7
$\begingroup$

For $\alpha,\beta\in\mathbb{C}$ and $\gamma\in\mathbb{C}\setminus\{0,-1,-2,\dotsc\}$, Gauss' hypergeometric function ${}_2F_1(\alpha,\beta;\gamma;z)$ can be defined by the series \begin{equation}\label{Gauss-HF-dfn} {}_2F_1(\alpha,\beta;\gamma;z)=\sum_{n=0}^{\infty}\frac{(\alpha)_n(\beta)_n}{(\gamma)_n}\frac{z^n}{n!},\quad |z|<1. \end{equation} The following special cases are well-known: \begin{align*} {}_2F_1(a,b;b;z)&=\frac{1}{(1-z)^a},\\ {}_2F_1(1,1;2;z)&=-\frac{\ln(1-z)}{z},\\ {}_2F_1\biggl(\frac12,1;\frac32;z^2\biggr)&=\frac1{2z}\ln\frac{1+z}{1-z},\\ {}_2F_1\biggl(\frac12,1;\frac32;-z^2\biggr)&=\frac{\arctan z}{z},\\ {}_2F_1\biggl(\frac12,\frac12;\frac32;z^2\biggr)&=\frac{\arcsin z}{z},\\ {}_2F_1\biggl(\frac12,\frac12;\frac32;-z^2\biggr)&=\frac{\ln\bigl(z+\sqrt{1+z^2}\bigr)}{z}. \end{align*} See Chapter 5 and page 109 in the book [1] below.

Lemma 2.6 in the paper [2] below reads that, for $0\ne|t|<1$ and $n=1,2,\dotsc$, \begin{equation*} {}_2F_1\biggl(\frac{1-n}{2}, \frac{2-n}{2};1-n;\frac1{t^2}\biggr) =\frac{t}{2^n\sqrt{t^2-1}\,} \biggl[\biggl(1+\frac{\sqrt{t^2-1}\,}{t}\biggr)^n -\biggl(1-\frac{\sqrt{t^2-1}\,}{t}\biggr)^n\biggr]. \end{equation*}

Corollary 4.1 in the paper [3] below states that, for $n=0,1,2,\dotsc$, \begin{multline}\label{Gauss-HF-Spec-Value} {}_2F_1\biggl(n+\frac{1}{2},n+1;n+\frac{3}{2};-1\biggr) =\frac{(2n+1)!!}{(2n)!!}\frac{\pi}{4}\\ +\frac{2n+1}{2^{2n}}\sum_{k=1}^{n} (-1)^{k} \binom{2n-k}{n} \frac{2^{k/2}}{k}\sin\frac{3k\pi}{4}. \end{multline}

My question is: can one find an elementary function $f(t)$ such that \begin{equation*} {}_2F_1\biggl(\frac{1}{2},\frac{1}{2};2;t\biggr)=f(t), \quad |t|\le1? \end{equation*} In other words, is the Gauss hypergeometric series $_2F_1\bigl(\frac{1}{2},\frac{1}{2};2;t\bigr)$ an elementary function?

References

  1. N. M. Temme, Special Functions: An Introduction to Classical Functions of Mathematical Physics, A Wiley-Interscience Publication, John Wiley & Sons, Inc., New York, 1996; available online at http://dx.doi.org/10.1002/9781118032572.
  2. Feng Qi, Qing Zou, and Bai-Ni Guo, The inverse of a triangular matrix and several identities of the Catalan numbers, Applicable Analysis and Discrete Mathematics 13 (2019), no. 2, 518--541; available online at https://doi.org/10.2298/AADM190118018Q.
  3. Feng Qi and Mark Daniel Ward, Closed-form formulas and properties of coefficients in Maclaurin's series expansion of Wilf's function composited by inverse tangent, square root, and exponential functions, arXiv (2022), available online at https://arxiv.org/abs/2110.08576v2.
$\endgroup$
4
  • $\begingroup$ The motivation and background of this problem can be found in the paper: Wei-Shih Du, Dongkyu Lim, and Feng Qi, Several recursive and closed-form formulas for some specific values of partial Bell polynomials, Advances in the Theory of Nonlinear Analysis and its Applications, vol. 6 (2022), no. 4, 528--537; available online at doi.org/10.31197/atnaa.1170948. $\endgroup$
    – qifeng618
    Commented Oct 11, 2022 at 1:50
  • 1
    $\begingroup$ I think it would help you most if you'd ask the second question separately. Since you've already accepted an answer for this one, it is unlikely the second question will garner a lot of attention $\endgroup$
    – Max Muller
    Commented Mar 29, 2023 at 10:35
  • 2
    $\begingroup$ @MaxMuller Thank you for your suggestion. I have asked the second question at math.stackexchange.com/q/4669567 separately. $\endgroup$
    – qifeng618
    Commented Mar 30, 2023 at 16:42
  • 1
    $\begingroup$ This question and some answers were mentioned in the paper "Yue-Wu Li and Feng Qi, A new closed-form formula of the Gauss hypergeometric function at specific arguments, Axioms 13 (2024), no. 5, 24 pages; available online at researchgate.net/publication/380431446." $\endgroup$
    – qifeng618
    Commented May 10 at 7:00

1 Answer 1

12
$\begingroup$

Maple does it in terms of complete elliptic integrals $\rm{K}$ and $\rm{E}$ ... $$ {\mbox{$_2$F$_1$}\left(\frac12,\frac12;\,2;\,t\right)}={\frac {4\left( t-1 \right){\rm K} \left( \sqrt {t} \right) +4{\rm E} \left( \sqrt {t} \right) }{t\pi}} \tag1$$ But that does not show it is elementary. In fact, I suspect it is not elementary.


Recall the known formulas: $$ \rm{K}\big(\sqrt{t}\big) = \frac{\pi}{2}\; {}_2F_1\left(\frac12 , \frac12 ; 1 ; t\right) , \\ \rm{E}\big(\sqrt{t}\big) = \frac{\pi}{2}\; {}_2F_1\left(-\frac12 , \frac12 ; 1 ; t\right) . $$ By themselves, they are not elementary. $(1)$ should follow from these two and a contiguous formula for the hypergeometrics. $$ {c\;\mbox{$_2$F$_1$}(a-1,b;\,c;\,x)} + \left( x-1 \right) c\;{\mbox{$_2$F$_1$}(a,b;\,c;\,x)} + \left( b-c \right) x\;{\mbox{$_2$F$_1$}(a,b;\,c+1;\,x)} =0 $$

$\endgroup$
2
  • 3
    $\begingroup$ Dear Peofessor Edgar, I recited your answer in Section 4.2 of my paper [Wei-Shih Du, Dongkyu Lim, and Feng Qi, Several recursive and closed-form formulas for some specific values of partial Bell polynomials, Advances in the Theory of Nonlinear Analysis and its Applications, vol. 6, no. 4, 528--537 (2022); available online at doi.org/10.31197/atnaa.1170948 ]. Thank you very much for your answer to my question. @GeraldEdgar $\endgroup$
    – qifeng618
    Commented Mar 30, 2023 at 17:02
  • $\begingroup$ This result can be directly obtained by differentiating $K\left( \sqrt{t} \right)$, expressed as an hypergeometric function, using DLMF 15.5.6 (dlmf.nist.gov/15.5.E6) with $a=b=/2,c=n=1$. $\endgroup$
    – Paul Enta
    Commented Jun 7 at 18:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.