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We know that the zeros of an analytic non-constant function are always isolated. A proof is here. Let $L(v)$ be an analytic function in $v$, where $v\in\mathbb{R}$.

Let us write $L(v) \equiv L(v,p)$ where $p \equiv p(v)$ is an analytic function. Moreover, it is injective (one-one) too.

Can we say that the zeros of $L(v,p)$ are also isolated ? I mean "isolated" on the $v$-$p$ plane, i.e, If $L(v_{0},p_{0}) = 0$, then $\exists$ $\epsilon > 0$ such that there is no other zero of $L(v,p)$ in a disk of radius $\epsilon$ centered at $(v_{0},p_{0})$.

I also have a followup question here.

Update: It is ok to assume that $L(v)$ is such that there exists an analytic function $p \equiv p(v)$ such that $L(v,p)$ is a bivariate polynomial.

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Does $L(v,p)$ depend on $p$? It doesn't appear to. –  Robert Israel Dec 10 '12 at 22:13
    
It does depend on p, but p is not an independent parameter here, because p = p(v). So, v is the only independent parameter. We can consider the form: L(v,p) = \sum_{i,j}a_{ij} v^{i} p^{j} where p = p(v) is some analytic function. I hope I am clear. –  Pavithran Iyer Dec 10 '12 at 22:31
    
If $L(v,p)$ is only defined on a curve $p=p(v)$ in the $(v,p)$ plane, then of course an isolated $v$ corresponds to an isolated $(v,p)$. –  Robert Israel Dec 10 '12 at 22:57
    
@RobertIsrael What I did was first to consider all the zeros of L(v,p) : these will, in general, be curves on the v-p plane. Then I consider the points at which the curve p=f(v) intersects these zero curves. These intersecting points would be zeros of L(v,p(v)). However, can it happen that p=f(v) itself is one of the zero curves ? (In which case the zeros of L(v,p(v)) would not be isolated.) Sorry if I seem to be confusing. –  Pavithran Iyer Dec 11 '12 at 20:20
    
Suppose L(v,p(v)) is a bivariate polynomial of degree n in variables {v,p(v)} , we know the number of zero curves of it are bounded by a quadratic function of its degree, say M(n). Can we use that information and claim to bound the number of zeros of L(v) also by M(n) ? –  Pavithran Iyer Dec 11 '12 at 20:24
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