mersenneforum.org Approximate derivative of x!
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 2020-05-10, 02:04 #1 othermath     May 2020 510 Posts Approximate derivative of x! Working experimentally, I found the approach $(x!)'=x!(\frac{x-2}{2(x-1)^2}+lnx)$ For x>2 the approach improves as the value of x increases. I don't know if this is true for very large values ​​of x, (x € R). Can you check it?
2020-05-10, 02:38   #2
Batalov

"Serge"
Mar 2008
Phi(4,2^7658614+1)/2

915810 Posts

Quote:
 Originally Posted by othermath I don't know if this is true for very large values ​​of x, (x € R).
What is x! for x € R ?
Define it and then we can talk about derivatives.

2020-05-10, 03:17   #3
othermath

May 2020

5 Posts

Quote:
 Originally Posted by Batalov What is x! for x € R ? Define it and then we can talk about derivatives.
I want to say that x is a real number.

 2020-05-10, 04:51 #4 VBCurtis     "Curtis" Feb 2005 Riverside, CA 2×2,239 Posts And what do you want to say the factorial of a real number is?
 2020-05-10, 05:16 #5 retina Undefined     "The unspeakable one" Jun 2006 My evil lair 133678 Posts Presumably n! = Γ(n + 1). For n >= 0 that would suffice, right? What did I miss?
2020-05-10, 06:11   #6
Batalov

"Serge"
Mar 2008
Phi(4,2^7658614+1)/2

2×19×241 Posts

Quote:
 Originally Posted by retina Presumably n! = Γ(n + 1). For n >= 0 that would suffice, right? What did I miss?
You are missing that n! = Γ(n+1) + sin(πn). Or Γ(n+1) + sin(π(n+1)); take your pick... Wikipedia is helpful to demonstrate that both are valid analytic continuations of the factorials to the non-integers.
Do Γ(n+1) and Γ(n+1) + sin(πn) have the same derivatives?

2020-05-10, 06:16   #7
othermath

May 2020

5 Posts

Quote:
 Originally Posted by VBCurtis And what do you want to say the factorial of a real number is?
I assume ...

2020-05-10, 06:31   #8
othermath

May 2020

5 Posts

Quote:
 Originally Posted by Batalov You are missing that n! = Γ(n+1) + sin(πn). Or Γ(n+1) + sin(π(n+1)); take your pick... Wikipedia is helpful to demonstrate that both are valid analytic continuations of the factorials to the non-integers. Do Γ(n+1) and Γ(n+1) + sin(πn) have the same derivatives?
Γ(n+1) requires the use of an integral. My relationship is simpler.

2020-05-10, 06:34   #9
othermath

May 2020

5 Posts

Quote:
 Originally Posted by retina Presumably n! = Γ(n + 1). For n >= 0 that would suffice, right? What did I miss?

2020-05-10, 06:40   #10
retina
Undefined

"The unspeakable one"
Jun 2006
My evil lair

10110111101112 Posts

Quote:
 Originally Posted by Batalov You are missing that n! = Γ(n+1) + sin(πn). Or Γ(n+1) + sin(π(n+1)); take your pick... Wikipedia is helpful to demonstrate that both are valid analytic continuations of the factorials to the non-integers. Do Γ(n+1) and Γ(n+1) + sin(πn) have the same derivatives?
Thanks. Makes sense.

 2020-05-10, 23:49 #11 Dr Sardonicus     Feb 2017 Nowhere EDA16 Posts There is a fine little book (good luck finding a copy!) entitled The Gamma Function by Emil Artin. In it he shows that the Gamma function is distinguished by being "log convex." As Retina has noted, x! = Γ(x+1) when x is a non-negative integer. As to the derivative: There is a well-known asymptotic expansion [Stirling's asymptotic series] for ln(Γ(z)), z a complex variable. Taking the derivative term by term gives an asymptotic series for Γ'(z)/Γ(z).

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