I'm doing a convex optimization problem. It requires me to fit a rational function to an exponential function. I assumed the original problem would be a quasiconvex optimization problem and based on this represented it via a family of convex functions and then solved it by bisection algorithm. But now, I cannot prove the quasiconvexity of the original problem :
$$minimize\quad \underset { i=1,...,k }{ max } \left| \frac { p({ t }_{ i }) }{ q({ t }_{ i }) } -{ y }_{ i } \right| $$
where: $$p(t)=a_0+a_1t+a_2t^2+...+a_mt^m,\quad q(t)=1+b_1t+...+b_nt^n$$
and the domain of the objective function is: $$D=\{ (a,b)\in \mathbf{R}^{m+1}\times \mathbf{R}^n | q(t)>0,\alpha <= t <= \beta \}$$
Can anyone give me some hint?
If we assume you can discretize the constraints as well (though we do not need to---see below), your original problem is $$\begin{array}{ll} \text{minimize}_{a,b} & \max_{i=1,2,\dots,k} \left|\frac{p(t_i)}{q(t_i)}-y_i\right| \\ \text{subject to} & q(t_i) > 0, ~ i=1,2,\dots, k \end{array}$$ A quasiconvex minimization problem has a quasiconvex objective and convex constraints. The constraints are linear in $b$, so all you need to do is prove the quasiconvexity of the objective. That, in turn, requires proving that the sublevel sets of the function are convex. That is, given any fixed $\delta$, prove that the set described by the inequality $$\max_{i=1,2,\dots,k} \left|\frac{p(t_i)}{q(t_i)}-y_i\right|\leq\delta$$ is a convex set. Since you asked for a hint, I will stop there. But hopefully you can see a path to proving convexity.
EDIT: Since the OP has indicated he "got it", I'll make sure the other readers "get it", too :-) Since our constraints ensure that $q(t)>0$, we can multiply through: $$\max_{i=1,2,\dots,k} \left|p(t_i)-y_iq(t_i)\right|\leq\delta q(t_i), \quad i=1,2,\dots,k$$ Since $p$ and $q$ are linear in $b$, these are $k$ convex inequalities in $a$ and $b$. If that isn't yet obvious to you, just convert them to $2k$ linear inequalities: $$-\delta q(t_i) \leq p(t_i)-y_iq(t_i)\leq\delta q(t_i), \quad i=1,2,\dots,k$$ So the sublevel set, being the intersection of convex inequalities, is convex. Hence the objective function is quasiconvex.
Now, keep in mind that discretization is just a practical choice. Even if you do not discretize, the model remains quasiconvex. $$\begin{array}{ll} \text{minimize}_{a,b} & \max_{t:~\alpha\leq t\leq\beta} \left|\frac{p(t)}{q(t)}-y(t)\right| \\ \text{subject to} & q(t) > 0, ~ \forall \alpha\leq t \leq \beta \end{array}$$ Convexity is preserved even with uncountable intersections of convex inequalities. So the constraint set remains convex, and the sublevel sets of the objective remain convex as well. The proof parallels what we did for the discretized problem. Alternatively, you could discretize just the objective, or just the constraints. In practice, you would probably end up discretizing the constraints more finely than the objective.