text: # What does normalized (k + 1)th divided difference at distinct nodes mean?
  
  1. Aug 28, 2011
  
  ### CantorSet
  
  Hi everyone,
  
  This is not a homework question but a clarification on the following proof:
  
  Suppose $h$ is an infinitely differentiable real-valued function defined on $/Re$ such that $h(1/n)=0$ for all $n \in N$. Then prove $h^{(k)}(0)=0$ for all $k \in$.
  
  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 $x_1,x_2,...,x_{k+1}$ as they tend to 0: $$h^{(k)}(0)=k! \lim_{x_1,x_2,...,x_{k+1} \rightarrow 0 } \nabla (x_1,...,x_{k+1})h$$
  
  Now, choosing $x_j := x_j(n) = \frac{1}{(n+j)}$ and letting $n \in \aleph$ tend to infinity, we see that $h^{k}(0) = 0$ for all $k \in N$.
  
  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. Sep 3, 2011
  
  ### Eynstone
  
  To rephrase the proof, we are using induction on k and Rolle's theorem.