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Science Forum Index » Math - Symbolic Forum » elliptic integral
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Message |
| dimitris |
 Posted: Fri May 11, 2007 10:26 am |
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Guest
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Hello to all!
I have the definite integral
o = HoldForm[Integrate[Sqrt[(1 - x)/((x - 2)*(x^2 - 2*x + 3))], {x, 1,
2}]]
Working with Mathematica...
Version 4. succeeds in getting a closed form result
(agreement with NIntegrate).
Version 5.2 returns the integral unevalueated.
Any ideas on how I could obtain an analytic result
(by CAS or hand)?
Thanks a lot!
Dimitris |
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| G. A. Edgar |
| Posted: Fri May 11, 2007 1:40 pm |
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Guest
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In article <1178897179.566923.120800@u30g2000hsc.googlegroups.com>,
dimitris <dimmechan@yahoo.com> wrote:
Quote: Hello to all!
I have the definite integral
o = HoldForm[Integrate[Sqrt[(1 - x)/((x - 2)*(x^2 - 2*x + 3))], {x, 1,
2}]]
Working with Mathematica...
Version 4. succeeds in getting a closed form result
(agreement with NIntegrate).
Version 5.2 returns the integral unevalueated.
Any ideas on how I could obtain an analytic result
(by CAS or hand)?
Thanks a lot!
Dimitris
Maple 10...
J1 := Int(sqrt((1-x)/(x-2)/(x^2-2*x+3)),x=1..2);
/2 (1/2)
| / 1 - x \
| |----------------------| dx
| | / 2 \|
/1 \(x - 2) \x - 2 x + 3//
evalf(J1);
0.9752615369
with(student):
changevar(y=x-1,J1,y):
simplify(value(simplify(%))):
lprint(%);
1/6*2^(1/4)*3^(1/4)*(-108*EllipticK(1/2*2^(1/2)/(3+3^(1/2)*2^(1/2))^(1/2)
)+44*3^(1/2)*2^(1/2)*EllipticK(1/2*2^(1/2)/(3+3^(1/2)*2^(1/2))^(1/2))-12
*EllipticPi(1/24*3^(1/2)*2^(1/2)*(2*3^(1/2)*2^(1/2)-5),
1/2*2^(1/2)/(3+3^(1/2)*2^(1/2))^(1/2))+5*EllipticPi(1/24*3^(1/2)*2^(1/2)
*(2*3^(1/2)*2^(1/2)-5),
1/2*2^(1/2)/(3+3^(1/2)*2^(1/2))^(1/2))*3^(1/2)*2^(1/2))/(2*3^(1/2)*2^(1/
2)-5)^2
evalf(%);
0.9752626767
--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/ |
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| David W. Cantrell |
| Posted: Fri May 11, 2007 6:22 pm |
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Guest
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dimitris <dimmechan@yahoo.com> wrote:
Quote: Hello to all!
I have the definite integral
o = HoldForm[Integrate[Sqrt[(1 - x)/((x - 2)*(x^2 - 2*x + 3))], {x, 1, 2}]]
Working with Mathematica...
Version 4. succeeds in getting a closed form result
(agreement with NIntegrate).
Version 5.2 returns the integral unevalueated.
Any ideas on how I could obtain an analytic result
(by CAS or hand)?
Huh? Didn't you _already_ get such a result using Mathematica 4?
But anyway, here's a way to get such a result using Mathematica 5.2:
In[4]:=
Integrate[(x - 1)/Sqrt[(1 - x)*(x - 2)*(x^2 - 2*x + 3)], {x, 1, 3/2}] +
Integrate[Sqrt[(1 - x)/((x - 2)*(x^2 - 2*x + 3))], {x, 3/2, 2}]
Out[4]=
EllipticPi[1 - I/Sqrt[2], ArcSin[Root[2 + 4*#1^2 + 3*#1^4 & , 3]],
(1/3)*(1 - 2*I*Sqrt[2])]*Root[8 + 8*#1^2 + 3*#1^4 & , 4] +
(EllipticF[ArcSin[Root[3 + 4*#1^2 + 2*#1^4 & , 3]], (1/3)*(1 -
2*I*Sqrt[2])] - EllipticPi[(1/3)*(2 - I*Sqrt[2]), ArcSin[Root[3 + 4*#1^2
+ 2*#1^4 & , 3]], (1/3)*(1 - 2*I*Sqrt[2])])*Root[8 + 8*#1^2 + 3*#1^4 & , 4]
In[5]:= Chop[N[%]]
Out[5]= 0.9752615369238598
Of course, I'm curious to know how Mathematica 6 does with this integral.
After all, writing it as I did in In[4] should not be necessary. We should
be able to ask straightforwardly for
Integrate[Sqrt[(1 - x)/((x - 2)*(x^2 - 2*x + 3))], {x, 1, 2}]
and get a result.
David |
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| dimitris |
| Posted: Sat May 12, 2007 1:56 am |
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Guest
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Hi David!
Your solution has a lot of magic; doesn't it?
Ok I even tried by myself to write the integrand as [(x - 1)/Sqrt[(1 -
x)*(x - 2)*(x^2 - 2*x + 3)]
but 3/2???
How you came to the conclusion and you broke the integral?
Do I miss something or it was a tactic based to your experience?
Dimitris
Ï/Ç David W. Cantrell Ýãñáøå:
Quote: dimitris <dimmechan@yahoo.com> wrote:
Hello to all!
I have the definite integral
o = HoldForm[Integrate[Sqrt[(1 - x)/((x - 2)*(x^2 - 2*x + 3))], {x, 1, 2}]]
Working with Mathematica...
Version 4. succeeds in getting a closed form result
(agreement with NIntegrate).
Version 5.2 returns the integral unevalueated.
Any ideas on how I could obtain an analytic result
(by CAS or hand)?
Huh? Didn't you _already_ get such a result using Mathematica 4?
But anyway, here's a way to get such a result using Mathematica 5.2:
In[4]:> Integrate[(x - 1)/Sqrt[(1 - x)*(x - 2)*(x^2 - 2*x + 3)], {x, 1, 3/2}] +
Integrate[Sqrt[(1 - x)/((x - 2)*(x^2 - 2*x + 3))], {x, 3/2, 2}]
Out[4]> EllipticPi[1 - I/Sqrt[2], ArcSin[Root[2 + 4*#1^2 + 3*#1^4 & , 3]],
(1/3)*(1 - 2*I*Sqrt[2])]*Root[8 + 8*#1^2 + 3*#1^4 & , 4] +
(EllipticF[ArcSin[Root[3 + 4*#1^2 + 2*#1^4 & , 3]], (1/3)*(1 -
2*I*Sqrt[2])] - EllipticPi[(1/3)*(2 - I*Sqrt[2]), ArcSin[Root[3 + 4*#1^2
+ 2*#1^4 & , 3]], (1/3)*(1 - 2*I*Sqrt[2])])*Root[8 + 8*#1^2 + 3*#1^4 & , 4]
In[5]:= Chop[N[%]]
Out[5]= 0.9752615369238598
Of course, I'm curious to know how Mathematica 6 does with this integral.
After all, writing it as I did in In[4] should not be necessary. We should
be able to ask straightforwardly for
Integrate[Sqrt[(1 - x)/((x - 2)*(x^2 - 2*x + 3))], {x, 1, 2}]
and get a result.
David |
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| dimitris |
| Posted: Sat May 12, 2007 2:13 am |
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Guest
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Thank you vry much!
BTW, in Mathematica 5.2 the substitution you proposed (y=x-1)
leads to a much simplify integral which however stays again
unevaluated!
Dimitris
Ï/Ç G. A. Edgar Ýãñáøå:
Quote: In article <1178897179.566923.120800@u30g2000hsc.googlegroups.com>,
dimitris <dimmechan@yahoo.com> wrote:
Hello to all!
I have the definite integral
o = HoldForm[Integrate[Sqrt[(1 - x)/((x - 2)*(x^2 - 2*x + 3))], {x, 1,
2}]]
Working with Mathematica...
Version 4. succeeds in getting a closed form result
(agreement with NIntegrate).
Version 5.2 returns the integral unevalueated.
Any ideas on how I could obtain an analytic result
(by CAS or hand)?
Thanks a lot!
Dimitris
Maple 10...
J1 := Int(sqrt((1-x)/(x-2)/(x^2-2*x+3)),x=1..2);
/2 (1/2)
| / 1 - x \
| |----------------------| dx
| | / 2 \|
/1 \(x - 2) \x - 2 x + 3//
evalf(J1);
0.9752615369
with(student):
changevar(y=x-1,J1,y):
simplify(value(simplify(%))):
lprint(%);
1/6*2^(1/4)*3^(1/4)*(-108*EllipticK(1/2*2^(1/2)/(3+3^(1/2)*2^(1/2))^(1/2)
)+44*3^(1/2)*2^(1/2)*EllipticK(1/2*2^(1/2)/(3+3^(1/2)*2^(1/2))^(1/2))-12
*EllipticPi(1/24*3^(1/2)*2^(1/2)*(2*3^(1/2)*2^(1/2)-5),
1/2*2^(1/2)/(3+3^(1/2)*2^(1/2))^(1/2))+5*EllipticPi(1/24*3^(1/2)*2^(1/2)
*(2*3^(1/2)*2^(1/2)-5),
1/2*2^(1/2)/(3+3^(1/2)*2^(1/2))^(1/2))*3^(1/2)*2^(1/2))/(2*3^(1/2)*2^(1/
2)-5)^2
evalf(%);
0.9752626767
--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/ |
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| CW |
| Posted: Sat May 12, 2007 6:09 am |
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Guest
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or
-1/3*(18-6*2^(1/2)*3^(1/2))^(1/2)*(44+18*2^(1/2)*3^(1/2))^(1/2)*EllipticK(1/6*(18-6*2^(1/2)*3^(1/2))^(1/2))+(1/12*(18-6*2^(1/2)*3^(1/2))^(1/2)*(44+18*2^(1/2)*3^(1/2))^(1/2)*2^(1/2)*3^(1/2)+1/6*(18-6*2^(1/2)*3^(1/2))^(1/2)*(44+18*2^(1/2)*3^(1/2))^(1/2))*EllipticPi(-5/24*2^(1/2)*3^(1/2)+1/2,1/6*(18-6*2^(1/2)*3^(1/2))^(1/2))
= .975261536923865519;
Chris
"G. A. Edgar" wrote:
Quote:
In article <1178897179.566923.120800@u30g2000hsc.googlegroups.com>,
dimitris <dimmechan@yahoo.com> wrote:
Hello to all!
I have the definite integral
o = HoldForm[Integrate[Sqrt[(1 - x)/((x - 2)*(x^2 - 2*x + 3))], {x, 1,
2}]]
Working with Mathematica...
Version 4. succeeds in getting a closed form result
(agreement with NIntegrate).
Version 5.2 returns the integral unevalueated.
Any ideas on how I could obtain an analytic result
(by CAS or hand)?
Thanks a lot!
Dimitris
Maple 10...
J1 := Int(sqrt((1-x)/(x-2)/(x^2-2*x+3)),x=1..2);
/2 (1/2)
| / 1 - x \
| |----------------------| dx
| | / 2 \|
/1 \(x - 2) \x - 2 x + 3//
evalf(J1);
0.9752615369
with(student):
changevar(y=x-1,J1,y):
simplify(value(simplify(%))):
lprint(%);
1/6*2^(1/4)*3^(1/4)*(-108*EllipticK(1/2*2^(1/2)/(3+3^(1/2)*2^(1/2))^(1/2)
)+44*3^(1/2)*2^(1/2)*EllipticK(1/2*2^(1/2)/(3+3^(1/2)*2^(1/2))^(1/2))-12
*EllipticPi(1/24*3^(1/2)*2^(1/2)*(2*3^(1/2)*2^(1/2)-5),
1/2*2^(1/2)/(3+3^(1/2)*2^(1/2))^(1/2))+5*EllipticPi(1/24*3^(1/2)*2^(1/2)
*(2*3^(1/2)*2^(1/2)-5),
1/2*2^(1/2)/(3+3^(1/2)*2^(1/2))^(1/2))*3^(1/2)*2^(1/2))/(2*3^(1/2)*2^(1/
2)-5)^2
evalf(%);
0.9752626767
--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/ |
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| David W. Cantrell |
| Posted: Sat May 12, 2007 9:25 am |
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Guest
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dimitris <dimmechan@yahoo.com> wrote:
Quote: Hi David!
Your solution has a lot of magic; doesn't it?
Ok I even tried by myself to write the integrand as
[(x - 1)/Sqrt[(1 - x)*(x - 2)*(x^2 - 2*x + 3)]
but 3/2???
How you came to the conclusion and you broke the integral?
Do I miss something or it was a tactic based to your experience?
OK, I'll try to explain briefly. All comments are based on Mathematica 5.2.
1) I noticed that the indefinite integral
Integrate[Sqrt[(1 - x)/((x - 2)*(x^2 - 2*x + 3))], x]
is evaluated. Although the resulting antiderivative cannot be evaluated
directly at x = 2, its limit as x -> 2 evaluates to 0. But regrettably,
neither can it be evaluated directly at x = 1 nor can its limit be
evaluated as x -> 1, for otherwise we would have been able to use
Newton-Leibniz.
2) The above led me to guess that the definite integral from a to 2 would
be evaluated if a were some specific number between 1 and 2, and that
turned out to be true. I chose a = 3/2 merely because it was the simplest
choice.
3) I then sought an antiderivative which would be better behaved at x = 1,
and that led me to try
Integrate[(x - 1)/Sqrt[(1 - x)*(x - 2)*(x^2 - 2*x + 3)], x]
Rather luckily, that succeeded. In fact, the resulting antiderivative can
be evaluated directly at x = 1. But now, neither can it be evaluated
directly at x = 2 nor can its limit be evaluated as x -> 2, for otherwise
we would have been able to use Newton-Leibniz here.
4) The above led me to guess that the corresponding definite integral from
1 to b would be evaluated if b were some specific number between 1 and 2.
Of course, I chose b = a as previously chosen.
5) Add the results form 2) and 4) to get the answer.
David
Quote: =CF/=C7 David W. Cantrell =DD=E3=F1=E1=F8=E5:
dimitris <dimmechan@yahoo.com> wrote:
Hello to all!
I have the definite integral
o = HoldForm[Integrate[Sqrt[(1 - x)/((x - 2)*(x^2 - 2*x + 3))],
{x, 1, 2}]]
Working with Mathematica...
Version 4. succeeds in getting a closed form result
(agreement with NIntegrate).
Version 5.2 returns the integral unevalueated.
Any ideas on how I could obtain an analytic result
(by CAS or hand)?
Huh? Didn't you _already_ get such a result using Mathematica 4?
But anyway, here's a way to get such a result using Mathematica 5.2:
In[4]:=
Integrate[(x - 1)/Sqrt[(1 - x)*(x - 2)*(x^2 - 2*x + 3)], {x, 1, 3/2}] +
Integrate[Sqrt[(1 - x)/((x - 2)*(x^2 - 2*x + 3))], {x, 3/2, 2}]
Out[4]=
EllipticPi[1 - I/Sqrt[2], ArcSin[Root[2 + 4*#1^2 + 3*#1^4 & , 3]],
(1/3)*(1 - 2*I*Sqrt[2])]*Root[8 + 8*#1^2 + 3*#1^4 & , 4] +
(EllipticF[ArcSin[Root[3 + 4*#1^2 + 2*#1^4 & , 3]], (1/3)*(1 -
2*I*Sqrt[2])] - EllipticPi[(1/3)*(2 - I*Sqrt[2]), ArcSin[Root[3 +
4*#1^2 + 2*#1^4 & , 3]], (1/3)*(1 - 2*I*Sqrt[2])])*Root[8 + 8*#1^2 +
3*#1^4 &, 4]
In[5]:= Chop[N[%]]
Out[5]= 0.9752615369238598
Of course, I'm curious to know how Mathematica 6 does with this
integral. After all, writing it as I did in In[4] should not be
necessary. We should be able to ask straightforwardly for
Integrate[Sqrt[(1 - x)/((x - 2)*(x^2 - 2*x + 3))], {x, 1, 2}]
and get a result.
David |
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