Q:

Shifting operator in Newton's forward and backward difference formula

Interpolation Newton's Forward Difference Formula
Let $y=f(x)$ be a function which takes values $f(x_0),f(x_0+h),\cdots$ corresponding to various equi-spaced values of $x$ with spacing $h$, say $x_0,x_0+h,\cdots$ 
  Suppose, we wish to evaluate the function $f(x)$ for a value $x_0+ph$, where $p$ is any real number, then for any real number $p$, we have the operator $E$ such that 
  \begin{align}
E^pf(x)&=f(x+ph)\\
f(x_0+ph)=E^pf(x_0)&=(1+\Delta)^pf(x_0)\\
&=\left[1+p\Delta+\frac{p(p-1)}{2!}\Delta^2+\cdots \right]f(x_0)
\end{align}
$$f(x_0+ph)=f(x_0)+p\Delta f(x_0)+\frac{p(p-1)}{2!}\Delta^2 f(x_0)+\cdots+\frac{p(p-1)(p-2)\cdots(p-n+1)}{n!}\Delta^n f(x_0)+\text{Error}$$
Interpolation Newton's Backward Difference Formula
Let $y=f(x)$ be a function which takes values $f(x_n),f(x_n-h),\cdots$ corresponding to various equi-spaced values of $x$ with spacing $h$, say $x_n,x_n-h,\cdots$ 
  Suppose, we wish to evaluate the function $f(x)$ for a value $x_n+ph$, where $p$ is any real number, then for any real number $p$, we have the shift operator $E$ such that 
  \begin{align}
f(x_n+ph)=E^pf(x_n)=(E^{-1})^{-p}&=(1-\Delta)^{-p}f(x_n)\\
&=\left[1+p\Delta+\frac{p(p+1)}{2!}\Delta^2+\cdots \right]f(x_n)
\end{align}
$$f(x_n+ph)=f(x_n)+p\Delta f(x_n)+\frac{p(p-1)}{2!}\Delta^2 f(x_n)+\cdots+\frac{p(p+1)(p+2)\cdots(p+n-1)}{n!}\Delta^n f(x_n)+\text{Error}$$

I really struggling with the shifting operator mentioned in this proof. How did it operate$?$Any help will be appreciated.
Thanks in advances.

A:

The shifting operator takes a function $f$ and returns a function $E[f]$. This new function has values taken from the old function per $E[f](x)=f(x+h)$. Repeated application gives $E^k[f](x)=f(x+kh)$. 
The difference operator is defined as $Δ=E-1$, $Δ[f](x)=f(x+h)-f(x)$, so that in return $E=1+Δ$. The remaining computation is the application of the binomial theorem
$$
E^p[f]=(1+Δ)^p[f]=f+\sum_{k=1}^p\binom{p}{k}Δ^k[f].
$$
The second formula uses a different difference operator $Δ_-=1-E^{-1}=E^{-1}Δ$, $Δ_-[f](x)=f(x)-f(x-h)$. Then indeed $E^{-1}=1-Δ_-$ and the binomial series can be (formally) applied
$$
E^p=(1-Δ_-)^{-p}=\sum_{k=0}^\infty\binom{-p}k(-Δ_-)^k
=\sum_{k=0}^\infty\binom{p+k-1}k(Δ_-)^k
$$

