[EM] Taylor or McLaurin polynomial for the complicated functinos would reduce the numerical work.
email9648742 at gmail.com
Fri Jul 20 23:01:10 PDT 2012
I spoke of using a polynomial approximation of G(q), the cumulative
state number,and differentiating it to get F(q), the
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.
More information about the Election-Methods