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Science Forum Index » Math - Symbolic Forum » An exact simplification challenge - 10
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Message |
| Vladimir Bondarenko |
 Posted: Sun Apr 22, 2007 4:50 am |
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Guest
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Hello all noble and valiant computer algebra fans,
A big surprise, none of modern generation computer
algebra systems is able to simplify this expression
directly
((Pi^2+Pi*(Pi^2-4)^(1/2)-4)^(1/2)
-(Pi*(Pi^2-4)^(1/2)-Pi^2+4)^(1/2))
/(Pi^2-4)^(3/4)
Is there a simplification-focused soul who can using
a sequence of a CAS commands get to the tiny exact
answer? (and show these commands... added for M ;)
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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| Peter Pein |
| Posted: Sun Apr 22, 2007 5:59 am |
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Guest
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Vladimir Bondarenko schrieb:
Quote: Hello all noble and valiant computer algebra fans,
A big surprise, none of modern generation computer
algebra systems is able to simplify this expression
directly
((Pi^2+Pi*(Pi^2-4)^(1/2)-4)^(1/2)
-(Pi*(Pi^2-4)^(1/2)-Pi^2+4)^(1/2))
/(Pi^2-4)^(3/4)
Is there a simplification-focused soul who can using
a sequence of a CAS commands get to the tiny exact
answer? (and show these commands... added for M ;)
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
quite easy...
In Mathematica 5.1:
In[1]:=
expr = ((Pi^2 + Pi*(Pi^2 - 4)^(1/2) - 4)^(1/2) -
(Pi*(Pi^2 - 4)^(1/2) - Pi^2 + 4)^(1/2)) / (Pi^2 - 4)^(3/4);
In[2]:=
possible = Simplify[Flatten[z /. Solve /@ (Pi == x /. Solve[z == expr /. Pi ->
x, x])]];
In[3]:=
Select[possible, N[#1] == N[expr] & ][[1]]
Out[3]=
Sqrt[2/(2 + Pi)]
Peter |
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| G. A. Edgar |
| Posted: Sun Apr 22, 2007 6:01 am |
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Guest
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In article <1177235428.120227.103470@y5g2000hsa.googlegroups.com>,
Vladimir Bondarenko <vb@cybertester.com> wrote:
Quote: Hello all noble and valiant computer algebra fans,
A big surprise, none of modern generation computer
algebra systems is able to simplify this expression
directly
((Pi^2+Pi*(Pi^2-4)^(1/2)-4)^(1/2)
-(Pi*(Pi^2-4)^(1/2)-Pi^2+4)^(1/2))
/(Pi^2-4)^(3/4)
((P^2+P*(P^2-4)^(1/2)-4)^(1/2)
-(P*(P^2-4)^(1/2)-P^2+4)^(1/2))
/(P^2-4)^(3/4)
= 2/(2*P+4)^(1/2)
of course P=Pi is not required.
Quote: Is there a simplification-focused soul who can using
a sequence of a CAS commands get to the tiny exact
answer? (and show these commands... added for M 
Left to M
--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/ |
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| Mate |
| Posted: Sun Apr 22, 2007 8:11 am |
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On Apr 22, 2:01 pm, "G. A. Edgar" <e...@math.ohio-state.edu.invalid>
wrote:
Quote: In article <1177235428.120227.103...@y5g2000hsa.googlegroups.com>,
Vladimir Bondarenko <v...@cybertester.com> wrote:
Hello all noble and valiant computer algebra fans,
A big surprise, none of modern generation computer
algebra systems is able to simplify this expression
directly
((Pi^2+Pi*(Pi^2-4)^(1/2)-4)^(1/2)
-(Pi*(Pi^2-4)^(1/2)-Pi^2+4)^(1/2))
/(Pi^2-4)^(3/4)
((P^2+P*(P^2-4)^(1/2)-4)^(1/2)
-(P*(P^2-4)^(1/2)-P^2+4)^(1/2))
/(P^2-4)^(3/4)
= 2/(2*P+4)^(1/2)
of course P=Pi is not required.
Is there a simplification-focused soul who can using
a sequence of a CAS commands get to the tiny exact
answer? (and show these commands... added for M ;)
Left to M
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
--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/
OK.
s:=solve(Groebner[Basis]([p^2-4-r^4,a^2-p^2-p*r^2+4,b^2-p*r^2+p^2-4,
f*r^3-a+b],plex(a,b,r,p,f))[1],f):
select(x->is(x>0),s) assuming p>2;
2
------------
1/2
(2 p + 4)
Mate |
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| Mate |
| Posted: Sun Apr 22, 2007 8:29 am |
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On Apr 22, 4:11 pm, Mate <mmat...@personal.ro> wrote:
Quote: On Apr 22, 2:01 pm, "G. A. Edgar" <e...@math.ohio-state.edu.invalid
wrote:
In article <1177235428.120227.103...@y5g2000hsa.googlegroups.com>,
Vladimir Bondarenko <v...@cybertester.com> wrote:
Hello all noble and valiant computer algebra fans,
A big surprise, none of modern generation computer
algebra systems is able to simplify this expression
directly
((Pi^2+Pi*(Pi^2-4)^(1/2)-4)^(1/2)
-(Pi*(Pi^2-4)^(1/2)-Pi^2+4)^(1/2))
/(Pi^2-4)^(3/4)
((P^2+P*(P^2-4)^(1/2)-4)^(1/2)
-(P*(P^2-4)^(1/2)-P^2+4)^(1/2))
/(P^2-4)^(3/4)
= 2/(2*P+4)^(1/2)
of course P=Pi is not required.
Is there a simplification-focused soul who can using
a sequence of a CAS commands get to the tiny exact
answer? (and show these commands... added for M ;)
Left to M
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
--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/
OK.
s:=solve(Groebner[Basis]([p^2-4-r^4,a^2-p^2-p*r^2+4,b^2-p*r^2+p^2-4,
f*r^3-a+b],plex(a,b,r,p,f))[1],f):
select(x->is(x>0),s) assuming p>2;
2
------------
1/2
(2 p + 4)
Mate
Correction:
select(x->is(limit(x,p=2)>0),[s]);
2
[------------]
1/2
(2 p + 4)
Mate |
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| Vladimir Bondarenko |
| Posted: Sun Apr 22, 2007 8:48 am |
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;)
http://upload.wikimedia.org/wikipedia/commons/thumb/b/b7/Simeon_Poisson.jpg/511px-Simeon_Poisson.jpg
Life is good for only two things,
discovering mathematics and teaching mathematics.
-- Simon-Denis Poisson
(There was no QA Engineering that time...)
On Apr 22, 6:29 am, Mate <mmat...@personal.ro> wrote:
Quote: On Apr 22, 4:11 pm, Mate <mmat...@personal.ro> wrote:
On Apr 22, 2:01 pm, "G. A. Edgar" <e...@math.ohio-state.edu.invalid
wrote:
In article <1177235428.120227.103...@y5g2000hsa.googlegroups.com>,
Vladimir Bondarenko <v...@cybertester.com> wrote:
Hello all noble and valiant computer algebra fans,
A big surprise, none of modern generation computer
algebra systems is able to simplify this expression
directly
((Pi^2+Pi*(Pi^2-4)^(1/2)-4)^(1/2)
-(Pi*(Pi^2-4)^(1/2)-Pi^2+4)^(1/2))
/(Pi^2-4)^(3/4)
((P^2+P*(P^2-4)^(1/2)-4)^(1/2)
-(P*(P^2-4)^(1/2)-P^2+4)^(1/2))
/(P^2-4)^(3/4)
= 2/(2*P+4)^(1/2)
of course P=Pi is not required.
Is there a simplification-focused soul who can using
a sequence of a CAS commands get to the tiny exact
answer? (and show these commands... added for M ;)
Left to M
Best wishes,
Vladimir Bondarenko
VM and GEMM architect
Co-founder, CEO, Mathematical Director
http://www.cybertester.com/CyberTester, LLC
http://maple.bug-list.org/Maple Bugs Encyclopaedia
http://www.CAS-testing.org/CASTesting
--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/
OK.
s:=solve(Groebner[Basis]([p^2-4-r^4,a^2-p^2-p*r^2+4,b^2-p*r^2+p^2-4,
f*r^3-a+b],plex(a,b,r,p,f))[1],f):
select(x->is(x>0),s) assuming p>2;
2
------------
1/2
(2 p + 4)
Mate
Correction:
select(x->is(limit(x,p=2)>0),[s]);
2
[------------]
1/2
(2 p + 4)
Mate- Hide quoted text -
- Show quoted text - |
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| Daniel Lichtblau |
| Posted: Sun Apr 22, 2007 9:09 am |
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On Apr 22, 5:59 am, Peter Pein <pet...@dordos.net> wrote:
Quote: Vladimir Bondarenko schrieb:
Hello all noble and valiant computer algebra fans,
A big surprise, none of modern generation computer
algebra systems is able to simplify this expression
directly
((Pi^2+Pi*(Pi^2-4)^(1/2)-4)^(1/2)
-(Pi*(Pi^2-4)^(1/2)-Pi^2+4)^(1/2))
/(Pi^2-4)^(3/4)
Is there a simplification-focused soul who can using
a sequence of a CAS commands get to the tiny exact
answer? (and show these commands... added for M ;)
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
quite easy...
In Mathematica 5.1:
In[1]:=
expr = ((Pi^2 + Pi*(Pi^2 - 4)^(1/2) - 4)^(1/2) -
(Pi*(Pi^2 - 4)^(1/2) - Pi^2 + 4)^(1/2)) / (Pi^2 - 4)^(3/4);
In[2]:=
possible = Simplify[Flatten[z /. Solve /@ (Pi == x /. Solve[z == expr /. Pi -
x, x])]];
In[3]:=
Select[possible, N[#1] == N[expr] & ][[1]]
Out[3]=
Sqrt[2/(2 + Pi)]
Peter- Hide quoted text -
- Show quoted text -
While I like this method I have a small preference for using
GroebnerBasis because I think it can be more reliable in explicitly
sorting out polynomial relations between variables.
ee = ((Pi^2 + Pi*(Pi^2 - 4)^(1/2) - 4)^(1/2) - (Pi*(Pi^2 - 4)^(1/2) -
Pi^2 + 4)^(1/2))/ (Pi^2 - 4)^(3/4);
In[32]:= InputForm[Select[
x /. (Solve[First[GroebnerBasis[ee - x /. Pi -> pi, {pi, x}]] == 0,
x] /.
pi -> Pi), Abs[N[# - ee]] < 10^(-3) &]]
Out[32]//InputForm=
{Sqrt[2/(2 + Pi)]}
This is similar to example 4 in the notebook at:
http://www.ima.umn.edu/2006-2007/seminars/index.html#lichtblau
A pdf version is also available there.
I think in some cases one will need to do as Mate did and make all the
intermediate polynomial relations explicit. This introduces
intermediate variables and thus allows for finer control over variable
ordering. I suspect that might sometimes be needed in order to cull
out the desired polynomial(s).
Daniel Lichtblau
Wolfram Research |
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| dimitris |
| Posted: Sun Apr 22, 2007 10:39 am |
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Rather trivial but here goes...
In[3]:eq = e1 == ((Pi^2 + Pi*(Pi^2 - 4)^(1/2) - 4)^(1/2) - (Pi*(Pi^2 -
4)^(1/2) - Pi^2 + 4)^(1/2))/(Pi^2 - 4)^(3/4)
Out[3]e1 == (-Sqrt[4 - Pi^2 + Pi*Sqrt[-4 + Pi^2]] + Sqrt[-4 + Pi^2 +
Pi*Sqrt[-4 + Pi^2]])/(-4 + Pi^2)^(3/4)
In[12]:(FullSimplify[Expand[#1^2]] & ) /@ eq
e1 /. ToRules[Reduce[% && e1 > 0, e1]]
Out[12]e1^2 == 2/(2 + Pi)
Out[13]Sqrt[2/(2 + Pi)]
Dimitris
/ Vladimir Bondarenko :
Quote: Hello all noble and valiant computer algebra fans,
A big surprise, none of modern generation computer
algebra systems is able to simplify this expression
directly
((Pi^2+Pi*(Pi^2-4)^(1/2)-4)^(1/2)
-(Pi*(Pi^2-4)^(1/2)-Pi^2+4)^(1/2))
/(Pi^2-4)^(3/4)
Is there a simplification-focused soul who can using
a sequence of a CAS commands get to the tiny exact
answer? (and show these commands... added for M ;)
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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| dimitris |
| Posted: Sun Apr 22, 2007 12:12 pm |
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Also even though it is not exactly what was asked...
In[1]:e1 = ((Pi^2 + Pi*(Pi^2 - 4)^(1/2) - 4)^(1/2) - (Pi*(Pi^2 - 4)^(1/2) -
Pi^2 + 4)^(1/2))/(Pi^2 - 4)^(3/4);
In[3]:Developer`ZeroQ[e1 - Sqrt[2/(2 + Pi)]]
Out[3]True
(Version->5.2)
Dimitris
Ο/Η dimitris έγραψε:
Quote: Rather trivial but here goes...
In[3]:> eq = e1 == ((Pi^2 + Pi*(Pi^2 - 4)^(1/2) - 4)^(1/2) - (Pi*(Pi^2 -
4)^(1/2) - Pi^2 + 4)^(1/2))/(Pi^2 - 4)^(3/4)
Out[3]> e1 == (-Sqrt[4 - Pi^2 + Pi*Sqrt[-4 + Pi^2]] + Sqrt[-4 + Pi^2 +
Pi*Sqrt[-4 + Pi^2]])/(-4 + Pi^2)^(3/4)
In[12]:> (FullSimplify[Expand[#1^2]] & ) /@ eq
e1 /. ToRules[Reduce[% && e1 > 0, e1]]
Out[12]> e1^2 == 2/(2 + Pi)
Out[13]> Sqrt[2/(2 + Pi)]
Dimitris
Ï/Ç Vladimir Bondarenko Ýãñáøå:
Hello all noble and valiant computer algebra fans,
A big surprise, none of modern generation computer
algebra systems is able to simplify this expression
directly
((Pi^2+Pi*(Pi^2-4)^(1/2)-4)^(1/2)
-(Pi*(Pi^2-4)^(1/2)-Pi^2+4)^(1/2))
/(Pi^2-4)^(3/4)
Is there a simplification-focused soul who can using
a sequence of a CAS commands get to the tiny exact
answer? (and show these commands... added for M ;)
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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| CW |
| Posted: Sun Apr 22, 2007 11:30 pm |
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Guest
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Consider this code :
A:=(RootOf(_Z^2-Pi^2-Pi*RootOf(_Z^2-Pi^2+4)+4)-1/2*RootOf(_Z^2-Pi^2-Pi*RootOf(_Z^2-Pi^2+4)+4)*Pi+1/2*RootOf(_Z^2-Pi^2-Pi*RootOf(_Z^2-Pi^2+4)+4)*RootOf(_Z^2-Pi^2+4))*RootOf(-RootOf(_Z^2-Pi^2+4)+_Z^2*(-8*Pi^2+Pi^4+16));
S:={allvalues}(A);
Is there a command that simplifies all elements of S ?
Chris
Vladimir Bondarenko wrote:
Quote:
Hello all noble and valiant computer algebra fans,
A big surprise, none of modern generation computer
algebra systems is able to simplify this expression
directly
((Pi^2+Pi*(Pi^2-4)^(1/2)-4)^(1/2)
-(Pi*(Pi^2-4)^(1/2)-Pi^2+4)^(1/2))
/(Pi^2-4)^(3/4)
Is there a simplification-focused soul who can using
a sequence of a CAS commands get to the tiny exact
answer? (and show these commands... added for M ;)
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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| Thomas Mautsch |
| Posted: Sun May 06, 2007 5:51 pm |
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In news:<462B8460.D27BCB26@ns.sympatico.ca>
schrieb CW <sylvester7@ns.sympatico.ca>:
Quote: Consider this code :
A:=(RootOf(_Z^2-Pi^2-Pi*RootOf(_Z^2-Pi^2+4)+4)-1/2*RootOf(_Z^2-Pi^2-Pi*RootOf(_Z^2-Pi^2+4)+4)*Pi+1/2*RootOf(_Z^2-Pi^2-Pi*RootOf(_Z^2-Pi^2+4)+4)*RootOf(_Z^2-Pi^2+4))*RootOf(-RootOf(_Z^2-Pi^2+4)+_Z^2*(-8*Pi^2+Pi^4+16));
S:={allvalues}(A);
Is there a command that simplifies all elements of S ?
Something like the following?
{seq(solve(y=i,{pi}),i=eval(S,Pi=pi))}:
eval({seq(solve(i[],y),i=%)},pi=Pi);
2 2
{- -------------, -------------}
1/2 1/2
(4 + 2 Pi) (4 + 2 Pi)
Problem is, I cannot garantee that no solutions may have been lost,
or additionally generated in this calculation.
Given the first, the latter option can be excluded, because of:
S_list := [S[]];
map(x->is(x>0),S_list);
[true, true, false, false, true, false, true, false]
However, according to Maple 8,
being '>0' does not seem to be the same as being 'positive':
map(is,S_list,positive);
[true, true, false, FAIL, FAIL, FAIL, FAIL, false]
??? |
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