Connection between Bernoulli polynomials and polygamma function - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T06:03:38Z http://mathoverflow.net/feeds/question/42696 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/42696/connection-between-bernoulli-polynomials-and-polygamma-function Connection between Bernoulli polynomials and polygamma function Anixx 2010-10-18T21:07:06Z 2012-06-17T23:08:33Z <p>There is an intricate connection between Hurwitz Zeta and the (traditional) polygamma function: </p> <p>$$\psi_n(z)=(-1)^{n+1}n!\zeta(n+1,z)$$</p> <p>If to use a generalization for Bernoulli numbers, this can be considered a formula, connecting polygamma and Bernoulli polynomials of negative order:</p> <p>$$\psi_n(z)=(-1)^{n+1}n!\frac{B_{-n}(x)}n$$ (1)</p> <p>As much as this equality is impressing, it is limited as it only holds for natural n.</p> <p>After a couple of unsuccessful attempts to find a more general formula, I encountered a more natural, "balanced" generalization of polygamma function explained in <a href="http://www.math.tulane.edu/~vhm/papers_html/genoff.pdf" rel="nofollow">this paper</a>. It turned out that while the old formula still holds for natural n (since the old polygamma and balanced polygamma coincide in integer positive orders), a completely new formula connecting this balanced polygamma with Zeta and Bernoulli numbers can be derived which holds for any z:</p> <p>$$\zeta(z,q)=\frac{\Gamma (1-z) \left(2^{-z} \left(\psi \left(z-1,\frac{q}{2}+\frac{1}{2}\right)+\psi \left(z-1,\frac{q}{2}\right)\right)-\psi(z-1,q)\right)}{\ln(2)}$$</p> <p>$$B_z(q) = -\frac{\Gamma (z+1) \left(2^{z-1} \left(\psi\left(-z,\frac{q}{2}+\frac{1}{2}\right)+\psi\left(-z,\frac{q}{2}\right)\right)-\psi(-z,q)\right)}{\ln (2)}$$</p> <p>Both of them can be expressed completely in terms of balanced polygamma and elementary functions if to notice that $\Gamma(x)=e^{\psi(-1,x)+\frac 12 \ln(2\pi)}$, which allows to get rid of the Gamma function.</p> <p>While the target was reached, these expressions still leave a bad impression. I cannot simplify it as no CAS system is capable of operations with the balanced polygamma. </p> <p>Hence I am asking for help on how to simplify the expressions so they could be easier to manage and use. It is also not evident how the letter formulas become the former ones at positive real z.</p> http://mathoverflow.net/questions/42696/connection-between-bernoulli-polynomials-and-polygamma-function/99862#99862 Answer by Hassan Jolany for Connection between Bernoulli polynomials and polygamma function Hassan Jolany 2012-06-17T23:08:33Z 2012-06-17T23:08:33Z <p>For finding such relations look at the generalized Bernoulli numbers whch is the poly- Bernoulli numbers and defined by Kaneko.In definition of Poly-Bernoulli numbers</p> <p>${Li_{k}(1-e^{-x}) \over 1-e^{-x}}=\sum_{n=0}^{\infty}B_{n}^{(k)}{x^{n}\over n!}$ </p> <p>he used of poly-logarithm which has some relationships with poly-gamma <a href="http://www.perfscipress.com/papers/UniversalTOC25.pdf" rel="nofollow">http://www.perfscipress.com/papers/UniversalTOC25.pdf</a> for more information see <a href="http://www2.math.kyushu-u.ac.jp/~mkaneko/papers/pB-MZV.pdf" rel="nofollow">http://www2.math.kyushu-u.ac.jp/~mkaneko/papers/pB-MZV.pdf</a> about poly-Bernoulli nubers</p>