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Vladimir Bondarenko science forum Guru
Joined: 23 Apr 2005
Posts: 601

Posted: Thu Mar 24, 2005 7:46 pm Post subject:
Re: An exact 1D integration challenge  2



"Icke Selba" <jumb...@aussiemail.com.au> write on
Feb 1, 4:13 pm
IS> As a variation, consider your integral with an arbitrary
IS> upper limit, 0<a<1, with b=f(a), a=erf(b):
Thank you for sending your detailed comments. Especially I
like the generalization.
As far as I can see from other your messages, you are partial
to analytic solutions? Oh I understand you!
By the way, Maple 9.5.2 gives up here immediately
Quote:  int(ln(RootOf(erf(_Z)+z)), z=0..1);
int(ln(RootOf(erf(_Z)z)),z = 0 .. 1) 
and Mathematica 5.0/5.1 invokes an invalid divergence
warning message
Integrate[Log[InverseErf[0, z]], {z, 0, 1}]
Integrate::idiv: Integral does not converge.
which is a bug.
Cheers,
Vladimir 

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Vladimir Bondarenko science forum Guru
Joined: 23 Apr 2005
Posts: 601

Posted: Thu Mar 24, 2005 7:46 pm Post subject:
Re: a limit wrong in Maple...



For persons still considering buying Maple, I'd just like
to add that here we deal with a fundamental flaw Maplesoft
did not bother to fix over a decade.
To get the same bug manifestation one can use BesselI,
factorial, GAMMA, StruveL, AiryAi etc
Quote:  limit(f(n)*exp(2*n), n= infinity);
limit(f(n)*exp(n^2), n= infinity);
limit(f(n)*exp(n^n), n= infinity);
limit(f(n)/(1+n)^n, n= infinity);
limit(f(n)/(1+n)^n, n= infinity);
limit(f(n)/n^n, n= infinity);
limit(f(n)/GAMMA(n), n= infinity);
limit(f(n)/n!, n= infinity);
limit(f(n)/BesselI(0,n), n= infinity);
limit(f(n)/StruveL(1/2,n), n= infinity);
limit(f(n)*AiryAi(n), n= infinity);
limit(f(n)/AiryBi(n), n= infinity);

0
0
0
0
0
0
0
0
0
0
0
0
And here we see a different specimen of ravings of
a madman
Quote:  limit(f(n)*exp(n), n= infinity);
signum(FAIL[1])*infinity 
Quote:  limit(f(n)*exp(n^2), n= infinity);
signum(FAIL[1])*infinity 
Quote:  limit(f(n)/BesselK(0,n), n= infinity);
signum(FAIL[1])*infinity 
Quote:  limit(f(n)*BesselI(0,n), n= infinity);
signum(FAIL[1])*infinity 
Quote:  limit(f(n)*StruveL(1/2,n), n= infinity);
signum(FAIL[1])*infinity 
Quote:  limit(f(n)/AiryAi(n), n= infinity);
signum(FAIL[1])*infinity 
etc
Best wishes,
Vladimir Bondarenko
http://www.cybertester.com/
http://maple.buglist.org/
http://www.CAStesting.org/ 

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Guest

Posted: Thu Mar 24, 2005 7:46 pm Post subject:
Re: Steps towards writing a computer algebra system



Richard J. Fateman wrote:
Quote:  )lisp
evaluates a lisp expression, so my guess is that Axiom could
(for example), gobble down all of Maxima's lisp code, using
it as a library. How about it, Tim?

Actually we plan to embed maxima for testing purposes.
It should be possible to load maxima into the lisp image (since both
use GCL) start axiom and, at the axiom command line run a maxima
expression with
axiom> )lisp (maxima)
maxima: run maxima expression
.....
exit maxima
axiom>
it would more useful to just create a maxima package that would
lift maxima up to the axiom command level so functions could
be invoked directly from the maxima package as in:
sqrt(x)$maxima
this wouldn't be hard. I'll look into it. 

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Guest

Posted: Thu Mar 24, 2005 7:46 pm Post subject:
Re: Steps towards writing a computer algebra system



Allan Adler wrote:
Quote:  g : LX := reduce(*, [X  rho for rho in r]) ;

If you got this far you have a fully working Axiom. At this point it is
just running the regression test suite.
Quote:  Question: if one wants to define something directly in Lisp and make
it
available in top level, where in the documentation is it explained
how to
do that?

There is a way to call lisp functions directly in Axiom by using
)lisp at the top level as a command. It is also possible to invoke
lisp functions directly from the algebra code (search the pamphlet
files for the string $Lisp). Note that algebra code is case sensitive
so lisp functions are uppercase. There is also a category (SEXCAT)
and two domains (SEXOF, SEX) which handle Sexpressions. The expression
axiom> (1+1)$Lisp
will call the '+' from Lisp and return an SExpression (SEX).
Alternatively you can modify a pamphlet file with your lisp code,
type make (which will recompile the modified file and remake the
system image) and then your function is now in the image.
So, technically you can invoke lisp from the command line, in
expressions, in algebra code, and as embedded lisp code.
Quote:  In my experience, this is the most poorly explained aspect of
every computer algebra system that depends on a Lisp substrate. Good
as it usually is, the user interface is actually a part of the
problem. 
Well it is also poorly explained in Axiom. But it needn't be.
Write a literate program and explain the use of $Lisp, add a
function and show how to call it, etc. Someone, probably me,
will eventually get around to writing a literate program that
explains how to do this. There is a multivolume set being written
(see http://arch.axiomdeveloper.org), the 3rd volume of which
is about programming. Authors are needed.
Of course, there is also the question of why a beginning user of
Axiom needs to use lisp code. I'd encourage you to look around at
what's available. Axiom is a very large system with a large number
of functions and data structures. There is very rarely a need to
reach for lisp.
Tim Daly 

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Bruno Salvy science forum beginner
Joined: 24 Mar 2005
Posts: 1

Posted: Thu Mar 24, 2005 7:46 pm Post subject:
Re: Calculation of Dfinite series.



In article <41faa871$1@nntp.onyx.net>,
"Bruce Westbury" <news@westbury.onyxnet.co.uk> wrote:
Quote:  I am studying a series which starts:
0 1 2 3 4 5 6 7 8 9 10 11 12
1 0 1 1 4 10 35 120 455 1792 7413 31780 140833
The definition of this sequence that is relevant here is that the kth term
is the coefficient of x^2y^2 in the Taylor series expansion of X^kY where X
and Y are the Laurent polynomials
X = 1 + x + 1/x + y + 1/y + xy + 1/(xy)
Y = x^2y^3  xy^3 + y^2/x  x^{2}y + x^{3}/y  x^{3}y^{2}
+ x^{2}y^{3}  y^{3}/x + xy^{2}  x^2y^{1} + x^3y  x^3y^2

I might be misunderstanding or mistyping something. When I try to get
your coefficients with your formulae they do not match:
Quote:  X := 1 + x + 1/x + y + 1/y + x*y + 1/(x*y):
Y := x^2*y^3 x*y^3 + y^2/x  x^(2)*y + x^(3)/y  x^(3)*y^(2)
+ x^(2)*y^(3)  y^(3)/x + x*y^(2)  x^2*y^(1) + x^3*y  x^3*y^2:
expand(X^3*Y):
coeff(%,x,2);
1 3 5 6 2 2 
  + y  2 y  2 y +  + 
y 3 4
y y
From your message, I expected 1. 

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Jens Axel Søgaard science forum beginner
Joined: 08 May 2005
Posts: 28

Posted: Thu Mar 24, 2005 7:46 pm Post subject:
Re: Calculation of Dfinite series.



Bruce Westbury wrote:
Quote:  I am studying a series which starts:
0 1 2 3 4 5 6 7 8 9 10 11 12
1 0 1 1 4 10 35 120 455 1792 7413 31780 140833
I would like some help with calculating this recurrence relation (or
differential equation).

The entry in "OnLine Encyclopedia of Integer Sequences"
contains a Maple snippet for its calculation:
<http://www.research.att.com/cgibin/access.cgi/as/njas/sequences/eismum.cgi>

Jens Axel Søgaard 

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Bruce Westbury science forum beginner
Joined: 21 Apr 2005
Posts: 2

Posted: Thu Mar 24, 2005 7:46 pm Post subject:
Re: Calculation of Dfinite series.



"Bruno Salvy" <Bruno.Salvy@inria.fr> wrote in message
news:Bruno.SalvyBDB80B.15521507022005@newsrocq1.inria.fr...
Quote:  In article <41faa871$1@nntp.onyx.net>,
"Bruce Westbury" <news@westbury.onyxnet.co.uk> wrote:
I am studying a series which starts:
0 1 2 3 4 5 6 7 8 9 10 11 12
1 0 1 1 4 10 35 120 455 1792 7413 31780 140833
The definition of this sequence that is relevant here is that the kth
term
is the coefficient of x^2y^2 in the Taylor series expansion of X^kY where
X
and Y are the Laurent polynomials
X = 1 + x + 1/x + y + 1/y + xy + 1/(xy)
Y = x^2y^3  xy^3 + y^2/x  x^{2}y + x^{3}/y  x^{3}y^{2}
+ x^{2}y^{3}  y^{3}/x + xy^{2}  x^2y^{1} + x^3y  x^3y^2
I might be misunderstanding or mistyping something. When I try to get
your coefficients with your formulae they do not match:
X := 1 + x + 1/x + y + 1/y + x*y + 1/(x*y):
Y := x^2*y^3 x*y^3 + y^2/x  x^(2)*y + x^(3)/y  x^(3)*y^(2)
+ x^(2)*y^(3)  y^(3)/x + x*y^(2)  x^2*y^(1) + x^3*y  x^3*y^2:
expand(X^3*Y):
coeff(%,x,2);
1 3 5 6 2 2
  + y  2 y  2 y +  + 
y 3 4
y y
coeff(%,y,2);
0
From your message, I expected 1.

My apologies. Take the coefficient of x*y in X^k*Y where
X := 1 + x + 1/x + x/y + y/x + x^2/y + y/x^2;
Y := x*y + 1/(x*y)  y^2/x  x/y^2 + y^3/x^4 + x^4/y^3  y^3/x^5  x^5/y^3 +
y^2/x^5 + x^5/y^2  y/x^4  x^4/y;
Just in case there is a mistake here, here is the picture you should draw to
see what is going on.
If I had a blackboard I would draw it for you.
First tile the plane with equilateral triangles in the usual way. Choose a
vertex and label it (0,0).
This vertex is connected to six other vertices which are the vertices of a
regular hexagon.
Go round these six vertices clockwise and label them in order
(1,0) (2,1) (1,1) (1,0) (2,1) (1,1)
This identifies the vertices of the tiling with Z + Z i.e. we now have
labeling of the vertices by
pairs of integers.
Associate with the point (r,s) the monomial x^r * y^s. Then X is the sum of
the monomials of (0,0)
and the monomials of the six neighbours.
Now take the group generated by reflection in the line x=0 and by reflection
in the line y=0.
This group has order 12 and the orbit of the point (1,1) has twelve points.
Then Y is the sum of the twelve monomials of these twelve points with a sign
which alternates as you move round the figure. More precisely the number of
reflections you can use to get from (1,1) to a particular point is not
welldefined but the parity is.

To clarify my question. Both Leonard and Jens have pointed out that this
sequence appears in the OnLine Encyclopedia of Integer Sequences (as
sequence A059710) and that the entry there includes Maple code for computing
the sequence. This Maple code is hard to read but I believe it is an
implementation of the above description of the sequence.
To explain my question; consider the wellknown sequence of Catalan numbers.
This sequence can be defined (and computed from) the definition that C(n) is
the coefficient of x in the expansion of (x1/x)*(x+1/x)^(2*n). This is the
analogue of the definition I have of the sequence I am interested in.
However the Catalan numbers can also be defined by
C(0) = 0 and for n+1>0, C(n+1) = C(n) * (4*n+2) /(n+2)
General theory says that the sequence I am looking at has a definition like
this (but not first order) but I have not succeeded in computing this.
The generating function of the Catalan sequence is also algebraic since it
satisfies a wellknown quadratic equation. However the generating function
of the sequence I am looking at will not be algebraic (although I do not
have a proof). 

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alain verghote science forum Guru Wannabe
Joined: 29 Apr 2005
Posts: 293

Posted: Thu Mar 24, 2005 7:46 pm Post subject:
Re: differential operator



Dear Amarpreet Rattan,
I prefer using the net for replying:
Carl Devoré gives a nice mapple program whih solve your
problem,try it: "nonpolynomial differential operator" 15 dec 2005
sci.math.symbolic .
You may also read "Differential Operators question " 21 jan 2005
alt.math.undergrad ,
Best Regards,
Alain. 

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Androcles science forum Guru
Joined: 03 Jun 2005
Posts: 2714

Posted: Thu Mar 24, 2005 7:46 pm Post subject:
Re: 1 QA engineer vs 7 users: opinions about the AXIOM computer algebra system



"Vladimir Bondarenko" <vb@cybertester.com> wrote blatant fucking
advertising in message
news:1107952475.042947.94730@c13g2000cwb.googlegroups.com...
Androcles. 

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Vladimir Bondarenko science forum Guru
Joined: 23 Apr 2005
Posts: 601

Posted: Thu Mar 24, 2005 7:46 pm Post subject:
Re: 1 QA engineer vs 7 users: opinions about the AXIOM computer algebra system



"Androcles" <Androcles@ MyPlace.org> wrote...
Gotcha!
Thomas De Quincey:
http://www.brainyquote.com/quotes/quotes/t/thomasdequ165064.html
TDQ> If once a man indulges himself in murder, very soon he
TDQ> comes to think little of robbing; and from robbing he
TDQ> comes next to drinking and Sabbathbreaking, and from
TDQ> that to procrastination and incivility. 

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Androcles science forum Guru
Joined: 03 Jun 2005
Posts: 2714

Posted: Thu Mar 24, 2005 7:46 pm Post subject:
Re: 1 QA engineer vs 7 users: opinions about the AXIOM computer algebra system



"Vladimir Bondarenko" <vb@cybertester.com> wrote in message
news:1108068797.683916.325490@c13g2000cwb.googlegroups.com...
Quote:  "Androcles" <Androcles@ MyPlace.org> wrote...
Gotcha!
Thomas De Quincey:
http://www.brainyquote.com/quotes/quotes/t/thomasdequ165064.html
TDQ> If once a man indulges himself in murder, very soon he
TDQ> comes to think little of robbing; and from robbing he
TDQ> comes next to drinking and Sabbathbreaking, and from
TDQ> that to procrastination and incivility.

Who the f*** is TDQuincy? (rhetorical question, I'm not interested)
Androcles. 

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Vladimir Bondarenko science forum Guru
Joined: 23 Apr 2005
Posts: 601

Posted: Thu Mar 24, 2005 7:46 pm Post subject:
Re: 1 QA engineer vs 7 users: opinions about the AXIOM computer algebra system



http://www.bcm.tmc.edu/neurol/struct/park/park6.html
Additional Info
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Fax: (718) 2799596
EMail: ts@tsausa.org
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What is Tourette's syndrome
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Prognosis
Many patients experience significant improvement in their late
teens or early 20s. Most people with Tourette's syndrome get
better as they mature. Some associated problems, such as
obsessivecompulsive disorder and attention problems, can
persist into adulthood and require longterm treatment. 

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Vladimir Bondarenko science forum Guru
Joined: 23 Apr 2005
Posts: 601

Posted: Thu Mar 24, 2005 7:46 pm Post subject:
Re: 1 QA engineer vs 7 users: opinions about the AXIOM computer algebra system



Date: Fri, 11 Feb 2005 23:27:33 +0200
From: Vladimir Bondarenko <vb@cybertester.com>
XMailer: The Bat! (v1.61) Personal
ReplyTo: Vladimir Bondarenko <vb@cybertester.com>
Organization: Cyber Tester, LLC
XPriority: 3 (Normal)
MessageID: <6752461535.20050211232733@cybertester.com>
ReturnReceiptTo: vb@cybertester.com
To: aard@spamcop.net
Subject: Please react to the postings by Androcles@ MyPlace.org
MIMEVersion: 1.0
ContentType: text/plain; charset=usascii
ContentTransferEncoding: 7bit
Good Day,
I have found your administrative email address via the
www.godaddy.com services and on behalf of the symbolic
computational community of many thousand readers ask you
to help us as it is you who is the person responsible
for www.myplace.org .
Please consider systematic attempts of disrupting our
normal placid discussions at sci.math.symbolic via
using obscene language by Androcles@MyPlace.org .
http://groupsbeta.google.com/group/sci.math.symbolic/msg/875d64ade2448fe0
http://groupsbeta.google.com/group/sci.math.symbolic/msg/9a92a7416dde1525
Such type of behavior is not new for Androcles@MyPlace.org
as you can see from
http://www.groupsrv.com/science/about791760asc60.html
http://www.groupsrv.com/science/post642900.html
http://www.groupsrv.com/science/post642898.html
http://www.groupsrv.com/science/post642532.html
http://www.groupsrv.com/science/post642748.html
http://www.groupsrv.com/science/post642881.html
http://www.groupsrv.com/science/about80960.html
http://www.groupsrv.com/science/about7917660.html
http://www.groupsrv.com/science/about7917645.html
http://www.groupsrv.com/science/post642743.html
and more, but I hope the quoted links will do.
Thank you in advance for your help.
Had you need any kind of support to drive the nail home
please don't hesitate to let me know and I guarantee you
the strongest backstop imaginable to stop the above
demonstrated disorderly conduct.
Best wishes,
Vladimir Bondarenko
GEMM architect
Cofounder, CEO, Mathematical Director
Cyber Tester, LLC
13 Dekabristov Str, Simferopol
Crimea 95000, Ukraine
tel: +38(0652)447325
tel: +38(0652)230243
tel: +38(0652)523144
fax: +38(0652)510700
http://www.cybertester.com/
http://maple.buglist.org/
http://www.CAStesting.org/ 

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Alec Mihailovs science forum addict
Joined: 28 Apr 2005
Posts: 95

Posted: Thu Mar 24, 2005 7:46 pm Post subject:
Re: Calculation of Dfinite series.



"Bruce Westbury" <news@westbury.onyxnet.co.uk> wrote in message
news:4208038d$1@nntp.onyx.net...
Quote: 
My apologies. Take the coefficient of x*y in X^k*Y where
X := 1 + x + 1/x + x/y + y/x + x^2/y + y/x^2;
Y := x*y + 1/(x*y)  y^2/x  x/y^2 + y^3/x^4 + x^4/y^3  y^3/x^5  x^5/y^3
+ y^2/x^5 + x^5/y^2  y/x^4  x^4/y;
To clarify my question. Both Leonard and Jens have pointed out that this
sequence appears in the OnLine Encyclopedia of Integer Sequences (as
sequence A059710) and that the entry there includes Maple code for
computing the sequence. This Maple code is hard to read but I believe it
is an implementation of the above description of the sequence.

No, it is not. I used Littelmann paths instead of Weyl's character formula.
Quote:  General theory says that the sequence I am looking at has a definition
like this (but not first order) but I have not succeeded in computing
this.

Here is the recurrence relation: a(0)=1, a(1)=0, a(2)=1, and for n>2
(n+5)(n+6)a(n)=2(n1)(2n+5)a(n1)+(n1)(19n+1a(n2)+14(n1)(n2)a(n3)
I just submitted it to the OEIS, so it should appear there next week. Also,
I changed my Maple program to
A059710:=rsolve({(n+5)*(n+6)*A(n)=2*(n1)*(2*n+5)*A(n1)+(n1)*(19*n+1*A(n2)+14*(n1)*(n2)*A(n3),A(0)=1,A(1)=0,A(2)=1},A(n),makeproc);
Alec Mihailovs
http://math.tntech.edu/alec/ 

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Vladimir Bondarenko science forum Guru
Joined: 23 Apr 2005
Posts: 601


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