Can Maple 11 calculate the TABLE integrals?
Enjoy yet another VM machine discovery.
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Mathematica 6.0.2
Integrate[1/z^2,{z,1,I Infinity}]
1
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Derive 6.1
INT(1/z^2,z,1,#i*inf)
1
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MuPAD 4.0.2
int(1/z^2, z = 1..I*infinity);
1
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Now, what about Maplesoft, after 25+ years of Maple development?
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Maple 11.02> int(1/z^2, z = 1..I*infinity);
int(1/z^2,z = 1 .. infinity*I)
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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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"We must understand that technologies
like these are the way of the future."
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1
int(1/(1+z)^2,z = 1 .. infinity*I)
Enjoy!
> http://www.cybertester.com/ Cyber Tester, LLChttp://maple.bug-list.org/ Maple Bugs Encyclopaediahttp://www.CAS-testing.org/ CAS Testing
One must be very careful when using such "complex infinities".
It is much better to transform first the integral e.g. over
[0,infinity)
DERIVE simply computes INT(f,a,b) = F(b) - F(a) where F is an
antiderivative,
even if a,b are "infinities". Do you think that this is a so smart
approach?
I agree. In this respect, Derive's approach is (too) simplistic.
But what you'd say about Mathematica 6 and MuPAD 4?
Just a remark on the ubiqiutous Maple bugs...
# Jacques Carette, Mar 24, 2008
Maple 11.02> assume(z>0);
Maple 11.02> is(sin(z)^2+cos(z)^2*z^2 > 0);
false (!) # Maple bugs strike again!
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So, what you'd say about Mathematica 6 and MuPAD 4?
...................................................
> On Mar 30, 6:59 am, Mate <mmat...@personal.ro> wrote:
> > approach?- Hide quoted text -
>
> - Show quoted text -
Just yet another ubiqiutous Maple 11 bug...
# Axel Vogt, Mar 28, 2008
Maple 11.02> Ei(1,1):
Maple 11.02> evalf(%=convert(%,Sum));
.2193839344 = 1.
honestly you should give reference to http://www.mapleprimes.com/tracker
So here is the link to a hot Axel Vogt's Ei-related bug
description a simplified form of which I presented:
http://www.mapleprimes.com/blog/axelvogt/errrexponentialintegralsum
Savour! ;)
I can only sing in sync with Herr Vogt, Arghrrrr! ;-)
A live discussion in Russian at http://forum.ru-board.com
http://forum.ru-board.com/topic.cgi?forum=5&topic=9935&start=180
http://forum.ru-board.com/topic.cgi?forum=5&topic=9935&start=200
I must say that I don't agree. It seems to me that Derive's approach here
is theoretically correct and as simple as possible.
> But what you'd say about Mathematica 6 and MuPAD 4?
>
> On Mar 30, 6:59=A0am, Mate <mmat...@personal.ro> wrote:
> > On Mar 30, 3:27 pm, Vladimir Bondarenko <v...@cybertester.com> wrote:
> >
> > > Maple 11.02> =A0int(1/(1+z)^2, z =3D 0..I*infinity);
> > > Maple 11.02> =A0int(1/(1+z)^2, z =3D 1..I*infinity);
> >
> > > 1
> > > int(1/(1+z)^2,z =3D 1 .. infinity*I)
> >
> > > Enjoy!
> >
> > One must be very careful when using such "complex infinities".
I'd be interested in seeing some examples showing why one must be very
careful when using such infinities.
> > It is much better to transform first the integral e.g. over
> > [0,infinity)
> > DERIVE simply computes INT(f,a,b) = F(b) - F(a) where F is an
> > antiderivative,
> > even if a,b are "infinities". Do you think that this is a so smart
> > approach?
The manual for Derive states that it computes definite integrals in
essentially that way. As a warning example, it mentions that
int(1/x^2, x, -1, 1)
yields -2, a result "which even Cauchy would call wrong." It then advises
"As is the case when using integral tables, it is your responsibility to
find singularities..." IMO, since the manual is adequately explicit about
this matter, I do not consider giving -2 for int(1/x^2, x, -1, 1) to be a
bug in the program.
David
The other day Mate said
M> One must be very careful when using such "complex infinities".
And I challenged him to show that both Wolfram Research and SciFace
GmbH engineers are wrong when they return 1 for this integral. But
he did not reply...
As as matter of fact I do not feel much that Albert Rich was wrong
returning 1 for this integral...
After all, we see that Maple returns a reasonable stuff for the
upper limit like I or 2*I... so why we cannot direct the upper
limit to I*infinity infinity and get this 1?