series
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in t so we look for a solution of the
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form x
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equals okay maybe a
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sub zero plus a sub
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1 1 times t
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plus a sub 2 times t squared plus
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3 dots carrying on in some fashion or
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another
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okay so that's that's interesting so
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what would happen if we just sort of did
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that
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if we sort of tried to fit a poly number
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um or on poly number solution to
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our original equation well
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one way of trying to make sense of this
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so one way of giving meaning
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to this is to restrict
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oneself to just terms
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of a certain fixed
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degree or less that is
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we ignore all terms beyond some point
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so for example let's say um
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let's say to be specific that we'll
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let x be
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we'll go a little bit more than
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quadratic maybe uh a sub
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3 t cubed plus
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a sub 4 t to the fourth plus
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a sub 5 t to the fifth
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and we're going to go only up to
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5
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or degree 5. let's say up to degree
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5 x is equal to a 0 plus a 1 t plus a
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two t
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squared plus a three t cubed plus a four
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to the fourth plus a
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fifth t to the fifth so in other words i
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e
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what we're doing is we're kind of set
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set
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t to the six equals zero
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okay that's sort of what we're thinking
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about
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we're trying to just
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approximately get a hold of something
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and so we're just going to agree to
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truncate our
15:50
our algebra and work in this fixed
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domain up to say degree 6 just to see
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what happens
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all right so let's plug in and see what
16:03
we get
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we get okay so our original expression
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was
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c0 plus c1 x
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plus c2x squared and
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we've already agreed that c0
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is going to be replaced with a slightly
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easier to deal with
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t okay and so now we're going to
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just let me put that t at the beginning
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there we're going to replace
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the x with
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with the expression that we have for it
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so c sub 1
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times x what is x x is this combination
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of powers of t
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and similarly this x is also going to be
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replaced
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with that particular combination so now
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the whole thing is a polynomial in t of
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degree
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- let's expand it out
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to see what it looks like and let's
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actually to make it a little bit
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easier to understand we're going to
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collect terms so
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organize it in terms of t
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all right there it is in decreasing
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powers of tt the 10 to
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the nine etcetera it down to to this
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thing here so let's have a look at uh
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at all this i
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want to use the fact that we are
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agreeing that t to the 6 is 0.
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so all of these terms from t to the 10
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down to the t to the 6
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can be removed all right
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so all of this is our original
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polynomial and the whole point is to try
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to set this equal
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to zero and solve for x in this case it
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means we have to solve for the
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the a's now
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this constant term the constant term
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gives us
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what's the constant term the constant
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term is this thing right here
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so if we factor this
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that's supposed to be equal to zero so
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we get two solutions there are two
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possibilities so two possibilities
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we could get a sub zero a
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sub zero equals zero
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and the other one is that a a
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sub zero equals
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uh i guess we that's just a linear
18:35
equation so that's minus
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c sub one divided by c
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sub two
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and recall that we've agreed that c sub
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2 is not
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zero so that quotient does make sense so
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these
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two possibilities now sort of provide a
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fork in the road there's going to be
18:54
two kinds of solutions that we're going
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to be able to generate one from each one
18:59
of them
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so let's look at the simpler one first
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okay
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so we'll start with
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a sub zero equals 0.
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and that's sort of a case so i'm going
19:12
to actually make a little
19:15
subsection here okay
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okay what happens if a sub zero equals
19:23
zero
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well that's just one of the terms so
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that constant term will be zero but we
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want this entire polynomial
19:31
in t to be zero so we also get
19:35
a number of conditions the
19:38
maybe i'll start with that one so a sub
19:40
0
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equals 0 so we're agreeing on that and
19:44
then
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the linear term here
19:51
that's going to be zero the quadratic
19:54
term
19:58
that's got to be zero
20:03
the cubic term that's going to be zero
20:08
quartic term that's going to be zero
20:12
and the degree five coefficient
20:18
all right let's stare at this for a
20:19
little while so what are we trying to do
20:20
we're trying to solve
20:22
for the a's in terms of the c's
20:29
okay uh we've got a zero here the second
20:33
equation
20:34
is a linear equation in a
20:37
one
20:42
a linear equation in a one involving a
20:46
zero okay but a0 we've said is 0. so
20:49
actually this
20:50
term here is going to be 0. so
20:54
let's simplify things all right i'm
20:55
going to take that term and i'm going to
20:57
kill it
20:57
because of our assumption that a 0 is 0.
21:00
similarly this thing here
21:02
will disappear anything else with a0 in
21:05
it
21:05
that will disappear
21:12
a0 so we also
21:16
get these relations
21:22
okay now we want to solve for
21:27
a1 a2 a3
21:30
a4 and a5 now scientific workplace
21:34
actually
21:35
is able in some cases to solve
21:39
a whole bunch of systems like this so
21:41
we're going to solve for a
21:43
sub 1 a sub 2
21:46
a sub 3 a sub 4 a
21:49
sub 5 maybe we'll put in a sub 0 just
21:52
to be complete even though we already
21:55
know what that one is
21:56
okay let's see if scientific workplace
21:59
will
22:00
figure out what the values are hey
22:03
it does so
22:06
we get
22:10
we get
22:17
these values for
22:23
our solution so what is this a zero is
22:26
zero
22:28
ie so now we can write down what x is x
22:30
is going to be
22:32
uh actually but the way i'm going to do
22:36
this
22:36
uh just to show you something else i'm
22:38
going to use a definition so i'm going
22:41
to
22:41
select this and i'm going to make this
22:42
definition up here okay that's a new
22:44
definition
22:45
and here are all the definitions there's
22:47
only one definition which is that a1
22:49
equals
22:49
minus 1 over c1
22:53
and then here um
22:57
i'm going to select this and make that
22:58
into a definition
23:03
make that into a definition
23:07
a4 into a definition
23:10
and a five in definition and i'll just
23:12
check my definition so i've got one for
23:14
each one of these
23:15
things a1 3a5 and so now i can go back
23:18
up to
23:19
x which is this thing here
23:31
and i can just evaluate this thing
23:36
okay that's kind of cool and a
23:40
zero um i didn't i should have also
23:43
selected that as something else so i'll
23:45
add that to the list
23:46
okay zero's on the bottom there so
23:48
that'll then
23:50
kill this thing and we'll get this as an
23:52
expression
23:53
for x
24:00
sorry try that again
24:04
display this
24:13
all right what have we got here it looks
24:15
like we've got something
24:17
that looks like a solution
24:20
to our original
24:24
quadratic expression or quadratic
24:27
formula
24:28
as a polynomial or poly number an
24:32
ongoing polynomial
24:33
in t and remember t was equal to
24:36
c zero where i remind you
24:40
t equals c zero
24:44
and there's uh notably there's powers
24:48
of c1 in the denominators
24:51
there are powers of c1
24:54
in the denominators increasing powers
24:58
so that means we better
25:02
ensure so c one
25:05
had better not equal zero
25:09
we better remind ourselves right we
25:11
can't allow multiplication
25:13
or division by zero and and otherwise
25:17
there are
25:18
some coefficients
25:21
with additional coefficients
25:28
one of these coefficients uh 1 1
25:31
ignore the minus sign 2 5 14.
25:39
now these are very famous numbers
25:43
these are very famous numbers
25:49
they are the first
25:54
five catalan numbers
26:01
so it looks like this elementary
26:05
kind of manipulation has allowed us to
26:08
find
26:09
a little formula at least the start of a
26:11
formula
26:13
for a solution to a quadratic equation
26:15
in which the
26:17
catalan numbers kind of miraculously
26:20
appear
26:21
and then there's then it going to be a
26:24
corresponding
26:25
other solution which this
26:29
uh the a0 will have to be replaced with
26:31
is more complicated minus c1 over
26:33
c2 okay but i think we can
26:36
already see here that we're kind of
26:38
interested we're getting somewhere here
26:40
right
26:40
we're making some progress where we're
26:43
seeing
26:44
the catalan numbers fall out it means
26:45
that we're actually
26:47
touching base with the classical
26:49
formulas because it turns out that the
26:50
catalan numbers are
26:52
very closely connected with the square
26:54
root of function like square root of one
26:56
plus four x or something has a has a
26:58
one over over that has a power
27:01
series involving catalan numbers so it
27:04
looks like this is maybe some kind of
27:06
algebraic way of
27:07
trying to you know get at the usual
27:10
square roots
27:11
so i think this is a good place to stop
27:13
um we've made an initial foray
27:15
into this and obviously we're
27:19
raring to go to see how we can go
27:21
further
27:22
in this direction so we'll be doing that
27:24
next time hope you'll join me for that
27:26
i'm norman wildberger thanks for listening