@Markiyan Hirnyk 

Thanks, this is correct

@Markiyan Hirnyk 

@Markiyan Hirnyk 

Hi,

The factor is:

-(1/4)*G/eta

@vv 

Thanks @w and thanks for all help me.

This i was what i wanted.

@vv 

Yes, is a error mine.

But if i wish have the same result with app Mathematica.

Operatintg Maple:

restart;

FM := proc (q, Jo, T) options operator, arrow; evalf(Int((1/2)*exp(-(1/2)*z^2)*Jo*sqrt(2)*(1-tanh(sqrt(q)*z/T)^2)/(sqrt(Pi)*T), z = -infinity .. infinity)) end proc;

FQ := proc (q, Jo, T) options operator, arrow; evalf(Int((1/2)*exp(-(1/2)*z^2)*sqrt(2)*tanh(sqrt(q)*z/T)^2/sqrt(Pi), z = -infinity .. infinity)) end proc;

Sol := proc (T) options operator, arrow; fsolve({FM(q, Jo, T) = 1, FQ(q, Jo, T) = q}, {Jo, q}) end proc; Sol(.3)

Sol(0.3);

{Jo = .3000000001, q = 0.}

When in Mathematica app is:

{Jo = 1.12479, q = 0.733283}

Regards.

@vv 

Yes, this is conditions

FM=1 and FQ=q

Then:

restart;

FM := proc (q, Jo, T) options operator, arrow; evalf(Int((1/2)*exp(-(1/2)*z^2)*Jo*sqrt(2)*(1-tanh(sqrt(q)*z/T)^2)/(sqrt(Pi)*T), z = -infinity .. infinity)) end proc;

FQ := proc (q, Jo, T) options operator, arrow; evalf(Int((1/2)*exp(-(1/2)*z^2)*sqrt(2)*tanh(sqrt(q)*z/T)^2/sqrt(Pi), z = -infinity .. infinity)) end proc;

Sol := proc (q) options operator, arrow; fsolve({FM(q, Jo, T)-1, FQ(q, Jo, T)-q}, {Jo, T}) end proc;

Sol(0.6);

But this does not work very well

@vv 

Thanks, why are "-1" and "-*2" in this? :

fsolve({FM(q, Jo, T) - 1, FQ(q, Jo, T) - 2}, {Jo, q}):

@AmusingYeti

Hi, I want to have a numerical solution of Jo and q for each value of T by solving the system of equations of the two integrals.

In other app that Mathemattica is made :

I want this in Maple.

Regards.

@AmusingYeti 

Hi, and use "roots" in this form near of q=1 and Jo=1?

Sol(.6) := roots({FM(q, Jo, T), FQ(q, Jo, T)}, {Jo = 1, q = 1})

But I dont known if this is possible..

Regardas.

@AmusingYeti 

Thanks,

Hi try solved parametres q and Jo for a T known. First, I define the equations of the integral numerical:

restart;

FM := proc (q, Jo, T) options operator, arrow; evalf(Int((1/2)*exp(-(1/2)*z^2)*Jo*sqrt(2)*(1-tanh(sqrt(q)*z/T)^2)/(sqrt(Pi)*T), z = -infinity .. infinity, method = _d01amc, methodoptions = [maxintervals = 1000])) end proc;

FQ := proc (q, Jo, T) options operator, arrow; evalf(Int((1/2)*exp(-(1/2)*z^2)*sqrt(2)*sqrt(q)^2*z^2/(T^2*sqrt(Pi)), z = -infinity .. infinity, method = _d01amc, methodoptions = [maxintervals = 1000])) end proc;

Sol := proc (T) options operator, arrow; solve({FM(q, Jo, T), FQ(q, Jo, T)}, {Jo, q}) end proc;

Sol(.6);

Here I would have to give the solutions of Jo and q for a given T in this case the value of T is 0.6. But this does not work.

How can i do?

Regards.

@tomleslie 

This equations are not recurrent, there are resolved for m<<1 and q finite, frist step expand m:

But integral m2 not resolve.

@vv Thanks, it is that i want!.

Regards!

Thanks for all,Expansion it is thay i want.

Best Regards!!

Laplace_Polar.mw

This is line right

g:= (x,y)-> `if`(y=sqrt(1-x^2), 100, `if`(x=sqrt(1-y^2), 100,`if`(y=0, 0, 'procname'(args))));

but error is the same:

x, y, w := algo_12_1([-1,1],[0,1],[6,3],1e-10,100):
Error, (in algo_12_1) unable to store '.2500000000*g(-1., HFloat(0.6666666666))+.2500000000*g(HFloat(-0.6666666667), 1.)' when datatype=float[8]

Regards.