| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 881.1 | Does anyone know where to begin? :-) | ZFC::DERAMO | I am, therefore I'll think. | Mon Jun 06 1988 23:43 | 12 |
| This isn't the solution in radicals, but it may help [?].
The five roots I found numerically were
-1.8420859661902543827111880601987
1.27289722392249919091427122056023 +/- 0.71979868148386138668102237400372 i
-0.35185424082737199955867719046086 +/- 1.70956104337032888201408195323997 i
For these values, the real and imaginary parts of z^5 - 5z + 12
were less than 10^-32.
Dan
|
| 881.2 | The water might be deeper than you think! | HPSTEK::XIA | | Tue Jun 07 1988 17:21 | 18 |
| > -< Does anyone know where to begin? :-) >-
Well, compute the Galois group. Since the originator of the problem
said that the equation is solvable, you can just do it by computing the
group. Let me warn you though. It is a BIG MESS.
> This isn't the solution in radicals, but it may help [?].
> The five roots I found numerically were
I don' think numerical solution will be of much help in this case.
Unless of course, you want to do it by trial and error.
a side note. I once corresponded with this Matt guy. He told me
that it was a big mess to do that type of problem. He used maxima
to help him out. (Do not mean to scare anybody from trying this
problem here).
Eugene
|
| 881.3 | do continued fraction on approximation? | VIDEO::OSMAN | type video::user$7:[osman]eric.vt240 | Tue Jun 07 1988 21:10 | 9 |
| Does "radical" mean square root ? Or are cube roots radicals too?
For square roots, one can take any numerical approximation and
do a continued fraction conversion to easily find what the radical
is.
(Does any similar method exist for cube roots conversion??)
/Eric
|
| 881.4 | sigh | ZFC::DERAMO | I am, therefore I'll think. | Wed Jun 08 1988 02:24 | 16 |
| To solve a polynomial equation in radicals includes n-th
degree roots for n = 2, 3, ... (you could limit yourself to
primes if you want :-). If the degree of the polynomial is
greater than four, such a solution does not in general
exist. The roots of the polynomial in .0 should involve
nesting square, cube, [fourth = nested square] and fifth
roots.
Only square roots can yield repeating continued fractions.
If that fact is part of the method of going from numerical
approximation to continued fraction to radical, then
it would seem to be a roadblock for higher radicals.
BTW, what's a Galois group and does one compute it? :-)
Dan
|
| 881.5 | How about rounding off a near miss? | POOL::HALLYB | Lookit all the happy creatures dancin' on the lawn | Wed Jun 08 1988 16:04 | 19 |
| How about calculating: (using Dan's real numeric solution in .1)
x^5 - 5x + 12
---------------------------------------
(x + 1.8420859661902543827111880601987)
rounding if possible, and calculating the solution to the general quartic
that results? (This should be nearly the same as multiplying the 4
complex answers in .1 and rounding off. Pick whatever seems easier.)
The notion being that if there is a "nice" solution in radicals, perhaps
this "roundabout" approach would yield something that is nearly nice,
say SQRT(42.00003) - CBRT(76.999998125) i. Which would then be a broad
hint that the real solution might be SQRT(42) - CBRT(77) i.
Of course there's no guarantee such will be the case, but from the problem
statement I would guess there are some nice numbers that fall out.
John
|
| 881.6 | | CLT::GILBERT | | Wed Jun 08 1988 16:06 | 2 |
| I saw two solutions fly by USEnet, but deleted them. Could someone
else post 'em? Thanks.
|
| 881.7 | | ZFC::DERAMO | I am, therefore I'll think. | Wed Jun 08 1988 16:18 | 26 |
| Re .5:
The following result was in a proof at home that there
is not, in general, a solutions in radicals to polynomials
(integral or rational coefficients) of degree five or more.
It was discussing finding the roots in radicals of
irreducible polynomials of odd prime degree. You start
with the rationals: the polynomial cannot be factored.
You add a p-th root for some prime p of a number in the
current field, to get the next field. Keep doing this
until the polynomial can factor. The result was that
it will not factor at all until an n-th root, where n
is the degree of the polynomial, is taken; and then,
it will factor completely into a product of linear terms.
So all five of the roots will involve that last fifth
root.
Re .6:
I have the USENET messages (didn't understand them --
see that title "sigh" a few notes back?). I will extract
and post them tonight.
Dan
|
| 881.8 | Just an idea for this problem, not a proposed theorem! | POOL::HALLYB | Lookit all the happy creatures dancin' on the lawn | Wed Jun 08 1988 17:01 | 25 |
| Re: .7
Dan, I don't see where you're coming from. Wasn't it Galois who
proved the general Nth degree polynomial is not solvable in terms
of radicals? Ahh, but we're given that this is solvable in terms
of radicals, hence at most 5th roots.
Well, by my calculations, -1.8420859661902543827111880601987 (the
real solution you found) is approximately -FIFTH(700/33). So divide
(algebraically) the polynomial by (x + FIFTH(700/33)), surely one
of our symbolic math packages can do that, and see if you get a
reasonable looking 4-th degree polynomial, or perhaps with a bit of
tweaking you end up with one.
I'm not suggesting this as a general equation-solving technique.
Rather, given that we have a polynomial that evidently reduces in
a "nice" manner, perhaps this approach yields polynomials with noise
terms (i.e., + 0.0000000037/x) that can be dropped in the search
for nice factors.
Then you plug in the nice factors and see if they multiply out to the
polynomial to be solved. If not, give up on this approach or go for
less nice factors. (If not FIFTH(700/33), maybe FIFTH(700)/SQRT(1000)).
John
|
| 881.9 | Well you ask for it :-) | HPSTEK::XIA | | Wed Jun 08 1988 19:04 | 96 |
| re. 4
I have to retract something I said in my previous note. Namely
that the numerical solution is not much of help. It turns out that
the numerical values given illustrates many features of the equation
that is of importance. One important feature is that there is no
trivial ways of solving it. For example, the problem will be easy
if one can factor the polynomial into products of polynomials of
lower degrees in a trivial way. In this case, the spliting poly
will be of degrees {1, 4} or {2, 3}. It is easy to compute
approximately the coeficient of such polynomials (just add or multiply
the roots). It turns out that these polynomials do not have
integer values. Ruling out the posibility that someone finding
out the solution by trial and error, one is left with the choice
of doing it the hard way (and I for sure ain't going to do it).
Now:
Abel proved that the roots of a polynomial of degree >= 5 can
not be solved in general by radicals. However, it was Galois who
associates each polynomial with a group, and give a criteron in
terms of some property of this group to determine whether a particular
polynomial is solvable by radicals. The idea is as follow:
Let Q denotes the field of the rationals. Let f be a polynomial
of Q, that is, f is in Q[x]. Obviously, in general, the roots of f will
not be in Q. However, they will be in some extention field F of Q.
Now let u be a root of f and u is not in Q. Then one can form
an extention field Q(u) that is the smallest field containing Q
and u. Note Q(u) forms a vector space over Q in terms of addition
and scaler multiplication (a scaler here is a rational number).
Definition of Galois group:
Let K be a field and F be an extention field of K. Then
The Galois group Aut(F,K) is the group of automorphisms (vector
space automorphism) of F fixing
K. In other word, if T is in Aut(F,K), then T(k) = k for all k
in K.
Now there are tons of theorems as to what type of properties Aut(F,K)
has. But the thing we are interested in is the solvability of
Aut(F,K).
Definition:
Let G be a group. Then the commutator of G, denoted by G' is the
group generated by the set {aba^(-1)b(-1): a, b are both in G}.
Let G be a group.
Now define a series of groups:
G(1), G(2), ..... such that G(1) = G and G(i) = G(i-1)'
Definition:
A group G is said to be solvable if G(n) is the trivial group <e>
for some integer n.
Now the extention field F = Q(u1, u2, .... un) such that u1, u2,....
are the roots of the above polynomial f.
Now for the BIG theorem.
a polynomial f is solvable by radicals if and only if
Aut(F,Q) is solvable as a group.
In case someone is still reading at this point, I will give some
theorems about some important features of Aut(F,Q) :-) :-).
If f is of degree n, then Aut(F,Q) is a subgroup of the group of
symmetry Sn. Now the reason f cannot be solved by radicals in general
is because Sn is not solvable for n > 4. However, for a particular
f, it is possible that the Aut(F,Q) associated with that f is a subgroup
of Sn and is solvable.
Finally, here is an interesting theorem:
Let p be a prime and f is an irreducible polynomial of degree p
over the field of rational numbers (meaning f cannot be factered
into polynomials of smaller degree with rational coefficients) which
has precisely two nonreal roots in the field of complex numbers,
then the Galois group of f is isomorphic to Sp.
In terms of polynomial of fifth degree, this above theorem says
that if you cannot factor it in the rationals, and it has exactly
two complex roots, then the polynomial is not solvable by radicals.
Now the problem is that the above theorem only tells whether a
particular polynomial is solvable by radicals, but does not tell
us how. Well, I am not aware of any particular method of doing
it, but the first thing I will do is computing the Aut(F,K) and
by doing so will learn a lot about the structure of this particular
polynomial.
I want to emphasize that there might be some elementary way of solving
this problem by algebraically munipulate the polynomial since the
problem comes out as an puzzle and in general there are tricks one
can play with puzzles that does not work in general cases.
Oh well I must have taken too many classes and do not know how to
follow my gut feelings any more (Read my comments on the problem
of the average series :-) :-).
Eugene
|
| 881.10 | I almost said "equally complex" :-) | ZFC::DERAMO | Daniel V. D'Eramo, VAX LISP developer | Wed Jun 08 1988 22:40 | 18 |
| Re .8
>> Re: .7
>>
>> Dan, I don't see where you're coming from. Wasn't it Galois who
>> proved the general Nth degree polynomial is not solvable in terms
>> of radicals? Ahh, but we're given that this is solvable in terms
>> of radicals, hence at most 5th roots.
I was just saying that in a sense all of the roots are
equally difficult. For this equation you can't divide out
one root and get a polynomial that has rational roots, or
roots in Q[sqrt(2),sqrt(3)], or something simple like that.
None of the roots will "drop out" as you add new radicals
until some point when they all drop out, and that step will
involve a fifth root (for x^5 - 5x + 12 = 0).
Dan
|
| 881.11 | more from USENET newsgroup sci.math | ZFC::DERAMO | Daniel V. D'Eramo, VAX LISP developer | Wed Jun 08 1988 22:52 | 174 |
| Here's what I got from USENET newsgroup sci.math, starting
with the original posting.
Dan
Newsgroups: sci.math
Path: decwrl!ucbvax!agate!garnet!weemba
Subject: A solvable quintic
Posted: 1 Jun 88 12:52:52 GMT
Organization: Brahms Gang Posting Central
Solve in radicals: x^5-5x+12=0. Have fun!
ucbvax!garnet!weemba Matthew P Wiener/Brahms Gang/Berkeley CA 94720
Newsgroups: sci.math
Path: decwrl!pyramid!prls!philabs!linus!bs
Subject: Re: A solvable quintic
Posted: 2 Jun 88 14:33:41 GMT
Organization: The MITRE Corporation, Bedford MA
In article <10450@agate.BERKELEY.EDU> weemba@garnet.berkeley.edu (Obnoxious Math Grad Student) writes:
>Solve in radicals: x^5-5x+12=0. Have fun!
How about a hint? What is it's Galois group???
Bob Silverman
========================================================================
Received: by decwrl.dec.com (5.54.4/4.7.34)
id AA24102; Fri, 3 Jun 88 07:21:57 PDT
Newsgroups: sci.math
Path: decwrl!ucbvax!agate!garnet!weemba
Subject: Re: A solvable quintic (small spoiler)
Posted: 3 Jun 88 15:53:33 GMT
Organization: Brahms Gang Posting Central
In-reply-to: bs@linus.UUCP (Robert D. Silverman)
>>Solve in radicals: x^5-5x+12=0. Have fun!
>How about a hint? What is its Galois group???
Surely you can solve this by metareasoning: D_5. Cyclic would have been
too easy!
But now that you mention it, when I solved it, I knew ahead of time that
it was D_5, as I challenged myself with this equation after seeing a J
Algebra paper a few years back that gave an appropriate criterion.
ucbvax!garnet!weemba Matthew P Wiener/Brahms Gang/Berkeley CA 94720
A mathematician's wife overhears her husband muttering the name 'Nancy'.
She wonders whether Nancy, the thing to which her husband referred, is
a woman or a Lie group. --Saul Kripke
Newsgroups: sci.math
Path: decwrl!purdue!gatech!udel!princeton!phoenix!mjlarsen
Subject: Re: A solvable quintic
Posted: 3 Jun 88 03:07:27 GMT
Organization: Princeton University, NJ
In article <10450@agate.BERKELEY.EDU> weemba@garnet.berkeley.edu (Obnoxious Math Grad Student) writes:
>Solve in radicals: x^5-5x+12=0. Have fun!
This kind of thing is not too difficult as long as you have a computer
and your coefficients are small. First we observe that the number of
solutions of this polynomial (mod p) is 0, 1, or 5 respectively 2/5,
1/2, and 1/10 of the time. Cebotarev's density theorem suggests that
the Galois group of the splitting field of the polynomial is the
dihedral group D_5. This is confirmed by the fact that the primes that
give rise to one solution are precisely the primes that are inert in
K = Q(\sqrt{-10}). So we expect the splitting field to be cyclic over K.
Now we apply Kummer theory. The idea is that if A_1, ..., A_p are
cyclically permuted by a field automorphism, \sigma,
that fixes w = e^{2\pi i / p}, then \sigma acts on
A = A_1 w + A_2 w^2 + ... + A_p w^p by multiplication by some power of
w. Thus A^p is \sigma-invariant. So we expect the fifth power of
some linear combination of the roots to lie in the ring generated over
Z by \sqrt{10} and the fifth roots of unity. Computing the roots
numerically, we discover that, in fact, this is so. The result is that
the roots are
(1/4)\sum_{i = 1}^4 w^{ik}(125((15+3r)-20w^i+(15-3r)w^{2i}+(-5+6r)w^{3i}
+(-5-6r)w^{4i}))^{1/5},
where r = \sqrt{-10}, w = (\sqrt{-5}-1)/2 + \sqrt{(-5-\sqrt{-5})/8}
is a fifth root of 1, and k = 0, 1, 2, 3, and 4 give the five roots of
the equation.
-Michael Larsen
Newsgroups: sci.math
Path: decwrl!hplabs!hpda!hpcupt1!mount
Subject: Re: A solvable quintic
Posted: 3 Jun 88 22:06:35 GMT
Organization: Hewlett Packard, Cupertino
> In article <10450@agate.BERKELEY.EDU> weemba@garnet.berkeley.edu (Obnoxious Math Grad Student) writes:
> >Solve in radicals: x^5-5x+12=0. Have fun!
>
>
> How about a hint? What is it's Galois group???
I worked out that this poly must have a Galois group of D5 (Dihedral group)
or A5 (Alternating group), so if it is indeed solvable its Galois group is D5.
John Mount
P.S. Anyone posting a solution please post what intermediate polynomials
they used to solve it.
Newsgroups: sci.math
Path: decwrl!hplabs!sdcrdcf!ucla-cs!julia!pmontgom
Subject: Re: A solvable quintic
Posted: 4 Jun 88 00:59:12 GMT
Organization: UCLA Mathematics Department
In article <10450@agate.BERKELEY.EDU> weemba@garnet.berkeley.edu (Obnoxious Math Grad Student) writes:
>Solve in radicals: x^5-5x+12=0. Have fun!
>
>ucbvax!garnet!weemba Matthew P Wiener/Brahms Gang/Berkeley CA 94720
program HadFun
implicit undefined(A-Z)
C This FORTRAN program computes the roots of
C X^5 - 5X + 12 = 0 using radicals, and prints them.
C Written by Peter Montgomery, graduate student, UCLA, June, 1988.
double precision COS72, SIN72, SQRT5, w1, w2, w3, w4, x
complex*16 root, u
real error
integer i
C Functions referenced
intrinsic ABS, DCMPLX, SIGN, SQRT
double precision FIFTH
C Statement function to compute fifth root of double precision value
FIFTH(x) = SIGN(ABS(x)**0.2D0, x)
C - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
SQRT5 = SQRT(5.0D0)
COS72 = (SQRT5 - 1.)/4.
SIN72 = SQRT(1. - COS72**2)
w1 = FIFTH(-1. + 0.4D0*SQRT5 - 0.6D0*SQRT(5. - 2.2D0*SQRT5))
w2 = FIFTH(-1. - 0.4D0*SQRT5 + 0.6D0*SQRT(5. + 2.2D0*SQRT5))
w3 = FIFTH(-1. - 0.4D0*SQRT5 - 0.6D0*SQRT(5. + 2.2D0*SQRT5))
w4 = FIFTH(-1. + 0.4D0*SQRT5 + 0.6D0*SQRT(5. - 2.2D0*SQRT5))
C Can use SQRT(5 + 2.2D0*SQRT5) = SIN72/(COS72*COS72*SQRT(10)) instead
do 100 i = 0, 4
u = DCMPLX(COS72, -SIN72)**i
root = w1*u + w2*u**2 + w3*u**3 + w4*u**4
error = ABS(root*(root**4 - 5.) + 12.)
print 50, root, error
50 format(' z = (',F19.15,',', F19.15,') |P(z)| = ', E12.5)
100 continue
end
Output (on a VAX 750, UNIX 4.3bsd)
z = ( -1.842085966190254, 0. ) |P(z)| = 0.21760e-13
z = ( 1.272897223922499, 0.719798681483861) |P(z)| = 0.14131e-13
z = ( -0.351854240827372, 1.709561043370329) |P(z)| = 0.18669e-13
z = ( -0.351854240827372, -1.709561043370329) |P(z)| = 0.20585e-13
z = ( 1.272897223922499, -0.719798681483861) |P(z)| = 0.13733e-13
--------
pmontgom@sdcrdcf.UUCP
pmontgom@julia.MATH.UCLA.EDU
|
| 881.12 | are we fooled.? Or is the 5th deg. equation solved | VIDEO::OSMAN | type video::user$7:[osman]eric.vt240 | Thu Jun 09 1988 17:51 | 12 |
| That Fortran program at the end merely verifies that the given
expressions are within 10**-13 of the exact answer. It doesn't
prove they are the answer.
In fact, it wouldn't be too hard to come up with (another?) phoney
set of radical expressions that are within 10**-13 of the root
of our equation.
Has anyone used MAPLE or some other symbolic expression package
to prove that the published radicals are indeed roots of the equation?
/Eric
|
| 881.13 | So *that*'s what they are! :-) | ZFC::DERAMO | Daniel V. D'Eramo, VAX LISP developer | Thu Jun 09 1988 22:37 | 6 |
| Re .9
Thanks for the definitions. Now it's time to reread
the USENET postings in .11 with these in mind.
Dan
|
| 881.14 | Lectures on the Icosahedron: F. Klein | CTCADM::ROTH | If you plant ice you'll harvest wind | Fri Jun 10 1988 11:13 | 11 |
| There was a book on solving equations of the 5-th degree (and a lot more)
by Felix Klein, which was reprinted by Dover. If I recall, there was
a straightforward way of using elliptic functions to generate a
solution if the 5-th degree polynomial was solvable in this case.
There was a very interesting connection with symmetry groups of the
icosahedron as well.
I haven't looked at it in a while, but have a photocopy somewhere.
My interest was in elliptic functions rather than Galios theory.
- Jim
|
| 881.15 | can we generate fifth-degree eq. from radicals? | VIDEO::OSMAN | type video::user$7:[osman]eric.vt240 | Fri Jun 10 1988 14:08 | 7 |
| Can we work backwards to generate difficult puzzles?
For example, can we take a complicated radical expression, and
then find a fifth degree equation for which it is a root, and then
pose the puzzle "solve the following equation in radicals...".
/Eric
|
| 881.16 | hmmmm.... | ZFC::DERAMO | Daniel V. D'Eramo, VAX LISP developer | Fri Jun 10 1988 23:05 | 25 |
| Re .15
>> Can we work backwards to generate difficult puzzles?
>>
>> For example, can we take a complicated radical expression, and
>> then find a fifth degree equation for which it is a root, and then
>> pose the puzzle "solve the following equation in radicals...".
>>
>> /Eric
I suppose you could try it, but unless there is some
theory on it that can be followed, you might stumble
over one of the following:
- You don't end up with a fifth (or other low) degree
polynomial.
- You don't end up with an irreducible polynomial
for your "root" but instead get one multiplied
by some other polynomial that is not solveable.
I don't know if just going for it in a straightforward
way will have these problems or not.
Dan
|
| 881.17 | It'll be f(ugly) | CHOVAX::YOUNG | Dumb, Expensive, Dumb ... (Pick Two) | Mon Jun 13 1988 02:33 | 8 |
| Re .16,.15:
Actually, I would think that the problem is that you are to likely
to end up with a valid 5th degree polynomial whose factors are 'uglier'
(more complicated) than the original root(s).
-- Barry
|
| 881.18 | another roadblock :-) | ZFC::DERAMO | Daniel V. D'Eramo, VAX LISP developer | Wed Jun 15 1988 00:24 | 4 |
| Does this Galois theory include a way of computing these
groups given the coefficients of the polynomial?
Dan
|
| 881.19 | Algorithm for Aut(F,K)? I don't think so, but... | HPSTEK::XIA | | Wed Jun 15 1988 15:32 | 25 |
| Re. 18
If you are asking for a step by step algorithm, I am not aware of
such things. However, I do know how to compute Aut(F,K) for
polynomials up to degree 4. The way you do it is to first prove
a few facts about the Galois group associated with degree 4 polynomial.
Then construct some intermediate fields between Q and F. Then
you constuct the resolvant cubic polynomial. Then you prove a few
more facts :-). The Galois group them can be determined with the
above facts :-). Big Mess. This is just for quartic. I don't
know how to do it with a quintic, but it involves doing things like
that. If you plan to compute some Galois groups for quintic, here
is a theorem that may be of some help:
Let f be in Q[x] (meaning the coefficient of f is in Q). Let
F = Q(u1, u2,...,un) where ui are the roots of f. If f is irreducible
in Q and separable (meaning not double roots), then G is isomorphic
to a transitive subgroup of Sn.
As Matt Wiener once put it. "If you want to find the roots of high
degree polynomials, Galois theory is less than useful :-)." However,
we both agreed that the theory is elegant :-). After all the the
guy (Galios) did it when he was a teenager (more :-). In any case,
Matt once told me that he once used maxima to compute the Galois
group of some quintic poly. and he said it was big mess (Don't quote
me on this :-).
Eugene
|
| 881.20 | more from USENET on the problem | ZFC::DERAMO | To err is human; to moo, bovine. | Thu Jun 30 1988 23:23 | 133 |
| Newsgroups: sci.math
Path: decwrl!sun!pitstop!sundc!seismo!uunet!mcvax!inria!morain
Subject: Re: A solvable quintic
Posted: 3 Jun 88 10:09:09 GMT
Organization: INRIA, Rocquencourt. France
In article <10450@agate.BERKELEY.EDU>, weemba@garnet.berkeley.edu (Obnoxious Math Grad Student) writes:
> Solve in radicals: x^5-5x+12=0. Have fun!
>
> ucbvax!garnet!weemba Matthew P Wiener/Brahms Gang/Berkeley CA 94720
First of all, is this equation solvable by radicals ? Using the program
`galois` of the share library of MAPLE 4.2, written by R. Sommeling, using
the ideas of McKay and al., we find that the galois group of
p(x) = x^5-5*x+12
is +D5, thus a subgroup of the metacyclic group F20 and thus is solvable.
Hence we can try to solve the equation by radicals.
|\^/|
._|\| |/|_. INRIA - Rocquencourt
\ MAPLE / Version 4.2 --- Dec 1987
<____ ____> For on-line help, type help();
|
> read galois;
> galois(x^5-5*x+12);
5
x - 5 x + 12
D = 8000^2
Possible groups: {+Z5, +D5, +A5}
p = 3 gives shape 2 2 1
Removing {+Z5}
Possible groups left: {+D5, +A5}
p = 7 gives shape 5
p = 11 gives shape 5
p = 13 gives shape 5
p = 17 gives shape 2 2 1
p = 19 gives shape 5
p = 23 gives shape 5
The Galois group is probably +D5
Using the orbit-length partition of 2-sets.
Calculating a resolvent polynomial...
Factoring the resolvent polynomial...
Orbit-length partition is 5 5
Removing {+A5}
Possible groups left: {+D5}
+D5, 10, {(1 2 3 4 5), (2 5)(3 4)}
# My personnal program to solve fifth degree equations by radicals whenever
# ... It is based on an article by G. P. Young: "Solvable quintic equations
# with commensurable coefficients", Am. J. of Math., X, 1888 (after correction
# of many typos).
> read solve5;
> solve5(x^5-5*x+12);
# first the real root
1 1 2 1 1/2 1/5
[(- 1 - 2 ------ + ((- 1 - 2 ------) - ------) )
1/2 1/2 5/2
5 5 5
1 1 2 1 1/2 1/5
+ (- 1 - 2 ------ - ((- 1 - 2 ------) - ------) )
1/2 1/2 5/2
5 5 5
1 1 2 1 1/2 1/5
+ (- 1 + 2 ------ + ((- 1 + 2 ------) + ------) )
1/2 1/2 5/2
5 5 5
1 1 2 1 1/2 1/5
+ (- 1 + 2 ------ - ((- 1 + 2 ------) + ------) ) ,
1/2 1/2 5/2
5 5 5
# then the four complex roots
(cos(2/5 Pi) - sin(2/5 Pi) I)
1 1 2 1 1/2 1/5
(- 1 - 2 ------ + ((- 1 - 2 ------) - ------) )
1/2 1/2 5/2
5 5 5
+ (- cos(1/5 Pi) - sin(1/5 Pi) I)
1 1 2 1 1/2 1/5
(- 1 - 2 ------ - ((- 1 - 2 ------) - ------) )
1/2 1/2 5/2
5 5 5
+ (- cos(1/5 Pi) + sin(1/5 Pi) I)
1 1 2 1 1/2 1/5
(- 1 + 2 ------ + ((- 1 + 2 ------) + ------) )
1/2 1/2 5/2
5 5 5
+ (cos(2/5 Pi) + sin(2/5 Pi) I)
1 1 2 1 1/2 1/5
(- 1 + 2 ------ - ((- 1 + 2 ------) + ------) ) ,
1/2 1/2 5/2
5 5 5
# etc...
# In order to handle these results with care, I give them in a readable format
# If
u1:=(-1-2*1/5**(1/2)+((-1-2*1/5**(1/2))**2-1/5**(5/2))**(1/2))**(1/5);
u2:=(-1-2*1/5**(1/2)-((-1-2*1/5**(1/2))**2-1/5**(5/2))**(1/2))**(1/5);
u3:=(-1+2*1/5**(1/2)+((-1+2*1/5**(1/2))**2+1/5**(5/2))**(1/2))**(1/5);
u4:=(-1+2*1/5**(1/2)-((-1+2*1/5**(1/2))**2+1/5**(5/2))**(1/2))**(1/5);
# and
z1:= u1+ u2+ u3+ u4;
z2:=r4*u1+r3*u2+r2*u3+r1*u4;
z3:=r3*u1+r1*u2+r4*u3+r2*u4;
z4:=r2*u1+r4*u2+r1*u3+r3*u4;
z5:=r1*u1+r2*u2+r3*u3+r4*u4;
# with r(i)=exp(2*i*I*Pi/5)
# then the 5 roots are
l:=[z1,z2,z3,z4,z5];
This yields the five roots of the equation.
F. Morain
|
| 881.21 | what the shape stuff of .-1 means | ZFC::DERAMO | To err is human; to moo, bovine. | Thu Jun 30 1988 23:46 | 39 |
| Re .20
>> p = 3 gives shape 2 2 1
This is a reference to factoring x^5 - 5x + 12 in the ring
of polynomials with coefficients over the integers mod p,
for various primes p.
For p = 3, x^5 - 5x + 12 = x^5 + x = x (x^4 + 1).
Now, over the usual integers x^4 + 4 = (x^2 + 2x + 2)(X^2 - 2x + 2).
But mod 3, x^4 + 1 = x^4 + 4, so we get x^5 - 5x + 12 =
x (x^2 + 2x + 2) (x^2 + x + 2). The shape 2 2 1 part refers
to the degrees of the factors, which here are two second
degree polynomials and one first degree polynomial.
Likewise, "p = 7 gives shape 5" must mean the over the
field of integers mod 7 the polynomial x^5 - 5x + 12 =
x^5 + 2x + 5 cannot be factored at all. Offhand I don't
know how to verify this, except to try the finitely many
possible factors (the lower degree polynomials).
For "p = 17 gives shape 2 2 1" one way to check this
is to verify that the polynomial only has one root mod 17.
I did this in VAX LISP, and verified that there were no
roots (which would have added a 1 to the shape list) for
some of the shape 5 primes listed.
The case p = 5 is probably ignored because the polynomial is
of degree 5. If we plugged it in we would get x^5 - 5x + 12
= x^5 + 2 = (x + 2)^5. Maybe Eugene HPSTEK:: Xia can
explain why p = 5 = degree of the original polynomial is
ignored?
How all of this rules out possibilities for the Galois group
and leads to "The Galois group is probably +D5" is still
beyond me (although I've been reading up on the stuff, so
watch out!).
Dan
|
| 881.22 | Sorry I don't really know. | HPSTEK::XIA | | Fri Jul 01 1988 20:58 | 32 |
| re .21
>The case p = 5 is probably ignored because the polynomial is
>of degree 5. If we plugged it in we would get x^5 - 5x + 12
>= x^5 + 2 = (x + 2)^5. Maybe Eugene HPSTEK:: Xia can
>explain why p = 5 = degree of the original polynomial is
>ignored?
To tell the truth, I do not know. However, I have some wild guesses.
I believe this has somethinig to do with cyclic and cyclotomic
extension. The idea is to look at the Galois group of the
polynomial of x^n - a = 0 (cyclic) and x^n - 1 = 0 (cyclotomic).
I will look more deeply into the stuff when I come back from this
long weekend (maybe :-). I am venturing into the grounds I am
not familiar with so what I said here should be taken with a grain
of salt.
>How all of this rules out possibilities for the Galois group
>and leads to "The Galois group is probably +D5" is still
>beyond me (although I've been reading up on the stuff, so
>watch out!).
I think the string of "p = 3 gives shape 2 2 1... " is the output
of a session of running Sommeling's program (Like the stuff you
see when you run an interactive program). It does not seem to mean
to make sense for humans :-). The best way to find out is perhaps
to get hold of the paper refered at the beginning of .20 (That is
if you are really into this stuff deep :-).
Eugene
|
| 881.23 | wow | ZFC::DERAMO | To err is human; to moo, bovine. | Tue Jul 05 1988 16:14 | 14 |
| I'm back (from July 4). About "p = 3 gives shape ...",
I find it incredible that the shapes are restricted the
way they seem to be. That is, for various odd primes p,
the field of integers mod p has either 0, 1, or 5 roots.
I realize that if it had 4, then dividing out by (x -
r1)(x - r2)(x - r3)(x - r4) would leave a linear term,
which would give a fifth root. But why not two roots
or three roots? If there is one root, why can't it occur
as a double root? There must be some very deep connection
here between the "shape" comments mod p and the Galois
group over the rationals that I haven't been able to
find in my references at home.
Dan
|
| 881.24 | x^6 + 2x^5 + 3x^4 + 4x^3 + 5x^2 + 6x + 7, PGL(2,5) | GUESS::DERAMO | Dan D'Eramo | Sat May 12 1990 14:53 | 27 |
| Path: shlump.nac.dec.com!e2big.dec.com!decuac!haven!aplcen!uunet!clyde.concordia.ca!concour.cs.concordia.ca!mckay
From: mckay@concour.cs.concordia.ca (MCKAY john)
Newsgroups: sci.math,sci.math.symbolic
Subject: Re: Galois groups (was platonic solids)
Message-ID: <2340@clyde.concordia.ca>
Date: 11 May 90 18:36:03 GMT
Sender: usenet@clyde.concordia.ca
Reply-To: mckay@concour.cs.concordia.ca (MCKAY john)
Followup-To: sci.math
Organization: cs
Lines: 14
Xref: shlump.nac.dec.com sci.math:11073 sci.math.symbolic:1375
In view of a recent remark and the comment on it by Richard Bumby, perhaps a bitof self-advertisement is in order. There is a program in maple written by Ron
Sommeling under my direction which, given f in Z[x], with deg f < 8, computes
Gal(f)/Q. This is an algorithm, not a heuristic.
This program is in the maple library.
As the original poster (I have lost the name) says, non-trivial examples of
Galois groups and their polynomials are hard to find in the books: here is
a pretty example : x**6+2x**5+3x**4+4x**3+5x**2+6x+7 .
Does anyone have a decent proof that its Galois group is PGL(2,5)?
This example is in need of generalization!
===============================================================================
Who said of whom: "Their men have no honour, their women no virtue, & their
flowers no perfume."?
|
| 881.25 | | GUESS::DERAMO | Dan D'Eramo | Sat May 12 1990 14:54 | 3 |
| What is the group PGL(2,5), and is it solveable? :-)
Dan
|
| 881.26 | | ALLVAX::JROTH | It's a bush recording... | Sun May 13 1990 04:47 | 14 |
| What is the group PGL(2,5), and is it solveable? :-)
Dan
It is the group of non singular 2 by 2 matrices
| a b |
| c d |
with a,b,c,d from the finite field of 5 elements, modulo multiples
of the identity matrix. From this you should be able to tell if it
has any self conjugate subgroups or not :-)
- Jim
|
| 881.27 | ... I think :-) | GUESS::DERAMO | Dan D'Eramo | Sun May 13 1990 15:27 | 3 |
| Thanks.
Dan
|
| 881.28 | Some finite matrix groups | ALLVAX::JROTH | It's a bush recording... | Mon May 14 1990 03:44 | 24 |
| GL(n,q) is the general linear group of n by n matrices with elements
from GF(q).
The order is (q^n-1)(q^n-q)...(q^n-q^(n-1))
PGL(n,p) is the projective linear group and is GL(n,p) modulo
scalar multiples of the identity matrix, where the scalar is a
primitive element of the multiplicative group of q.
The order of this subgroup is 1/(q-1) of the order of GL(n,q).
Other subgroups of interest are SL(n,q), the special linear group
of matrices of determinant 1 and PSL(n,q) the projective special
linear group.
These groups can be defined in terms of generators and relations.
As for PGL(2,5), there are 5^4 = 625 possible matrices in the algebra of
2 by 2 matrices over GF(5), but only (5^2-1)(5^2-5) = 480 are
nonsingular; they fall into 4 classes so the order of PGL(2,5) is 120.
It happens to be isomorphic to S5 [I think], so would not be solvable.
- Jim
|