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## Proof: (related to "Nth Difference Operator")

• By hypothesis, $f:D\to\mathbb R$ is a function with $D\subseteq\mathbb R.$ Moreover, let $n\in\mathbb N,$ $x, x+1,\ldots, x+n\in D.$
• The statement can be proved by induction.
• Base case $n=1$
• Induction step $n\to n+1$
• Inductin premise: Assume $n\ge 1$ and let the formula $$\Delta^m f(x)=\sum_{k=0}^m f(x+k)\cdot (-1)^{m-k}\cdot\binom mk$$ be true for all $m\le n.$
• By the induction step premise, and basic calculations involving the difference operator (no. 1 and no. 2) it follows \begin{align}\Delta^{n+1} f(x)&=\Delta (\Delta^n f(n))\\ &=\Delta\left( \sum_{k=0}^n f(x+k)\cdot (-1)^{n-k}\cdot\binom nk\right)\\ &=\sum_{k=0}^n(f(x+k+1)-f(x+k))\cdot (-1)^{n-k}\cdot\binom nk\\ &=\sum_{k=0}^n f(x+k+1)\cdot (-1)^{n-k}\cdot\binom nk-\sum_{k=0}^n f(x+k)\cdot (-1)^{n-k}\cdot\binom nk\\ &=\sum_{k=0}^n f(x+k+1)\cdot (-1)^{n-k}\cdot\binom nk+\sum_{k=0}^n f(x+k)\cdot (-1)^{n+1-k}\cdot\binom nk\\ \end{align}
• Changing the index $k+1\to k$ in the left sum allows combining both sums together:
\begin{align} &=\sum_{k=1}^{n+1} f(x+k)\cdot (-1)^{n-(k-1)}\cdot\binom n{k-1}+\sum_{k=0}^n f(x+k)\cdot (-1)^{n+1-k}\cdot\binom nk\\ &=f(x+n+1)+\sum_{k=1}^{n} f(x+k)\cdot (-1)^{(n+1)-k}\cdot\binom n{k-1}\cdot \binom nk+f(x)\cdot (-1)^{n+1}\\ \end{align}
• With the recursive formula for the binomial coefficient, it follows
\begin{align}&=f(x+n+1)+\sum_{k=1}^{n} f(x+k)\cdot (-1)^{(n+1)-k}\cdot\binom {n+1}{k}+f(x)\cdot (-1)^{n+1}\\ &=\sum_{k=0}^{n+1} f(x+k)\cdot (-1)^{(n+1)-k}\cdot\binom {n+1}{k}. \end{align}
q.e.d

| | | | created: 2020-03-25 13:46:36 | modified: 2020-03-25 13:46:45 | by: | references: [1591]

### Bibliography (further reading)

[1591] Piotrowski, Andreas: “bookofproofs.org”, Own Work, 2014