2008-01-08, 05:53 | #1 |
Nov 2005
181 Posts |
Derivative of Gamma
If the convention is used that x!=Gamma(x+1), then the factorial function can have all real numbers except negative integers as inputs. Thinking about this lead me to a path close to the reasoning behind Stirling's approximation of factorial. He used ln 0 + ln 1 + ln 2 + ln 3 + ln 4 + ... to derive his approximation. I was looking at functions like x^x, and trying to create a sequence in place of the power: x^g_n(x) where g_0(x) = x and each next g(x) gives a closer approximation.
The derivative is x!*Digamma(x+1). In other words: Dx x! = Gamma(x+1) * Digamma(x+1) x! * Digamma(x+1) = x!* Dx Ln(x!) Dx Digamma(x+1) = Zeta(2, x+1) 3rd derivative of x! = x!*(Digamma(x+1)^3 + 3*Zeta(2, x+1)*Digamma(x+1) - 2*Zeta(3, x+1)) According to Derive, this is the derivative of that previous line without the x! in the product: 3·Zeta(2, x + 1)^2·Digamma(x + 1) - 6·Zeta(3, x + 1)·Digamma(x + 1) + 6·Zeta(4, x + 1) + 3·Zeta(2, x + 1) There seems to be a chain: Gamma -> Digamma -> Zeta I'm wondering if there's a formula to get the n'th derivative of x!, past the first one containing the Zeta function. Note: I'm using the convention that allows factorial to take all real numbers except negative integers as inputs. ('dependant variables' if I remember right) This problem is connected to the third one I'm thinking of. I need a derivative of (-1)^x and x! before I can continue. |
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