I spoke of using a polynomial approximation of G(q), the cumulative state number,and differentiating it to get F(q), the probability-density.
I'd like to add that a Taylor or McLaurin polynomial approximation of a complicated function could be used. ...after you've determined, by whatever method, exactly what the complicated function is to be, and what its constants should be. That could allow a completely analytical solution for R, the rounding point in a particular interval. But very likely not. That Taylor or McLaurin polynomial approximation of the complicated function will probably have a non-zero constant term. When that term is divided by q, as it will be, when that F(q) approximation is multiplied by s/q, there will then be a term with 1/q. Then, when the result of that multiplication of F(q) by s/q is integrated, the term with q in the denominator will result in a term that is logarithmic in q. But the other terms will, when integrated, result in terms that are algebraic functions of q. In evaluating a definite integral, the antiderivative is evaluated with q taking on the values of the limits of integration. In both of the integrations, from a to R, and from R to b, then, q has the value of R. Resulting in two terms logarithmic in R. And they don't cancel out, because the two integrations are of functions in which s/q has a different value of s. That means that a term logarithmic in R will remain. ...along with the other terms of the integration-result, which are polynomial functions of q. The resulting equation, then, doesn't sound like a promising candidate for an analytical solution. It will probably need an iterative equation-solving method, such as Newton's method, or Regula Falsi, or Bisection. Newton's method is popular as the first one to try, because it converges rapidly.Bisection is recommended for when Newton doesn't converge, or converges too slowly, because Bisection reliably converges at a good, if not as great, rate. Even though that equation would need a numerical solution, the problem would still be a lot easier than if you'd done as I described earlier, and done a numerical integration as part of each iteration of the equation-solving method, because this way the only numerical task is the equation-solving itself, making for less numerical work. Mike Ossipoff ---- Election-Methods mailing list - see http://electorama.com/em for list info
