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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?

gen-ℤ ready to perish
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2 Answers2

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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 $n$:

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

leonbloy
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  • So when you say “cannot,” do you mean it is *literally* impossible? – gen-ℤ ready to perish Aug 20 '17 at 05:33
  • That word in bold doesn't mean anything, unless you define exactly what we mean by "closed form". See eg https://math.stackexchange.com/questions/155/how-can-you-prove-that-a-function-has-no-closed-form-integral – leonbloy Aug 20 '17 at 14:51
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    Your interpretation of the superscript notation is indeed correct as far as WA goes. – Simply Beautiful Art Aug 20 '17 at 22:50
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    Interestingly, $R(n,x)$ is not so ugly if you allow $\Gamma^{(n)}(x)$ in it. However, $\Gamma^{(n)}(x)$ cannot be written nicely in terms of polygamma functions, which is somewhat unfortunate. – Simply Beautiful Art Aug 20 '17 at 23:15
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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,

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

Simply Beautiful Art
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