Logarithm of a hypergeometric series - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T07:26:05Z http://mathoverflow.net/feeds/question/80323 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/80323/logarithm-of-a-hypergeometric-series Logarithm of a hypergeometric series Max Karev 2011-11-07T19:21:43Z 2011-11-08T00:34:56Z <p>I am sorry if the answer to my question is well-known. I am quite new in this topic, so it will be also nice to have a reference, if it exists.</p> <p>I was wondered if there exists a nice closed formula for a logarithm of an arbitrary hypergeometric series in the terms of, say, a linear combination of some other hypergeometric series. The reason that makes me believe in the existence of such a formula is the following.</p> <p>It is an easy exercise to show that the derivative of a hypergeometric series can be expressed as follows: $\frac{d}{dx} {}_nF_m (a_1,\ldots, a_n;b_1\ldots b_m; x) = \frac {a_1\cdots a_n}{b_1\cdots b_m} {}_nF_m (a_1+1,\ldots, a_n+1;b_1+1\ldots b_m+1; x)$. From the other hand, for an arbitrary function $G(x)$ we have $(\log G(x))' = \frac {G'(x)}{G(x)}$.</p> <p>It follows that it suffices to find a ratio of two hypergeometric series to find a logarithmic derivative. In some cases this ratio is known to be a hypergeometric series again. So after the integration we'll obtain the desired result.</p> <p>Thank you in advance for any help.</p> http://mathoverflow.net/questions/80323/logarithm-of-a-hypergeometric-series/80330#80330 Answer by Igor Rivin for Logarithm of a hypergeometric series Igor Rivin 2011-11-07T20:10:59Z 2011-11-07T20:10:59Z <p>Check out this very cool paper (in Proceedings of the National Academy of the US&lt; so freely available, if you care):</p> <p>Classification of hypergeometric identities for Ï€ and other logarithms of algebraicâ€‰numbers D. V. Chudnovsky* and G. V. Chudnovsky</p> <p>I am not sure it answers your question, but it comes very close.</p> http://mathoverflow.net/questions/80323/logarithm-of-a-hypergeometric-series/80339#80339 Answer by Jacques Carette for Logarithm of a hypergeometric series Jacques Carette 2011-11-07T22:56:21Z 2011-11-08T00:27:49Z <p>No. This is because all hypergeometrics are holonomic, and holonomic functions can only have a finite number of singularities, which themselves can only be of certain types. If the logarithm of all hypergeometrics could be so expressed, then you could have a holonomic function with a $\ln \ln (x)$ singularity, which is not possible.</p> <p>I find the paper <a href="http://algo.inria.fr/flajolet/Publications/FlGeSa05.pdf" rel="nofollow">On the non-holonomic character of logarithms, powers, and the nth prime function </a> by Flajolet, Gerhold and Salvy (<em>The Electronic Journal of Combinatorics</em>, 2005, vol. 11) to be a wonderful compendium of useful tools for disproving holonomicity. Searching through the literature to find these tools is tedious, and so these authors ought to be commended for assembling so many into one pleasant paper.</p>