Originally Posted by

**ColinJ** I've taken my project to the next step using the equation you provided earlier and my next hurdle occurs with the addition of a second variable r2 and substituting the denominator of the original equation as the d variable in your solution. The latex probably explains better what I’m chasing;

$\displaystyle \lim_{x \to \infty} \sum_{n = 0}^x (1/x) \frac { d-r_{1}cos(2\pi n/x)} { ((d-r_{1} \cos(2\pi n/x))^2 – r_{2}^2)^{(3/2)}}$

Is this still a Riemann sum?

If I've got this right this should also converge to around 0.06 for d=4 and r = 1.

Yes it is still a Riemann sum, and the limit is equal to the integral

$\displaystyle \int_0^1\!\!\frac{d-r_1\cos(2\pi x)}{\bigl((d-r_1\cos(2\pi x))^2 - r_2^2\bigr)^{3/2}}\,dx.$

That is a far more elaborate integral than the one in the previous problem, and I doubt whether it is possible to give the answer in a closed form. For particular values of d, r_1 and r_2, you can presumably get Mathematica or Wolfram to churn out an answer.