In the comments for the following linked question Descartes rule of sign with positive real exponents, the following was stated:
" The positive roots depend continuously on the exponents. This holds true, except in the case of a double (or higher order) positive root. But in that case a perturbation of the exponents either changes the double root to a pair of real roots, or eliminates the real roots altogether - so the number of positive roots either stays the same or goes down, meaning the rule of signs still applies."
Is it saying that there is an upper bound for the number of positive roots of a perturbed polynomial in terms of the original? In other words,
Suppose that $$q(t) = a_nx^n + \cdots + a_kx^k + \cdots +a_0 = (x-r_1) \cdots (x-r_k)\tilde q(t)$$
where $r_1,\cdots,r_k>0$ and that $\tilde q(t)$ has no positive roots.
Now define $$p(t) = a_nx^n + \cdots + a_kx^{k+\alpha} + \cdots +a_0 $$
where $\alpha>0$ such that $k + \alpha <n$. Can we say that $p(t)$ has at most $k$ positive roots. If so how would I prove that? A reference would be helpful too!
In the case the statement is not true, is there a counter example?