Solving Polynomial Equations 6: On to the quartic equation!
Wild Egg Maths
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Having the beginnings of a tentative solution to a cubic equation, can we extend the story to the general quartic equation? Having a good understanding now of the recursive approach to finding a power/rational series to fit a given equation (after all, a standard procedure in physics and engineering) it is not hard to push things to the degree four situation.
The results are a very pleasant surprise!
00:01
hello everyone i'm norman wahlberger and
00:02
thank you for joining me on this
00:04
adventure
00:05
towards solving polynomial equations so
00:08
we've just had a string of good luck i
00:11
think
00:12
and found an interesting new kind of
00:15
formula for the
00:16
cubic equation actually sort of a
00:18
general cubic equation
00:19
so here is sort of a summary of what we
00:22
have accomplished that
00:23
if we're looking at c zero plus c one x
00:26
plus c two x squared plus c
00:27
three x cubed equals zero then at least
00:30
up to a certain point so up to say
00:32
degree seven in
00:34
in the variable c zero we get some
00:36
expression like this
00:37
x equals this expression it's like a
00:39
power series but it's not really a power
00:41
series because
00:42
it actually has
00:45
terms in the denominator so it's kind of
00:48
a
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has a mixed aspect say it's not just
00:51
polynomial it's a it's more like a
00:53
rational function that's going on
00:55
and and there's some rhyme and reason to
00:58
these uh
00:58
terms and we can summarize like this
01:01
that we have powers of c1 in the
01:03
denominator
01:04
at powers of c0 c2 and c3 in the
01:06
numerator
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and the sign depends on the parity of
01:09
the power of c3 that appears
01:11
here are all the bunch of minus ones
01:13
because there's um
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c3 is appearing to the zero power here
01:17
c3 is appearing to the first power
01:19
here's c3 appearing to second power here
01:22
c3 appearing to the
01:24
third power and so the kind of alternate
01:26
side and
01:27
in general uh the terms are all balanced
01:30
that is they have total degree
01:31
zero so actually if we look at both
01:34
the numerator and the denominator so
01:36
it's not just a polynomial now it's like
01:38
a rational function
01:40
and then the degree should be calculated
01:42
how i'll say something like this right
01:44
there's a c 0 to the fourth in the
01:47
numerator a c
01:48
2 cubed we would normally say that
01:50
numerator has degree 7
01:52
but there's a something of degree seven
01:54
on the on the bottom of
01:55
the denominator so all together this
01:57
really has degree
01:58
seven minus seven so we say it has
02:00
degree zero total degree zero
02:02
and they all have that property okay
02:05
that's nice
02:06
and they're all of total index degree
02:08
minus one
02:09
let me remind you about the what we call
02:11
the index degree so
02:13
let's say this term here to calculate
02:14
the index degree what we do is we look
02:16
at the sum of all the products of the
02:18
exponents
02:19
times the indices so here we have four
02:21
times zero
02:23
three times two and actually
02:26
minus seven times one so we get zero
02:29
plus six plus minus seven that's minus
02:32
one
02:34
let's have a look and say this one here
02:36
uh 84
02:38
index degree 6 times 0 that's 0. 3 times
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2 is 6
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and 1 times 3 is 9
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well that's 1 times 3 is 3 but
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altogether that's 9 in the numerator
02:49
and then in the denominator we have a
02:52
- so the
02:52
the index degree would be minus 10
02:56
times 1. okay so altogether
02:59
9 minus 10 it's a total index degree of
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minus
03:03
- so that's kind of a pleasant way of
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sort of restating what we did
03:07
earlier but with the bigger picture
03:10
sort of including this rational function
03:13
aspect to the terms that are appearing
03:15
here
03:15
and the general term which was of this
03:18
form
03:19
seems to have a coefficient which is uh
03:21
the power of the c1 in the bottom k
03:24
minus one of the factorial and a
03:26
factorial b factorial c
03:28
factorial if those are the exponents of
03:29
c zero c two and c c3 in the numerator
03:32
and there's all the sine of minus 1 to
03:34
the c plus 1.
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oh great great okay so this of course
03:38
was only up to degree 7 so you know
03:40
there's there's more to be done
03:41
in the cubic case and of course we want
03:43
to really get at these numbers that are
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appearing here and it looks quite
03:46
interesting and
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so there's probably a lot more that we
03:48
can do
03:50
in the cubic case and we we need to go
03:52
back and do that but
03:54
now we are so excited by the possibility
03:56
that
03:57
this seems to be a somewhat general
04:00
formula and so
04:02
we're immediately interested in whether
04:04
it extends to the degree four situation
04:07
okay that would be really exciting and
04:10
if not we want to know
04:11
what additional complexities can we
04:14
expect as we go from degree three up to
04:16
degree four
04:17
okay so going from degree two degree
04:19
three there was a big jump in complexity
04:21
basically this first line here sort of
04:23
captured the degree
04:25
two the quadratic situation right those
04:27
familiar kind of catalan numbers
04:29
now we're extending that catalan story
04:31
to this broader thing
04:32
and now we want to know okay so we go
04:34
now from degree three degree four
04:36
how much more complexity is there going
04:37
to be so let's press on right now with
04:39
this
04:40
this overwhelmingly important question
04:45
okay so let's start with our cubic
04:47
equation now
04:48
now there's an additional term c
04:51
for x to the 4 equals 0. and let's
04:55
set x equal to a familiar power series
04:58
in t where we've set c zero
05:01
equal to t for convenience
05:05
so we'll go up to degree eight um that's
05:07
rather arbitrary but the idea is we're
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going to sort of cut
05:10
off at degree 8 and ignore all terms
05:13
of degree 9 and higher in t
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okay and the the point is that we're
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going to substitute this into our
05:20
basic quartic equation and try to solve
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for
05:24
a0 a1 up to a8 in terms of
05:27
the variables c0 c1 c2 c3
05:30
and c4 okay so let's um
05:34
see how to do that so maybe we could
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start by taking
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this vowel this variable x here and
05:41
defining
05:42
x to be this expression here all right
05:45
so we'll set that equal to that
05:47
just check our definitions okay there is
05:49
x
05:51
and now we'll substitute into
05:54
sub into
05:58
this quartic equation
06:02
and we'll rewrite that replacing the
06:06
c0 with t as we've agreed okay
06:10
so we're going to now wanting to set
06:12
this equal to zero
06:14
okay so let's put a zero equals in the
06:16
front so
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this is going to be a big polynomial in
06:20
in t that we're going to set
06:21
equal to 0. and now let's see what we
06:24
actually get
06:25
okay so there's t plus c 1 x plus c 2 x
06:28
squared plus c 3 x cubed plus c 4
06:32
x to the 4. okay now what we want to do
06:36
is we want
06:37
to expand this out
06:40
and sort of collect terms involving t
06:44
but we want to do that carefully because
06:45
if we just expand it out as it is we're
06:47
going to get all kinds of
06:49
terms additional terms that we don't
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want which are degree
06:53
9 or higher remember we're ignoring all
06:55
of those
06:56
so the way we can do that is to use our
06:59
power series capability here so we're
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going to evaluate this as a power series
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the number of terms say nine in terms of
07:09
t okay
07:13
okay we get this expression here and
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the a constant term is this one here
07:20
okay so the constant term
07:24
the term gives
07:28
this thing equals zero but if we just
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take that and factor it as we've done
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before
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we get a zero times this
07:39
expression equals zero
07:44
okay so this is the very first of our
07:47
terms it has to be 0 and what can we
07:49
conclude from this
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well we can conclude something about a0
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either a0 is equal to 0
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or it satisfies this cubic equation
07:58
so the simplest possibility is if we
08:02
set a0 equal to zero so
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set a sub zero
08:08
equal to zero okay
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and we'll define that
08:15
and then the the remaining
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terms will simplify because a0 being
08:22
zero means that this
08:23
expression will simplify dramatically
08:26
then we get
08:27
a simplification of the remainder
08:33
say
08:36
so we'll take this again all the way
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from here
08:40
down to down to there
08:46
and let's simplify that
08:56
maybe we'll expand it
09:01
and get rid of the o t sub 9 that's just
09:04
representing the additional terms that
09:05
we're setting equal to zero
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and let's now rewrite this in powers of
09:11
t
09:13
polynomial collect t
09:21
all right so i can dispense with this
09:23
other stuff that we had really it
09:24
amounted to this
09:27
this expression um is a polynomial
09:31
in t and we want to set it equal to zero
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so there's a linear part quadratic part
09:36
cubic part
09:37
etc okay so then we get
09:43
a certain number of equations the linear
09:46
part
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has to be zero so that expression is
09:49
going to be
09:50
- that will mean that we'll be able to
09:52
solve for a1
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and then the quadratic part is this
09:56
thing here
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set that equal to 0 once we know a1 then
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we'll be able to solve for a2
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again rather simply because it's just
10:06
linear in a2
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and then the cubic that one there
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that'll solve
10:13
for a3 and the next one over here
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oops where are we that one there
10:24
that will give us a four and maybe we'll
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go one more
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a5 okay
10:31
actually we don't really need these
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brackets let's get rid of them
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or because it's getting a little bit
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complicated let's first just calculate
10:41
a 1 through a 5 using these five
10:44
equations and then we'll go
10:46
on to a6a78 subsequently okay
10:49
so this is a straightforward now
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exercise which our computer
10:52
hopefully can do very quickly so we're
10:54
going to solve
10:55
exact for a sub 1 a
10:59
sub 2 a sub 3 a sub 4
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and a sub 5.
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great okay so we get some solutions so
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let's grab those solutions
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don't need that anymore let's arrange
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them in a
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more readable way there's a one
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there's a two
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three
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a4 and a5
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[Music]
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okay so those are the values of a1 a2 a3
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a4 a5
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in terms of the original coefficients
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c0 c1 c2 c3 and c4
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now we recognize that there's something
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familiar here we've seen
11:53
some of these things before we recognize
11:54
some catalan
11:56
numbers ones twos five fourteen
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and however we do also recognize
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something that's
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new that we didn't have before which is
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this term here which involves
12:09
the c4 that's the additional term
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so it's kind of nice it's suggesting
12:14
that what we're getting now is an
12:16
expansion
12:17
of what we got for the cubic situation
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right we're taking the cubic solution
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and we're adding to it
12:24
that's what it looks like it's not that
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we are sort of starting from scratch
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and that the quartic equation solution
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is completely different from the
12:32
cubic one no uh and this is a really
12:36
nice aspect that there's the potential
12:38
that we're building up as we go
12:41
solutions to higher and higher degree
12:43
equations one on top of the other
12:46
not independently not separate equations
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think how different that is from the
12:51
usual radical story
12:54
okay so what we're gonna do is we're
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going to get our computer to remember
12:58
these things
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so we'll make definition for a1
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for a2 for a3 for a4
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and for a5 and we'll see where we are
13:10
good we've got them all there
13:14
okay now we can carry on now carrying on
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we can take what we had before
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which was this stuff up to say t
13:26
to the sixth
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[Music]
13:31
and we're going to want to evaluate that
13:34
to be a zero and that will give us some
13:36
additional equations
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okay so we're getting um there's some
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additional equations now involving a6 so
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here's the coefficient of a6 i'll just
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grab the whole thing
13:49
and maybe i will put zero equals okay
13:52
that way you can
13:55
maybe not have to see the entire
13:57
equation but we know the entire equation
13:59
is there
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and then the coefficient of t to the 7
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is all of this we want that
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to be equal to 0 as well and there's
14:07
also the coefficient of
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t 8 which is all of that
14:13
so i'm just going to do some line breaks
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here so that you can see
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at least what these things look like
14:20
at least partially maybe i need another
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line break here
14:28
all right so there there are the three
14:30
new additional equations that
14:32
hopefully will allow us to calculate a
14:34
six a seven and
14:35
a eight all right so i'm going to go
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back and remove those line breaks
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because
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i need to do that um before
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i get
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the computer to solve
14:54
and we had a line break also
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here okay
15:01
all right so these were the additional
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equations involving a6
15:04
a7 a8 so now let us get the computer to
15:07
solve what our a6 a7
15:09
and a8 solve
15:12
exact a sub 6
15:15
a sub 7 a sub 8
15:18
okay good it does something okay let's
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grab what it does
15:24
we get and let's write this out
15:30
there's an a6
15:34
there's an a7 and there's an a8
15:43
okay good and now i'm going to
15:47
remember those values to add them to the
15:49
ones that we already have
15:55
and let's just check that i have all of
15:57
them i've got a6 a7 a8 added to the a0
16:00
up to a5 that i had previously so we've
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solved for all of the coefficients
16:04
a 0 up to a8 under the assumption that
16:07
a0
16:08
was 0 that's the starting point and then
16:10
the rest of it is forced
16:11
we get these expressions involving c0 c1
16:14
c2 c3 and c4 and we see that
16:18
you know it's very parallel seemingly
16:22
close to
16:22
a cubic case but more complicated
16:26
all right good so now we have all of the
16:29
the a i's and so now let's look at the
16:32
final you know
16:33
what we've got we've got an expression
16:34
for x and let's have a look at it
16:37
and we'll do the same kind of thing that
16:39
we did in the cubic
16:40
write everything in terms of the
16:42
original coefficient
16:44
c0 up to c to the 4.
16:48
all right so let's let t equals c sub
16:51
zero because remember we were using
16:52
t throughout a lot of this um so we'll
16:56
we'll make that definition and then then
16:59
we get
17:00
the following expression for x x
17:07
and maybe we'll expand that out
17:13
okay get rid of that earlier thing all
17:18
right so let's look at this
17:19
expression we want to try to explain
17:21
various aspects of it
17:22
let's consider the kinds of terms that
17:25
are appearing for example let's say this
17:27
one
17:27
here okay that term there
17:32
what is its total degree the degree is 8
17:36
plus 3 plus 2 that's 13
17:39
minus the 13 on the bottom so that has
17:41
degree 0.
17:43
how about this one here total degree six
17:47
plus one plus one minus eight that's
17:49
total degree
17:50
zero how about this one over here eight
17:52
plus four
17:53
plus one that's thirteen minus thirteen
17:55
that's degree zero
17:56
how about this one here eight plus seven
17:58
minus 15 that's also degree 0. it looks
18:00
like they all have degree 0
18:02
just as in the cubic case what about the
18:04
index degree so if we go back here what
18:06
about the index degree
18:08
so remember that means we have to take
18:09
products of exponents
18:12
and an indices
18:14
a times zero that's zero plus three
18:17
times two that's six plus two times
18:19
three that's another six
18:21
and on the bottom minus thirteen times
18:24
one because that
18:24
exponent is really minus thirteen so the
18:27
index
18:28
degree of that is minus 13 so we get 6
18:30
plus 6 minus 13 that's minus 1.
18:32
that's good that's exactly the same kind
18:34
of thing we were getting before
18:36
how about this last term here index
18:38
degree is 0 plus 4
18:40
plus 3 plus 4 4 plus 3 plus 4 that's uh
18:44
11 minus 12 is minus 1.
18:49
how about over here
18:52
0 plus 2 plus 3 plus 4 that's 9 minus 10
18:56
that's again minus 1.
18:59
so that same thing that we saw in the
19:02
cubic case
19:03
where the degrees are all homogeneously
19:06
zero and the index degrees are all
19:08
homogeneously minus one it's happening
19:10
here and it's independent it's not
19:12
doesn't it's not changing
19:14
the cubic values are exactly the same as
19:17
the quartic values
19:20
that's cool what about the sign of the
19:22
expression can we figure out what that
19:23
is so there's pluses and minuses when do
19:25
we get
19:26
a minus sign so here we're getting a
19:27
minus sign
19:29
here we're getting a plus sign what's
19:32
different
19:36
there's a c3 term here
19:40
that's uh odd but here it's even
19:43
here there's an odd here there's an even
19:47
c3 an even c3
19:51
and even c3 even c3 even c3 even c3
19:56
even yeah uh oh this one here has a
19:59
has an odd odd c3 because that term it
20:03
does have a plus sign
20:05
here there's an even c3 with a minus
20:07
sign
20:09
here there's an odd c3 with a
20:12
plus sign so it looks like it's c3 that
20:15
determines
20:16
the sign just as it did in the previous
20:18
case
20:20
what about the coefficients
20:23
so let's grab one of them let's say say
20:26
this one here
20:30
and we want to uh
20:33
have a look at what was the kind of
20:35
formula that we were looking at we were
20:36
looking at a fraction
20:38
involving some some factorials divided
20:41
by a factorial if we're doing exactly
20:42
the
20:42
corresponding thing that we did in the
20:44
cubic case we would put one less than
20:46
than the denominator here so we put an
20:48
11 factorial
20:49
at the top and on the bottom we would
20:51
have a product of the various
20:54
exponents or powers that are appearing
20:55
in the numerator so it'd be a 1
20:57
factorial
20:59
corresponding to the c3 term a 4
21:02
factorial corresponding to the c
21:04
2 term and a 7 factorial corresponding
21:07
to the
21:07
c 0 term and let's see what that thing
21:09
evaluates it evaluates to 330.
21:14
cool let's choose another one how about
21:15
this one here
21:17
here there's um
21:23
also a c4 term let's look at that
21:27
okay so um so let me
21:31
explain the coefficient here
21:35
is found by that formula okay now we
21:38
hopefully want to find a similar
21:40
justification for the coefficient here
21:42
the 495
21:44
so hopefully we'll still have an 11
21:45
factorial in the numerator corresponding
21:48
to one less than the power of c1
21:50
down here and then in the denominator
21:53
we should have a one factorial another
21:56
one factorial
21:58
a 2 factorial and a
22:02
8 factorial corresponding to various
22:05
exponents here for c0 c2 c3 and c4
22:10
fabulous all right so um
22:13
let's see what that thing is let's
22:15
evaluate that
22:17
- oh that's so excellent
22:20
um should we try one more how about this
22:22
990 let's try that one
22:24
okay
22:29
and uh so hopefully something like that
22:32
will work
22:35
the coefficient here is found by so we
22:37
should take the 13 subtract 1 that's 12
22:39
factorial in the top
22:41
and then there's an 8 factorial a 3
22:44
factorial
22:45
a 2 factorial and a
22:49
like a 0 factorial and that amounts to
22:52
990.
22:56
so we could go through the rest of these
22:57
things and we could
23:00
you know become very confident that
23:03
we're on the right track here
23:04
and note that
23:07
you know it really looks like the same
23:10
general procedure that we found in the
23:12
cubic case is working here in this
23:14
what in this quarter case right so this
23:17
is very exciting it means that
23:18
we perhaps have captured the general
23:21
situation
23:22
it's not impossible now that this
23:24
general framework that we've found
23:26
will extend because it's gone smoothly
23:29
from two to three and also from three to
23:31
four
23:32
although we haven't checked all the
23:33
coefficients but you know you can check
23:35
some more
23:37
and then it's likely that it carries on
23:39
to even higher power so that
23:41
in some sense we we might be looking at
23:43
right at this stage here the possibility
23:46
of a general solution to a general nth
23:49
order equation
23:52
okay that's obviously very exciting so
23:54
probably here's a good time to
23:56
stop and rest and and and
23:59
be happy with the work that we've done
24:01
so far okay very exciting
24:03
now we want to you know have a closer
24:05
look at
24:06
you know does this really work what are
24:09
these these coefficients
24:10
can we write down some nice formulas to
24:12
to really
24:14
come up with some succinct theorems that
24:16
capture what's going on in
24:18
rather great generality it's all very
24:21
exciting