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Prove that the set of those $c$ is countable

by zeraoulia
Tags: analytic function, real and complex
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zeraoulia
#1
Apr27-13, 12:35 PM
P: 12
The **mean values theorem** says that there exists a $c∈(u,v)$ such that $$f(v)-f(u)=f′(c)(v-u)$$ My question is: Assume that $u$ is a root of $f$, hence we obtain $$f(v)=f′(c)(v-u)$$ Assume that $f$ is a non-zero analytic function in the whole real line. My interest is about the real $c∈(u,v)$. I believe that the set of those $c$ is contable, otherwise we conclud that $f′(c)$ is constant (in an open set containing $c$) since $v$ is a constant and hence $f$ is identically zero. Also, I think that the set of those $c$ is countable (for analytic functions) but I am not able to prove that.
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HS-Scientist
#2
Apr27-13, 01:18 PM
P: 183
For all c that satisfy that equation, [itex] f'(c)(v-u)-f(v)=0 [/itex]. The LHS is an analytic function in c (v and u are constant) and is not identically zero unless f is linear. That any non-zero analytic function has countable zeroes is well known. (http://planetmath.org/ZeroesOfAnalyt...onsAreIsolated is a proof for complex functions, but the idea is the same)

Edit: My link only proves that the zeroes are isolated. Here is how you would show that the set of zeroes in countable: Let a be a zero. If all other zeroes are separated from a by at least ε, let q be a rational number which is no more than ε/3 away from a. Each rational number can be associated with a zero in this way. The cardinality of the set of zeroes is the same as for this set of rational numbers and rational numbers are countable.
Fredrik
#3
Apr28-13, 01:44 PM
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