Question

What is the derivative of 1/x! and why?

Answer 1

The derivative of \(\dfrac{1}{x}\) or \(\dfrac{1}{x!}\)? I'm assuming it's the first one. :)

Answer 2

No it's the second

Answer 3

one over x factorial.

Answer 4

So then we're defining factorial for non-natural numbers with the Gamma function evaluated at a corresponding value?

Answer 5

yes, right.

Answer 6

So we're \(\dfrac{d}{dx}\dfrac{1}{\Gamma (x+1)}\).

Answer 7

ok, I get that much.

Answer 8

Daniel, you got me interested, but unfortunately I don't know if I can be of help here.
I hope someone with more background can jump in and help.
With that being said though, have you tried applying the Leibniz Integral Rule to the Gamma function to find its derivative?

Answer 9

I looked at that but it seems to be about understanding the Clausen Function.

Answer 10

While nobody else is posting:
We have \(\displaystyle \Gamma(x) = \int_0^\infty e^{-t}t^{x-1} \, dt\). For the partial derivative of the integrand, we have \[ \frac{\partial}{\partial x} (t^{x-1}e^{-t}) = e^{-t}t^{x-1}\log t,\] (Where \(\log\) is the natural log.) Therefore \[\dfrac{d}{dx}\Gamma (x) = \int_0^\infty e^{-t}t^{x-1}\log t \, dt.\]
So if that's right, then by the Chain Rule \[\dfrac{d}{dx}\dfrac{1}{\Gamma (x+1)} = \dfrac{-1}{(\Gamma(x))^2}\int_0^\infty e^{-t}t^{x-1}\log t \, dt.\]

Answer 11

Not entirely. I forgot about the \(x+1\), so the \(x\) on the RHS should be shifted. \[ \dfrac{d}{dx}\dfrac{1}{\Gamma (x+1)} = \dfrac{-1}{(\Gamma(x+1))^2}\int_0^\infty e^{-t}t^{x}\log t \, dt. \]
Like I said, I hope someone with more background will take a look at this thread and give us their thoughts

Answer 12

It's interesting, wolfram used the polygamma function. http://www.wolframalpha.com/input/?i=derive+1%2Fx!

Is your form equivalent? Might be worth a check.

Answer 13

Yeah, it's pretty interesting. I also had Wolfram Alpha give me the derivative of the Gamma function itself yesterday to see what it said. I haven't studied the polygamma function (as I said, I'm sort of in unfamiliar territory here), but I looked specifically at the Wikipedia page for the Digamma function:
https://en.wikipedia.org/wiki/Digamma_function
It is defined to be the logarithmic derivative of the Gamma function.

Typing the result I wrote above for the derivative of the Gamma function into Wolfram Alpha,
http://www.wolframalpha.com/input/?i= \int_0^\infty+e^{-t}t^{x-1}\log+t+\,+dt
we get the digamma function times the gamma function as the result (at least for \(\text{Re}(x) > 0\), Wolfram Alpha says), i.e., the derivative of the gamma function, so that checks out.

The rest is an application of the Chain Rule, so the final result should be equivalent as well.

Answer 14

The answer looks to be a product of the quotient rule but because of the nature of the gamma function it's difficult to see how. Do you really know how the chain rule or the quotient rule may be applied in this situation?

Answer 15

There is nothing special about this situation which affects how the Chain Rule should be applied. We simply have a composition of functions: Let \(g(x)=\dfrac{1}{x}\) and \(f(x) = \Gamma(x+1)\). Then your function \(\dfrac{1}{x!}\) is \(g(f(x))\), and its derivative is \(g'(f(x))f'(x)\), which is \(\dfrac{-1}{(\Gamma(x+1))^2} \Gamma'(x+1)\).

Answer 16

I'm not convinced. Can you show your steps?

Answer 17

I just showed my steps in the last post. Which part is not convincing?

Answer 18

Last part.

Answer 19

I can't tell what you mean by "last part". I just applied the Chain Rule and I showed you how I decomposed the function so that there would be no confusion. Please include the specific expression or claim which I need to explain further.

Answer 20

The last step you did, can you explain why -1?

Answer 21

Because the derivative of \(\dfrac{1}{x}\) is \(\dfrac{-1}{x^2}\).