Approximate derivative of x!

[othermath]

Working experimentally, I found the approach



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?

[Batalov]

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.

[othermath]

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.

[VBCurtis]

And what do you want to say the factorial of a real number is?

[retina]

Presumably n! = Γ(n + 1). For n >= 0 that would suffice, right? What did I miss?

[Batalov]

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?

[othermath]

Originally Posted by VBCurtis

And what do you want to say the factorial of a real number is?

I assume ...

[othermath]

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.

[othermath]

Originally Posted by retina

Presumably n! = Γ(n + 1). For n >= 0 that would suffice, right? What did I miss?

That I answered Batalov.

[retina]

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.

[Dr Sardonicus]

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).