An algorithm for checking if a nonlinear function f is always positive

Question

Is there an algorithm to check if a given (possibly nonlinear) function f is always positive?

The idea that I currently have is to find the roots of the function (using newton-raphson algorithm or similar techniques, see http://en.wikipedia.org/wiki/Root-finding_algorithm) and check for derivatives, or finding the minimum of the f, but they don't seems to be the best solutions to this problem, also there are a lot of convergence issues with root finding algorithms.

For example, in Maple, function verify can do this, but I need to implement it in my own program. Maple Help on verify: http://www.maplesoft.com/support/help/Maple/view.aspx?path=verify/function_shells Maple example: assume(x,'real'); verify(x^2+1,0,'greater_than' ); --> returns true, since for every x we have x^2+1 > 0

[edit] Some background on the question: The function $f$ is the right hand-side differential nonlinear model for a circuit. A nonlinear circuit can be modeled as a set of ordinary differential equations by applying modified nodal analysis (MNA), for sake of simplicity, let's consider only systems with 1 dimension, so $x' = f(x)$ where $f$ describes the circuit, for example $f$ can be $f(x) = 10x - 100x^2 + 200x^3 - 300x^4 + 100x^5$ ( A model for nonlinear tunnel-diode) or $f=10 - 2sin(4x)+ 3x$ (A model for josephson junction).

$x$ is bounded and $f$ is only defined in interval $[a,b] \in R$. $f$ is continuous. I can also make an assumption that $f$ is Lipschitz with Lipschitz constant L>0, but I don't want to unless I have to.

Answer 1

If I understand your problem correctly, it boils down to counting the number of (real) roots in an interval without necessarily identifying them. In fact, you don't even need to get the exact number, just whether or not it's equal to zero.

If your function is a polynomial, I think that Sturm's theorem may be applicable. The Wikipedia article claims two other procedures are preferred, so you might want to check those out, too. I'm not sure if Descartes' rule of signs works on an interval, but Budan's theorem does appear to.