The special case of Pochhammer Symbol at Zero?

Solution 1:

(EDIT)

I would start with the 'extended' definition for any real value of $n$ : $$\tag{1}x^{\bar{n}}=\frac{\Gamma(x+n)}{\Gamma(x)}$$ which should provide directly $x^{\bar{0}}=1$ (since $\Gamma(x)\not =0$ for $x\in \mathbb{R}$ and by analytic continuation for non positive integers as $\Gamma(x)$ becomes infinite)

Let's use $(1)$ too for $x^{\bar{n}}$ as $\,x\to 0$.

For $n>0$ the limit should give us $0^{\bar{n}}=0$ (because $\dfrac 1{\Gamma(x)}=x+\gamma x^2+O(x^3)\;$ to multiply by nearly $\Gamma(n)$)

For $n<0$ the results are more interesting :

• for $n$ not integer the results are $0$ again (as for $n>0$)
• for $n$ negative integer we get $\dfrac {(-1)^n}{(-n)!}$ (factorizing the numerator)

Should we consider the limit as $n\to 0$ then the result would be undefined since the limit would be between $1$ and $0$ depending of $x$ or $n$ going faster to $0$ (as $x\to 0$, $\dfrac 1{\Gamma(x)}\sim x$ so that forcing $n:=x$ would for example generate the limit $\dfrac 12$).
This written and if we set $n=0$ (instead of taking the limit as $n\to 0$) then we may apply our $x^{\bar{0}}=1$ result and obtain $0^{\bar{0}}=1$ (corresponding to our formula for negative integers) similar to the result for the ordinary power $0^{0}=1$ (and with the same reservations...).

Solution 2:

Fix $n>0$ a natural number. Then by your definition $a^\bar{n}$ is a polynomial in $a$ with a root at $a=0$ coming from the first factor, and hence $0^\bar{n}=0$. (Note that the expression is not a polynomial if $n$ varies).

For the case $n=0$ by your definition we have $0^{\bar{0}} =\prod_{k=1}^0(k-1)$ and this empty product is equal to 1. However I would want to see how this matches up with other definitions before making a definitive statement here.