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## Proposition: Nth Difference Operator

Let $f:D\to\mathbb R$ be a function with $D\subseteq\mathbb R$ and let $n\in\mathbb N,$ $x, x+1,\ldots, x+n\in D.$ The recursively defined nth difference operator $$\Delta^n f(x):=\Delta(\Delta^{n-1}f(x))$$ can be calculated explicitely using the binomial coefficients as follows:

$$\Delta^n f(x)=\sum_{k=0}^n f(x+k)\cdot (-1)^{n-k}\cdot\binom nk.$$

| | | | | created: 2020-03-25 13:44:04 | modified: 2020-03-25 13:46:05 | by: bookofproofs | references: [1591]