1
$\begingroup$

After playing around with transforms of a certain parametric integral, I am inclined to think that the linear combination $$f(n):=\dfrac1{n-2}\left({\,}_2F_1(\dfrac{n-2}{4n},\dfrac12;\dfrac{5n-2}{4n};-1)\right)+\dfrac1{n+2}\left({\,}_2F_1(\dfrac{n+2}{4n},\dfrac12;\dfrac{5n+2}{4n};-1)\right)$$ has a closed form for integer $n$. I know for example that $f(3)=\dfrac{1}{12^{3/4}}\dfrac{\Gamma(\frac14)^2}{\sqrt{\pi}}$. Any ideas?

Edit: putting $a:=\frac14-\frac1{2n}$, we can define $g(a):= \frac{8}{1-4a}f(\frac{8}{1-4a})$ to get arguments closer to the "standard" notation used in formula collections. The question then becomes:

Which rational values of $a$, other than $a= \frac1{12}$, allow a closed form for$$g(a)=\dfrac1{a}\left({\,}_2F_1(a,\frac12;a+1;-1)\right)+\dfrac1{\frac{1}{2}-a}\left({\,}_2F_1(\dfrac{1}{2}-a,\frac12;\frac32-a;-1)\right)?$$

$\endgroup$
  • $\begingroup$ Can you include that "parametric integral"? I might be a clue to why the $n$ in my answer has closed-forms. $\endgroup$ – Tito Piezas III Jan 28 at 15:52
  • $\begingroup$ @TitoPiezasIII oh sorry, that was such a long time ago that I don't remember at all what that was and where it occurred! Nothing for f(24)? And it is surprising that nothing came up for f(10). $\endgroup$ – Wolfgang Jan 28 at 17:05
  • $\begingroup$ Actually, "apparently" nothing for $n<48$, though I did use a hundred decimal digits. $\endgroup$ – Tito Piezas III Jan 28 at 17:22
  • $\begingroup$ I guess you have already raised $f(n)/K(k_{whatever})$ to the 4th power to increase your chances? $\endgroup$ – Wolfgang Jan 28 at 17:56
  • $\begingroup$ Yes, and I assume that the $d$ of $K(k_d)$ divides $n$. So it was natural to check $f(5)/K(k_5)$ and $f(7)/K(k_7)$. Why the former is algebraic, while the latter apparently is not is intriguing. P.S. However, I've found that certain patterns using eta quotients may involve only $p = 2,3,5$. $\endgroup$ – Tito Piezas III Jan 28 at 18:12
1
$\begingroup$

If we use the original function,

$$f(n):=\tfrac1{n-2}\left({\,}_2F_1\big(\tfrac{n-2}{4n},\tfrac12;\tfrac{5n-2}{4n};-1\big)\right)+\tfrac1{n+2}\left({\,}_2F_1\big(\tfrac{n+2}{4n},\tfrac12;\tfrac{5n+2}{4n};-1\big)\right)$$

then there are several other $f(n)$ with a closed form. Let $K(k_d)$ be an elliptic integral singular value. Then,

$$f(3) = \sqrt{\frac{2}{3\sqrt3}}\; K(k_1)$$ $$f(4) = \sqrt{\frac{-1+\sqrt2}{2\sqrt2}}\; K(k_2)$$ $$f(5) = \frac{\sqrt2}{(\sqrt5\phi)^{5/4}}\; K(k_5)$$ $$f(6) = \sqrt{\frac{1}{12\sqrt3}}\; K(k_3)$$ $$f(12) = \sqrt{\alpha}\; K(k_6)$$

where $\phi$ is the golden ratio and $\alpha$ is a quartic root.

P.S. Try as I might, I couldn't find anything else with $2<n<20$ which begs the question what makes these $n$ so special.

$\endgroup$
0
$\begingroup$

I have found two instances of $_2F_1(a,b;c;z), z=-1$ in

A. Erdelyi, Higher Transcendental Functions, Vol. 1 (and particularly, Sec. 2.8), Krieger Publishing.

that may be of help to you.

$$ _2F_1(a,b;1+a-b;-1)=2^{-a}\frac{\Gamma(1+a-b)\Gamma(1/2)}{\Gamma(1-b+a/2)\Gamma(1/2+a/2)}, \quad 1+a-b\ne0,-1,-2,... $$ $$ (a+1)_2F_1(-a,1;b+2;-1)+(b+1)_2F_1(-b,1;a+2;-1)=2^{a+b+1}\frac{\Gamma(a+2)\Gamma(b+2)}{\Gamma(a+b+2)}, \quad a,b\ne-2,-3,-4,... $$ And this one from the NIST Handbook of Mathematical Functions, which is a variation of the first one above

$$ _2F_1(a,b;1+a-b;-1)=\frac{\Gamma(1+a-b)\Gamma(a/2+1)}{\Gamma(1-b+a/2)\Gamma(1+a)} $$

There was one additional relation for $_2F_1(a,b;c;z), z=-1$, but it was in terms of the digamma function.

$\endgroup$
  • $\begingroup$ Thank you for searching, but this helps only for $_2F_1(a,1/2;a+1/2;-1)$, What I need is $_2F_1(a,1/2;a+1;-1)$. And the digamma one (15.4.27) has 1 not 1/2 as argument. $\endgroup$ – Wolfgang Jul 24 '17 at 6:51
  • $\begingroup$ @spanferkel I haven't had a chance to see if the case for $n=3$ you presented fits any of the solutions I found. I did, however, check that solution numerically. At any rate, my point is that there are only a limited number of known solutions for $z=-1$ and if these don't do the trick it's unlikely that you'll find a general solution for arbitrary $n$. $\endgroup$ – Cye Waldman Jul 24 '17 at 21:15
  • $\begingroup$ You say "it's unlikely that you'll find a general solution", and of course I'd agree if it was just about one of the two terms. But here is a (conveniently weighted) sum. I remember having seen once a certain weighted sum of two order $n$ digamma functions with arguments $\frac{1\pm\sqrt{5}}2$ which had a closed form... $\endgroup$ – Wolfgang Jul 25 '17 at 8:38
  • $\begingroup$ @spanferkel Apologies. I didn't mean to disparage your efforts, but merely to point out that such solutions are rare and limited, as evidenced by what I turned up in my own references. $\endgroup$ – Cye Waldman Jul 25 '17 at 15:52
  • $\begingroup$ Sure, I am just searching a needle in a hay stick... :) $\endgroup$ – Wolfgang Jul 25 '17 at 16:00

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.