# Lower-bounding the distance between zeros of a continuous function

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 $L(v_{\star} + \epsilon) \neq 0$ ?

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Would it not be easier (and avoid a square root) to solve for $v$ first instead of $p$, then substitute that into $L$? –  Antonio Vargas Dec 5 '12 at 7:39
You're absolutely right in the sense the equation becomes simpler. But, may I still insist on L being a function of v ? This is because in my context v is the only tunable parameter. Sorry for not saying that. Thanks. –  Pavithran Iyer Dec 5 '12 at 15:25