## Logarithm of a hypergeometric series

### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

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.

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.

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 $(\log G(x))' = \frac {G'(x)}{G(x)}$.

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.

Thank you in advance for any help.

-

## 2 Answers

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.

I find the paper On the non-holonomic character of logarithms, powers, and the nth prime function by Flajolet, Gerhold and Salvy (The Electronic Journal of Combinatorics, 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.

-
Ah, great. But, if I am not mistaken, this argument seems to work in the case n≥m+1. Is anything known for the case n≤m, when the corresponding hypergeometric series is an entire function? – Max Karev Nov 8 2011 at 21:35

1
@Max: I am not aware of any specific results in that area. And clearly there are hypergeometric functions whose logarithm is also hypergeometric/holonomic. I would try to figure out what conditions to impose on the ODE satisfied by G so that ln(G) also satisfies an ODE (with polynomial coefficients), and see where that leads. The Weierstrass and Hadamard factorization theorems are likely to be useful, as well as the Structure Theorem (from paper cited above). – Jacques Carette Nov 8 2011 at 22:20

### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.

Check out this very cool paper (in Proceedings of the National Academy of the US< so freely available, if you care):

Classification of hypergeometric identities for π and other logarithms of algebraic numbers D. V. Chudnovsky* and G. V. Chudnovsky

I am not sure it answers your question, but it comes very close.

-

1
the link to the paper: pnas.org/content/95/6/2744.full – jc Nov 7 2011 at 21:48
Thank you for the link. – Max Karev Nov 8 2011 at 21:28
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 4, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9366285800933838, "perplexity_flag": "head"}