Difference Calculus of the Factorial Function for Negative Integer Powers of $x$

Question

Under the difference calculus, we have that

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

We also have factorial polynomials (sometimes referred to the rising and or falling factorials). The text I am reading defines the factorial function as

$$x^{(n)}=x(x-1)(x-2)...(x-(n-1))$$

with $x^{(0)}=1$ by definition. Applying Delta to the factorial function gives

$$\Delta x^{(n)}=(x+1)^{(n)}-x^{(n)}$$ $$=(x+1)[x(x-1)...(x-(n-2))-[x(x-1)...(x-(n-2))](x-(n-1))$$ $$=[(x+1)-(x-n+1)][x(x-1)...(x-(n-2))]$$ $$=nx^{(n-1)}$$

and this holds for positive $n$. It is then noted that

$$x^{(m+n)}=x(x-1)...(x-(m-1))\cdot(x-m)(x-m-1)...(x-m-(n-1))$$

and so

$$x^{(m+n)}=x^{(m)}\cdot (x-m)^{(n)}$$

Now by the above, if we let $m+n=0$, then $m=-n$. So we have

$$1=x^{(0)}=x^{(-n+n)}=x^{(-n)}(x-(-n))^{(n)}=x^{(-n)}(x+n)^{(n)}$$

So since $1=x^{(-n)}(x+n)^{(n)}$, we have that

$$x^{(-n)}=\frac{1}{(x+n)^{(n)}}$$

Now I am in the game to show that

$$\Delta x^{(n)}=nx^{(n-1)}$$

for negative integers. Which I interpret as trying to find

$$\Delta x^{(-n)}$$ So when I attempt the calculation, I get

$$\Delta x^{(-n)}=(x+1)^{(-n)}-x^{(n)}$$ $$=\frac{1}{(x+1+n)^{(n)}}-\frac{1}{(x+n)^{(n)}}$$ $$=\frac{1}{(x+1+n)(x+n)...(x+2)}-\frac{1}{(x+n)(x+(n-1))...(x+1)}$$ $$=\frac{1}{(x+1+n)[(x+n)...(x+2)]}-\frac{1}{[(x+n)...(x+2)](x+1)}$$ $$=\frac{1}{x+1+n}\frac{1}{(x+n)...(x+2)}-\frac{1}{x+1}\frac{1}{(x+n)...(x+2)}$$ $$=\left[\frac{1}{x+1+n}-\frac{1}{x+1}\right]\frac{1}{(x+n)...(x+2)}$$ $$=\frac{(x+1)-(x+1+n)}{(x+1)(x+1+n)}\frac{1}{(x+n)^{(n-1)}}$$

And this is where i get stuck...

Answer 1

Everything looks fine: We obtain \begin{align*} \color{blue}{\Delta x^{(-n)}}&=(x+1)^{(-n)}-x^{(n)}\\ &=\cdots\\ &=\frac{(x+1)-(x+1+n)}{(x+1)(x+1+n)}\cdot\frac{1}{(x+n)^{(n-1)}}\\ &=-n\cdot\frac{1}{(x+n+1)\cdot (x+n)^{(n-1)}\cdot (x+1)}\\ &=-n\cdot\frac{1}{(x+n+1)^{(n+1)}}\\ &=\color{blue}{-nx^{(-n-1)}} \end{align*} and we conclude the rule is valid for negative $n$ as well.

Hint: You might be interested in the calculation following (2.52) in Concrete Mathematics by R.L. Graham, D.E. Knuth and O. Patashnik.