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Science Forum Index » Math - Symbolic Forum » An exact simplification challenge - 9
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| Vladimir Bondarenko |
 Posted: Sat Apr 21, 2007 12:21 am |
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Hello all computer algebra buffs,
None of the current generation computer algebra systems can
simplify this expression directly
6*sin(Pi/14)*(1+sqrt(14-3*sqrt(21))
*cos(arccot(3*sqrt(3))/6)-sqrt(14+3*sqrt(21))
*sin(arccot(3*sqrt(3))/6))/(15+2*sqrt(14-3*sqrt(21))
*cos(arccot(3*sqrt(3))/6)-3*sqrt(21)
*cos(arccot(3*sqrt(3))/3)-2*sqrt(14+3*sqrt(21))
*sin(arccot(3*sqrt(3))/6)-sqrt(7)
*sin(arccot(3*sqrt(3))/3))
Is there a soul who can using a sequence of a CAS commands
get to the nice exact answer?
Best wishes,
Vladimir Bondarenko
VM and GEMM architect
Co-founder, CEO, Mathematical Director
http://www.cybertester.com/ Cyber Tester, LLC
http://maple.bug-list.org/ Maple Bugs Encyclopaedia
http://www.CAS-testing.org/ CAS Testing |
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| Mate |
| Posted: Sat Apr 21, 2007 9:13 am |
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On Apr 21, 8:21 am, Vladimir Bondarenko <v...@cybertester.com> wrote:
Quote: Hello all computer algebra buffs,
None of the current generation computer algebra systems can
simplify this expression directly
6*sin(Pi/14)*(1+sqrt(14-3*sqrt(21))
*cos(arccot(3*sqrt(3))/6)-sqrt(14+3*sqrt(21))
*sin(arccot(3*sqrt(3))/6))/(15+2*sqrt(14-3*sqrt(21))
*cos(arccot(3*sqrt(3))/6)-3*sqrt(21)
*cos(arccot(3*sqrt(3))/3)-2*sqrt(14+3*sqrt(21))
*sin(arccot(3*sqrt(3))/6)-sqrt(7)
*sin(arccot(3*sqrt(3))/3))
Hint:
Substitute each sin(...) cos(...) by RootOfs, e.g.
cos(1/6*arccot(3*3^(1/2)))=RootOf(28672*_Z^12-86016*_Z^10+96768*_Z^8-50176*_Z^6+11760*_Z^4-1008*_Z^2+1,1);
Then simplify.
Mate |
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| CW |
| Posted: Sat Apr 21, 2007 9:25 am |
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{cos(1/14*Pi) = 1/2*RootOf(_Z^6-7*_Z^4+14*_Z^2-7), sin(1/14*Pi) =
5/2-5/2*RootOf(_Z^6-7*_Z^4+14*_Z^2-7)^2+1/2*RootOf(_Z^6-7*_Z^4+14*_Z^2-7)^4,
sin(1/6*arccot(3*3^(1/2))) =
RootOf(RootOf(7*_Z^6-42*_Z^4+63*_Z^2-14-20*RootOf(7*_Z^4-13*_Z^2+7)+7*RootOf(7*_Z^4-13*_Z^2+7)^3)^2+4*_Z^2-4),
cos(1/6*arccot(3*3^(1/2))) =
1/2*RootOf(7*_Z^6-42*_Z^4+63*_Z^2-14-20*RootOf(7*_Z^4-13*_Z^2+7)+7*RootOf(7*_Z^4-13*_Z^2+7)^3),
cos(1/3*arccot(3*3^(1/2))) =
1/2*RootOf(7*_Z^3+7*RootOf(7*_Z^4-13*_Z^2+7)^3-20*RootOf(7*_Z^4-13*_Z^2+7)-21*_Z),
sin(1/3*arccot(3*3^(1/2))) =
RootOf(RootOf(7*_Z^3+7*RootOf(7*_Z^4-13*_Z^2+7)^3-20*RootOf(7*_Z^4-13*_Z^2+7)-21*_Z)^2+4*_Z^2-4)}
to be more precise.
Mate wrote:
Quote:
On Apr 21, 8:21 am, Vladimir Bondarenko <v...@cybertester.com> wrote:
Hello all computer algebra buffs,
None of the current generation computer algebra systems can
simplify this expression directly
6*sin(Pi/14)*(1+sqrt(14-3*sqrt(21))
*cos(arccot(3*sqrt(3))/6)-sqrt(14+3*sqrt(21))
*sin(arccot(3*sqrt(3))/6))/(15+2*sqrt(14-3*sqrt(21))
*cos(arccot(3*sqrt(3))/6)-3*sqrt(21)
*cos(arccot(3*sqrt(3))/3)-2*sqrt(14+3*sqrt(21))
*sin(arccot(3*sqrt(3))/6)-sqrt(7)
*sin(arccot(3*sqrt(3))/3))
Hint:
Substitute each sin(...) cos(...) by RootOfs, e.g.
cos(1/6*arccot(3*3^(1/2)))=RootOf(28672*_Z^12-86016*_Z^10+96768*_Z^8-50176*_Z^6+11760*_Z^4-1008*_Z^2+1,1);
Then simplify.
Mate |
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| Vladimir Bondarenko |
| Posted: Sat Apr 21, 2007 10:21 am |
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On Apr 21, 7:13 am, Mate <mmat...@personal.ro> wrote:
M> Substitute each sin(...) cos(...) by RootOfs, e.g.
Name me a little green man or a thinking cuttlefish
had the original challenge wasn't
"using a sequence of a CAS commands get to the
nice exact answer"
However, I admit that I really didn't use "and display
these commands" wording explicitly...
.... while you ARE a genuine math man ;)
Quote: On Apr 21, 8:21 am, Vladimir Bondarenko <v...@cybertester.com> wrote:
Hello all computer algebra buffs,
None of the current generation computer algebra systems can
simplify this expression directly
6*sin(Pi/14)*(1+sqrt(14-3*sqrt(21))
*cos(arccot(3*sqrt(3))/6)-sqrt(14+3*sqrt(21))
*sin(arccot(3*sqrt(3))/6))/(15+2*sqrt(14-3*sqrt(21))
*cos(arccot(3*sqrt(3))/6)-3*sqrt(21)
*cos(arccot(3*sqrt(3))/3)-2*sqrt(14+3*sqrt(21))
*sin(arccot(3*sqrt(3))/6)-sqrt(7)
*sin(arccot(3*sqrt(3))/3))
Hint:
Substitute each sin(...) cos(...) by RootOfs, e.g.
cos(1/6*arccot(3*3^(1/2)))=RootOf(28672*_Z^12-86016*_Z^10+96768*_Z^8-50176*_Z^6+11760*_Z^4-1008*_Z^2+1,1);
Then simplify.
Mate |
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| dimitris |
| Posted: Sat Apr 21, 2007 1:37 pm |
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In the same vein...
In[11]:ex(6*Sin[Pi/14]*(1 + Sqrt[14 - 3*Sqrt[21]]*Cos[(1/6)*ArcCot[3*Sqrt[3]]]
- Sqrt[14 + 3*Sqrt[21]]*Sin[(1/6)*ArcCot[3*Sqrt[3]]]))/(15 + 2*Sqrt[14
- 3*Sqrt[21]]*Cos[(1/6)*ArcCot[3*Sqrt[3]]] -
3*Sqrt[21]*Cos[(1/3)*ArcCot[3*Sqrt[3]]] -
2*Sqrt[14 + 3*Sqrt[21]]*Sin[(1/6)*ArcCot[3*Sqrt[3]]] -
Sqrt[7]*Sin[(1/3)*ArcCot[3*Sqrt[3]]]);
In[12]:Timing[FullSimplify[ex1, TransformationFunctions -> {Automatic,
TrigToExp, RootReduce}]]
Out[12]{41.827999999999996*Second, 1}
(*Note that Simplify can't be used instead of FullSimplify in In[12]*)
Dimitris
Ï/Ç Vladimir Bondarenko Ýãñáøå:
Quote: Hello all computer algebra buffs,
None of the current generation computer algebra systems can
simplify this expression directly
6*sin(Pi/14)*(1+sqrt(14-3*sqrt(21))
*cos(arccot(3*sqrt(3))/6)-sqrt(14+3*sqrt(21))
*sin(arccot(3*sqrt(3))/6))/(15+2*sqrt(14-3*sqrt(21))
*cos(arccot(3*sqrt(3))/6)-3*sqrt(21)
*cos(arccot(3*sqrt(3))/3)-2*sqrt(14+3*sqrt(21))
*sin(arccot(3*sqrt(3))/6)-sqrt(7)
*sin(arccot(3*sqrt(3))/3))
Is there a soul who can using a sequence of a CAS commands
get to the nice exact answer?
Best wishes,
Vladimir Bondarenko
VM and GEMM architect
Co-founder, CEO, Mathematical Director
http://www.cybertester.com/ Cyber Tester, LLC
http://maple.bug-list.org/ Maple Bugs Encyclopaedia
http://www.CAS-testing.org/ CAS Testing |
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| G. A. Edgar |
| Posted: Sat Apr 21, 2007 2:00 pm |
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In article <1177171466.927649.192630@b75g2000hsg.googlegroups.com>,
Daniel Lichtblau <danl@wolfram.com> wrote:
Quote:
ee = 6*Sin [
Pi/14]*(1 + Sqrt [14 - 3*Sqrt [21]]*Cos [ArcCot [3*Sqrt [3]]/6] -
Sqrt [14 + 3*Sqrt [21]]*Sin [ArcCot [3*Sqrt [3]]/6])/(15 +
2*Sqrt [14 - 3*Sqrt [21]]*Cos [ArcCot [3*Sqrt [3]]/6] -
3*Sqrt [21]*Cos [ArcCot [3*Sqrt [3]]/3] -
2*Sqrt [14 + 3*Sqrt [21]]*Sin [ArcCot [3*Sqrt [3]]/6] -
Sqrt [7]*Sin [ArcCot [3*Sqrt [3]]/3]);
This would be a set of commands corresponding to what previous
respondents indicated.
In[10]:= RootReduce[TrigToExp[ee]]
Out[10]= 1
Daniel Lichtblau
Wolfram Research
Copying this in Maple...
e :=
6*sin(Pi/14)*(1+sqrt(14-3*sqrt(21))
*cos(arccot(3*sqrt(3))/6)-sqrt(14+3*sqrt(21))
*sin(arccot(3*sqrt(3))/6))/(15+2*sqrt(14-3*sqrt(21))
*cos(arccot(3*sqrt(3))/6)-3*sqrt(21)
*cos(arccot(3*sqrt(3))/3)-2*sqrt(14+3*sqrt(21))
*sin(arccot(3*sqrt(3))/6)-sqrt(7)
*sin(arccot(3*sqrt(3))/3)):
simplify(convert(ee,exp));
1
The time was 161 seconds on my Mac, nearly 3 minutes.
--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/ |
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| CW |
| Posted: Sat Apr 21, 2007 2:37 pm |
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A better description :
alias(alpha[1] =
RootOf(2*_Z^2-1+RootOf(28*_Z^2-27-12*RootOf(8*_Z^3-4*_Z^2-4*_Z+1,.2225209340)+36*RootOf(8*_Z^3-4*_Z^2-4*_Z+1,.2225209340)^2,.9979924640),.3168229788e-1));
alias(alpha[2] =
RootOf(28*_Z^2-1+12*RootOf(8*_Z^3-4*_Z^2-4*_Z+1,.2225209340)-36*RootOf(8*_Z^3-4*_Z^2-4*_Z+1,.2225209340)^2,.6333278610e-1));
alias(alpha[3] = RootOf(8*_Z^3-4*_Z^2-4*_Z+1,.2225209340));
alias(alpha[4] =
RootOf(28*_Z^2-27-12*alpha[3]+36*alpha[3]^2,.9979924640));
[cos(1/3*arccot(3*3^(1/2))) = alpha[4], sin(1/14*Pi) = alpha[3],
sin(1/6*arccot(3*3^(1/2))) = alpha[1], sin(1/3*arccot(3*3^(1/2))) =
alpha[2], cos(1/6*arccot(3*3^(1/2))) =
244*alpha[1]*alpha[4]*alpha[2]+120*alpha[1]*alpha[4]*alpha[3]*alpha[2]-432*alpha[1]*alpha[4]*alpha[3]^2*alpha[2]+244*alpha[1]*alpha[2]+120*alpha[1]*alpha[3]*alpha[2]-432*alpha[1]*alpha[3]^2*alpha[2]];
Chris |
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| CW |
| Posted: Sat May 05, 2007 5:11 pm |
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These substitiutions work best :
A:=6*sin(Pi/14)*(1+sqrt(14-3*sqrt(21))
*cos(arccot(3*sqrt(3))/6)-sqrt(14+3*sqrt(21))
*sin(arccot(3*sqrt(3))/6))/(15+2*sqrt(14-3*sqrt(21))
*cos(arccot(3*sqrt(3))/6)-3*sqrt(21)
*cos(arccot(3*sqrt(3))/3)-2*sqrt(14+3*sqrt(21))
*sin(arccot(3*sqrt(3))/6)-sqrt(7)
*sin(arccot(3*sqrt(3))/3));
sub := {7^(1/2) = r1, sin(1/3*arccot(3*3^(1/2))) = -1/14*r1+3/7*r1*r3,
(14-3*21^(1/2))^(1/2) = r2, sin(1/14*Pi) = r3,
cos(1/3*arccot(3*3^(1/2))) = 3/14*r4+1/7*r4*r3-4/7*r4*r3^2, 21^(1/2) =
r4, (14+3*21^(1/2))^(1/2) = 2*r1*r2+3/7*r4*r1*r2,
cos(1/6*arccot(3*3^(1/2))) =
13/14*r2-4/7*r4*r3^2*r2+1/7*r4*r2*r3+6/7*r2*r3-18/7*r3^2*r2+3/14*r4*r2,
sin(1/6*arccot(3*3^(1/2))) =
1/14*r1*r2-4/49*r1*r4*r3*r2-2/49*r1*r4*r3^2*r2-3/7*r1*r3*r2+3/98*r4*r1*r2},
[r1 = RootOf(_Z^2-7), r2 = RootOf(_Z^2-14+3*RootOf(_Z^2-21)), r3 =
RootOf(8*_Z^3-4*_Z^2-4*_Z+1,.2225209340), r4 = RootOf(_Z^2-21)];
(evala@Expand@subs)(sub,A);
A :=
6*sin(1/14*Pi)*(1+(14-3*21^(1/2))^(1/2)*cos(1/6*arccot(3*3^(1/2)))-(14+3*21^(1/2))^(1/2)*sin(1/6*arccot(3*3^(1/2))))/(15+2*(14-3*21^(1/2))^(1/2)*cos(1/6*arccot(3*3^(1/2)))-3*21^(1/2)*cos(1/3*arccot(3*3^(1/2)))-2*(14+3*21^(1/2))^(1/2)*sin(1/6*arccot(3*3^(1/2)))-7^(1/2)*sin(1/3*arccot(3*3^(1/2))))
sub := {7^(1/2) = r1, sin(1/3*arccot(3*3^(1/2))) = -1/14*r1+3/7*r1*r3,
(14-3*21^(1/2))^(1/2) = r2, sin(1/14*Pi) = r3,
cos(1/3*arccot(3*3^(1/2))) = 3/14*r4+1/7*r4*r3-4/7*r4*r3^2, 21^(1/2) =
r4, (14+3*21^(1/2))^(1/2) = 2*r1*r2+3/7*r4*r1*r2,
cos(1/6*arccot(3*3^(1/2))) =
13/14*r2-4/7*r4*r3^2*r2+1/7*r4*r2*r3+6/7*r2*r3-18/7*r3^2*r2+3/14*r4*r2,
sin(1/6*arccot(3*3^(1/2))) =
1/14*r1*r2-4/49*r1*r4*r3*r2-2/49*r1*r4*r3^2*r2-3/7*r1*r3*r2+3/98*r4*r1*r2},
[r1 = RootOf(_Z^2-7), r2 = RootOf(_Z^2-14+3*RootOf(_Z^2-21)), r3 =
RootOf(8*_Z^3-4*_Z^2-4*_Z+1,.2225209340), r4 = RootOf(_Z^2-21)]
1
Chris
Mate wrote:
Quote:
On Apr 21, 8:21 am, Vladimir Bondarenko <v...@cybertester.com> wrote:
Hello all computer algebra buffs,
None of the current generation computer algebra systems can
simplify this expression directly
6*sin(Pi/14)*(1+sqrt(14-3*sqrt(21))
*cos(arccot(3*sqrt(3))/6)-sqrt(14+3*sqrt(21))
*sin(arccot(3*sqrt(3))/6))/(15+2*sqrt(14-3*sqrt(21))
*cos(arccot(3*sqrt(3))/6)-3*sqrt(21)
*cos(arccot(3*sqrt(3))/3)-2*sqrt(14+3*sqrt(21))
*sin(arccot(3*sqrt(3))/6)-sqrt(7)
*sin(arccot(3*sqrt(3))/3))
Hint:
Substitute each sin(...) cos(...) by RootOfs, e.g.
cos(1/6*arccot(3*3^(1/2)))=RootOf(28672*_Z^12-86016*_Z^10+96768*_Z^8-50176*_Z^6+11760*_Z^4-1008*_Z^2+1,1);
Then simplify.
Mate |
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| Thomas Mautsch |
| Posted: Sun May 06, 2007 2:42 pm |
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Quote: On Apr 21, 8:21 am, Vladimir Bondarenko wrote:
None of the current generation computer algebra systems can
simplify this expression directly
6*sin(Pi/14)*(1+sqrt(14-3*sqrt(21))
*cos(arccot(3*sqrt(3))/6)-sqrt(14+3*sqrt(21))
*sin(arccot(3*sqrt(3))/6))/(15+2*sqrt(14-3*sqrt(21))
*cos(arccot(3*sqrt(3))/6)-3*sqrt(21)
*cos(arccot(3*sqrt(3))/3)-2*sqrt(14+3*sqrt(21))
*sin(arccot(3*sqrt(3))/6)-sqrt(7)
*sin(arccot(3*sqrt(3))/3))
Mate wrote:
Hint:
Substitute each sin(...) cos(...) by RootOfs,
[ ... ]
Then simplify.
In news:<463D012B.4D593337@ns.sympatico.ca> schrieb CW ("Chris"):
Quote: These substitiutions work best :
[ ... ]
Why so complicated??
simplify(combine(%));
is more than sufficient. - Result: 1
It is only a pity that this way
one cannot see that the formula rests
on the fact that sin(Pi/14) is the root of a cubic equation
and on a solution formula for such equations. |
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| CW |
| Posted: Mon May 07, 2007 12:44 am |
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Quote:
Why so complicated??
simplify(combine(%));
is more than sufficient. - Result: 1
That code produces no simplifications of Maple V R4...
Converting to algebraic field extension does however.
Chris |
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