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| M.A.Fajjal... |
 Posted: Sun Oct 18, 2009 10:18 pm |
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| Can MAPLE or any other software solve all solvable sextics and higher order solvable polynomials by radical |
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| M.A.Fajjal... |
| Posted: Thu Oct 22, 2009 10:17 am |
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[quote]Can MAPLE or any other software solve all solvable
sextics and higher order solvable polynomials by
radical
[/quote]
I think I succeed to design this software up to degree 9 using MAPLE
Can anyone double check the output results of one root for
X^8+3*X^7+20*X^4+18*X^3-18*X^2-8*X+14 |
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| Posted: Fri Oct 23, 2009 7:54 pm |
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"M.A.Fajjal" schrieb:
[quote]
Can MAPLE or any other software solve all solvable
sextics and higher order solvable polynomials by
radical
I think I succeed to design this software up to degree 9 using MAPLE
Can anyone double check the output results of one root for
X^8+3*X^7+20*X^4+18*X^3-18*X^2-8*X+14
[/quote]
I don't think my Derive 6.10 is able to verify your root symbolically; I
didn't wait long enough to see what happens (either memory overflow, or
some transformed - or unchanged - expression whose equivalence to zero
remains undecided).
A numerical verification of your root meets with two problems.
First, your expression is about 35 kilobyte in size; so, to be on the
safe side, it should be verified using significantly more than 35,000
decimal digits. I haven't attempted to do that.
Second, numerical evaluation of your expression randomly returns any of
the following sixteen values:
-1.491265654 - 1.462163491*#i
-1.491265654 - 1.233645780*#i
-1.491265654 - 0.6631721784*#i
-1.491265654 - 0.6157694271*#i
-1.491265654 - 0.4346544675*#i
-1.491265654 - 0.4117395969*#i
-1.491265654 - 0.3872517162*#i
-1.491265654 - 0.1832218861*#i
-1.491265654 + 0.1832218861*#i
-1.491265654 + 0.3872517162*#i
-1.491265654 + 0.4117395969*#i
-1.491265654 + 0.4346544675*#i
-1.491265654 + 0.6157694271*#i
-1.491265654 + 0.6631721784*#i
-1.491265654 + 1.233645780*#i
-1.491265654 + 1.462163491*#i
These are pairwise complex conjugates, but only the 8th and 9th are true
roots of the given octic. Surely, this happens because certain
intermediate radicands end up on one side or the other of a branch cut
in dependence on numerical cancellation. You should therefore
reformulate your expression such that roots of negative real numbers do
not appear.
Martin. |
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| Gerry... |
| Posted: Fri Oct 23, 2009 9:49 pm |
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On Oct 22, 10:17 pm, "M.A.Fajjal" <h2... at (no spam) yahoo.com> wrote:
[quote] octic168.txt
35KViewDownload
Can MAPLE or any other software solve all solvable
sextics and higher order solvable polynomials by
radical
I think I succeed to design this software up to degree 9 using MAPLE
Can anyone double check the output results of one root for
X^8+3*X^7+20*X^4+18*X^3-18*X^2-8*X+14
[/quote]
Hi
i pasted your amazing r1 string into Pari-GP and played a little with
the precision, i assume your root expression is correct although with
a set precision of 100 digits the polynomial doesn't evaluate to 0.
With 1000 digits it does become zero again.
Regards
Gerry |
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| Peter Pein... |
| Posted: Fri Oct 23, 2009 10:16 pm |
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On my old-fashioned deterministic finite state machine I get only one value
for 100 evaluations of the same input:
Union[Table[N[r1], {100}]] --> {-1.491265654272576 - 0.1832218861092626*I}
and
poly/.x->N[r1,10^5] --> 0.*10^-99998+0.*10^-99998 I
and
N[poly/.x->r1,10^5] --> 0.*10^-100045+0.*10^-100045 I
where poly = 14 + x (-8 + x (-18 + x (18 + x (20 + x^3 (3 + x)))))
Peter
clicliclic at (no spam) freenet.de schrieb:
[quote]"M.A.Fajjal" schrieb:
Can MAPLE or any other software solve all solvable
sextics and higher order solvable polynomials by
radical
I think I succeed to design this software up to degree 9 using MAPLE
Can anyone double check the output results of one root for
X^8+3*X^7+20*X^4+18*X^3-18*X^2-8*X+14
I don't think my Derive 6.10 is able to verify your root symbolically; I
didn't wait long enough to see what happens (either memory overflow, or
some transformed - or unchanged - expression whose equivalence to zero
remains undecided).
A numerical verification of your root meets with two problems.
First, your expression is about 35 kilobyte in size; so, to be on the
safe side, it should be verified using significantly more than 35,000
decimal digits. I haven't attempted to do that.
Second, numerical evaluation of your expression randomly returns any of
the following sixteen values:
-1.491265654 - 1.462163491*#i
-1.491265654 - 1.233645780*#i
-1.491265654 - 0.6631721784*#i
-1.491265654 - 0.6157694271*#i
-1.491265654 - 0.4346544675*#i
-1.491265654 - 0.4117395969*#i
-1.491265654 - 0.3872517162*#i
-1.491265654 - 0.1832218861*#i
-1.491265654 + 0.1832218861*#i
-1.491265654 + 0.3872517162*#i
-1.491265654 + 0.4117395969*#i
-1.491265654 + 0.4346544675*#i
-1.491265654 + 0.6157694271*#i
-1.491265654 + 0.6631721784*#i
-1.491265654 + 1.233645780*#i
-1.491265654 + 1.462163491*#i
These are pairwise complex conjugates, but only the 8th and 9th are true
roots of the given octic. Surely, this happens because certain
intermediate radicands end up on one side or the other of a branch cut
in dependence on numerical cancellation. You should therefore
reformulate your expression such that roots of negative real numbers do
not appear.
Martin.[/quote] |
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| Posted: Sat Oct 24, 2009 12:52 am |
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Peter Pein schrieb:
[quote]
On my old-fashioned deterministic finite state machine I get only one value
for 100 evaluations of the same input:
Union[Table[N[r1], {100}]] --> {-1.491265654272576 - 0.1832218861092626*I}
and
poly/.x->N[r1,10^5] --> 0.*10^-99998+0.*10^-99998 I
and
N[poly/.x->r1,10^5] --> 0.*10^-100045+0.*10^-100045 I
where poly = 14 + x (-8 + x (-18 + x (18 + x (20 + x^3 (3 + x)))))
Peter
clicliclic at (no spam) freenet.de schrieb:
"M.A.Fajjal" schrieb:
Can MAPLE or any other software solve all solvable
sextics and higher order solvable polynomials by
radical
I think I succeed to design this software up to degree 9 using MAPLE
Can anyone double check the output results of one root for
X^8+3*X^7+20*X^4+18*X^3-18*X^2-8*X+14
I don't think my Derive 6.10 is able to verify your root symbolically; I
didn't wait long enough to see what happens (either memory overflow, or
some transformed - or unchanged - expression whose equivalence to zero
remains undecided).
A numerical verification of your root meets with two problems.
First, your expression is about 35 kilobyte in size; so, to be on the
safe side, it should be verified using significantly more than 35,000
decimal digits. I haven't attempted to do that.
Second, numerical evaluation of your expression randomly returns any of
the following sixteen values:
-1.491265654 - 1.462163491*#i
-1.491265654 - 1.233645780*#i
-1.491265654 - 0.6631721784*#i
-1.491265654 - 0.6157694271*#i
-1.491265654 - 0.4346544675*#i
-1.491265654 - 0.4117395969*#i
-1.491265654 - 0.3872517162*#i
-1.491265654 - 0.1832218861*#i
-1.491265654 + 0.1832218861*#i
-1.491265654 + 0.3872517162*#i
-1.491265654 + 0.4117395969*#i
-1.491265654 + 0.4346544675*#i
-1.491265654 + 0.6157694271*#i
-1.491265654 + 0.6631721784*#i
-1.491265654 + 1.233645780*#i
-1.491265654 + 1.462163491*#i
These are pairwise complex conjugates, but only the 8th and 9th are true
roots of the given octic. Surely, this happens because certain
intermediate radicands end up on one side or the other of a branch cut
in dependence on numerical cancellation. You should therefore
reformulate your expression such that roots of negative real numbers do
not appear.
Martin.
[/quote]
Lucky Peter! I forgot to mention: Mine is one of those modern quantum
models. Unfortunately, this doesn't mean that software bugs can't get
into it.
Martin. |
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| TPiezas... |
| Posted: Mon Oct 26, 2009 10:28 pm |
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On Oct 24, 1:49 am, Gerry <gerry... at (no spam) gmail.com> wrote:
[quote]On Oct 22, 10:17 pm, "M.A.Fajjal" <h2... at (no spam) yahoo.com> wrote:
octic168.txt
35KViewDownload
Can MAPLE or any other software solve all solvable
sextics and higher order solvable polynomials by
radical
I think I succeed to design this software up to degree 9 using MAPLE
Can anyone double check the output results of one root for
X^8+3*X^7+20*X^4+18*X^3-18*X^2-8*X+14
Hi
i pasted your amazing r1 string into Pari-GP and played a little with
the precision, i assume your root expression is correct although with
a set precision of 100 digits the polynomial doesn't evaluate to 0.
With 1000 digits it does become zero again.
Regards
Gerry
[/quote]
Hello Fajjal,
For nonics, test it with:
x^9-2x^8+8x^7-8x^6+20x^5-8x^4+4x^3-24x^2+23x-6 = 0.
This will not factor over a cubic subfield, but over a _sextic_, so
will be tricky. This has transitive group 9T9. Go to this website:
http://www.math.uni-duesseldorf.de/~klueners/minimum/minimum.html
In the page for 9th deg eqns, you will find 34 transitive groups. Only
the first 13 (with size < 60) are solvable. If you are using the
method I think you are using, namely "polynomial decomposition", and
factoring the nonic into three cubics whose coefficients are
determined by an nth-deg eqn with rational coefficients, the lower
groups are easy because n = 3. But for 9T9 and some of the higher ones
(I forgot which as my notes are in another country), it may be n 4,6, or even 12.
Yes, trying to factor some solvable nonics into three cubics may
involve solving a 12-deg eqn with rational coefficients. (Its 84-deg
resolvent has a 12-deg as the smallest factor.)
Update us with which of the 13 nonic groups your algorithm can't
solve.
- Titus |
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| M.A.Fajjal... |
| Posted: Tue Oct 27, 2009 3:07 am |
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Guest
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[quote]
Hello Fajjal,
For nonics, test it with:
x^9-2x^8+8x^7-8x^6+20x^5-8x^4+4x^3-24x^2+23x-6 = 0.
This will not factor over a cubic subfield, but over
a _sextic_, so
will be tricky. This has transitive group 9T9. Go to
this website:
http://www.math.uni-duesseldorf.de/~klueners/minimum/m
inimum.html
In the page for 9th deg eqns, you will find 34
transitive groups. Only
the first 13 (with size < 60) are solvable. If you
are using the
method I think you are using, namely "polynomial
decomposition", and
factoring the nonic into three cubics whose
coefficients are
determined by an nth-deg eqn with rational
coefficients, the lower
groups are easy because n = 3. But for 9T9 and some
of the higher ones
(I forgot which as my notes are in another country),
it may be n =
4,6, or even 12.
Yes, trying to factor some solvable nonics into three
cubics may
involve solving a 12-deg eqn with rational
coefficients. (Its 84-deg
resolvent has a 12-deg as the smallest factor.)
Update us with which of the 13 nonic groups your
algorithm can't
solve.
- Titus
Hello Titus[/quote]
The solution for x^9-2*x^8+8*x^7-8*x^6+20*x^5-8*x^4+4*x^3-24*x^2+23*x-6 = 0 shall be
[quote]nonic(1, -2, 8, -8, 20, -8, 4, -24, 23, -6);
x^3+RootOf(_Z^3+2*_Z^2+(13+7*2^(1/2))*_Z+20+14*2^(1/2))*x^2+((10/7)*RootOf(_Z^3+2*_Z^2+(13+7*2^(1/2))*_Z+20+14*2^(1/2))+15/7-(9/7)*2^(1/2)-(6/7)*RootOf(_Z^3+2*_Z^2+(13+7*2^(1/2))*_Z+20+14*2^(1/2))*2^(1/2)-(1/7)*RootOf(_Z^3+2*_Z^2+(13+7*2^(1/2))*_Z+20+14*2^(1/2))^2*2^(1/2)+(4/7)*RootOf(_Z^3+2*_Z^2+(13+7*2^(1/2))*_Z+20+14*2^(1/2))^2)*x-2*(115455+17846*RootOf(_Z^3+2*_Z^2+(13+7*2^(1/2))*_Z+20+14*2^(1/2))^2*2^(1/2)+79090*RootOf(_Z^3+2*_Z^2+(13+7*2^(1/2))*_Z+20+14*2^(1/2))*2^(1/2)+102814*RootOf(_Z^3+2*_Z^2+(13+7*2^(1/2))*_Z+20+14*2^(1/2))+88563*2^(1/2)+18503*RootOf(_Z^3+2*_Z^2+(13+7*2^(1/2))*_Z+20+14*2^(1/2))^2)/(69831+589*RootOf(_Z^3+2*_Z^2+(13+7*2^(1/2))*_Z+20+14*2^(1/2))^2*2^(1/2)+35692*RootOf(_Z^3+2*_Z^2+(13+7*2^(1/2))*_Z+20+14*2^(1/2))*2^(1/2)+37006*RootOf(_Z^3+2*_Z^2+(13+7*2^(1/2))*_Z+20+14*2^(1/2))+56744*2^(1/2)-5387*RootOf(_Z^3+2*_Z^2+(13+7*2^(1/2))*_Z+20+14*2^(1/2))^2)[/quote]
Elapsed time is, 47.077 seconds
For nonic only 9T27,9T32,9T33,9T34 are not solvable. All other nonic transitive groups are solvable. My algorithm can solve all of them
Example 9T26
x^9 - x^7 - 5*x^6 + x^5 + 2*x^4 + 4*x^3 - 3*x^2 - x + 1
[quote]nonic(1, 0, -1, -5, 1, 2, 4, -3, -1, 1);
x = RootOf(4576305062*_Z^3+4576305062*u*_Z^2+(-12987403928*u^5-8336700754-3043300196*u^4+3693046449*u^7+83289548*u^11-7272554149*u-330564483*u^6+17506845797*u^2-57123865*u^10+1053902165*u^8-247258826*u^9+14492023447*u^3)*_Z+95464235652*u^3-95341462800*u^5+24153988612*u^7-1781675484*u^9-8411266464*u^6+7816610996*u^8-511218930*u^10+70387077978*u^2-11852788124*u^4-36456880970-51016723702*u+584710268*u^11)[/quote]
u = -RootOf(401*(_Z-v)^3+1203*v*(_Z-v)^2+(1489*v^2-24*v^3-456-847*v)*(_Z-v)+473*v^2+423*v^3-384-2465*v)
v = RootOf(_Z^4-11*_Z^3+55*_Z^2-67*_Z-16)
Elapsed time is, 49.351 seconds |
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| TPiezas... |
| Posted: Tue Oct 27, 2009 5:42 am |
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On Oct 27, 7:07 am, "M.A.Fajjal" <h2... at (no spam) yahoo.com> wrote:
[quote]Hello Fajjal,
For nonics, test it with:
x^9-2x^8+8x^7-8x^6+20x^5-8x^4+4x^3-24x^2+23x-6 = 0.
This will not factor over a cubic subfield, but over
a _sextic_, so
will be tricky. This has transitive group 9T9. Go to
this website:
http://www.math.uni-duesseldorf.de/~klueners/minimum/m
inimum.html
In the page for 9th deg eqns, you will find 34
transitive groups. Only
the first 13 (with size < 60) are solvable. If you
are using the
method I think you are using, namely "polynomial
decomposition", and
factoring the nonic into three cubics whose
coefficients are
determined by an nth-deg eqn with rational
coefficients, the lower
groups are easy because n = 3. But for 9T9 and some
of the higher ones
(I forgot which as my notes are in another country),
it may be n > > 4,6, or even 12.
Yes, trying to factor some solvable nonics into three
cubics may
involve solving a 12-deg eqn with rational
coefficients. (Its 84-deg
resolvent has a 12-deg as the smallest factor.)
Update us with which of the 13 nonic groups your
algorithm can't
solve.
- Titus
Hello Titus
The solution for x^9-2*x^8+8*x^7-8*x^6+20*x^5-8*x^4+4*x^3-24*x^2+23*x-6 = 0 shall be
nonic(1, -2, 8, -8, 20, -8, 4, -24, 23, -6);
x^3+RootOf(_Z^3+2*_Z^2+(13+7*2^(1/2))*_Z+20+14*2^(1/2))*x^2+((10/7)*RootOf(_Z^3+2*_Z^2+(13+7*2^(1/2))*_Z+20+14*2^(1/2))+15/7-(9/7)*2^(1/2)-(6/7)*RootOf(_Z^3+2*_Z^2+(13+7*2^(1/2))*_Z+20+14*2^(1/2))*2^(1/2)-(1/7)*RootOf(_Z^3+2*_Z^2+(13+7*2^(1/2))*_Z+20+14*2^(1/2))^2*2^(1/2)+(4/7)*RootOf(_Z^3+2*_Z^2+(13+7*2^(1/2))*_Z+20+14*2^(1/2))^2)*x-2*(115455+17846*RootOf(_Z^3+2*_Z^2+(13+7*2^(1/2))*_Z+20+14*2^(1/2))^2*2^(1/2)+79090*RootOf(_Z^3+2*_Z^2+(13+7*2^(1/2))*_Z+20+14*2^(1/2))*2^(1/2)+102814*RootOf(_Z^3+2*_Z^2+(13+7*2^(1/2))*_Z+20+14*2^(1/2))+88563*2^(1/2)+18503*RootOf(_Z^3+2*_Z^2+(13+7*2^(1/2))*_Z+20+14*2^(1/2))^2)/(69831+589*RootOf(_Z^3+2*_Z^2+(13+7*2^(1/2))*_Z+20+14*2^(1/2))^2*2^(1/2)+35692*RootOf(_Z^3+2*_Z^2+(13+7*2^(1/2))*_Z+20+14*2^(1/2))*2^(1/2)+37006*RootOf(_Z^3+2*_Z^2+(13+7*2^(1/2))*_Z+20+14*2^(1/2))+56744*2^(1/2)-5387*RootOf(_Z^3+2*_Z^2+(13+7*2^(1/2))*_Z+20+14*2^(1/2))^2)
Elapsed time is, 47.077 seconds
For nonic only 9T27,9T32,9T33,9T34 are not solvable. All other nonic transitive groups are solvable. My algorithm can solve all of them
Example 9T26
x^9 - x^7 - 5*x^6 + x^5 + 2*x^4 + 4*x^3 - 3*x^2 - x + 1
nonic(1, 0, -1, -5, 1, 2, 4, -3, -1, 1);
x = RootOf(4576305062*_Z^3+4576305062*u*_Z^2+(-12987403928*u^5-8336700754-3043300196*u^4+3693046449*u^7+83289548*u^11-7272554149*u-330564483*u^6+17506845797*u^2-57123865*u^10+1053902165*u^8-247258826*u^9+14492023447*u^3)*_Z+95464235652*u^3-95341462800*u^5+24153988612*u^7-1781675484*u^9-8411266464*u^6+7816610996*u^8-511218930*u^10+70387077978*u^2-11852788124*u^4-36456880970-51016723702*u+584710268*u^11)
u = -RootOf(401*(_Z-v)^3+1203*v*(_Z-v)^2+(1489*v^2-24*v^3-456-847*v)*(_Z-v)+473*v^2+423*v^3-384-2465*v)
v = RootOf(_Z^4-11*_Z^3+55*_Z^2-67*_Z-16)
Elapsed time is, 49.351 seconds- Hide quoted text -
- Show quoted text -
[/quote]
Hello Fajjal,
Impressive work! Oops, regarding the solvable nonic groups, my
mistake about the size < 60 thing. That was for quintics since only
5T1, 5T2, and 5T3 are solvable and they have size < 60. Sigh, I don't
have my notes here, but I did have the complete list of solvable nonic
groups before. (When I made the post, I noticed that 13 solvable nonic
groups seemed to be too few.)
I. Regarding the 9T9 example I gave, namely,
x^9-2*x^8+8*x^7-8*x^6+20*x^5-8*x^4+4*x^3-24*x^2+23*x-6 = 0 (eq.1)
it can be decomposed as three cubics x^3+F(m)_1*x^2+F(m)_2*x+F(m)_3 0 where F(m)_i are roots of sextics. This sextic factors over a Sqrt
[2] extension. Specifically, eq.1 decomposes as,
x^3+mx^2+(am^2+b-1)x-(am^2+m+b-2) = 0
where m is a root of,
m^3+2m^2-m+b = 0
and {a,b} = {2-Sqrt[2], 2*Sqrt[2]}. I gave this nonic as an example in
my paper "Solving Solvable Sextics Using Polynomial Decomposition".
II. Regarding the 9T26 example, is that the one where the F(m)_i are
roots of 12-deg eqns? (I remember one nonic I solved decomposed into 3
cubics with coefficients that involved a 12-deg eqn. The 12-deg could
then be decomposed into 4 cubics with coefficients that involve a
quartic. Since you ended up with a quartic, do I presume this is the
same group?) Could you clarify the soln of the 9T26? The variables
make it a little confusing.
Thanks.
Sincerely,
- Titus |
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| M.A.Fajjal... |
| Posted: Tue Oct 27, 2009 8:33 am |
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Guest
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[quote]Could you clarify the soln of the 9T26?
The variables
make it a little confusing.
[/quote]
The solution can be explained as following
f3(u)*x^3 + f2(u)*x^2 + f1(u)*x + f0(u) = 0
f3(u) = 4576305062
f2(u) = 4576305062*u
f1(u) = 83289548*u^11 - 57123865*u^10 - 247258826*u^9 + 1053902165*u^8 + 3693046449*u^7 - 330564483*u^6 - 12987403928*u^5 - 3043300196*u^4 + 14492023447*u^3 + 17506845797*u^2 - 7272554149*u - 8336700754
f0(u) = 584710268*u^11 - 511218930*u^10 - 1781675484*u^9 + 7816610996*u^8 + 24153988612*u^7 - 8411266464*u^6 - 95341462800*u^5 - 11852788124*u^4 + 95464235652*u^3 + 70387077978*u^2 - 51016723702*u - 36456880970
Where
g3(v)*u^3 + g2(v)*u^2 + g1(v)*u + g0(v)=0
g3(v) = 401
g2(v) = 0
g1(v) = -24*v^3+286*v^2-847*v-456
g0(v) = -24*v^4+384+264*v^3-1320*v^2+2009*v
Where
v^4 - 11*v^3 + 55*v^2 - 67*v - 16 = 0
v has 4 different values take any one of them
u has 3 values
x has 3*3 values |
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| TPiezas... |
| Posted: Tue Oct 27, 2009 1:44 pm |
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On Oct 27, 12:33 pm, "M.A.Fajjal" <h2... at (no spam) yahoo.com> wrote:
[quote]Could you clarify the soln of the 9T26?
The variables
make it a little confusing.
The solution can be explained as following
f3(u)*x^3 + f2(u)*x^2 + f1(u)*x + f0(u) = 0
f3(u) = 4576305062
f2(u) = 4576305062*u
f1(u) = 83289548*u^11 - 57123865*u^10 - 247258826*u^9 + 1053902165*u^8 + 3693046449*u^7 - 330564483*u^6 - 12987403928*u^5 - 3043300196*u^4 + 14492023447*u^3 + 17506845797*u^2 - 7272554149*u - 8336700754
f0(u) = 584710268*u^11 - 511218930*u^10 - 1781675484*u^9 + 7816610996*u^8 + 24153988612*u^7 - 8411266464*u^6 - 95341462800*u^5 - 11852788124*u^4 + 95464235652*u^3 + 70387077978*u^2 - 51016723702*u - 36456880970
Where
g3(v)*u^3 + g2(v)*u^2 + g1(v)*u + g0(v)=0
g3(v) = 401
g2(v) = 0
g1(v) = -24*v^3+286*v^2-847*v-456
g0(v) = -24*v^4+384+264*v^3-1320*v^2+2009*v
Where
v^4 - 11*v^3 + 55*v^2 - 67*v - 16 = 0
v has 4 different values take any one of them
u has 3 values
x has 3*3 values
[/quote]
Hello Fajjal,
I. Beautiful! Ah, so 9T26 is the group where you need to solve a 12-
deg. I checked your soln with Mathematica and it's correct:
Factor[Resultant[
401*u^3 + (-24*v^3 + 286*v^2 - 847*v - 456)*
u + (-24*v^4 + 384 + 264*v^3 - 1320*v^2 + 2009*v),
v^4 - 11*v^3 + 55*v^2 - 67*v - 16, v]]
Eliminating v, this gives a 12-deg in u:
-16-56u+8u^2+155u^3+71u^4-119u^5-132u^6+30u^7+51u^8+11u^9-3u^10+u^12 0
Factor[Resultant[-16-56u
+8u^2+155u^3+71u^4-119u^5-132u^6+30u^7+51u^8+11u^9-3u^10+u^12,
4576305062*x^3 + 4576305062*u*x^2
+ (83289548*u^11 - 57123865*u^10 - 247258826*u^9 + 1053902165*u^8 +
3693046449*u^7 - 330564483*u^6 - 12987403928*u^5 -
3043300196*u^4 +
14492023447*u^3 + 17506845797*u^2 - 7272554149*u - 8336700754)
*
x + (584710268*u^11 - 511218930*u^10 - 1781675484*u^9 +
7816610996*u^8 +
24153988612*u^7 - 8411266464*u^6 - 95341462800*u^5 -
11852788124*u^4 +
95464235652*u^3 + 70387077978*u^2 - 51016723702*u -
36456880970), u]]
Eliminating u, this gives a 3*12 = 36-deg eqn in x which is just the
original nonic repeated 4 times. Very nice!
II. I'm sure you can extend your algorithm to 12-deg. (I stopped at 10-
deg.) Have you tried checking its transitive groups (there's 301) to
see which are solvable? I believe those with size divisible by 5 are
unsolvable. Anyway, can you tell us which group holds a SURPRISE,
just like 9T26? To explain further, unlike the 9-deg which can be
decomposed only as a cubic, the 12-deg can be decomposed into a
solvable form in two ways:
1. a cubic (with coefficients determined by an eqn of deg m)
2. a quartic (with coefficients determined by an eqn of deg n)
One will expect m = 4 and n = 3. But, as shown by 9T26, one can always
be surprised, so I'm sure there will be groups such that {m,n} > 4.
Can you tell us what solvable group for 12-deg is the most difficult?
Thanks.
- Titus |
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| M.A.Fajjal... |
| Posted: Tue Oct 27, 2009 5:30 pm |
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Guest
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[quote]Can you tell us what solvable group for 12-deg is the
most difficult?
Thanks.
This may takes long time to check all groups. I will try.[/quote] |
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| TPiezas... |
| Posted: Tue Oct 27, 2009 6:33 pm |
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Guest
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On Oct 27, 9:30 pm, "M.A.Fajjal" <h2... at (no spam) yahoo.com> wrote:
[quote]Can you tell us what solvable group for 12-deg is the
most difficult?
Thanks.
This may takes long time to check all groups. I will try.
[/quote]
Thanks. I remember now, ANY transitive group with a size less than 60
is solvable, so the first 32 for deg-12 will pass, though I don't know
if any will be tricky. For size 60 and beyond, you can eliminate any
that is div by 5, 7, or 11 since a solvable and irreducible 12-deg eqn
with rational coefficients does not use the 5th, 7th, or 11th root of
unity. That eliminates 36 groups already. The rest will have to be
checked individually (too bad Klueners didn't put a column indicating
if it is solvable or not).
Some of the solvable groups will be routine, decomposing easily, but
wouldn't it be nice to know if there is a 12-deg transitive group that
needs solving a 9-deg, another 12-deg, or even an 18-deg?
Then there are the 15-deg transitive groups, and Klueners stops here.
But who knows what surprises there are for the higher 16-deg groups
since it is a square as well as a power of 2. I recall (I hope I
remember my notes accurately) there are some solvable 16-deg groups
that need solving a QUINTIC. (How a 5th root of unity can squeeze into
a power of 2 is remarkable!) Then for the next square group, the 25-
deg, some solvable groups need a CUBE ROOT, even though solvable
quintics never use the cube root of unity. Ramanujan gave a very
beautiful and simple soln for this.
Playing around with solvable eqns is quite fun.
- Titus |
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| M.A.Fajjal... |
| Posted: Tue Oct 27, 2009 9:39 pm |
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Guest
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[quote]On Oct 27, 9:30Â pm, "M.A.Fajjal" <h2... at (no spam) yahoo.com
wrote:
Can you tell us what solvable group for 12-deg is
the
most difficult?
Thanks.
 This may takes long time to check all groups. I
will try.
Thanks. I remember now, ANY transitive group with a
size less than 60
is solvable, so the first 32 for deg-12 will pass,
though I don't know
if any will be tricky. For size 60 and beyond, you
can eliminate any
that is div by 5, 7, or 11 since a solvable and
irreducible 12-deg eqn
with rational coefficients does not use the 5th, 7th,
or 11th root of
unity. That eliminates 36 groups already. The rest
will have to be
checked individually (too bad Klueners didn't put a
column indicating
if it is solvable or not).
Some of the solvable groups will be routine,
decomposing easily, but
wouldn't it be nice to know if there is a 12-deg
transitive group that
needs solving a 9-deg, another 12-deg, or even an
18-deg?
Then there are the 15-deg transitive groups, and
Klueners stops here.
But who knows what surprises there are for the higher
16-deg groups
since it is a square as well as a power of 2. I
recall (I hope I
remember my notes accurately) there are some solvable
16-deg groups
that need solving a QUINTIC. (How a 5th root of unity
can squeeze into
a power of 2 is remarkable!) Then for the next
square group, the 25-
deg, some solvable groups need a CUBE ROOT, even
though solvable
quintics never use the cube root of unity. Ramanujan
gave a very
beautiful and simple soln for this.
Playing around with solvable eqns is quite fun.
- Titus
[/quote]
As I remember (may any one correct me if I am wrong), all groups of order less than 100 are solvable except 60 for quintic and sextic
But for degree 20 , 20T14 group order 60 is solvable and you may see cube root of unity
For degree 16 you may see 5th root of unity for group order 80,160,320
May see both cube root of unity and 5th root of unity for group order 240,480,960
May see 7th root of unity severals
For nonic you shall find surprises for
9T14,9T15,9T19,9T23,9T26
Half surprises for
9T9 and 9T16
For Degree 12
All groubs are solvable except which are divided by 10 or 11
i.e 36 nonsolvable groups. The rest are solvable
You may face a lot of surprises |
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| TPiezas... |
| Posted: Wed Oct 28, 2009 5:17 am |
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Guest
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On Oct 28, 1:39 am, "M.A.Fajjal" <h2... at (no spam) yahoo.com> wrote:
[quote]On Oct 27, 9:30 pm, "M.A.Fajjal" <h2... at (no spam) yahoo.com
wrote:
Can you tell us what solvable group for 12-deg is
the
most difficult?
Thanks.
This may takes long time to check all groups. I
will try.
Thanks. I remember now, ANY transitive group with a
size less than 60
is solvable, so the first 32 for deg-12 will pass,
though I don't know
if any will be tricky. For size 60 and beyond, you
can eliminate any
that is div by 5, 7, or 11 since a solvable and
irreducible 12-deg eqn
with rational coefficients does not use the 5th, 7th,
or 11th root of
unity. That eliminates 36 groups already. The rest
will have to be
checked individually (too bad Klueners didn't put a
column indicating
if it is solvable or not).
Some of the solvable groups will be routine,
decomposing easily, but
wouldn't it be nice to know if there is a 12-deg
transitive group that
needs solving a 9-deg, another 12-deg, or even an
18-deg?
Then there are the 15-deg transitive groups, and
Klueners stops here.
But who knows what surprises there are for the higher
16-deg groups
since it is a square as well as a power of 2. I
recall (I hope I
remember my notes accurately) there are some solvable
16-deg groups
that need solving a QUINTIC. (How a 5th root of unity
can squeeze into
a power of 2 is remarkable!) Then for the next
square group, the 25-
deg, some solvable groups need a CUBE ROOT, even
though solvable
quintics never use the cube root of unity. Ramanujan
gave a very
beautiful and simple soln for this.
Playing around with solvable eqns is quite fun.
- Titus
As I remember (may any one correct me if I am wrong), all groups of order less than 100 are solvable except 60 for quintic and sextic
But for degree 20 , 20T14 group order 60 is solvable and you may see cube root of unity
For degree 16 you may see 5th root of unity for group order 80,160,320
May see both cube root of unity and 5th root of unity for group order 240,480,960
May see 7th root of unity severals
For nonic you shall find surprises for
9T14,9T15,9T19,9T23,9T26
Half surprises for
9T9 and 9T16
For Degree 12
All groubs are solvable except which are divided by 10 or 11
i.e 36 nonsolvable groups. The rest are solvable
You may face a lot of surprises- Hide quoted text -
- Show quoted text -
[/quote]
For deg 20, a cube root of unity may be expected for some solvable
groups since one can always "compose" an eqn of the form x^4+P(m)
_1*x^3+P(m)_2*x^2+P(m)_3*x+P(m)_4 = 0, where the P(m)_i are
polynomials in terms of a root m of a solvable quintic F(m) = 0.
Eliminating m between the two will, of course, result in a deg 20
solvable eqn in x.
It is harder for deg 25 since, using the same solvable quintic F(m) 0, if you use it in the composition of x^5+P(m)_*x^4 + .... = 0, there
is no guarantee the resulting 25-deg is solvable. But Ramanujan
managed to find a solvable 25-deg parametrization that used the CUBE
roots of unity (as you know, solvable quintics don't need cube roots).
To top it off, he found solvable 49-deg eqns as well. (It's in one of
his Notebooks, the third one, I think.)
For deg 16, why there is 5th root of unity needed in some solvable
groups may be explained. Simply form the expression, Product[y+/-(Sqrt
[x1]+/-Sqrt[x2]+/-Sqrt[x3]+/-Sqrt[x4]+/-Sqrt[x5])] where the x_i are
the 5 roots of a solvable quintic. The resulting 2^5 = 32 deg eqn in y
has only even exponents, so can be reduced to a solvable 16-deg.
Going higher, if you use a special kind of solvable septic, the
resulting 2^7 = 128 deg eqn may have an octic FACTOR, an example of
which is the one you gave at the start of this thread, as well as the
beautiful solvable octic x^8-x^7+29x^2+29 = 0. I don't know if this
128 deg can ever have a 16-deg factor.
For solvable eqns of degree prime p or 2p, their behaviour is rather
predictable; it's the composite degrees not 2p that can be tricky. And
the first one after the 8th and 9th is the 12 deg, and I'm curious to
know how difficult some of its solvable groups can get. Update us,
will you?
Thanks.
- Titus |
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