Andrew Carmichael

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14 years, 228 days

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Dear Folks, I find this interesting. I tried this integral using the CauchyPrincipalValue command and got one result. I also tried doing the principal value manually (integrating up to the pole then from the pole to infinity and adding the two) and ended up with something different. Any thoughts? Could somebody try it on Mathematica (I don't have it on my computer)? Int(omega*sqrt(x)/(x*(omega-x)),x=0..infinity)=int(omega*sqrt(x)/(x*(omega-x)),x=0..infinity,CauchyPrincipalValue);
Dear All, On doing the following integral I get different results analytically and numerically. # Define the integrand P3 := proc (x, m) options operator, arrow; -(1/2)*(-1+4*x^2*m^2+8*x^3*m+2*x*m-3*x^2+m^2)/((2*x^2+1+2*sqrt((x+m)^2/((x+m+1)*(x+m-1)))*x^2)*(x+m+1)*(x+m-1)*sqrt(x)) end proc; IntegralP3 := proc (m) options operator, arrow; Int(P3(x, m), x = 0 .. infinity) end proc; evalf(IntegralP3(1.0)); -2.262741700 int(P3(x,1),x=0..infinity); 0 evalf(-8*sqrt(2)/5); -2.262741699
Dear all, I recently had to solve the following integral: A:=(x,m)->(-m^2-2*x*m+1)/(sqrt(x)*sqrt((m+x)^2-1)*(x+sqrt((m+x)^2-1))); IntegralA:=(m)->Int(A(x,m),x=0..infinity); When I used evalf(IntegralA(1)); I get -2.828427125 But when I use int(A(x,1),x=0..infinity); to get a symbolic answer I get a complicated expression involving the function MeijerG function. Mathematica, on the other hand gives -2sqrt(2) exactly for the symbolic value. Does anybody know why? I'm curious even though the decimal values are the same. I was using Maple 11 and Mathemtica (I think) 5. Regards, Drew
Hi folks, When I use a command like: implicitplot3d(f(x,y,z)=0,x=...,y=...,z=...) obviously maple has to compute all the points in the given range which satisfy the equation. Does anybody know which numerical method it uses to do this? Regards, Drew
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