Understanding the Derivative of x: A Scientist's Perspective

[Emmanuel_Euler]

what is the derivative of x! ??

 

[pwsnafu]

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

 

[Emmanuel_Euler]

thank you for help.

 

[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.

 

[certainly]

but x! is only defined on the natural numbers

The previous response is quite correct

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......

 

[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 $\mathbb {R}^-$.

 

[micromass]

In-fact the value of the gamma function can also be computed for all negative integers

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 $\mathbb{C}$).

But I believe it is not continuous in $\mathbb {R}^-$.

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.

 

[pwsnafu]

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......

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.

 

[certainly]

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.

Sorry....that was a silly error on my part.....

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 C\mathbb{C}).

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".]

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.

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......

 

[pwsnafu]

[It's sort of like an extended gamma function, obtained by defining the gamma function in terms of a "neutrix".]

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.

 

[Emmanuel_Euler]

Thanks to all who helped me.

but please can you give me the derivative formula??

 

[lonely_nucleus]

Thanks to all who helped me.

but please can you give me the derivative formula??

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

 

[FactChecker]

This is not correct...
The gamma function is defined for all $x$ .

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.

 

[riemannsigma]

Not continuous

 

[WWGD]

what is the derivative of x! ??

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

 

[phion]

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

 

[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.

 

[SnotRocketSci]

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.

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

 

[Emmanuel_Euler]

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

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

 

[Emmanuel_Euler]

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

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.

 

[phion]

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.

I was hoping it would, but it does not.

 

[WWGD]

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

 

[phion]

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

Oh really?

 

[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 ;-).

 

[WWGD]

Oh really?

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

 

[phion]

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

We're reduced to finding the factorial of a fraction with the gamma function I think.

 

[WWGD]

We're reduced to finding the factorial of a fraction with the gamma function I think.

But then, like it has been posted, it is not the $x!$ function.

 

[phion]

But then, like it has been posted, it is not the #x!# function.

I still don't see how what Wolfram Alpha stated was actually the derivative of the gamma function.

 

[pwsnafu]

I still don't see how what Wolfram Alpha stated was actually the derivative of the gamma function.

The digamma is the logarithmic derivative of the gamma. If you multiply by the gamma what do you get?

 

[phion]

The digamma is the logarithmic derivative of the gamma. If you multiply by the gamma what do you get?

Got it.