I ran $\frac{d^n}{dx^n}[(x!)!]$ through the calculator but don’t understand the $R^{(0,1)}$ and $R(n,x)$ in the output

Question

I ran $\frac{d^n}{dx^n}[(x!)!]$ through Wolfram|Alpha, which returned

$$\frac{\partial^n(x!)!}{\partial x^n} = \Gamma(1+x!)\,R(n,1+x!)$$ for

• $R(n,x)=\psi(x)\,R(-1+n,x)+R^{(0,1)}(-1+n,x)$

• $R(0,x)=1$

• $n\in\mathbb{Z}$

• $n>0$

where $\psi^{(n)}(x)$ is the $n$^th derivative of the digamma function

They define the polygamma function as $$\psi^{(n)}(x)=\frac{d^{n+1}}{dx^{n+1}}\ln[\Gamma(x)]$$

What on Earth is $R^{(0,1)}$, and how can I make sense of this $R(n,x)$ business?

Answer 1

Define $f_n(x)$ as follows and show that$$f_n(x)=\Gamma(x)R(n,x)=\frac d{dx}\Gamma(x)R(n-1,x)=f'_{n-1}(x)\\f_0(x)=\Gamma(x)$$

From this, you can see that we actually have

$$f_n(x)=\Gamma^{(n)}(x)$$

And particularly,

$$R(n,x)=\frac{f_n(x)}{\Gamma(x)}=\frac{\Gamma^{(n)}(x)}{\Gamma(x)}$$

Answer 2

The superscript notation means (I think) the derivative: $$R^{(0,1)}(−1+n,x)=\frac{\partial R(n,x)}{\partial x}\Bigr|_{(−1+n,x)}$$

$R(n,x)$ is so ugly, it seems, that it cannot be expressed in some closed form but only as a recursion in $n$:

$$R(n,x)=\frac{d \ln[\Gamma(x)]}{dx} \,R(-1+n,x)+\frac{\partial R(n,x)}{\partial x}\Bigr|_{(−1+n,x)}$$