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What does $\Delta ^{k}$ mean?

For example in this Newton Series:

$\displaystyle f(x)=\sum _{k=0}^{\infty }{\frac {\Delta ^{k}[f](a)}{k!}}~(x-a)_{k}=\sum _{k=0}^{\infty }{x-a \choose k}~\Delta ^{k}[f](a)~$

I'm not clear on whether it's the difference between the $k^{th}$ term and the $0^{th}$ term, or the difference between the $k^{th}$ term and the $(k-1)^{th}$ term, or the difference between the $(k+1)^{th}$ term and the $k^{th}$ term.

I'm inclined to interpret it

$\displaystyle\Delta ^{k}[f](a)=f^{k}(a)-f^{k-1}(a)$

In which case if $f(x)=2x+1$ then $\displaystyle\Delta ^{3}[f](a)=(8x+7)-(4x+3)=4x+4$

but not sure.

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    $\begingroup$ It is the $\Delta$ operator, iterated $k$ times. $\Delta f=f_{k+1}-f_k$, then $\Delta^2 f=\Delta(f_{k+1}-f_k)=f_{k+2}-2f_{k+1}+f_k$... $\endgroup$ – Yves Daoust Mar 11 '17 at 15:32
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The delta operator is defined as

$$\Delta f_k:=f_{k+1}-f_k$$

where $f_k:=f(k)$, and $$\Delta ^j f_k=\Delta^{j-1}(\Delta f_k)=\Delta^{j-1}(f_{k+1}-f_k)=\Delta^{j-2}(\Delta f_{k+1}-\Delta f_k)$$

because the operator is linear. Using induction you get the general formula

$$\Delta^j f_k:=\sum_{h=0}^j(-1)^{j-h}\binom{j}{h}f_{k+h}$$

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