Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

What is ${\partial\over \partial x_i}(x_i !)$ where $x_i$ is a discrete variable?

Do e consider $(x_i!)=(x_i)(x_i-1)...1$ and do product rule on each term, or something else? THanks.

share|improve this question
    
What do you mean by the 'derivative'? Since you're working with discrete things, do you want the forward difference or something like that? –  Antonio Vargas Feb 11 '13 at 20:37
1  
Writing an expression with a variable number of terms/factors and treating it as if it were fixed formulas is a very bad idea in doing differentation. You will find examples under the tag (fake-proofs) on this site. –  Marc van Leeuwen Aug 27 '13 at 8:55
add comment

4 Answers

The derivative of a function of a discrete variable doesn't really make sense in the typical calculus setting. However, there is a continuous variant of the factorial function called the Gamma function, for which you can take derivatives and evaluate the derivative at integer values.

In particular, since $n!=\Gamma(n+1)$, there is a nice formula for $\Gamma^\prime$ at integer values: $$ \Gamma^\prime(n+1)=n!\left(-\gamma+\sum_{k=1}^n\frac{1}{k}\right) $$ where $\gamma$ is the Euler-Mascheroni constant.

share|improve this answer
    
THank you, icurays1! –  Grizzly Feb 11 '13 at 20:25
1  
It might be good to observe that ther are other differentiable (and even analytic) functions that restrict to the factorial functions on the natural numbers, and that they have different derivatives; the question, even with a liberal interpretation of what it is asking, really has no definite answer. –  Marc van Leeuwen Aug 27 '13 at 8:50
add comment

$x!$ is usually defined only for nonnegative integer $x$. However, there is an extension to non-integers, given by the Gamma function: $x! = \Gamma(x+1)$, and the derivative of this is $\Psi(x+1) \Gamma(x+1)$ where $\Psi$ is the Digamma function. The values of this derivative at $x=0,1,\ldots,10$ are $-\gamma,1-\gamma,3-2\,\gamma,11-6\,\gamma,50-24\,\gamma,274-120\, \gamma,1764-720\,\gamma,13068-5040\,\gamma,109584-40320\,\gamma, 1026576-362880\,\gamma,10628640-3628800\,\gamma$ where $\gamma$ is Euler's constant.

share|improve this answer
    
Thank you, Robert! –  Grizzly Feb 11 '13 at 20:29
add comment

As has been mentioned, the Gamma function $\Gamma(x)$ is the way to go.

Integration by parts yields $$ \begin{align} \Gamma(x) &=\int_0^\infty e^{-t}t^{x-1}\,\mathrm{d}t\\ &=(x-1)\int_0^\infty e^{-t}t^{x-2}\,\mathrm{d}t\\ &=(x-1)\Gamma(x-1) \end{align} $$ Taking the derivative of the logarithm of $\Gamma(x)$ gives $$ \frac{\Gamma'(x)}{\Gamma(x)}=\frac1{x-1}+\frac{\Gamma'(x-1)}{\Gamma(x-1)} $$ Because $\Gamma(x)$ is log-connvex and $$ \lim_{x\to\infty}\frac{\Gamma'(x)}{\Gamma(x)}-\log(x)=0 $$ we get that $$ \frac{\Gamma'(x)}{\Gamma(x)}=-\gamma+\sum_{k=1}^\infty\left(\frac1k-\frac1{k+x-1}\right) $$ For integer $n$, $n!=\Gamma(n+1)$, so the derivative is $$ \begin{align} \Gamma'(n+1) &=\Gamma(n+1)\left(-\gamma+\sum_{k=1}^\infty\frac{n}{k(k+n)}\right)\\ &=n!(-\gamma+H_n) \end{align} $$ where $H_n$ is the $n^\text{th}$ Harmonic Number (with the convention that $H_0=0$).

share|improve this answer
    
Thank you, robjohn! –  Grizzly Feb 12 '13 at 2:33
add comment

It's probably best to use an analytic continuation of the factorial function, rather than the factorial itself. Consider the gamma function:

$\Gamma(x) = \int_{0}^{\infty}x^{t}e^{-t}dt$

Obviously, $\Gamma(0) = 1$, and we also have:

$$\begin{align} \Gamma(x+1) &= \int_{0}^{\infty}x^{t+1}e^{-x}dt\\ &=[t^{x+1}e^{-t}]_{0}^{\infty} + (x+1)\int^{\infty}_{0}t^{x}e^{-t}dt\\ &=(x+1)\Gamma(x) \end{align}$$

So, $\Gamma(x) = (x-1)!$. So, just freely take derivatives now.

share|improve this answer
    
Thank you, Jerry! –  Grizzly Feb 11 '13 at 20:30
1  
There is no such thing as an analytic continuation of the factorial funcion on$~\Bbb N$. The function $x\mapsto\Gamma(x+1)$ is an analytic extension (or maybe interpolation or extrapolation is a better term), but it is not the only one that exists. Other extensions have different derivatives of course. –  Marc van Leeuwen Aug 27 '13 at 8:46
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.