Consider a continuous function of the form:
$L(v) = \sum_{i = 0}^{m}[vA_{i} - B_{i}]p^{i}$
where $p$ is the root of the polynomial equation: $vf(p) - g(p) = 0$ with $f(p)$ and $g(p)$ being two $n-$degree polynomials. Suppose we express the root as: $p = F(v)$, then:
$L(v) = \sum_{i = 0}^{m}(vA_{i} - B_{i})[F(v)]^{i}$
$L(v)$ is a continuous function and moreover, let us say its a non-constant function of $v$. Hence if it has a zero for some $v = v_{\star}$, $\exists \epsilon > 0$ such that $L(v_{\star} + \epsilon) \neq 0$.
I would like to know, whether given an interval, is there a general way to bound the distance between the zeros of $L(v)$. In otherwords, I wanted to know a systematic method to find such an $\epsilon$ where $L(v_{\star}+\epsilon) \neq 0$.
Let us consider an example:
Let: $n = 2, m = 3$ and:
$f(p) = 3p^{2} + 4p + 2$
$g(p) = 8p^2 + p + 7$
$L(v,p) = (7v-5)p^{3} + (v-7)p^{2} + (3v-1)p + (7v-9)$
Hence $p$ is the root of the polynomial equation: $p^{2}(3v - 8) + p(4v - 1) + (2v - 7) = 0$.
On solving for $p$, we find one of the solutions as:
$p = F(v) = \dfrac{-(4v-1) + \sqrt{(4v-1)^{2} - 4(3v-8)(2v-7)}}{2(3v-8)}$
On substituting this in $L(v,p)$, we can express $L(v,p) \equiv L(v)$ as:
$L(v) = (7v-5) \left[\dfrac{-(4v-1) + \sqrt{(4v-1)^{2} - 4(3v-8)(2v-7)}}{2(3v-8)}\right]^{3} + (v-7)\left[\dfrac{-(4v-1) + \sqrt{(4v-1)^{2} - 4(3v-8)(2v-7)}}{2(3v-8)}\right]^{2} + (3v-1)\left[\dfrac{-(4v-1) + \sqrt{(4v-1)^{2} - 4(3v-8)(2v-7)}}{2(3v-8)}\right] + (7v-9)$
Could anyone give hint to me how I could think about lowerbounding the distance between the zero's of such a function ? Suppose that I know it has a zero at $v_{\star} \in (0,1)$, how can I find an $\epsilon > 0$ such that $L(v_{\star} + \epsilon) \neq 0$ ?