Mariusz Iwaniuk

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The short answer is that generally no, for your matrix B:=Matrix([[a, -b, e], [c, l, d], [-e, -k, m]], simplify command can't  simplify it. Mathematica also can't.


@Preben Alsholm 

Thanks for clarification.

Intial condition: D[1](T)(x, 0) = 0 it's seems to be wrong ?

If I change to D[2](T)(x, 10) = 0(I assume),pdsolve(numeric) give me: Error, (in pdsolve/numeric) unable to handle elliptic PDEs.

Likely, that means Maple doesn't know a closed-form solution, if it exists.


"What do you mean by "trivial" corrections?"


I mean for a "trivial": unimportant fixes.

"Do you have an example and do you expect a large jump with 2019?"


I do not have it, and I do not know anything about it.

"Can you provide a list of a few "weaknesses" of int, that are going to be fixed?"


I do not have it, and I do not know anything about it.


English is not my native language,sorry for misunderstanding.


See Table 1. and another Table 1 (A very slow progress)

My opinion People from Maple Company are too busy(In introduction of new features,fixing bugs) and do not have time for such trivial corrections in int function.

"Error, (in dsolve/numeric/BVPSolve) matrix is singular" it's probaly means No solution or  infinitely many

Almost the same Question here.


The question concerned to integral : Int(cos(2*x)/(1+2*sin(3*x)^2), x = 0 .. Pi) not for Int(cos(2*x)/(1+2*sin(3*x)^2), x = 0 .. Pi/2) ? My answer is correct and Maple  returns 0.

J := Int(convert(cos(2*x)/(1+2*sin(3*x)^2), exp), x = 0 .. Pi); evalf(J) = evalf(value(J));

# 2.065847493*10^(-13) = 0.



@Preben Alsholm 


I do not care what the method Maple uses. In this example,Zero is the correct answer.


sol := pdsolve([pde, ic], w(x, t)) assuming t>0,x>0;

#sol := w(x, t) = f(-3*t^2*(1/2)+x)*exp(-t)


With no error massages:

sol := pdsolve([pde, ic], w(x, t), HINT = `*`);

#sol := w(x, t) = exp(t)*invfourier(fourier(f(x), x, s)*exp(-(1/2)*(3*I)*t^2*s), s, x)


Thanks for information.


Thanks for quick fix.

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