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What does normalized (k + 1)th divided difference at distinct nodes mean?

  1. Aug 28, 2011 #1
    Hi everyone,

    This is not a homework question but a clarification on the following proof:

    Suppose [itex]h[/itex] is an infinitely differentiable real-valued function defined on [itex]/Re[/itex] such that [itex]h(1/n)=0[/itex] for all [itex] n \in N [/itex]. Then prove [itex]h^{(k)}(0)=0[/itex] for all [itex] k \in [/itex].

    Proof: Since h is infinitely differentiable in a neighborhood of 0, the kth derivative of h at 0 is the limit of its normalized (k+1)th divided difference at distinct nodes [itex]x_1,x_2,...,x_{k+1}[/itex] as they tend to 0: [tex] h^{(k)}(0)=k! \lim_{x_1,x_2,...,x_{k+1} \rightarrow 0 } \nabla (x_1,...,x_{k+1})h [/tex]

    Now, choosing [itex] x_j := x_j(n) = \frac{1}{(n+j)}[/itex] and letting [itex] n \in \aleph [/itex] tend to infinity, we see that [itex] h^{k}(0) = 0 [/itex] for all [itex] k \in N [/itex].

    I don't understand the proof above since I don't know what normalized (k+1)th divided difference at distinct nodes means. Does anyone know?
  2. jcsd
  3. Sep 3, 2011 #2
    To rephrase the proof, we are using induction on k and Rolle's theorem.
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