Hi David.
I think I have an answer to your problem.
Consider the incomplete integral of the third kind in Mathematica.
In[2]:=
Integrate[1/((1 - n*Sin[o]^2)*Sqrt[1 - m*Sin[o]^2]), {o, 0, e},
GenerateConditions -> False]
Out[2]=
EllipticPi[n,e,m]
Go to Maple now and load the package MmaTranslator execute the command
FromMmaToMaple and in the resulting window asking for translation of
the Mathematica
expression EllipticPi[n,e,m]. Then the Maple equivalent is
EllipticPi(sin(e),n,sqrt(m)).
with(MmaTranslator):
MmaToMaple();
Let's check it!
Mathematica
In[25]:=
NIntegrate[1/((1 - (1/3)*Sin[o]^2)*Sqrt[1 - (1/2)*Sin[o]^2]), {o, 0,
Pi/2},WorkingPrecision->30,PrecisionGoal->20]
EllipticPi[n, e, m] /. {e -> Pi/2, n -> 1/3, m -> 1/2}
N[%,20]
Out[25]=
2.31079499075428181544
Out[26]=
EllipticPi[1/3, 1/2]
Out[27]=
2.3107949907542818154
subs(n=1/3,m=1/2,e=Pi/2,EllipticPi(sin(e),n,sqrt(m)));
1/2
Pi 2
EllipticPi(sin(----), 1/3, ----)
2 2
evalf(%,20);
2.3107949907542818154
Perfectly!
The same procedure can be used for the other elliptic integrals too!
Dimitris
/ David W. Cantrell :
dimitris <dimmechan@yahoo.com> wrote:
Show
Sqrt[2] (
(1 + I) EllipticPi[1/2 - I/2, 1/2]
- (1 - I) EllipticPi[1/2 + I/2, 1/2]
- (1 + I) Sqrt[2] EllipticPi[1 - I, 2]
+ (1 - I) Sqrt[2] EllipticPi[1 + I, 2] )
equal to Pi in Mathematica?
This is indeed a true challenge Vladimir!
But it should be noted that it's not really the same challenge as the one
given using Maple's conventions. That challenge involved two incomplete and
two complete elliptic integrals of the third kind. In contrast, the newly
stated challenge using Mathematica's conventions involves four complete
integrals. Thus, when VB restated the challenge, he did _more_ than merely
converting it from one convention to the other.
What is the general conversion between Maple's convention for an incomplete
elliptic integral of the third kind and Mathematica's convention for an
incomplete elliptic integral of the third kind?
David
/ Vladimir Bondarenko :
On Jun 2, 9:54 am, David W. Cantrell <DWCantr...@sigmaxi.net> writes:
DWC> you need either to state the challenge
DWC> also using Mathematica's conventions
Here you are,
Sqrt[2] (
(1 + I) EllipticPi[1/2 - I/2, 1/2]
- (1 - I) EllipticPi[1/2 + I/2, 1/2]
- (1 + I) Sqrt[2] EllipticPi[1 - I, 2]
+ (1 - I) Sqrt[2] EllipticPi[1 + I, 2]
)
Also, a question for Maple wizards...
Could you come up with a more short solution
than that invented by Wizard (pluralistically
Professor) Edgar?
http://groups.google.com/group/sci.math.symbolic/msg/a7a6ffe9fcc17efe?h
l=en&
(Beware, this could be not easy!)
Vladimir Bondarenko <v...@cybertester.com> wrote:
On Jun 2, 7:20 am, David W. Cantrell <DWCantr...@sigmaxi.net
writes:
DWC> Clearly, you need to specify the convention
DWC> you have in mind
Yes you are absolutely right. ALL Cyber Tester's
challenges use either Maple or Mathematica syntax.
So I assume then that your challenge was stated using Maple's
conventions for elliptic integrals. If you want Mathematica users to
consider your challenge, you need either to state the challenge also
using Mathematica's conventions or else to give a reference which
would inform Mathematica users how to convert from Maple's
conventions to Mathematica's.
David
As of today, in terms of computational powers, say,
minimally, ability to solve math textbooks problems,
there are only two REAL competitors to compare.
Frankly, all the rest CASs are too weak now.
Vladimir Bondarenko <v...@cybertester.com> wrote:
Hello oh you computer algebra adorers,
Is there a CAS lady or a CAS gentleman who
can show in steps how to turn this nightmare
sqrt(2)*(
(-1-I)*EllipticPi(sqrt(2), (1-I)/2, 1/sqrt(2))
+( 1-I)*EllipticPi(sqrt(2), (1+I)/2, 1/sqrt(2))
+(-1+I)*EllipticPi((1+I)/2, 1/sqrt(2))
+( 1+I)*EllipticPi((1-I)/2, 1/sqrt(2))
)
into a Beauty of bright mathematician's eyes?
Anyone asking such questions should know enough to realize that
they are ambiguous unless a notational convention is _specified_!
(This comment also applies to your previous simplification
challenge.) Even for complete elliptic integrals of the first and
second kinds, there are two commonly used conventions, one of
which happens to be used by Mathematica and the other by Maple.
(Example: If the argument of EllipticE is 1/2, the numerical
value which Mathematica gets is 1.350... while Maple would, IIRC,
get 1.237...) And the ambiguity is worse for complete elliptic
integrals of the third kind. And the ambiguity is worse still if
the integrals are incomplete! Clearly, you need to specify the
convention you have in mind; otherwise, your challenge is
unclear.
David
