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Derivative at a is 0 for any order n

  1. Apr 2, 2009 #1
    1. The problem statement, all variables and given/known data

    Let U be an open interval in R, let f : U [tex]\rightarrow[/tex] R be infinitely differentiable and let a [tex]\in[/tex] U. Prove that if f has a zero arbitrarily close to a then f[tex]^{(n)}[/tex](a) = 0 for all n [tex]\geq[/tex] 0.

    2. Relevant equations



    3. The attempt at a solution

    f is differentiable so f is continuous. f is continuous so for any [tex]\epsilon[/tex] > 0 there is a [tex]\delta[/tex] > 0 such that when | x - a | < [tex]\delta[/tex] then | f(x) - f(a) | < [tex]\epsilon[/tex]. there is an x where | x - a | < [tex]\delta[/tex] such that f(x) = 0. Therefore | f(a) | < [tex]\epsilon[/tex]. But epsilon is abitrary so f(a) = 0.

    I'm not sure how to approach proving that the nth derivative of f is 0 at a. Any thoughts?
     
  2. jcsd
  3. Apr 2, 2009 #2

    Dick

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    Science Advisor
    Homework Helper

    You could start by showing that f'(a)=0 using pretty much the same argument with the difference quotient. To get to higher derivatives, start using the mean value theorem.
     
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