# What is the derivative of x! ?

1. Apr 14, 2015

### Emmanuel_Euler

what is the derivative of x! ??

2. Apr 14, 2015

### pwsnafu

Very easy: doesn't exist. Any differentiable function is necessarily continuous, but x! is only defined on the natural numbers, and not continuous.

3. Apr 14, 2015

### Emmanuel_Euler

thank you for help.

4. May 8, 2015

### daverusin

The previous response is quite correct. However, there is a "natural" continuous extension of the factorial function -- a way to "connect the dots", if you like. Look up the "Bohr–Mollerup theorem" and learn about the Gamma function. For integers x we have x! = Gamma(x+1), so perhaps you would like to know about Gamma'(x+1)? That would be equal to (x!) ( 1 + 1/2 + ... + 1/x - gamma ) where gamma=0.577... is Euler's constant.

It's also worth learning Stirling's Approximation: x! is roughly (x/e)^x sqrt(2 pi x ) . That's a formula you can differentiate using calculus-1 tools.

5. May 8, 2015

### certainly

This is not correct................
The gamma function is defined for all $x$ and values of the function can be found for infinitely many rational arguments by the identity $\Gamma(\frac{1}{2})=\sqrt{\pi}$
Also note that this means that $\Big(-\frac{1}{2}\Big)!=\sqrt{\pi}$. Values of the function exist for other rational numbers as well but they are somewhat complicated.
Also the gamma function is continuous in the domain $\mathbb{R}^+$
See this stackexchange thread...................

Last edited: May 8, 2015
6. May 8, 2015

### certainly

In-fact the value of the gamma function can also be computed for all negative integers.............
See this paper by Fischer and Kilicman.............................
But I believe it is not continuous in $\mathbb {R}^-$.

7. May 8, 2015

### micromass

The gamma function has poles at the negative integers. It is not defined there. And if it were to be defined, it would be $\infty$ (in the one-point compactification of $\mathbb{C}$).

It is. It is meromorphic, which means that it induces a continuous (even holomorphic) function $\Gamma:\mathbb{C}\rightarrow \mathbb{C}_\infty$, where $\mathbb{C}_\infty$ is the Riemann sphere.

8. May 8, 2015

### pwsnafu

If you define the factorial as the gamma sure. Most texts don't. They define the factorial as a product, and show that the gamma function is an extension of the factorial.

9. May 9, 2015

### certainly

Sorry.............that was a silly error on my part..............
Perhaps I should have been a little more careful.
What you said is perfectly true, in the classical sense the gamma function is not defined for negative integers.....................
Please have a look at the link i provided ...............
[It's sort of like an extended gamma function, obtained by defining the gamma function in terms of a "neutrix".]
While teaching this approach is more intuitive and often historically accurate, however, in practice, the general version (what maybe called the extended version) is always preferred as a definition. So perhaps my reply should have been something like, now that you know about the gamma function the answer will be the derivative of the gamma function........................

10. May 9, 2015

### pwsnafu

For people who are interested the paper uses neutrix calculus, which concerns taking diverging limits/series/integrals and removing the "infinite part" of it. The idea is that $G$ is a group of functions under addition and define $N$ a subgroup such that the only constant function contained is the zero function. N is the called the "neutrix" and the elements of N are called "negligible". The limits are then calculated modulo N.

Neurtix calculus sees a lot of use in generalized functions (because it was Hadamard's work that started this technique) and you see it when defining intrinsic product of Schwartz distributions. I'm told QFT uses it, but I'm not a physicist so I can't say anything about that.

Last edited: May 9, 2015
11. May 11, 2015

### Emmanuel_Euler

Thanks to all who helped me.

but please can you give me the derivative formula??

12. Jun 19, 2015

### lonely_nucleus

do the exponent short cut or you can find the limit of the function as Δx→0.
da/dh=

13. Jun 19, 2015

### FactChecker

You can not say that the gamma function and factorial are the same function. They are not defined on the same domain or in the same way. Just because gamma is an extension of factorial does not mean that factorial is more than its original, simple definition.

14. Jun 20, 2015

### riemannsigma

Not continuous

15. Jun 21, 2015

### WWGD

Still, think of the differential quotient. If x! is defined only on the integers, what is then $f(x+h) , h \rightarrow 0$ ?

16. Jun 24, 2015

### phion

Wolfram Alpha states the derivative of $x!$ is $\Gamma (x+1) \psi^{0}(x+1)$.

17. Jun 24, 2015

### HallsofIvy

I'm sorry to hear that! Yes, for n an integer, $\Gamma(n)= (n- 1)!$ but I would NOT agree that the factorial is the gamma function. As FactChecker said, they are different functions that happen to have a simple relationship.

18. Jun 25, 2015

### SnotRocketSci

For small x it can be a poor approximation. I never thought about this being differentiable. Cool thought.

19. Jun 26, 2015

### Emmanuel_Euler

Friend it is not that easy to solve.
But i will try.

20. Jun 26, 2015

### Emmanuel_Euler

Does the wolfram alpha show the steps??
It usually does, but i do not think that it will show me the steps only the formula.
Again i want to thank you all.

21. Jun 26, 2015

### phion

I was hoping it would, but it does not.

22. Jun 26, 2015

### WWGD

I think Wolfram is computing the derivative of the Gamma function, not of x!

23. Jun 26, 2015

### phion

Oh really?

24. Jun 26, 2015

### my2cts

A pedestrian approach is (x+1)!-x!=x+1, which gives xx! or x!-(x-1)! which gives (x-1)(x-1)!
The truth must be somewhere near the middle.
The derivative grows even faster with x than the original which makes my estimate very unreliable ;-).

25. Jun 26, 2015

### WWGD

Well, if this is the standard derivative of x!, defined for integers, what is then $f(a+h):=(a+h)!$ , for $0<h<1$ ?