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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?
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?