Hi Benorin, Hi Hurkyl,
meanwhile I solved the problem using mathematica instead of maple which I found out is much more performant numerically. I still don't know if it works other way, but thank you guys, anyway.
Hendrik
Thanks for the answer, but no, I meant:
\psi_n(z) = \int_0^{2\pi}\int_0^1 \frac{ (z-\frac{1}{2}) \cdot (r \cos(\theta) + \frac{1}{2})^n \cdot r} {\sqrt{4z^2+4r^2+4r\cos(\theta)+2-4z}^{2n+3}}\,dr\,d\theta
...sorry for the latex trouble. Taking out the constants is a good idea and it might...
Hi,
I am new here, but apparently there are some decent mathematicians around, so I would like to confront you with a double integral problem.
Consider
\psi_n(z) = \int_0^{2\pi}\int_0^1 \frac{ (z-\frac{1}{2}) \cdot (r \cos(\theta) + \frac{1}{2})^n \cdot r}...
Hi,
I am new here, but apparently there are some decent mathematicians around, so I would like to confront you with a double integral problem.
Consider
\psi_n(z) = \int_0^{2\pi}\int_0^1 \frac{ (z-\frac{1}{2}) \cdot (r \cos(\theta) + \frac{1}{2})^n \cdot r}...
Hi,
I am new here, but apparently there are some decent mathematicians around, so I would like to confront you with a double integral problem.
Consider
\psi_n(z) = \int_0^{2\pi}\int_0^1 \frac{ (z-\frac{1}{2}) \cdot (r \cos(\theta) + \frac{1}{2})^n \cdot r}...
Hi,
I am new here, but apparently there are some decent mathematicians around, so I would like to confront you with a double integral problem.
Consider
\psi_n(z) = \int_0^{2\pi}\int_0^1 \frac{ (z-\frac{1}{2}) \cdot (r \cos(\theta) + \frac{1}{2})^n \cdot r} { \sqrt{...