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$$ 143\,\sqrt {3}\;{\mbox{$_2$F$_1$}\left(\frac{1}{2},\frac{1}{2};\,1;\,{\frac {3087}{8000}}\right)}= 40\,\sqrt {5}\; {\mbox{$_2$F$_1$}\left(\frac{1}{3},\frac{2}{3};\,1;\,{\frac {2923235}{2924207}}\right)} $$ While working on another problem, I found two different ways of expressing my solution. Plugging in a value, I get the above equation.

Are there methods/references for proving something like this?

Maple does not know them, it seems.

added

More generally, I have $$ \frac{\displaystyle 3\,\sqrt {2}\; {\mbox{$_2$F$_1$}\Bigg(\frac{1}{2},\frac{1}{2};\,1;\,{\frac { 64\left( 1-y \right) ^{3}}{ \left( -{y}^{3/2}\sqrt {y+8}-8\,\sqrt {y}\sqrt {y+8}+{y}^{2}-20\,y-8 \right) ^{2}}}}\Bigg)} {{\sqrt {-3\,{y}^{2}+60\,y+24+3\,{y}^{3/2}\sqrt {y+8}+24\, \sqrt {y}\sqrt {y+8}}}} = \frac{\displaystyle {\mbox{$_2$F$_1$}\bigg(\frac{1}{3},\frac{2}{3};\,1;\,{\frac { \left( y+8 \right) ^{2} \left( 1-y \right) }{ \left( 4-y \right) ^{3}}}\bigg)}} {( 4 - y)} $$ for $0 < y < 4$ ... and for the above problem I found a value of $y$ so that both $y$ and $y+8$ were perfect squares.

But I was hoping someone knew methods to do it other than the round-about way I came to it.

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    $\begingroup$ Both of these expressions can be expressed in terms of LegendreP functions: $$143\,\sqrt {3}{\it LegendreP} \left( -1/2,{\frac {913}{4000}} \right) ,\ 40\,\sqrt {5}{\it LegendreP} \left( -1/3,-{\frac {2922263}{ 2924207}} \right) $$ But I haven't found any general relation between $LegendreP(-1/2,\cdot)$ and $LegendreP(-1/3,\cdot)$. Both can be written as integrals over $[0,1]$, but there's no obvious (to me) relation between the integrands. You'd probably have better luck with the expressions before plugging in a value. $\endgroup$ – Robert Israel Sep 12 '14 at 17:37
  • $\begingroup$ I agree with Robert Israel; often the first step in proving something like this is to introduce an extra parameter, to enable WZ-like methods to be applied. It sounds like you already have an extra parameter, so it might help to say explicitly what it is. $\endgroup$ – Timothy Chow Sep 12 '14 at 22:26
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These formulas originate from modular parametrizations of the underlying hypergeometric functions and there are many such appearances in the literature (a collection of references can be found in http://arxiv.org/abs/1302.0548 where such algebraic transformations are used in a related context). But the most classical source is Goursat's treatment of such transformations in his thesis: http://www.numdam.org/item?id=ASENS_1881_2_10__S3_0.

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  • $\begingroup$ @GeraldEdgar: Or even after two years. :) $\endgroup$ – Tito Piezas III Dec 28 '16 at 17:12
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The OP's general identity is only half equivalent to the one on page 258 of Ramanujan's second notebook. (Also Theorem 5.6 of Berndt's Ramanujan's Notebooks Vol 5, p.112.) It needs a second one, namely Theorem 6.1 on p.116 to complete the puzzle pieces.

I. Piece 1: Theorem 5.6 If

$$u_1 = \frac{p^3(2+p)}{1+2p},\quad\quad v_1 = \frac{27p^2(1+p)^2}{4(1+p+p^2)^3}$$ then for $0\leq p <1$, $$(1+p+p^2)\,_2F_1\big(\tfrac12,\tfrac12;1;u_1\big)=\sqrt{1+2p}\;_2F_1\big(\tfrac13,\tfrac23;1;v_1\big)\tag1$$

II. Piece 2: Theorem 6.1 If

$$u_2 = \frac{q(3+q)^2}{2(1+q)^3},\quad\quad v_2 = \frac{q^2(3+q)}{4}$$ then for $0\leq q <1$, $$\,_2F_1\big(\tfrac13,\tfrac23;1;u_2\big)=(1+q)\;_2F_1\big(\tfrac13,\tfrac23;1;v_2\big)\tag2$$

III. Piece 3: $v_1=v_2$

It turns out the OP's was the special case $v_1=v_2$, hence,

$$\frac{27p^2(1+p)^2}{4(1+p+p^2)^3} = \frac{q^2(3+q)}{4}$$

While a cubic in $q$, it conveniently factors linearly. Choosing the correct factor such that $0\leq p,q <1$, we can aesthetically express the OP's general identity as,

$$\large \tfrac{(1+p+p^2)(1+q)}{\sqrt{1+2p}}\,_2F_1\Big(\tfrac12,\tfrac12;1;\tfrac{p^3(2+p)}{1+2p}\Big)=\;_2F_1\Big(\tfrac13,\tfrac23;1;\tfrac{q(3+q)^2}{2(1+q)^3}\Big)\tag{3a}$$

where,

$$q = \frac{3p}{1+p+p^2}\tag{3b}$$

For example, the OP used $y=\tfrac1{36}$. But using $p=\tfrac7{10}$ on $(3)$, we recover the original equality,

$$\tfrac{143\sqrt3}{40\sqrt5}\;{\mbox{$_2$F$_1$}\left(\tfrac{1}{2},\tfrac{1}{2};\,1;\,{\tfrac {3087}{8000}}\right)}={\mbox{$_2$F$_1$}\left(\tfrac{1}{3},\tfrac{2}{3};\,1;\,{\tfrac {2923235}{2924207}}\right)}$$

and puzzle solved.

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    $\begingroup$ This is great. Thanks. It is too bad I cannot accept more than one answer. I wonder what will happen after 3 years... $\endgroup$ – Gerald Edgar Dec 28 '16 at 17:20
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This should be related to Ramanujan's theory of cubic elliptic functions. Presumably your identity is equivalent to one occurring on page 258 of Ramanujan's second notebook, see Thm. 5.6 in Berndt, Bhargava and Garvan, Ramanujan's theory of elliptic functions to alternative bases, Trans. Amer. Math. Soc. 347 (1995), 4163-4244. You can download a pdf from Frank Garvan's homepage.

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