Axel Vogt

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20 years, 225 days
Munich, Bavaria, Germany

MaplePrimes Activity


These are replies submitted by Axel Vogt

Correct, and I do have a citation at hand. Though it is enough that the very machine or program can not decide it ( = practically).

As vv pointed out: just use the more expensive setting for Testzero. Det is certainly not the right way.

If you want to report a bug: consider Rank. Maple says Rank(M) = 5, while it is 4 (numerically and symbolically).

Anyway: if b is undecidable to be zero or not (theoretically or practically) then the 1-dim matrix b will be returned as 1/b.

No bug, as I said.

Your personal opinion may be different.

This is not a bug.

Only the help should be more precise, something like "If A is recognized as singular matrix, an error message is returned."

@vv yes, my setting is WIN 7 SP 1 64 Bit Intel, 32 Bit Maple on 64 Bit Windows

Unfortunately I get the same error (I have has no ini-file that test), using your posted sheet in the application area), executing MultiIntPoly(1, rels, [x,y]) - the last example in that sheet.

@_Maxim_ , thank you.

@Markiyan Hirnyk 

It was a rhetorical question, yes.

But I think it is a bit "asymmetric" and not polite to demand "code comments" if the result of MMA is just cited and MMA neither exhibits its code nor comments on it.

@maxim: would you mind to explain it a bit?

Either I reset the variable before the call or would use

x:=m;
m:='m'; # forget m
return x;

How MMA computes it?

@ecterrab 

Plotting Re and Im shows that there is a problem for z < 0 as well, I would say (black = the line from convert):


 

restart: interface(version);
MeijerG([[a], [b]],[[c], [d]],z):
Re(%), Im(%), convert(%, hypergeom):
eval(%, [a=1/2,d=1, b=2, c=1]);
plot([%], z=-4 .. 4, thickness=3, color=[red, blue, grey]);

`Classic Worksheet Interface, Maple 2017.2, Windows, Jul 19 2017, Build ID 1247392`

Re(MeijerG([[1/2], [2]], [[1], [1]], z)), Im(MeijerG([[1/2], [2]], [[1], [1]], z)), Pi^(1/2)*z/2

 

MeijerG([[a], [b]],[[c], [d]],z):
eval(%, [a=1/2,d=1, b=2, c=1]); eval(%, z=-2);
evalf(%);

MeijerG([[1/2], [2]], [[1], [1]], z)

MeijerG([[1/2], [2]], [[1], [1]], -2)

-0.551653471779524e-16+0.915439115453814e-1*I

(1)


 

Download tmp.mws


 

MeijerG([[a], [b]],[[c], [d]],z):
%=convert(%, hypergeom);
eval(%, [a=1/2,d=1]):
EQ:=eval(%, [b=2, c=1]);
[op(%)]: plot(%, color=[red, blue]);

MeijerG([[a], [b]], [[c], [d]], z) = GAMMA(1-a+c)*z^c*hypergeom([1-a+c, 1-b+c], [1-d+c], z)/(GAMMA(b-c)*GAMMA(1-d+c))

EQ := MeijerG([[1/2], [2]], [[1], [1]], z) = Pi^(1/2)*z/2

 

 


 

Download tmp.mws

Yes, upload your very task.

I think that is essentially a bi-variate normal over 1 quadrant (formulae
are in Abramowitz & Stegun, with some minor typos if I remember). For the
following I have not cared too much about all cases for B and G.

P := (x,y)->(1/2)*exp(-(1/2)*(x^2+G*y^2-2*B*x*y)/(-B^2+G))/(Pi*sqrt(-B^2+G));

Int(Int(P(x,y),x=0..infinity),y=-infinity..0);
PDEtools:-dchange({x=-sqrt(G)*xx,y=yy},%,[xx,yy]): subs(xx=x,yy=y,%):
combine(%):
eval(%, B=-2*R*sqrt(G)/2);
factor(%);

Now compare with the integrand for the bivariate normal

Int(Int(exp(-(xi^2-2*rho*xi*eta+eta^2)/2/(1-rho^2))/2/Pi/sqrt(1-rho^2),
  eta = -infinity .. y),xi = -infinity .. x);
subs(x=0, y=0, %);
combine(%): factor(%);

# subs(rho=R, xi=x,eta=y, %);

Either provide the code as text or upload you worksheet using the "green arrow" which you have in the editor if you post.

And you may say what you get and what you expect to get

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