# A Feynmanistic Derivation

The operator $\Delta$ introduced in my previous post has many properties. The most notable ones are:

(1) $\Delta c = 0, c$ is a constant

(2) $\Delta(c\cdot f(x)) = c\cdot \Delta f(x)$

(3) $\Delta (f_1(x) + f_2(x) + ... + f_n(x) ) = \Delta f_1(x) + \Delta f_2(x) + ... + \Delta f_n(x)$

(4) $\Delta(f(x)\cdot g(x)) = (f(x)+1)\Delta g(x) + g(x) \Delta f(x)$

(5) $\Delta\frac{f(x)}{g(x)} = \frac{g(x)\Delta f(x) - f(x)\Delta g(x)}{g(x)(g(x)+1)}$

(6) $\Delta^{p} x^m = 0, p > m$

To derive (6), let us consider

$\Delta^{k-1} x^m, k=2,3,...m$.

When $k=2$,

$\Delta^{k-1} x^m = \Delta^{2-1}x^m=\Delta x^m$

$= (x+1)^m-x^m$

$= mx^{m-1} + \binom{m}{2}x^{m-1} +...$

$= mx^{m-1} + \frac{m(m-1)}{2!}x^{m-2} + ...$

$= mx^{m-1} + 1\cdot \frac{m(m-1)}{2!}x^{m-2} + ...$

$=m x^{m-(2-1)} + (2-1)\cdot \frac{m(m-1)}{2!}x^{m-2}+ ...$

$= (m-k+2) x^{m-(k-1)} + (k-1)\cdot \frac{m(m-k+1)}{2!}x^{m-k} + ...$.

When $k=3$,

$\Delta^{k-1} x^m = \Delta^{3-1} x^m = \Delta^2 x^m = \Delta (\Delta x^m)$

$= m((m-1)x^{m-2} + \binom{m-1}{2}x^{m-3} + ... ) + \binom{m}{2}((m-2) x^{m-3} + ... )$

$= m(m-1)x^{m-2} + m(\binom{m-1}{2} + \binom{m}{2}(m-2))x^{m-3} + ...$

$= m(m-1)x^{m-2} +(\frac{m(m-1)(m-2)}{2!} + \frac{m(m-1)(m-2)}{2!})x^{m-3} + ...$

$= m(m-1)x^{m-2} + 2\cdot \frac{m(m-1)(m-2)}{2!}x^{m-3} + ...$

$= m(m-1)x^{m-(3-1)} + (3-1) \cdot \frac{m(m-1)(m-2)}{2!}x^{m-3} + ...$

$= m(m-k+2)x^{m-(k-1)} + (k-1) \cdot \frac{m(m-1)(m-k+1)}{2!} x^{m-k}+ ...$.

When $k = 4$,

$\Delta^{k-1} x^m = \Delta^{4-1}x^m = \Delta^3 x^m = \Delta(\Delta^2 x^m)$

$=m(m-1)((m-2)x^{m-3} + \binom{m-2}{2}x^{m-4} + ...) + 2\cdot\frac{m(m-1)(m-2)}{2!}((m-3)x^{m-4}+...)$

$=m(m-1)(m-2)x^{m-3} + \frac{m(m-1)(m-2)(m-3)}{2!}x^{m-4}+2\cdot \frac{m(m-1)(m-2)(m-3)}{2!}x^{m-4} + ...$

$= m(m-1)(m-2)x^{m-3} + 3\cdot \frac{m(m-1)(m-2)(m-3)}{2!}x^{m-4} + ...$

$= m(m-1)(m-2)x^{m-(4-1)} + (4-1) \cdot \frac{m(m-1)(m-2)(m-3)}{2!}x^{m-4} + ...$

$= m(m-1)(m-k+2)x^{m-(k-1)} + (k-1) \cdot \frac{m(m-1)(m-2)(m-k+1)}{2!} x^{m-k} + ...$.

When $k = 5$,

$\Delta^{k-1}x^m = \Delta^{5-1} x^m = \Delta^4 x^m = \Delta(\Delta^3 x^m)$

$= m(m-1)(m-2)((m-3)x^{m-4} + \binom{m-3}{2} x^{m-5}+ ...)$

$+ 3\cdot \frac{m(m-1)(m-2)(m-3)}{2!}((m-4)x^{m-5} +...)$

$= m(m-1)(m-2)(m-3)x^{m-4} +\frac{m(m-1)(m-2)(m-3)(m-4)}{2!}x^{m-5}$

$+ 3\cdot\frac{m(m-1)(m-2)(m-3)(m-4)}{2!}x^{m-5} + ...$

$= m(m-1)(m-2)(m-3)x^{m-4} + 4\cdot\frac{m(m-1)(m-2)(m-3)(m-4)}{2!} x^{m-5} + ...$

$= m(m-1)(m-2)(m-3)x^{m-(5-1)} + (5-1)\cdot\frac{m(m-1)(m-2)(m-3)(m-4)}{2!}x^{m-5} + ...$

$= m(m-1)(m-2)(m-k+2)x^{m-(k-1)} + (k-1)\cdot\frac{m(m-1)(m-2)(m-3)(m-k+1)}{2!}x^{m-k} + ...$

$\vdots$

When $k=m$,

$\Delta^{k-1} x^m = \Delta^{m-1}x^m$

$= m(m-1)(m-2)(m-3) \cdots 2 x + (m-1)\cdot \frac{m(m-1)(m-2)(m-3) \cdots 1}{2!} + 0$

$= m! x + (m-1)\cdot \frac{m!}{2!}$

$= m! x + (m-1)\cdot \frac{m!}{2}$

$= m!(x+\frac{m-1}{2})$.

Therefore,

$\Delta^m x^m = \Delta(\Delta^{m-1}x^m)=\Delta(m!(x+\frac{m-1}{2})= m! \Delta(x+\frac{m+1}{2}))= m! (\Delta x + \Delta(\frac{m-1}{2}))= m! (1+0)= m!.$

It follows that

$p>m \implies p-m>0 \implies p-m \ge 1 \implies \Delta^p x^m = \Delta^{p-m}(\Delta^{m}x^m) = \Delta^{p-m}m!= 0.$

# 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 $r=k$,

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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