# Simple calculus question

Definition: $\Delta f(x) = f(x + \Delta x) - f(x)$. Find $\Delta^nf(x)$. My guess is

$$\Delta^nf(x) = f(x + n\Delta x) + \sum_{k=0}^{n}(-1)^kkf(x + (n-k)\Delta x) + (-1)^nf(x)$$

Is this guess correct?

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Did you mean $\,\Delta^n f(x)\,$? – DonAntonio Oct 17 '12 at 4:40
You may want to check N-th finite difference formula in: en.wikipedia.org/wiki/Finite_difference – NoChance Oct 17 '12 at 5:13

Calculate the first few values:

\begin{align*} \Delta^2f(x)&=\Delta\left(f(x+\Delta x)-f(x)\right)\\ &=\Big(f(x+2\Delta x)-f(x+\Delta x)\Big)-\Big(f(x+\Delta x)-f(x)\Big)\\ &=f(x+2\Delta x)-2f(x+\Delta x)+f(x)\;, \end{align*}

\begin{align*} \Delta^3f(x)&=\Delta\left(\Delta^2f(x)\right)\\ &=\Delta\Big(f(x+2\Delta x)-2f(x+\Delta x)+f(x)\Big)\\ &=\Big(f(x+3\Delta x)-2f(x+2\Delta x)+f(x+\Delta x)\Big)\\ &\qquad\quad-\Big(f(x+2\Delta x)-2f(x+\Delta x)+f(x)\Big)\\ &=f(x+3\Delta x)-3f(x+2\Delta x)+3f(x+\Delta x)-f(x)\;, \end{align*}

and

\begin{align*} \Delta^4f(x)&=\Delta\left(\Delta^3f(x)\right)\\ &=\Delta\Big(f(x+3\Delta x)-3f(x+2\Delta x)+3f(x+\Delta x)-f(x)\Big)\\ &=\Big(f(x+4\Delta x)-3f(x+3\Delta x)+3f(x+2\Delta x)-f(x+\Delta x)\Big)\\ &\qquad\quad-\Big(f(x+3\Delta x)-3f(x+2\Delta x)+3f(x+\Delta x)-f(x)\Big)\\ &=f(x+4\Delta x)-4f(x+3\Delta x)+6f(x+2\Delta x)-4f(x+\Delta x)+f(x)\;. \end{align*}

Now look at those sequences of coefficients:

\begin{align*} &1,-2,1\\ &1,-3,3,-1\\ &1,-4,6,-4,1 \end{align*}

1. Do they match your formula?

2. Are they familiar from some other context?

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Your formula gives $\Delta f(x) = f(x+\Delta x)-2 f(x)$, so it can't be right. Try computing the formula for $\Delta^2 f, \Delta^3 f$ and see if you can spot the pattern. Then verify by induction.

Here is the correct formula:

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

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Did something change? I don't care about the points, but why did you 'unaccept' after two years? – copper.hat Dec 2 '14 at 16:52