Numerical solution of quintic equations using bootstrapping
Wild Egg Maths
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The problem of solving quintic equations has generated considerable interest over the centuries. From the point of view of Galois theory, it is generally thought that solutions of such equations are intrinsically different from lower degree equations, at least in terms of finding formulas using expressions with radicals.
In particular, theoretical procedures for generating even approximate solutions using numerical methods are not known, or at least not widely known.
However in this series Solving Polynomial Equations, we have developed a powerful new algebraic way of solving general equations using a massive generalization of the Catalan story, involving roofed subdivided polygons (RSP's) and associated power series as suitable generating functions. Let's now see if we can use this novel technology to get approximate numerical solutions to explicit quintic equations!
The method of bootstrapping will be particularly useful here; this is a procedure outlined in the previous videos for using approximated solutions to generate more accurate ones. We will be happily harnessing the power of Wolfram Mathematica to help us!
The ideas here generate a wide range of questions and additional avenues for exploration this is fertile ground for keen amateur explorers of mathematics, and fits nicely into our general plan of laying our fruitful directions to Explore Research Level Mathematics.
Be sure to join our Wild Egg Maths channel as a Member, and get the entire series of videos on this topic: the full Playlist for Solving Polynomial Equations is at https://www.youtube.com/playlist?list.... However as a Member you access not only this content, but much more!
Screenshot PDFs for my videos are available at the website http://wildegg.com. These give you a concise overview of the contents of the lectures for various Playlists: great for review, study and summary.
My research papers can be found at my Research Gate page, at https://www.researchgate.net/profile/...
My blog is at http://njwildberger.com/, where I will discuss lots of foundational issues, along with other things.
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00:01
hello everyone i'm norman wahlberger and
00:02
thanks for joining me on this adventure
00:04
towards solving general polynomial
00:06
equations
00:07
today numerical solutions of quintic
00:10
equations
00:10
using bootstrapping so first we'll just
00:14
review
00:15
the theoretical setup and then we'll
00:17
turn to mathematica
00:19
and see that these theoretical
00:22
algebraic expressions actually provide a
00:25
very powerful and direct way
00:28
of obtaining numerical solutions for
00:30
these quintic equations
00:33
but they don't automatically work so we
00:35
have to adjust
00:37
our approach depending on the nature of
00:40
the coefficients
00:41
and that's where this bootstrapping idea
00:44
comes in
00:45
okay so let's review it's very parallel
00:48
to what we've already done in the
00:49
quadratic
00:50
and the cubic cases so i hope you've
00:52
been following that along and you'll be
00:53
able to
00:54
relatively quickly understand that what
00:56
i'm doing now is just a
00:58
direct generalization of those earlier
01:00
examples
01:01
so first of all our rsp numbers these
01:03
are these coefficients which play the
01:05
crucial role
01:07
and they're defined in the usual way and
01:09
then
01:10
we start with the geometric cubic that
01:12
has a special form 1 minus
01:14
x and so on and here is our formula for
01:18
it which is beautifully just a power
01:20
series in
01:21
the coefficients t2 up to t5
01:24
whose coefficients are these are numbers
01:28
these rsp numbers
01:30
now first order of business is to move
01:34
to the
01:34
more general equation where we have
01:37
something of the form
01:38
c0 minus c 1 y plus c 2 y squared and so
01:42
on equals 0.
01:43
so how do we transition from this
01:45
equation which we're calling equation 1
01:47
to this equation which we're calling
01:48
equation 2.
01:50
okay we're making changes of variables
01:53
first of all the x is going to be
01:54
replaced with c 1
01:56
over c 0 y
01:59
and we're going to replace the t 2 t 3
02:03
t 5 variables with these expressions
02:06
have a look at this
02:08
t 2 is c 0 c 2 over c 1 squared
02:12
t 3 is c 0 squared c 3 over c 1 cubed
02:15
t sub 4 is c 0 cubed c 4 over c 1 to the
02:18
fourth
02:19
c sub 5 equals c 0 to the 4th c sub 5
02:22
over c 1 to the 5th
02:24
okay now to explain where we got that
02:26
from well you should go back to the
02:27
cubic and
02:28
look at that very carefully okay and
02:31
when we do
02:32
all these substitutions into here what
02:35
does it become
02:37
well the 1 minus x there is the new x c
02:40
1 over c 0 y and we've replaced
02:42
all of these x's with that extreme
02:44
expression and then we plug in these t
02:46
so there's t2 there's t3 and there's t4
02:49
and there's t5
02:51
and what happens when we simplify this
02:53
well if you look at the power of
02:55
c1 in one of these expressions lay this
02:57
one here there's a c1 cubed
02:59
and there's a c1 cubed in the
03:00
denominator so they all cancel
03:02
and the c zeros well there's a c zero
03:05
squared and there's a c
03:06
zero cubed in the denominator so
03:09
altogether that's like one over
03:10
c zero so there will be a one over c
03:12
zero in each one of these expressions so
03:15
if we multiply everything through by
03:17
c zero we'll just get this c zero and
03:20
then other terms which don't have any c
03:22
zeros and if you look at see what the
03:24
other terms are well after we've
03:25
multiplied through by c zero we're gonna
03:27
have minus
03:27
c one y and over here all what's going
03:30
to remain is
03:31
the c2 that's the only thing that hasn't
03:34
canceled
03:34
over here the only thing that's going to
03:35
remain is the c3 here's c4
03:38
and here's e5 so yes indeed we
03:40
definitely get equation 2
03:42
after we've made these substitutions
03:44
into equation one
03:46
well that means that we can read off a
03:49
solution
03:50
to this general equation from the
03:52
solution of
03:53
the simpler geometric form
03:57
okay we just carry on
04:00
with the substitutions so the solution
04:02
is let's scroll down here
04:06
where's our cursor
04:09
there okay um so the solution is
04:12
here it is right here okay so what do we
04:15
get
04:16
well we just get this thing here where
04:18
we have to replace the t2
04:19
up to t5 with the corresponding
04:21
expressions here so there's t2
04:23
there's t3 there's t4 there's t5 raised
04:26
to the
04:26
appropriate powers then there's a c0
04:29
over c1
04:30
in front and the coefficient is as usual
04:33
the rsp number r
04:37
conclusion then is we get what we might
04:39
call
04:40
the mega theorem for the quintic okay
04:43
that the solution of this quintic in
04:46
this form
04:49
is given by this expression y is a
04:52
certain
04:53
function of the coefficients c0 to c5
04:58
and if we look to see how many c zeros
05:00
and everything else there are
05:02
the total number of c zeros in the
05:04
numerator is one
05:06
from this term plus m two plus 2m3
05:10
plus 3m4 plus 4m5
05:13
that's the reason why that exponent is
05:15
the way it is
05:16
and the c2 c3 c4 and c5 terms they're
05:20
just directly just the
05:22
associated powers and the c1 well
05:25
there's one of them down here in this
05:26
denominator and then there's two m2
05:28
3 m3 4 m4 and 5 m5 that explains
05:32
that exponent there and we're going to
05:35
call this big sum
05:38
k okay so it's an expression involving
05:41
the coefficients
05:42
so this is like the holy grail right we
05:44
want a
05:45
formula for the solution in terms of the
05:48
coefficients
05:50
in a very very general way so it just
05:51
depends on the coefficients
05:56
now the question is you know what about
05:58
the interpretation well
05:59
i've already explained to you how we can
06:01
interpret this kind of thing
06:03
algebraically
06:04
now we want to show that it also can be
06:06
used for
06:07
numerical calculations of
06:11
approximate zeros
06:16
okay so here is mathematica okay
06:19
and we're going to apply the formulas
06:21
that we've
06:23
um laid out here and just to start with
06:26
i'm just putting this little blurb here
06:28
this is just a little sequence of things
06:31
that
06:32
that get mathematica to agree
06:35
that zero to the zero is equal to one
06:39
i want that because in certain
06:41
situations when we're evaluating some
06:43
things we have
06:43
certain things raised to certain powers
06:46
and
06:46
the powers are sometimes zero and
06:50
the things that we're going to plug in
06:52
can also be sometimes 0 and i want
06:54
mathematica to evaluate that as 0 so
06:56
this explicitly uh
06:58
tells that it tells of that
07:04
oh see so here i haven't actually run it
07:06
so it says you know if you if you just
07:08
evaluate this thing here the
07:10
indeterminate expression encountered
07:12
so now i'm going to go here and i'm
07:13
going to run this thing
07:15
and now i'm going to run this and now
07:17
mathematica agrees that
07:18
that's equal to 1. that's the effect of
07:20
this little
07:22
sequence of commands okay so let's get
07:25
to it
07:25
first of all first order of business
07:27
defining the rsp numbers are
07:30
so this is direct thing here you will
07:32
note that i'm not using subscripts
07:34
maybe just say a word about that so
07:36
subscripts
07:37
are nice to look at but generally when
07:41
you're working with mathematica it's
07:43
probably better
07:44
to not use subscripts to stick with
07:48
variables have sort of simple names just
07:50
like m2 m3
07:51
even though it doesn't quite look as
07:53
pretty okay you're less likely to run
07:56
into
07:56
problems so that's what we're going to
07:58
do so here's our
08:00
and it's defined in the way that you're
08:02
all familiar there's the factorial
08:04
for the numerator the one in the
08:05
denominator consisting of this thing and
08:07
the factorials of the individual
08:08
m's from m2 to m5 okay let's run that
08:12
so we know what r is let's just check
08:15
okay r of something one three two
08:19
six what is that okay it's some big
08:22
number excellent
08:23
okay so it knows what r is
08:26
so let's delete that okay so now let's
08:29
go to f
08:30
so here is the f function remember that
08:33
was the
08:34
solution to the to the geometric
08:38
quintic okay the one involving ts so
08:41
it's a sum
08:42
over r m2 m3 m4 m5
08:45
and it's a pure power series in t2 t3
08:48
t4 and t5 and these brackets here
08:51
just tell mathematica the range of
08:55
summation values for each of these
08:57
indices so this really is a
08:59
quadruple sum because there's four
09:01
things that are being
09:03
summed over m2 m3 m4 and m5 so let us
09:09
ctrl enter that great and
09:12
the more general solution involved that
09:16
k function remember that involved all
09:18
the coefficients of the general
09:20
thing so that's a function of c0 up to
09:23
c5
09:25
defined in a similar kind of way
09:28
c0 to the 1 plus m2 plus 2 m3 et cetera
09:31
the c2 c3
09:33
c4 and c5 terms appearing with the usual
09:36
factorials
09:36
and the c one to the one plus two m two
09:39
plus three and three plus four m four
09:40
plus
09:41
five m five okay that's what we're
09:43
summing and again we'll let uh
09:45
m2 m3 m4 and m5 range say from
09:48
zero to six okay we happy with that
09:52
so boom
09:55
okay so now we have the basic
09:59
numerical engines ready to go
10:02
let's see if we can solve some quintix
10:05
okay
10:06
so some numerical examples of the
10:08
geometric quintic so let's start with
10:11
a geometric quantix so that's a
10:12
relatively simple
10:14
so solve so the equation is something
10:16
like 1 minus
10:18
x i'll say plus sub 2 x
10:21
squared minus x cubed
10:25
plus 3 x to the 4
10:30
plus x to the five
10:34
okay we might solve that for x
10:39
but i want a numerical solution so i'm
10:42
going to put
10:43
n solve here okay so let's see what
10:46
happens
10:48
okay i get some solutions i get some
10:50
complex solutions
10:52
actually oh
10:55
yeah four of them and one
10:58
approximate rational solution try to
11:01
avoid saying real solution it's not a
11:02
real number it's just a
11:03
finite decimal which is an approximate
11:05
solution to this thing
11:07
okay so let's see if our method
11:10
might have gotten that how would we try
11:12
to get that all right so let's see if
11:15
our basic formula
11:16
will come up with this so remember for
11:19
the f function we're just inputting the
11:21
coefficients of x squared up to x to the
11:24
- that's 2
11:26
minus one three and one
11:32
okay that is a big number that is
11:35
nowhere close
11:36
to what we're what we're interested in
11:39
okay and of course we're not surprised
11:41
because our series is a power series in
11:45
in in these things here okay and
11:48
that thing is not likely to be
11:50
converging
11:52
to anything useful unless these numbers
11:54
here are small
11:56
okay so this is not not close
12:00
so let's um let me i'll keep this thing
12:02
here and we'll just look at
12:04
another variant okay let's do the same
12:07
thing but i'm going to modify the
12:09
coefficients now
12:10
so i'm going to put in some fractions
12:12
one say
12:13
seventh
12:18
fraction one quarter
12:23
fraction one third
12:26
fraction one one
12:33
say ten okay
12:36
and let's solve that now okay so now
12:39
let's uh
12:40
calculate the answer from the formula
12:43
so that involves the coefficients of
12:47
1 7 and then
12:50
we had minus one quarter
12:54
and then um
12:57
one third and then
13:00
oops sorry one-third and then
13:03
1 10.
13:08
okay and i want to evaluate that
13:10
numerically so
13:11
numerical evaluation is like an n so i'm
13:13
going to put an
13:14
n in front of that
13:17
and evaluate numerically so that's
13:19
that's a very big thing
13:21
okay that's still uh our method is still
13:24
not
13:24
working and probably it's because these
13:28
numbers are not yet
13:29
small enough to get convergence so i'm
13:30
going to just modify things up here
13:34
so i just want to see if this thing is
13:35
converging to something
13:38
if the coefficients are sort of say
13:40
suitable so i'm just going to multiply
13:42
all of these by 10.
13:43
and i'm going to reevaluate this
13:46
okay now the solution is minus 4.91
13:50
and some other stuff and over here now i
13:52
can quickly just go over here and modify
13:55
these i'm just multiplying them all by
13:56
- and i'm going to rerun this
14:02
and i'm getting um
14:08
1.03821 oh actually but wait that is
14:11
here's one of the solutions so the thing
14:12
has changed it's not just this solution
14:14
there's also this one over here 1.03821
14:18
so we're actually getting that
14:20
and now i'm kind of interested in how
14:22
close are we okay because it's all just
14:24
approximate so let's try to
14:27
involve some precision the way we do
14:29
that is we
14:30
in the end solve we just add an extra
14:35
factor which is the number of
14:39
digits in the answer so now we have the
14:41
answer
14:42
to this quintic
14:46
to 10 decimal places
14:50
and now let's calculate this i can do
14:52
correspondingly just put a comma
14:54
and a 10 there and hopefully it will
14:56
calculate that and what do i see
14:59
one point zero three eight two one one
15:02
five eight five three zero nine so we're
15:04
within uh
15:05
we have six decimal places of accuracy
15:09
great so i think we can
15:13
we could conclude that our f function is
15:15
is working okay
15:17
so now i want to go to the the general
15:20
case and look at a more
15:21
general quintic
15:25
okay so now let's have a look at the
15:26
general quintic
15:28
okay where we have rather arbitrary uh
15:31
coefficients maybe i'll make this one
15:34
three three minus eight x plus x
15:36
squared minus two x cubed plus x to the
15:38
fourth plus three x to the fifth
15:39
equals zero we wanna solve that for x
15:42
numerically
15:44
okay and we see we have um
15:47
three three uh zeros
15:51
okay so let's have a look um first of
15:53
all can we
15:55
can we find this using our k function
15:58
okay
15:59
so let's look at n that's
16:02
evaluation then k and then the
16:05
coefficients are going to be 3
16:07
eight i have to ignore the minus sign
16:09
there and then one
16:11
and then minus two and then one and then
16:14
three
16:17
all right and one more bracket
16:21
so let's evaluate that we get 0.385194
16:25
excellent okay that's good agreement
16:28
with with this
16:29
this value here in fact if i wanted to
16:31
see how good agreement it is i could go
16:33
up to say 10
16:35
places and i could also go up to 10
16:37
places here
16:39
and now how's the agreement 3 8 5
16:42
1 9 0 so up to
16:46
up to about five decimal places
16:50
okay that's good now
16:53
what about well first of all one of the
16:56
reasons why this thing worked is because
16:58
this this eight here
16:59
right is a little bit bigger than the
17:01
other ones
17:02
so that's i've arranged that
17:07
so i wanted to check whether this thing
17:10
is going to
17:11
give me the answer that i know it should
17:13
give me if
17:15
the coefficients are kind of well chosen
17:17
for our method
17:18
so what happens if that's not the case
17:20
so let's have another
17:21
look modify this we'll make this
17:25
eight much more modest say maybe just um
17:29
four now it's comparable with
17:32
the other ones let's see if we still get
17:34
something
17:35
so first of all we'll solve it and then
17:38
we'll
17:42
we'll have a look at this we can change
17:44
this 8 to a4
17:46
okay and see what we get we get
17:48
something 10 to the minus
17:50
10 to the 14. so obviously that's far
17:52
off
17:53
all right so how could we modify
17:57
our our strategy now to try to get at a
18:00
zero in this more general situation
18:02
where we do not
18:03
have the favorable situation of having a
18:06
big
18:06
c1 term compared to the others well
18:09
bootstrapping is is one way of thinking
18:11
about this okay
18:12
now what we should do is we should first
18:14
look for an approximate
18:17
zero and then then move towards that
18:21
okay so an approximate zero is say minus
18:24
1.5 so let's define
18:27
say a function g of x
18:32
g of x will be equal to
18:36
and this is what we're going to take all
18:38
right
18:39
that's our function
18:43
let's define that and now we're going to
18:48
look at expanding a translate of that so
18:52
we're going to translate that function
18:54
by an approximate zero so we let's look
18:56
at the approximate zero minus 1.5 that's
18:59
an approximate zero to g
19:02
okay so let's just check that with g of
19:05
minus 1.5
19:09
that is 0.28125 okay you know
19:12
it's pretty close to zero okay fine so
19:15
what i'm going to do now is
19:16
instead of calculating that i'm going to
19:18
calculate g of x
19:21
minus 1.5
19:24
okay and i'm going well let me just do
19:27
that if i just do that
19:28
i get this thing here but i want to
19:31
actually expand that so let me
19:32
expand it right off the bat
19:36
expand okay
19:41
so have a look i'm getting a new quintic
19:44
but
19:45
crucially you see this constant term was
19:48
the
19:48
value of g of minus 1.5 that we just
19:50
calculated 0.281
19:52
and that's not really small but it's
19:55
smaller than the other ones
19:56
so there's a good chance that that's
19:58
going to help us
20:00
in having a series which is
20:03
kind of converging to something so let's
20:07
go here and let's go down and let's
20:10
now replace these things here with
20:13
say 0.28 125
20:18
and then 41. i better put a minus sign
20:21
here
20:22
minus 41.9375 we probably don't need all
20:27
these digits but anyway
20:29
minus 77.75
20:32
and then 59.5
20:36
and 21.5
20:40
and finally three do i have them all
20:43
right
20:44
that one that one that one that one that
20:46
what oh there's a minus sign there sorry
20:49
minus 21.5 and three all right let's
20:52
evaluate that thing
20:55
minus point zero zero six two four
20:58
six three now what am i going to do with
21:01
that
21:02
well i'm going to add that to minus 1.5
21:06
okay so if i take minus 1.5
21:09
1.5 and i add what i just got
21:14
i get minus 1.5062
21:18
which is um what i'm getting here
21:22
to five decimal places
21:25
so this is a demonstration of how
21:28
bootstrapping
21:29
can get you closer to a zero
21:32
once you know you have an approximate
21:36
zero
21:38
and this might also be useful i mean if
21:40
we had a an example where we had
21:42
say more than just one rational or
21:46
approximate rational solution let's say
21:47
let's say we have about five approximate
21:49
rational solutions so we might be able
21:51
to obtain
21:53
more than just one of them by by by
21:55
starting
21:56
with giving one of these seed values
21:58
which is close to
22:00
to one of these and then hopefully it
22:01
will it will converge to um
22:05
to the one that we're interested in so
22:08
in summary i think
22:09
this shows that our formulas are
22:11
essentially correct
22:12
uh there's still a very big question of
22:15
you know how to efficiently
22:17
um create a technology that actually
22:20
helps us
22:21
solve numerical solutions in general
22:24
right
22:24
you know so what do we do depending on
22:27
the
22:28
nature of the coefficients but overall
22:30
we have
22:32
now a mega solution that covers both the
22:35
algebraic
22:36
side of things and is pretty effective
22:39
on the numerical side
22:41
so i think that's a cause for um
22:44
for celebration