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)}

We will add to this list an additonal property:

(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.

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