Content deleted Content added
Undid revision 756936983 by 86.138.203.45 (talk) |
Reciprocist (talk | contribs) No edit summary |
||
Line 23:
:<math>\begin{align}
\psi(x) &= \psi(0,x)
\psi^{(n)}(x)&=\psi(n,x) \qquad n\in\mathbb{N}
\Gamma(x)&=\exp\left( \psi(-1,x)+\tfrac12 \ln 2\pi \right)
\zeta(z,q)&=\frac{\Gamma (1-z)}{\ln 2} \left(2^{-z} \psi \left(z-1,\frac{q+1}{2}\right)+2^{-z} \psi \left(z-1,\frac{q}{2}\right)-\psi(z-1,q)\right)
\zeta'(-1,x)&=\psi(-2, x) + \frac{x^2}2 - \frac{x}2 + \frac1{12}
B_n(q) &= -\frac{\Gamma (n+1)}{\ln 2} \left(2^{n-1} \psi\left(-n,\frac{q+1}{2}\right)+2^{n-1} \psi\left(-n,\frac{q}{2}\right)-\psi(-n,q)\right)
\end{align}</math>
Line 40:
The balanced polygamma function can be expressed in a closed form at certain points (where {{mvar|A}} is the [[Glaisher constant]] and {{mvar|G}} is the [[Catalan constant]]):
:<math>\begin{align}
\psi\left(-2,\tfrac14\right)&=\tfrac18\ln 2\pi+\tfrac98\ln A+\frac{G}{4\pi} &&
\psi\left(-2,\tfrac12\right)&=\tfrac14\ln\pi+\tfrac32\ln A+\tfrac5{24}\ln2 &
\psi\left(-3,\tfrac12\right)&=\tfrac1{16}\ln 2\pi+\tfrac12\ln A+\frac{7\zeta(3)}{32\pi^2}
\psi(-2,1)&=\tfrac12\ln 2\pi &
\psi(-3,1)&=\tfrac14\ln 2\pi+\ln A
\psi(-2,2)&=\ln 2\pi-1 &
\psi(-3,2)&=\ln 2\pi+2\ln A-\tfrac34 \end{align}</math>
| |||