Recurrence Relation for Digamma Function
Jump to navigation
Jump to search
Theorem
- $\map \psi {z + 1} = \map \psi z + \dfrac 1 z$
where:
- $\psi$ denotes the digamma function
- $z \in \C \setminus \Z_{\le 0}$.
Proof
| \(\ds \map \Gamma {z + 1}\) | \(=\) | \(\ds z \map \Gamma z\) | Gamma Difference Equation | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \map \ln {\map \Gamma {z + 1} }\) | \(=\) | \(\ds \map \ln {z \map \Gamma z}\) | applying $\ln$ on both sides | ||||||||||
| \(\ds \) | \(=\) | \(\ds \ln z + \map \ln {\map \Gamma z}\) | Sum of Logarithms | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \dfrac \d {\d z} \map \ln {\map \Gamma {z + 1} }\) | \(=\) | \(\ds \dfrac \d {\d z} \ln z + \dfrac \d {\d z} \map \ln {\map \Gamma z}\) | differentiation with respect to $z$ | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \dfrac {\map {\Gamma'} {z + 1} } {\map \Gamma {z + 1} }\) | \(=\) | \(\ds \dfrac 1 z + \dfrac {\map {\Gamma'} z} {\map \Gamma z}\) | Derivative of Natural Logarithm Function, Chain Rule for Derivatives | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \map \psi {z + 1}\) | \(=\) | \(\ds \map \psi z + \dfrac 1 z\) | Definition of Digamma Function |
$\blacksquare$
Also see
Sources
- Weisstein, Eric W. "Digamma Function." From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/DigammaFunction.html