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| M.A.Fajjal... |
 Posted: Fri Oct 30, 2009 1:05 am |
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[quote]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
[/quote]
My PC RAM is half GB. I need to upgrade it to at least 4GB to check degree 12
The algorthim can be extended to any degree
By the way can Mathematica or any other software find radical expression for cos(2pi/47) |
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| TPiezas... |
| Posted: Fri Oct 30, 2009 5:04 am |
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On Oct 30, 5:05 am, "M.A.Fajjal" <h2... at (no spam) yahoo.com> wrote:
[quote]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
My PC RAM is half GB. I need to upgrade it to at least 4GB to check degree 12
The algorthim can be extended to any degree
By the way can Mathematica or any other software find radical expression for cos(2pi/47)
My PC RAM is half GB. I need to upgrade it to at least 4GB to check degree 12
The algorthim can be extended to any degree
[/quote]
Well, maybe I can suggest something? As the problem is merely to see
which of the solvable groups of the 12-deg are "exceptional" (like
the nonics 9T26), perhaps all we need is to calculate resolvents, and
not generate the actual radical soln of the 12-deg itself. Is this
easily done? (I'm not much of a programmer.) The algorithm would test
the examples in Klueners' site with the ff steps:
1. Check if it factors over a square root extension Sqrt[d], where d
is the discriminant. If not, go to next step.
2. Go through all Binomial[12,3]= 220 combinations of y1 = (x1+x2+x3)
to see if it is the EXACT root of a quartic. (Or generate the 220-deg
polynomial and see if it has a quartic factor, or a sextic or less.)
If not, go to next step.
3. Go through all Binomial[12,4]= 495 combinations of z1 (x1+x2+x3+x4) to see if it is the EXACT root of a cubic.
If a solvable 12-deg fails any of the 3 steps, then it is
"exceptional" and will be harder to solve than the routine ones.
(Mathematica has a feature called Recognize[x,n,t], such that it gives
an equation in t of deg n such that x is approximately a root. Of
course, one can then test if it is exactly a root.)
[quote]By the way can Mathematica or any other software find radical expression for cos(2pi/47)
[/quote]
I don't think it is practical even for the newest release, Mathematica
7. Note that cos(2pi/23) already takes up 500 MB of space. See here
http://mathworld.wolfram.com/TrigonometryAnglesPi23.html
P.S. I'm really curious how high the MINIMUM deg of y1 or z1 of the
exceptional cases can get. I'm assuming, from the factors of the group
size, there may solvable 12-deg that need solving a 18-deg, but that's
just a guess.
- Titus |
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| M.A.Fajjal... |
| Posted: Fri Oct 30, 2009 6:57 am |
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Guest
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[quote]
P.S. I'm really curious how high the MINIMUM deg of
y1 or z1 of the
exceptional cases can get. I'm assuming, from the
factors of the group
size, there may solvable 12-deg that need solving a
18-deg, but that's
just a guess.
- Titus
I have checked randomly the following groups[/quote]
27,35,41,44,62,80,96,104,115,127,141,159,173,201,204,225,238,251,268,277,292
No surprises at all
All can be reduced to form
x^3+b2*x^2+b1*x+b0
where b0, b1 are linear functions of b2
b2 is a quartic equation |
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| M.A.Fajjal... |
| Posted: Fri Oct 30, 2009 7:28 am |
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Guest
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[quote]
P.S. I'm really curious how high the MINIMUM deg
of
y1 or z1 of the
exceptional cases can get. I'm assuming, from the
factors of the group
size, there may solvable 12-deg that need solving
a
18-deg, but that's
just a guess.
- Titus
I have checked randomly the following groups
27,35,41,44,62,80,96,104,115,127,141,159,173,201,204,2
25,238,251,268,277,292
No surprises at all
All can be reduced to form
x^3+b2*x^2+b1*x+b0
where b0, b1 are linear functions of b2
b2 is a quartic equation
[/quote]
I think these are expected results
I have noticed that for degree n polynomial, the surprises are for groups of the form n(n-1) and their multipliers as group 42, 56, 168, 72, 144, ...,110, 156,....
For degree 12 there is no group 132, so no surprises
Same for degrees 6, 10, 14, 15 |
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| Nasser M. Abbasi... |
| Posted: Fri Oct 30, 2009 12:44 pm |
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"M.A.Fajjal" <h2maf at (no spam) yahoo.com> wrote in message
[quote]
By the way can Mathematica or any other software find radical expression
for cos(2pi/47)
[/quote]
using Mathematica 7:
When I do
ToRadicals[Cos[2*(Pi/47)]]
(-(1/2))*(-1)^(45/47)*(1 + (-1)^(4/47))
It looks like it uses the algebraic formula as shown on this page
?http://www.mathpages.com/home/kmath186.htm
(although a negative sign pops up at front of 1/2 here compared to the
page?)
But it does tell one that Cos[2*(Pi/47)] is the 23 third root of the
following polynomial:
In[85]:= RootReduce[Cos[2*(Pi/47)]];
p=First[%][x]
Out[86]= 8388608*x^23 + 4194304*x^22 -
46137344*x^21 - 22020096*x^20 +
110100480*x^19 + 49807360*x^18 -
149422080*x^17 - 63504384*x^16 +
127008768*x^15 + 50135040*x^14 -
70189056*x^13 - 25346048*x^12 +
25346048*x^11 + 8200192*x^10 -
5857280*x^9 - 1647360*x^8 + 823680*x^7 +
192192*x^6 - 64064*x^5 - 11440*x^4 +
2288*x^3 + 264*x^2 - 24*x - 1
And it is indeed the case by comparing the 23 roots and the numerical value
of Cos[2*(Pi/47)]
N[Cos[2*(Pi/47)]]
0.9910774881547801
NSolve[p, x][[23]]
{x -> 0.9910774881547794}
There might be other commands or options to play with, but this is not my
area, and the above is all what I can get for now.
--Nasser |
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| TPiezas... |
| Posted: Fri Oct 30, 2009 3:14 pm |
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On Oct 30, 10:57 am, "M.A.Fajjal" <h2... at (no spam) yahoo.com> wrote:
[quote]P.S. I'm really curious how high the MINIMUM deg of
y1 or z1 of the
exceptional cases can get. I'm assuming, from the
factors of the group
size, there may solvable 12-deg that need solving a
18-deg, but that's
just a guess.
- Titus
I have checked randomly the following groups
27,35,41,44,62,80,96,104,115,127,141,159,173,201,204,225,238,251,268,277,292
No surprises at all
All can be reduced to form
x^3+b2*x^2+b1*x+b0
where b0, b1 are linear functions of b2
b2 is a quartic equation
[/quote]
I. Hm, that is strange. For solvable sextics, they fall into 3
classes:
a) they decompose ONLY as x^2+b1x+b0 = 0, where b0 is a polynomial
function of b1, and b1 is a cubic eqn
b) they decompose ONLY as x^3+b2x^2+b1x+b0 = 0, where b0, b1 are
polynomial functions of b2, and b2 is a quadratic.
c) they decompose BOTH as (a) or (b)
For solvable 12-deg, you'd expect there may be several classes as
well:
a) as ONLY x^2+b1x+b0 = 0, b1 is a sextic
b) as ONLY x^3+b2x^2+... = 0, b2 is a quartic
c) as ONLY x^4+b3x^3+... = 0, b3 is a cubic
d) as ONLY x^6+b5x^5+... = 0, b5 is a quadratic
e) as ANY combination of {a,b,c,d,e}
f) and (perhaps) none of the above
Maybe I'll just email Klueners and ask him about it.
II. Regarding your observation about n(n-1), I had conjectured before
that the deg 2^p after a Mersenne prime 2^p-1 would be "exceptional".
Thus, degs 4,8, need the 3rd and 7th roots of unity, respectively. If
the conjecture is true, that would mean there are 32-degs that need
the 31st root of unity, and so on. (But I wonder if instead it covers
ALL degs 2^m? While some deg-16 needs a 3rd and 5th root of unity,
would some deg 2^9 = 512 = 7*73 need a 73rd root of unity? Or maybe it
is just when 2^p-1 is a prime?)
So many questions, and so little time and RAM. :-)
- Titus |
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| TPiezas... |
| Posted: Sat Oct 31, 2009 5:42 am |
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On Oct 30, 7:14 pm, TPiezas <tpie... at (no spam) gmail.com> wrote:
[quote]On Oct 30, 10:57 am, "M.A.Fajjal" <h2... at (no spam) yahoo.com> wrote:
P.S. I'm really curious how high the MINIMUM deg of
y1 or z1 of the
exceptional cases can get. I'm assuming, from the
factors of the group
size, there may solvable 12-deg that need solving a
18-deg, but that's
just a guess.
- Titus
I have checked randomly the following groups
27,35,41,44,62,80,96,104,115,127,141,159,173,201,204,225,238,251,268,277,292
No surprises at all
All can be reduced to form
x^3+b2*x^2+b1*x+b0
where b0, b1 are linear functions of b2
b2 is a quartic equation
I. Hm, that is strange. For solvable sextics, they fall into 3
classes:
a) they decompose ONLY as x^2+b1x+b0 = 0, where b0 is a polynomial
function of b1, and b1 is a cubic eqn
b) they decompose ONLY as x^3+b2x^2+b1x+b0 = 0, where b0, b1 are
polynomial functions of b2, and b2 is a quadratic.
c) they decompose BOTH as (a) or (b)
For solvable 12-deg, you'd expect there may be several classes as
well:
a) as ONLY x^2+b1x+b0 = 0, b1 is a sextic
b) as ONLY x^3+b2x^2+... = 0, b2 is a quartic
c) as ONLY x^4+b3x^3+... = 0, b3 is a cubic
d) as ONLY x^6+b5x^5+... = 0, b5 is a quadratic
e) as ANY combination of {a,b,c,d,e}
f) and (perhaps) none of the above
Maybe I'll just email Klueners and ask him about it.
II. Regarding your observation about n(n-1), I had conjectured before
that the deg 2^p after a Mersenne prime 2^p-1 would be "exceptional".
Thus, degs 4,8, need the 3rd and 7th roots of unity, respectively. If
the conjecture is true, that would mean there are 32-degs that need
the 31st root of unity, and so on. (But I wonder if instead it covers
ALL degs 2^m? While some deg-16 needs a 3rd and 5th root of unity,
would some deg 2^9 = 512 = 7*73 need a 73rd root of unity? Or maybe it
is just when 2^p-1 is a prime?)
So many questions, and so little time and RAM. :-)
- Titus- Hide quoted text -
- Show quoted text -
[/quote]
Hello Fajjal,
I've contrived a solvable 12-deg:
Resultant[x^4+mx^3+(m-1)x+1, m^3-m+1, m]
Eliminating m gives:
x^12-x^10-4x^9+x^8-2x^7+x^6-7x^5+x^4-x^3+2x^2-3x+1 = 0
Can you try to decompose this into a cubic of form:
x^3+f2x^2+f1x+f0 = 0
where f2 is the root of a quartic and f1,f0 are polynomial functions
of f2. If it is possible, this _random_ example would strongly
indicate that 12-degs can be decomposable in two or more ways.
P.S. By the way, in my previous post, I meant to say if deg 2^9 = 512
eqn would need a 73rd root of unity since 2^9-1 = 7*73. (I made a typo
and equated 2^9 = 512 = 7*73. Was sleepy.) Though I believe it will
be only if 2^p-1 is prime.
- Titus |
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| M.A.Fajjal... |
| Posted: Sat Oct 31, 2009 7:13 am |
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[quote]
Hello Fajjal,
I've contrived a solvable 12-deg:
Resultant[x^4+mx^3+(m-1)x+1, m^3-m+1, m]
Eliminating m gives:
x^12-x^10-4x^9+x^8-2x^7+x^6-7x^5+x^4-x^3+2x^2-3x+1 =
0
Can you try to decompose this into a cubic of form:
x^3+f2x^2+f1x+f0 = 0
where f2 is the root of a quartic and f1,f0 are
polynomial functions
of f2. If it is possible, this _random_ example
would strongly
indicate that 12-degs can be decomposable in two or
more ways.
[/quote]
Good Example!
x^12-x^10-4*x^9+x^8-2*x^7+x^6-7*x^5+x^4-x^3+2*x^2-3*x+1
If x^3+f2x^2+f1x+f0 = 0
Then
f2^12-3*f2^10+10*f2^8+12*f2^7+6*f2^6+9*f2^5-13*f2^4-21*f2^3+12*f2^2-5
But If x^4+b2x^3+b1x2+b0
Then b2^3-b2-1 =0 |
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| TPiezas... |
| Posted: Sat Oct 31, 2009 10:39 am |
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On Oct 31, 11:13 am, "M.A.Fajjal" <h2... at (no spam) yahoo.com> wrote:
[quote]Hello Fajjal,
I've contrived a solvable 12-deg:
Resultant[x^4+mx^3+(m-1)x+1, m^3-m+1, m]
Eliminating m gives:
x^12-x^10-4x^9+x^8-2x^7+x^6-7x^5+x^4-x^3+2x^2-3x+1 > > 0
Can you try to decompose this into a cubic of form:
x^3+f2x^2+f1x+f0 = 0
where f2 is the root of a quartic and f1,f0 are
polynomial functions
of f2. If it is possible, this _random_ example
would strongly
indicate that 12-degs can be decomposable in two or
more ways.
Good Example!
x^12-x^10-4*x^9+x^8-2*x^7+x^6-7*x^5+x^4-x^3+2*x^2-3*x+1
If x^3+f2x^2+f1x+f0 = 0
Then
f2^12-3*f2^10+10*f2^8+12*f2^7+6*f2^6+9*f2^5-13*f2^4-21*f2^3+12*f2^2-5
But If x^4+b2x^3+b1x2+b0
Then b2^3-b2-1 =0- Hide quoted text -
- Show quoted text -
[/quote]
But the f2 is not a quartic. So there are classes to the 12-deg, after
all!
Solving the composite degree pq, its "ideal" decomposition is an eqn
of degree p with coefficients determined by an eqn of degree q, where
the latter has rational coefficients. For some "exceptional" solvable
groups (like 9T26), no ideal decomposition is possible.
However, some groups can be ideally solved in more than one way. For
example, for the sextic x^6-2x^5-2x^4+4x^3-4x+4 = 0, this is either,
a) x^2+mx+(m^2+2m-2) = 0; where m is any root of m^3+2m^2-4m+2 = 0, or
b) x^3+nx^2-nx-2 = 0; where n is any root of n^2+2n-4 = 0
So there are some 12-degs that can be ideally decomposed ONLY as {p,q}
= {4,3}, and some you tested as {3,4}. I believe some will only be
{6,2}, or {2,6}, and some will be in more than one way. Or none at
all. I think there is still a chance there is a solvable 12-deg with
no ideal decomposition.
- Titus |
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| ... |
| Posted: Sun Nov 01, 2009 12:39 pm |
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"Nasser M. Abbasi" schrieb:
[quote]
"M.A.Fajjal" <h2maf at (no spam) yahoo.com> wrote in message
By the way can Mathematica or any other software find radical expression
for cos(2pi/47)
using Mathematica 7:
[...]
But it does tell one that Cos[2*(Pi/47)] is the 23 third root of the
following polynomial:
In[85]:= RootReduce[Cos[2*(Pi/47)]];
p=First[%][x]
Out[86]= 8388608*x^23 + 4194304*x^22 -
46137344*x^21 - 22020096*x^20 +
110100480*x^19 + 49807360*x^18 -
149422080*x^17 - 63504384*x^16 +
127008768*x^15 + 50135040*x^14 -
70189056*x^13 - 25346048*x^12 +
25346048*x^11 + 8200192*x^10 -
5857280*x^9 - 1647360*x^8 + 823680*x^7 +
192192*x^6 - 64064*x^5 - 11440*x^4 +
2288*x^3 + 264*x^2 - 24*x - 1
[/quote]
According to a Derive procedure of mine that checks whether a given
polynomial represents a hypergeometric function, your specimen equals:
= -2F1(-23, 24, 1/2, (1+x)/2)
= 47*2F1(-23, 24, 3/2, (1-x)/2)
= CHEBYCHEV_U(46, sqrt((1+x)/2))
= -CHEBYCHEV_T(47, sqrt((1-x)/2)) / sqrt((1-x)/2)
Martin. |
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| Axel Vogt... |
| Posted: Sun Nov 01, 2009 12:47 pm |
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clicliclic at (no spam) freenet.de wrote:
[quote]"Nasser M. Abbasi" schrieb:
"M.A.Fajjal" <h2maf at (no spam) yahoo.com> wrote in message
By the way can Mathematica or any other software find radical expression
for cos(2pi/47)
using Mathematica 7:
[...]
But it does tell one that Cos[2*(Pi/47)] is the 23 third root of the
following polynomial:
In[85]:= RootReduce[Cos[2*(Pi/47)]];
p=First[%][x]
Out[86]= 8388608*x^23 + 4194304*x^22 -
46137344*x^21 - 22020096*x^20 +
110100480*x^19 + 49807360*x^18 -
149422080*x^17 - 63504384*x^16 +
127008768*x^15 + 50135040*x^14 -
70189056*x^13 - 25346048*x^12 +
25346048*x^11 + 8200192*x^10 -
5857280*x^9 - 1647360*x^8 + 823680*x^7 +
192192*x^6 - 64064*x^5 - 11440*x^4 +
2288*x^3 + 264*x^2 - 24*x - 1
According to a Derive procedure of mine that checks whether a given
polynomial represents a hypergeometric function, your specimen equals:
= -2F1(-23, 24, 1/2, (1+x)/2)
= 47*2F1(-23, 24, 3/2, (1-x)/2)
= CHEBYCHEV_U(46, sqrt((1+x)/2))
= -CHEBYCHEV_T(47, sqrt((1-x)/2)) / sqrt((1-x)/2)
Martin.
[/quote]
Do you do that check by looking whether the coefficients satisfy
the conditions of a hypergeometric series (around 0)? |
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| Posted: Mon Nov 02, 2009 3:28 am |
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Axel Vogt schrieb:
[quote]
clicliclic at (no spam) freenet.de wrote:
"Nasser M. Abbasi" schrieb:
"M.A.Fajjal" <h2maf at (no spam) yahoo.com> wrote in message
By the way can Mathematica or any other software find radical expression
for cos(2pi/47)
using Mathematica 7:
[...]
But it does tell one that Cos[2*(Pi/47)] is the 23 third root of the
following polynomial:
In[85]:= RootReduce[Cos[2*(Pi/47)]];
p=First[%][x]
Out[86]= 8388608*x^23 + 4194304*x^22 -
46137344*x^21 - 22020096*x^20 +
110100480*x^19 + 49807360*x^18 -
149422080*x^17 - 63504384*x^16 +
127008768*x^15 + 50135040*x^14 -
70189056*x^13 - 25346048*x^12 +
25346048*x^11 + 8200192*x^10 -
5857280*x^9 - 1647360*x^8 + 823680*x^7 +
192192*x^6 - 64064*x^5 - 11440*x^4 +
2288*x^3 + 264*x^2 - 24*x - 1
According to a Derive procedure of mine that checks whether a given
polynomial represents a hypergeometric function, your specimen equals:
= -2F1(-23, 24, 1/2, (1+x)/2)
= 47*2F1(-23, 24, 3/2, (1-x)/2)
= CHEBYCHEV_U(46, sqrt((1+x)/2))
= -CHEBYCHEV_T(47, sqrt((1-x)/2)) / sqrt((1-x)/2)
Do you do that check by looking whether the coefficients satisfy
the conditions of a hypergeometric series (around 0)?
[/quote]
Yes, with a linear transformation of the argument thrown in. It's no big
deal, I'm simply relying on Derive's equation solver. Here's the
procedure code, call, and result:
poly_gauss(p,x,n,v,s):=PROG(n:=POLY_DEGREE(p,x),v:=VECTOR(POLY_C~
OEFF(SUBST(p,x,(x-q_)/p_),x,i_),i_,0,n),s:=SOLUTIONS(VECTOR(((b_~
+i_-1)*(i_-n-1))*v SUB i_=(c_+i_-1)*i_*v SUB (i_+1),i_,1,n),[b_,~
c_,p_,q_]),RETURN(VECTOR(SUBST(v SUB 1*'GAUSS_SERIES(-n,b_,c_,p_~
*x+q_,n),[b_,c_,p_,q_],t_),t_,s)))
poly_gauss(8388608*x^23+4194304*x^22-46137344*x^21-22020096*x^20~
+110100480*x^19+49807360*x^18-149422080*x^17-63504384*x^16+12700~
8768*x^15+50135040*x^14-70189056*x^13-25346048*x^12+25346048*x^1~
1+8200192*x^10-5857280*x^9-1647360*x^8+823680*x^7+192192*x^6-640~
64*x^5-11440*x^4+2288*x^3+264*x^2-24*x-1,x)
[-GAUSS_SERIES(-23,24,1/2,x/2+1/2,23),47*GAUSS_SERIES(-23,24,3/2~
,1/2-x/2,23)]
My Chebychev equivalents were based on handbook formulae. So far I was
to lazy to generalize this procedure to arbitrary pFq; instead, I have
variants for 1F1, 3F2, etc.
Nasser's polynomial doesn't seem to decompose over the rationals.
Martin.
PS: Have you tried the new two dimensional integration problem? Is it as
bad as last year's problem? Really impossible on Maple after one year of
subconscious processing? |
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| Axel Vogt... |
| Posted: Mon Nov 02, 2009 10:54 am |
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Guest
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clicliclic at (no spam) freenet.de wrote:
[quote]Axel Vogt schrieb:
clicliclic at (no spam) freenet.de wrote:
"Nasser M. Abbasi" schrieb:
"M.A.Fajjal" <h2maf at (no spam) yahoo.com> wrote in message
By the way can Mathematica or any other software find radical expression
for cos(2pi/47)
using Mathematica 7:
[...]
But it does tell one that Cos[2*(Pi/47)] is the 23 third root of the
following polynomial:
In[85]:= RootReduce[Cos[2*(Pi/47)]];
p=First[%][x]
Out[86]= 8388608*x^23 + 4194304*x^22 -
46137344*x^21 - 22020096*x^20 +
110100480*x^19 + 49807360*x^18 -
149422080*x^17 - 63504384*x^16 +
127008768*x^15 + 50135040*x^14 -
70189056*x^13 - 25346048*x^12 +
25346048*x^11 + 8200192*x^10 -
5857280*x^9 - 1647360*x^8 + 823680*x^7 +
192192*x^6 - 64064*x^5 - 11440*x^4 +
2288*x^3 + 264*x^2 - 24*x - 1
According to a Derive procedure of mine that checks whether a given
polynomial represents a hypergeometric function, your specimen equals:
= -2F1(-23, 24, 1/2, (1+x)/2)
= 47*2F1(-23, 24, 3/2, (1-x)/2)
= CHEBYCHEV_U(46, sqrt((1+x)/2))
= -CHEBYCHEV_T(47, sqrt((1-x)/2)) / sqrt((1-x)/2)
Do you do that check by looking whether the coefficients satisfy
the conditions of a hypergeometric series (around 0)?
Yes, with a linear transformation of the argument thrown in. It's no big
deal, I'm simply relying on Derive's equation solver. Here's the
procedure code, call, and result:
poly_gauss(p,x,n,v,s):=PROG(n:=POLY_DEGREE(p,x),v:=VECTOR(POLY_C~
OEFF(SUBST(p,x,(x-q_)/p_),x,i_),i_,0,n),s:=SOLUTIONS(VECTOR(((b_~
+i_-1)*(i_-n-1))*v SUB i_=(c_+i_-1)*i_*v SUB (i_+1),i_,1,n),[b_,~
c_,p_,q_]),RETURN(VECTOR(SUBST(v SUB 1*'GAUSS_SERIES(-n,b_,c_,p_~
*x+q_,n),[b_,c_,p_,q_],t_),t_,s)))
poly_gauss(8388608*x^23+4194304*x^22-46137344*x^21-22020096*x^20~
+110100480*x^19+49807360*x^18-149422080*x^17-63504384*x^16+12700~
8768*x^15+50135040*x^14-70189056*x^13-25346048*x^12+25346048*x^1~
1+8200192*x^10-5857280*x^9-1647360*x^8+823680*x^7+192192*x^6-640~
64*x^5-11440*x^4+2288*x^3+264*x^2-24*x-1,x)
[-GAUSS_SERIES(-23,24,1/2,x/2+1/2,23),47*GAUSS_SERIES(-23,24,3/2~
,1/2-x/2,23)]
My Chebychev equivalents were based on handbook formulae. So far I was
to lazy to generalize this procedure to arbitrary pFq; instead, I have
variants for 1F1, 3F2, etc.
Nasser's polynomial doesn't seem to decompose over the rationals.
Martin.
PS: Have you tried the new two dimensional integration problem? Is it as
bad as last year's problem? Really impossible on Maple after one year of
subconscious processing?
[/quote]
a) Identifying: thanks, may be one should have such in Maple as well,
there is only a easy way to convert to the usual orthogonal polynomials
b) 2-dim integration: I have had a look at your without success and
have to admit, that I did not re-read the older thread do get some
reasonable idea how to do it with Maple 12.
Axel |
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| TPiezas... |
| Posted: Mon Nov 02, 2009 5:24 pm |
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Guest
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On Oct 31, 2:39 pm, TPiezas <tpie... at (no spam) gmail.com> wrote:
[quote]On Oct 31, 11:13 am, "M.A.Fajjal" <h2... at (no spam) yahoo.com> wrote:
Hello Fajjal,
I've contrived a solvable 12-deg:
Resultant[x^4+mx^3+(m-1)x+1, m^3-m+1, m]
Eliminating m gives:
x^12-x^10-4x^9+x^8-2x^7+x^6-7x^5+x^4-x^3+2x^2-3x+1 > > > 0
Can you try to decompose this into a cubic of form:
x^3+f2x^2+f1x+f0 = 0
where f2 is the root of a quartic and f1,f0 are
polynomial functions
of f2. If it is possible, this _random_ example
would strongly
indicate that 12-degs can be decomposable in two or
more ways.
Good Example!
x^12-x^10-4*x^9+x^8-2*x^7+x^6-7*x^5+x^4-x^3+2*x^2-3*x+1
If x^3+f2x^2+f1x+f0 = 0
Then
f2^12-3*f2^10+10*f2^8+12*f2^7+6*f2^6+9*f2^5-13*f2^4-21*f2^3+12*f2^2-5
But If x^4+b2x^3+b1x2+b0
Then b2^3-b2-1 =0- Hide quoted text -
- Show quoted text -
But the f2 is not a quartic. So there are classes to the 12-deg, after
all!
Solving the composite degree pq, its "ideal" decomposition is an eqn
of degree p with coefficients determined by an eqn of degree q, where
the latter has rational coefficients. For some "exceptional" solvable
groups (like 9T26), no ideal decomposition is possible.
However, some groups can be ideally solved in more than one way. For
example, for the sextic x^6-2x^5-2x^4+4x^3-4x+4 = 0, this is either,
a) x^2+mx+(m^2+2m-2) = 0; where m is any root of m^3+2m^2-4m+2 = 0, or
b) x^3+nx^2-nx-2 = 0; where n is any root of n^2+2n-4 = 0
So there are some 12-degs that can be ideally decomposed ONLY as {p,q}
= {4,3}, and some you tested as {3,4}. I believe some will only be
{6,2}, or {2,6}, and some will be in more than one way. Or none at
all. I think there is still a chance there is a solvable 12-deg with
no ideal decomposition.
- Titus- Hide quoted text -
- Show quoted text -
[/quote]
I just emailed Klueners and he was kind enough to answer that when the
eqn is irreducible but solvable, the exceptional cases happen ONLY
when the deg n is a power of a PRIME. (By the O'Nan-Scott theorem.)
Thus, n = 8,9,16,25,(etc) as we discussed, but none for n = 12. Good
to finally know the answer.
- Tito |
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| ... |
| Posted: Tue Nov 03, 2009 1:05 pm |
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Guest
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Axel Vogt schrieb:
[quote]clicliclic at (no spam) freenet.de wrote:
Axel Vogt schrieb:
clicliclic at (no spam) freenet.de wrote:
"Nasser M. Abbasi" schrieb:
"M.A.Fajjal" <h2maf at (no spam) yahoo.com> wrote in message
By the way can Mathematica or any other software find radical expression
for cos(2pi/47)
using Mathematica 7:
[...]
But it does tell one that Cos[2*(Pi/47)] is the 23 third root of the
following polynomial:
In[85]:= RootReduce[Cos[2*(Pi/47)]];
p=First[%][x]
Out[86]= 8388608*x^23 + 4194304*x^22 -
46137344*x^21 - 22020096*x^20 +
110100480*x^19 + 49807360*x^18 -
149422080*x^17 - 63504384*x^16 +
127008768*x^15 + 50135040*x^14 -
70189056*x^13 - 25346048*x^12 +
25346048*x^11 + 8200192*x^10 -
5857280*x^9 - 1647360*x^8 + 823680*x^7 +
192192*x^6 - 64064*x^5 - 11440*x^4 +
2288*x^3 + 264*x^2 - 24*x - 1
According to a Derive procedure of mine that checks whether a given
polynomial represents a hypergeometric function, your specimen equals:
= -2F1(-23, 24, 1/2, (1+x)/2)
= 47*2F1(-23, 24, 3/2, (1-x)/2)
= CHEBYCHEV_U(46, sqrt((1+x)/2))
= -CHEBYCHEV_T(47, sqrt((1-x)/2)) / sqrt((1-x)/2)
Do you do that check by looking whether the coefficients satisfy
the conditions of a hypergeometric series (around 0)?
Yes, with a linear transformation of the argument thrown in. It's no big
deal, I'm simply relying on Derive's equation solver. Here's the
procedure code, call, and result:
poly_gauss(p,x,n,v,s):=PROG(n:=POLY_DEGREE(p,x),v:=VECTOR(POLY_C~
OEFF(SUBST(p,x,(x-q_)/p_),x,i_),i_,0,n),s:=SOLUTIONS(VECTOR(((b_~
+i_-1)*(i_-n-1))*v SUB i_=(c_+i_-1)*i_*v SUB (i_+1),i_,1,n),[b_,~
c_,p_,q_]),RETURN(VECTOR(SUBST(v SUB 1*'GAUSS_SERIES(-n,b_,c_,p_~
*x+q_,n),[b_,c_,p_,q_],t_),t_,s)))
poly_gauss(8388608*x^23+4194304*x^22-46137344*x^21-22020096*x^20~
+110100480*x^19+49807360*x^18-149422080*x^17-63504384*x^16+12700~
8768*x^15+50135040*x^14-70189056*x^13-25346048*x^12+25346048*x^1~
1+8200192*x^10-5857280*x^9-1647360*x^8+823680*x^7+192192*x^6-640~
64*x^5-11440*x^4+2288*x^3+264*x^2-24*x-1,x)
[-GAUSS_SERIES(-23,24,1/2,x/2+1/2,23),47*GAUSS_SERIES(-23,24,3/2~
,1/2-x/2,23)]
My Chebychev equivalents were based on handbook formulae. So far I was
to lazy to generalize this procedure to arbitrary pFq; instead, I have
variants for 1F1, 3F2, etc.
Nasser's polynomial doesn't seem to decompose over the rationals.
a) Identifying: thanks, may be one should have such in Maple as well,
there is only a easy way to convert to the usual orthogonal polynomials
[/quote]
Then, Maple should have been able to "identify" a Jacobi-polynomial
here. A very similar procedure can be used to "identify" any
hypergeometric function from its Taylor expansion, as Martin Rubey has
pointed out a while ago. Note that the equation solver in the procedure
must be able to handle polynomial equations.
Martin. |
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