Solving Polynomial Equations 5: Cracking the pattern with the Cubic

Wild Egg Maths

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We have a good look at the interesting numbers appearing in our proposed power series solution of the general cubic equation (actually it is not exactly a power series since negative powers of one of the variables occurs). The Catalan part of the pattern is particularly helpful as it gives us a handle on the other less familiar terms.

This is mathematical exploration in action: not just austere and perhaps unmotivated final results. We want to get some experience with thinking actively about mathematics --common sense and a scientific attitude count for more than quick thinking.

00:01

hello everyone i'm norman wahlberger

00:03

and thank you for joining me on this

00:04

little journey towards trying to

00:07

get at polynomial equations and how to

00:09

solve them

00:10

so i left you hanging last time

00:13

we had gotten somewhere in terms of a

00:17

general solution to a cubic equation so

00:19

let me remind you that we started with

00:21

this cubic equation here

00:23

c 0 plus c 1 x plus c 2 x squared plus c

00:25

3 x cubed equals 0.

00:26

we replaced c 0 with t and then looked

00:29

for a solution

00:30

x which is a power series or sort of

00:32

ongoing

00:33

series in t with coefficients a0 a1 a2

00:37

etcetera to be determined

00:39

we found that one possibility was that

00:41

a0 equals zero and then

00:43

that generated these subsequent

00:47

values for the ais in terms of the

00:49

original

00:50

c's so we got a1 a2 up to a7

00:53

and the expressions involve certain

00:56

terms involving the c's and we

00:58

sort of got our head more or less around

01:00

you know which

01:01

kinds of terms we think are going to

01:03

appear

01:04

and we also looked at the signs plus or

01:07

minus that occur

01:08

in front of these terms and now the main

01:10

challenge is to try to get at the

01:12

numbers these coefficients that are in

01:14

front of these things

01:15

and uh so last time i said okay maybe we

01:19

can just look at the numbers themselves

01:21

and i put them in a matrix

01:23

and we had something that looked like

01:25

this and

01:26

then i factored those numbers so we

01:28

could see their prime power

01:29

decomposition to give us a bit more of

01:31

an insight into the nature

01:33

of the formulas that are behind this

01:37

so what we want to do is is we don't

01:39

want to necessarily assume that there's

01:41

going to be nice formulas but we want to

01:43

kind of

01:44

at least start with that general

01:46

orientation

01:48

okay so often in mathematics we want to

01:50

be looking for the simplest

01:53

the easiest kinds of solutions first

01:56

it's a kind of a natural evolution we

01:58

start with the simplest possibilities

02:00

and then move up increasingly

02:01

complicated ones by necessity

02:04

ruling out the simpler cases first but

02:06

generally we like to have this

02:08

orientation that mathematics is

02:10

ultimately quite beautiful

02:11

and that for some miraculous reason

02:14

things often work out really pleasantly

02:16

at least once you look at things in the

02:18

right way

02:19

okay so we look for simple explanations

02:22

first now i left you with this

02:26

and uh and maybe you've had a go

02:29

and probably you've found it challenging

02:31

okay so it's

02:32

it's not so easy to figure out you know

02:34

which numbers which products of small

02:36

primes are appearing in front of these

02:38

terms and why

02:40

so in this kind of situation what do we

02:43

do

02:44

well we can you know make it

02:47

inspired guesses but there's another

02:51

kind of general strategy that i want to

02:52

share with you which is to

02:54

to sort of go back a little bit and look

02:57

at what we're doing and saying

02:58

are we seeing the entire picture here

03:00

you know the way we're looking at things

03:04

we want it to be as as as broad and

03:08

encompassing as we can given the

03:10

circumstances of what we're looking at

03:13

let me illustrate that in this case here

03:15

so

03:16

what we're really doing is we're looking

03:18

to see you know what this solution

03:20

x looks like and i'm going to

03:23

um do that in in another way

03:27

to so we can actually see what's what's

03:29

going on so let me actually let me just

03:31

check my definitions up here

03:33

i don't have any definitions i've

03:34

cleared them previously let me

03:36

define all these a's so a sub 0

03:40

a sub 1 a sub 2

03:44

a sub 3 a sub 4 a sub 5

03:48

a sub 6 a sub 7 i'll check that i have

03:51

them all there a 0 1

03:52

2 3 4 5 6 7 yes they're all there

03:56

and i'm also going to

04:00

actually go back

04:03

and replace the t with c0 because

04:06

remember we were

04:07

we were using t in a special kind of way

04:10

here because it's simpler to deal with

04:11

because it doesn't have the index

04:13

but actually the t just represents c0

04:16

and so strictly speaking

04:18

to to be as sort of encompassing and as

04:21

natural as possible we should

04:22

replace the t with what it really is so

04:25

let me also

04:27

set t equal to c zero okay

04:30

and now let's have another look at this

04:33

solution

04:34

that we have so far found up to degree

04:38

seven okay there it is

04:42

and let me see what that actually is

04:45

okay

04:46

so it's got lots of terms

04:50

each of the terms is an expression

04:53

actually a rational

04:54

function in the seas now and because

04:58

there's

04:58

some brackets let's see if we can just

05:00

make things simpler by just expanding

05:02

everything

05:04

expand

05:08

what did that happen now let's put the

05:10

thing there let's try that again

05:12

expand there we go

05:22

now

05:24

it's it's this expression that i really

05:27

wanted to um

05:30

to look at because now what we're doing

05:32

is we're doing something a little bit

05:33

more general than we were before

05:35

so before what we had done is we had

05:38

just looked at essentially the

05:40

numerators here the things in brackets

05:43

okay

05:44

but we had not incorporated the power of

05:47

c1 in the denominator and we had also

05:50

not incorporated

05:51

the power of t or c 0

05:54

which is also actually in the numerator

05:56

multiplying the a's

05:58

so a 3 has to really be multiplied by c

06:01

0

06:01

cubed and that's what we're doing here

06:03

so over here now we've

06:04

expressed all these terms purely in

06:07

terms of

06:08

the c's c0 c1 c2

06:12

and nc3 okay

06:15

now what do we know

06:18

well we know that the

06:22

the first column of this this table here

06:26

and that corresponds to the

06:30

um the terms where

06:33

there is no c3 involved

06:36

okay so um

06:40

1 1 2 5 14 42 and so on

06:44

how can i organize those at the front so

06:47

let me organize this thing

06:49

compute i'm going to use this polynomial

06:52

collecting this organ allows me to

06:54

organize a polynomial or

06:55

an expression in terms of powers

06:59

of one particular variable and

07:02

i'm going to use c3

07:10

okay so this one here that you see here

07:13

this last thing here

07:14

these are this of the constant term with

07:16

respect to c3

07:17

so these are the the terms which don't

07:19

involve

07:20

any of the of the c3 these correspond

07:23

exactly to this first

07:24

column and we do recognize the catalan

07:26

numbers appearing here

07:28

so i want to just stare at this just for

07:31

a second

07:32

so let's just consider

07:35

the catalan terms

07:40

okay because we know what what they are

07:42

we know what

07:44

the actual values of these things are

07:46

these are these catalan numbers

07:48

which are you know the central binomial

07:50

coefficients

07:51

2n choose n and divided by

07:55

n plus 1. so for example this 42 let's

07:59

have a look at it here

08:00

so 42 was the fifth catalan number

08:05

which was i remind you uh

08:08

2 times 5 so that's 10 i don't need the

08:11

brackets 10

08:12

factorial over 5 factorial

08:15

times 5 factorial that's the central

08:19

binomial coefficient and then 1 over 6

08:21

times that

08:23

let me just check that that's equal yes

08:24

that's correct

08:26

okay so another way of saying what this

08:28

is i'm going to rewrite this just a

08:29

little bit

08:31

it's equal to instead of the 6 times 5

08:34

factorial i'll just make it a 6

08:35

factorial

08:36

and get rid of this 1 6.

08:40

can make this a little bit bigger maybe

08:42

you having trouble seeing it

08:44

let's go to 195.

08:54

all right so we're looking at this term

08:56

here

08:57

and we're trying to say you know where

09:00

how could we predict this 42 just from

09:02

the nature of things here

09:05

well there's a c 0 to the 6th on top

09:07

there's a c2 to the 5th

09:09

there's a c1 to the 11 on the bottom

09:13

okay that's interesting because there's

09:14

a 6 factorial on the bottom here there's

09:16

a 5 factorial on the bottom

09:17

corresponding to these two

09:19

powers here and there's well a 10

09:21

factorial on the top

09:23

instead of well there's an 11 as a power

09:26

of c1 in the

09:27

in the numerator in the denominator

09:30

let's

09:30

see how that combines say compares with

09:33

say 132.

09:35

so 132

09:39

is the next one so that's c sub 5

09:42

so that should be equal to

09:43

correspondingly

09:47

sort of c sub 6 sorry that should be 12

09:50

factorial

09:52

over 7 factorial times

09:55

6 factorial let's just see what that is

09:58

that is 132 right

10:00

okay so how could we have figured this

10:02

thing out or determined it from the

10:04

nature of the terms

10:06

here involving c0 c1 c2 and c3

10:09

well there's no c3 but there's a c0 to

10:12

the 7 that corresponds to that

10:14

factorial in the denominator there's a

10:16

c2 to the 6 that corresponds to that

10:18

6 factorial in the denominator and

10:21

there's this 13

10:23

okay which sort of perhaps is

10:26

responsible for the 12 factorial

10:28

even though 12 is 1 less than 13. but

10:30

that's the same kind of thing that we

10:32

had appearing here so what we're seeing

10:35

is that

10:36

when we when we rewrite things by

10:39

putting

10:39

all the terms in so we've got c0 c1 c2

10:43

and c3

10:44

and we stare at all of things and then

10:45

we concentrate on

10:47

that part of it that we do actually know

10:49

about namely the

10:50

the terms involving the catalan numbers

10:53

then we see that there's a kind of an

10:55

explanation

10:56

so at the level of just those ones we

10:58

can kind of

10:59

guess what the coefficients would be

11:01

once we've sort of established this

11:03

pattern

11:03

there seems to be a pattern here okay

11:06

so now obviously let's have a look at

11:09

some of the other terms

11:10

and see you know if it may be the same

11:12

kind of thing works so if we go back

11:14

up here let's have a look at this thing

11:16

here let's just grab this term

11:18

all right that's just an individual term

11:21

uh

11:21

actually that's times c3 okay all of

11:24

this times c3 so i better put the c3

11:26

there

11:27

so i also need to put a c3 here

11:30

okay that's the corresponding that so

11:32

what would be perhaps the corresponding

11:34

expression

11:37

to you know to explain this 84 if we

11:39

were going to do exactly the same kind

11:41

of thing or a parallel kind of thing

11:43

to what we've just seen happens to the

11:46

catalan

11:46

story okay well we would

11:50

do a ratio of

11:54

factorials and the powers in the

11:58

numerator

11:58

seem to end up corresponding to the

12:02

factorials in the denominator so what

12:04

powers are there in the numerator

12:06

there's a c 0 with a 6 a c

12:07

2 cubed and a c 3 to the 1. so maybe we

12:11

have 1 factorial

12:13

times maybe 3 factorial

12:16

times 6 factorial okay and what should

12:19

be

12:19

in the numerator well if it was the same

12:22

thing that was going on here we should

12:24

take this power of c1 which is in the

12:26

denominator namely 10

12:27

and maybe reduce it by one to nine

12:32

let's try it what is that equal to oh

12:34

it's equal to 84.

12:37

good okay let's try it again how about

12:39

this next one right beside it

12:41

330

12:45

330 okay remember we have to include

12:47

this term

12:48

c3 here so put that there okay we want

12:52

to explain that 330.

12:54

so let's grab the

12:57

um this thing and modified

13:01

so in the numerator now we have a

13:02

there's a seven factorial

13:04

it's going to be a 4 factorial 4

13:08

and it's still a 1 factorial for the c3

13:11

and in the numerator there's a 12 down

13:14

here so probably we should put 11.

13:18

what does that equal to oh it equals

13:19

330.

13:24

this is very cool okay so we have this

13:27

start of a conjecture

13:30

we start a conjecture that the

13:33

coefficient of

13:36

the the

13:40

the terms involving

13:44

the c's c sub 0 c sub 1

13:48

c sub 2 c sub 3

13:52

are obtained as quotients

13:56

of products of factorials

14:01

the numerator

14:04

has a factorial for each

14:08

of the sorry not the numerator let's say

14:11

let's start with the denominator

14:13

the denominator denominator

14:16

has a factorial for each of the powers

14:20

of well c0

14:24

c1 c2 and not c1 because c1 is in the

14:28

numerator

14:29

in the denominator c0 c2

14:32

and c3 in the

14:36

numerator

14:40

of the term okay so what i'm talking

14:43

about here

14:44

um this denominator like what this

14:46

number that we're getting

14:48

has a denominator of 1 factorial 4

14:49

factorial 7 factorial

14:51

and that corresponds to the powers of c0

14:55

c2 and c3 in the numerator of the of the

14:57

term involving the c's

15:00

while the numerator of the coefficient

15:04

maybe i should put that make it clear

15:06

the denominator of the coefficient

15:10

and while the numerator of the

15:11

coefficient

15:13

is a factorial

15:18

which is one less

15:23

well one less than the power

15:27

of c1

15:31

in the denominator

15:36

how does that sound something like that

15:39

so let's let's

15:40

now go back up here let's grab this

15:42

thing again

15:44

uh here here's what we want to explain

15:46

right this thing here

15:47

was what we wanted to explain so let's

15:49

have a more careful look at it

15:51

um i guess we can get rid of this we

15:53

don't need that anymore

15:55

okay so let's check further

16:03

okay we don't just want to do uh you

16:05

know two or three and say oh now we have

16:06

the

16:07

conjecture now we want to systematically

16:08

look um

16:10

if this works in fact if i was doing

16:12

this you know

16:14

on my own i would check each one of

16:16

these terms just to make sure that each

16:18

one of them

16:18

was was working but let's do a couple

16:21

and i'll leave the rest to you so how

16:23

about this 180 have we talked

16:24

about that one yet so let's look at this

16:27

term so that's got

16:28

180 and there's also this

16:31

power of c 3 which we better not forget

16:34

okay so i'll say this has a coefficient

16:40

maybe if our conjecture is right

16:46

of what let me just grab copy the form

16:50

of it

16:51

and modify it okay so in the

16:54

denominator here we should get a seven

16:55

we already have a seven and then it

16:57

should be a two

16:58

and then another two

17:02

and there's an a power of 11 down here

17:04

c1 to the 11

17:06

so we should get maybe a 10 up here

17:10

okay hopefully if i evaluate this i get

17:13

180 and i do

17:15

that's great let's try another one how

17:18

about this one out here that has a c

17:22

3 cubed so that's going to have a

17:25

coefficient

17:26

of

17:30

okay um in the denominator

17:35

there's there's no c2 actually so

17:38

there's

17:38

there's we'll just have two terms

17:40

there's a a seven

17:42

corresponding to the c zero to the seven

17:44

and there's a 3

17:46

corresponding to the c 3 but there's no

17:50

power of c 2 or if you like it it's c

17:54

2 to the 0 so you could put a 0

17:56

factorial there

17:57

but that's equal to 1. okay

18:01

so and then what about the denominator

18:04

uh well the numerator of this thing

18:05

which is the denominator of that here

18:07

there's a 10

18:08

and so we should put one less than that

18:10

which is a 9.

18:12

and what's that equal to 12 that's the

18:15

same

18:15

thing okay so this is uh

18:19

very exciting okay it means that

18:22

potentially

18:23

now we have um

18:27

a kind of a process that allows us to

18:30

predict

18:31

what the coefficients are going to be

18:34

for

18:37

these terms which are made out of the

18:39

c's

18:40

the coefficients of the original

18:44

equation that we were going to to solve