I stumbled upon the following identity, which I have not tried to prove, but seems true: the function $$f(t):={}_2F_1(1/2,2t;1-t;4)$$ is periodic of period 1, and more precisely $$f(t)=\dfrac{1+2e^{-2\pi it}}{3}\;.$$ This begs several questions:

(1). Is this true? (2). Are there other (infinitely many?) formulas of this type (of course outside of trivial manipulations of this) ? (3) More generally, call a function $f(t)$ scale-periodic if there exists $A>0$ such that $A^{-t}f(t)$ is $1$-periodic (so periodic if $A=1$). I have found a number of such $f(t)$ of the form ${}_2F_1(a(t),b(t);c(t);z)$: is there some way to find them? (this last question was inspired by a paper of Beukers and Forsgard arXiv:2004.08117).

  • 1
    $\begingroup$ How can the equality hold if the lhs is real for real $t$ and the rhs is not?? $\endgroup$ Jul 7, 2021 at 19:25
  • 1
    $\begingroup$ Since the argument is $z=4$, this is the standard analytic continuation, which is indeed complex for $z>1$. $\endgroup$ Jul 7, 2021 at 19:49
  • $\begingroup$ @Cohen With Mathematica only the real parts coincide. This is a summary of all my previous comments. Usually there is a cut on the real axis for the implementation of the hypergeometric function. $\endgroup$
    – juan
    Jul 7, 2021 at 20:02
  • $\begingroup$ $\Im f(1/5)$ appear to be a solution of the equation $81x^4-45x^2+5=0$ and the imaginary part of the expression with exponential appear to be a root of $81x^4-585x^2+5=0$. All this as given in Mathematica and recognising the algebraic numbers. $\endgroup$
    – juan
    Jul 7, 2021 at 20:15
  • 1
    $\begingroup$ In principle question (1) can be answered by finding the (unique) coefficients of the contiguous relation which gives ${{}_2F_1}(a,b+2;c-1;z)$ as $p\; {{}_2F_1}(a,b;c;z) + q\;{{}_2F_1}(a+1,b;c;z)$ where $p,q$ are rational functions in $a,b,c,z$; then test whether $q=0$ for your parameters. Question (2) is easily answered in the affirmative, and question (3) is answered by applying the same technique for arbitrary shifts and getting solutions whenever the parameters are chosen to set $q=0$. arXiv:math/0109222 claims to give an algorithm, but I'm only getting tautologies for $Q(2,0,0)$. $\endgroup$ Jul 7, 2021 at 20:25

1 Answer 1


Gauss' contiguous relations provide a basis for finding a linear relationship between three functions of the form ${}_2F_1(a+k, b+l, c+m, z)$, henceforth $\mathbf{F}\left(\begin{matrix}a+k, b+l \\ c+m\end{matrix}\right)$. Various papers have been published on calculating these relationships; I'm using the notation of Vidūnas's paper Contiguous relations of hypergeometric series:

$$\textbf{F}\left(\begin{matrix}a+k, b+l \\ c+m\end{matrix}\right) = \textbf{P}(k,l,m) \textbf{F}\left(\begin{matrix}a, b \\ c\end{matrix}\right) + \textbf{Q}(k,l,m) \textbf{F}\left(\begin{matrix}a+1, b \\ c\end{matrix}\right)$$

Vidūnas recommends finding the coefficients by first chaining up $k$ with his formula (9), but a naïve approach runs into bootstrap problems (as I discovered by trying it). Once $\textrm{Q}(2,0,0)$ is found, I think the approach works, but here we have such small shifts that we can do directly.

Rephrasing his (2), (3), (4) in terms of $\textbf{Q}$ we obtain

$$\mathbf{Q}(k+1,l,m) = \tfrac{(2a+2k-c-m+(b+l-a-k)z)}{(a+k)(1-z)}\mathbf{Q}(k,l,m) + \tfrac{(c+m-a-k)}{(a+k)(1-z)}\mathbf{Q}(k-1,l,m)$$ $$\mathbf{Q}(k,l,m-1) = \tfrac{a+k}{c+m-1} \mathbf{Q}(k+1,l,m) + \tfrac{c+m-a-k-1}{c+m-1} \mathbf{Q}(k,l,m)$$ $$\mathbf{Q}(k,l+1,m) = \tfrac{a+k}{b+l} \mathbf{Q}(k+1,l,m) + \tfrac{b+l-a-k}{b+l}\mathbf{Q}(k,l,m)$$

and by repeated application of these I calculate that

$$\mathbf{Q}(0,2,-1) = a \tfrac{(a+b-c+1)(a+b-c+2) + (c-a-1)(2a+b-c+1)(1-z) + (b-a+1)(c-a-1)(1-z)^2}{(c-1)b(b+1)(1-z)^2} \\ $$

$$\mathbf{P}(0,2,-1) = (c-a-1) \tfrac{a(a+b-c+2) + a(-2a+b+c-1)(1-z) + (b-a)(b-a+1)(1-z)^2}{(c-1)b(b+1)(1-z)^2}$$

Substituting your parameters $a=\tfrac12$, $b=2t$, $c=1-t$, $z=4$ I get respectively $0$ and $1$, which establishes the periodicity of your $f$.

To search systematically for scale-periodic hypergeometric functions, I propose the following approach:

  1. Pick $k$, $l$, $m \in \mathbb{Z}$ (taking into account symmetries for efficiency).
  2. Compute $\mathbf{Q}(k,l,m)$ and $\mathbf{P}(k,l,m)$.
  3. Substitute $a = a_0 + kt$, $b = b_0 + lt$, $c = c_0 + mt$ into $\mathbf{P}(k,l,m) = A$ and $\mathbf{Q}(k,l,m) = 0$. Equate coefficients of $t$ in each to get a collection of polynomial constraints over $a_0,b_0,c_0,z,A$. Use Gröbner bases or other techniques to solve these constraints.

To take the obvious example of $k=0$, $l=2$, $m=-1$, and (for simplicity) applying the substitution $x = 1-z$, $\textbf{Q}(0,2,-1) = 0$ gives

$$ a_0 (9-3x-2x^2) = 0 \\ 6a_0+6b_0-6c_0+9+(-5a_0-b_0+4c_0-4)x+(-a_0-b_0+2c_0-3)x^2 = 0 \\ (a_0+b_0-c_0+1)(a_0+b_0-c_0+2) + (-a_0+c_0-1)(2a_0+b_0-c_0+1)x + (-a_0+b_0+1)(-a_0+c_0-1)x^2 = 0 $$

Ignoring the trivial $a_0 = 0$, we see that $x \in \{-3, \tfrac32\}$.

Case $x=-3$

$a_0 = \tfrac12$ drops out of the next constraint, leaving the final constraint to reduce to $(b_0+2c_0-3)(b_0+2c_0-2) = 0$. If $b_0+2c_0-2 = 0$ then we're going to recover your $f$, so consider instead $b_0+2c_0-3 = 0$. This turns out to have no solutions: substituting into $\mathbf{P}$ and expanding we get a contradiction in the coefficients of $t^0$.

Case $x = \tfrac32$

Comparing coefficients for $t^3$ in $\mathbf{P}$ gives $A = 1$; subsequently comparing coefficients for $t^2$ in $\mathbf{P}$ gives $a_0 = 0$, so there is no non-trivial solution.

I therefore withdraw my blithe assertion in an earlier comment that question (2) is easily answered in the affirmative; it remains plausible, but with larger shifts we're going to have more coefficients of $t$ to impose constraints on a fixed number of variables and so there might not be very many sets of shifts which admit solutions.

  • 1
    $\begingroup$ Very nice, thank you. I will try to implement this to find more examples, unless of course you beat me to it! $\endgroup$ Jul 8, 2021 at 15:55
  • $\begingroup$ @HenriCohen, I wasn't planning to take it further, since you are undoubtedly more proficient with symbolic algebra than I, but I changed my mind. gist.github.com/pjt33/345b7d70a320887e850b456c9fe04557 has my Sage code and some preliminary classification of results. $\endgroup$ Jul 15, 2021 at 11:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.