# Introducing Operator Delta

The $r^{th}$ order finite difference of function$f(x)$ is defined by

$\Delta^r f(x) = \begin{cases} f(x+1)-f(x), r=1\\ \Delta(\Delta^{r-1}f(x)), r > 1\end{cases}$

From this definition, we have

$\Delta f(x) = \Delta^1 f(x) = f(x+1)-f(x)$

and,

$\Delta^2 f(x) = \Delta (\Delta^{2-1} f(x))$

$= \Delta (\Delta f(x))$

$= \Delta( f(x+1)-f(x))$

$= (f(x+2)-f(x+1)) - (f(x+1)-f(x))$

$= f(x+2)-2f(x)+f(x+1)$

as well as

$\Delta^3 f(x) = \Delta (\Delta^2 f(x))$

$= \Delta (f(x+2)-2f(x)+f(x+1))$

$= (f(x+3)-2f(x+1)+f(x+2)) - (f(x+2)-2f(x)+f(x+1))$

$= f(x+3)-3f(x+2)+3f(x+1)-f(x)$

The function shown below generates $\Delta^r f(x), r:1\rightarrow 5$ (see Fig. 1).

delta_(g, n) := block(
local(f),

define(f[1](x),
g(x+1)-g(x)),

for i : 2 thru n do (
define(f[i](x),
f[i-1](x+1)-f[i-1](x))
),

return(f[n])
);


Fig. 1

Compare to the result of expanding $(f(x)-1)^r=\sum\limits_{i=0}^r(-1)^i \binom{r}{i} f(x)^{r-i}, r:1\rightarrow 5$ (see Fig. 2)

Fig. 2

It seems that

$\Delta^r f(x) = \sum\limits_{i=0}^r(-1)^i \binom{r}{i} f(x+r-i)\quad\quad\quad(1)$

Lets prove it!

We have already shown that (1) is true for $r= 1, 2, 3$.

Assuming (1) is true when $r=k-1 \ge 4$:

$\Delta^{k-1} f(x) = \sum\limits_{i=0}^{k-1}(-1)^i \binom{r}{i} f(x+k-1-i)\quad\quad\quad(2)$

When $r=k$,

$\Delta^k f(x) = \Delta(\Delta^{k-1} f(x))$

$\overset{(2)}{=}\Delta (\sum\limits_{i=0}^{k-1}(-1)^i \binom{k-1}{i}f(x+k-1-i))$

$=\sum\limits_{i=0}^{k-1}(-1)^i \binom{k-1}{i}f(x+1+k-1-i)-\sum\limits_{i=0}^{k-1}(-1)^i \binom{k-1}{i}f(x+k-1-i)$

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

$+\sum\limits_{i=1}^{k-1}(-1)^i \binom{k-1}{i}f(x+k-i)-\sum\limits_{i=0}^{k-2}(-1)^i \binom{k-1}{i}f(x+k-1-i)$

$-(-1)^{k-1}\binom{k-1}{k-1}f(x+k-1-(k-1))$

$\overset{\binom{k-1}{0} = \binom{k-1}{k-1}=1}{=}$

$f(x+k)+ \sum\limits_{i=1}^{k-1}(-1)^i \binom{k-1}{i}f(x+k-i)-\sum\limits_{i=0}^{k-2}(-1)^i \binom{k-1}{i}f(x+k-1-i) -(-1)^{k-1}f(x)$

$=f(x+k)+ \sum\limits_{i=1}^{k-1}(-1)^i \binom{k-1}{i}f(x+k-i)+\sum\limits_{i=0}^{k-2}(-1)^{i+1}\binom{k-1}{i}f(x+k-1-i) -(-1)^{k-1}f(x)$

$\overset{j=i+1, i:0 \rightarrow k-2\implies j:1 \rightarrow k-1}{=}$

$f(x+k)+ \sum\limits_{i=1}^{k-1}(-1)^i\binom{k-1}{i}f(x+k-i) + \sum\limits_{j=1}^{k-1}(-1)^j \binom{k-1}{j-1}f(x+k-j)-(-1)^{k-1}f(x)$

$= f(x+k)+ \sum\limits_{i=1}^{k-1}(-1)^i\binom{k-1}{i}f(x+k-i) + \sum\limits_{i=1}^{k-1}(-1)^i\binom{k-1}{i-1}f(x+k-i)+(-1)^k f(x)$

$= f(x+k) + \sum\limits_{i=1}^{k-1}(-1)^i f(x+k-i) (\binom{k-1}{i} + \binom{k-1}{i-1})+(-1)^k f(x)$

$\overset{\binom{k-1}{i} + \binom{k-1}{i-1}=\binom{k}{i}}{=}$

$f(x+k)+ \sum\limits_{i=1}^{k-1}(-1)^i \binom{k}{i} f(x+k-i)+(-1)^k f(x)$

$= (-1)^0 \binom{k}{0}f(x+k-0)+\sum\limits_{i=1}^{k-1}(-1)^i \binom{k}{i} f(x+k-i)+(-1)^k \binom{k}{k} f(x+k-k)$

$= \sum\limits_{i=0}^{k}(-1)^i \binom{k}{i} f(x+k-i)$