To continue our adventures (which started here) with the hypergeometric function we are going to establish the following identity
If $ a + b + (1/2)$ is neither zero nor a negative integer and if $ |x| < 1$ and $ |4x(1 - x)| < 1$, then $$F\left(a, b; a + b + \frac{1}{2}; 4x(1 - x)\right) = F\left(2a, 2b; a + b + \frac{1}{2}; x\right)$$
A curious thing to observe here is that the term on the left does not change if we replace $ x$ by $ (1 - x)$, but the right term does change its value. If we restrict ourselves to real positive values of $ x$ then we need to understand that in the right term the symbol $ x$ should actually be replaced by minimum of $ x$ and $ 1 - x$ or we can say that the identity as it is holds only when $ 0 \leq x \leq 1/2$.
Again to establish the identity we start with function $ y = F(a, b; a + b + 1/2; z)$ which satisfies the differential equation $$z(1 - z)\frac{d^{2}y}{dz^{2}} + \left(a + b + \frac{1}{2} - (a + b + 1)z\right)\frac{dy}{dz} - aby = 0$$ Putting $ z = 4x(1 - x)$ we get $$\frac{dy}{dz} = \dfrac{\dfrac{dy}{dx}}{\dfrac{dz}{dx}} = \frac{1}{4(1 - 2x)}\frac{dy}{dx}$$ and \begin{align} \frac{d^{2}y}{dz^{2}} &= \dfrac{\dfrac{d}{dx}\left(\dfrac{1}{4(1 - 2x)}\dfrac{dy}{dx}\right)}{\dfrac{dz}{dx}}\notag\\ &= \frac{1}{16(1 - 2x)}\left(\frac{1}{1 - 2x}\frac{d^{2}y}{dx^{2}} + \frac{2}{(1 - 2x)^{2}}\frac{dy}{dx}\right)\notag\\ &= \frac{1}{16(1 - 2x)^{2}}\left(\frac{d^{2}y}{dx^{2}} + \frac{2}{1 - 2x}\frac{dy}{dx}\right)\notag\end{align} and after some symbolic manipulations we arrive at the following differential equation $$x(1 - x)\frac{d^{2}y}{dx^{2}} + \left(a + b + \frac{1}{2} - (2a + 2b + 1)x\right)\frac{dy}{dx} - 4aby = 0$$ This is clearly satisfied by $ y = F(2a, 2b; a + b + 1/2; x)$ and hence the identity follows. On putting $ a = b = 1/4, x = k^{2}$ we see that $ 4x(1 - x) = 4k^{2}(1 - k^{2}) = (2kk')^{2}$ and therefore we obtain the formula $$K = \frac{\pi}{2}F\left(\frac{1}{4},\frac{1}{4}; 1; (2kk')^{2}\right)$$ or in simpler and expanded form $$\frac{2K}{\pi} = 1 + \left(\frac{1}{4}\right)^{2}(2kk')^{2} + \left(\frac{1\cdot 5}{4\cdot 8}\right)^{2}(2kk')^{4} + \left(\frac{1\cdot 5\cdot 9}{4\cdot 8\cdot 12}\right)^{2}(2kk')^{6} + \cdots$$ and this is valid for $ 0 \leq k \leq 1/\sqrt{2}$.
The equation satisfied by the series on the right side is \begin{align}&z^{2}(z - 1)y''' - 3z\left(a + b + \frac{1}{2} - (a + b + 1)z\right)y''\notag\\ &\,\,\,\,+[\{2(a^{2} + b^{2} + 4ab) + 3(a + b) + 1\}z - (a + b)(2a + 2b + 1)]y'\notag\\ &\,\,\,\, + 4ab(a + b)y = 0\notag\end{align} On the other hand the function $ v = F(a, b; a + b + 1/2; z)$ satisfies the differential equation $$z(z - 1)v'' + \{(a + b + 1)z - a - b - 1/2\}v' + abv = 0$$ or $$(z^{3} - z^{2})v'' + \{(a + b + 1)z^{2} - (a + b + 1/2)z\}v' + abzv = 0\tag{1}$$ Differentiating with respect to $ z$ we get \begin{align}&(z^{3} - z^{2})v''' + \{(a + b + 4)z^{2} - (a + b + 5/2)z\}v''\notag\\ &\,\,\,\,+\{(2a + 2b + ab + 2)z - (a + b + 1/2)\}v' + abv = 0\tag{2}\end{align} Let's put $ y = v^{2}$ so that $$ y' = 2vv', y'' = 2v'^{2} + 2vv'', y''' = 6v'v'' + 2vv'''$$ Now mulplying $(1)$ by $ 6v'$ and $(2)$ by $ 2v$ and adding the equations we get \begin{align}&(z^{3} - z^{2})(6v'v'' + 2vv''') + 6\{(a + b + 1)z^{2} - (a + b + 1/2)z\}v'^{2}\notag\\ &\,\,\,\, + 6abzvv' + 2\{(a + b + 4)z^{2} - (a + b + 5/2)z\}vv''\notag\\ &\,\,\,\, + 2\{(2a + 2b + ab + 2)z - (a + b + 1/2)\}vv' + 2abv^{2} = 0\notag\end{align} or \begin{align}& (z^{3} - z^{2})(6v'v'' + 2vv''') + 3z\{(a + b + 1)z - (a + b + 1/2)\}(2v'^{2} + 2vv'')\notag\\ &\,\,\,\,+ 2z(z - 1)(-2a - 2b + 1)vv'' + \{2z(a + b + 2ab + 1) - (a + b + 1/2)\}(2vv')\notag\\ &\,\,\,\, + 2abv^{2} = 0\notag\end{align} or \begin{align}&(z^{3} - z^{2})(6v'v'' + 2vv''') + 3z\{(a + b + 1)z - (a + b + 1/2)\}(2v'^{2} + 2vv'')\notag\\ &\,\,\,\, + 2(-2a - 2b + 1)[z(z - 1)vv'' + \{(a + b + 1)z - (a + b + 1/2)\}vv' + abv^{2}]\notag\\ &\,\,\,\, + 2(2a + 2b - 1)[\{(a + b + 1)z - (a + b + 1/2)\}vv' + abv^{2}]\notag\\ &\,\,\,\, + \{2z(a + b + 2ab + 1) - (a + b + 1/2)\}(2vv') + 2abv^{2} = 0\notag\end{align} or \begin{align}&(z^{3} - z^{2})(6v'v'' + 2vv''') + 3z\{(a + b + 1)z - (a + b + 1/2)\}(2v'^{2} + 2vv'')\notag\\ &\,\,\,\, + 2(a + b - 1/2)\{(a + b + 1)z - (a + b + 1/2)\}(2vv')\notag\\ &\,\,\,\, + \{2z(a + b + 2ab + 1) - (a + b + 1/2)\}(2vv') + 4ab(a + b)v^{2}= 0\notag\end{align} and finally \begin{align}&(z^{3} - z^{2})(6v'v'' + 2vv''') + 3z\{(a + b + 1)z - (a + b + 1/2)\}(2v'^{2} + 2vv'')\notag\\ &\,\,\,\, + [\{2(a^{2} + b^{2} + 4ab) + 3(a + b) + 1\}z - (a + b)(2a + 2b + 1)](2vv')\notag\\ &\,\,\,\, + 4ab(a + b)v^{2}= 0\notag\end{align} Thus $ y = v^{2} = (F(a, b; a + b + 1/2; z))^{2}$ satisfies the differential equation \begin{align}&z^{2}(z - 1)y''' - 3z\left(a + b + \frac{1}{2} - (a + b + 1)z\right)y''\notag\\ &\,\,\,\, + [\{2(a^{2} + b^{2} + 4ab) + 3(a + b) + 1\}z - (a + b)(2a + 2b + 1)]y'\notag\\ &\,\,\,\, + 4ab(a + b)y = 0\notag\end{align} This establishes the Clausen's Formula. Immediate application of this formula in the context of elliptic integrals is obtained by putting $ a = b = 1/4, z = (2kk')^{2}$ as follows: $$\left(\frac{2K}{\pi}\right)^{2} = 1 + \left(\frac{1}{2}\right)^{3}(2kk')^{2} + \left(\frac{1\cdot 3}{2\cdot 4}\right)^{3}(2kk')^{4} + \left(\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\right)^{3}(2kk')^{6} + \cdots$$ or in the more striking form as \begin{align}&\left(1 + \left(\frac{1}{2}\right)^{2}k^{2} + \left(\frac{1\cdot 3}{2\cdot 4}\right)^{2}k^{4} + \left(\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\right)^{2}k^{6} + \cdots\right)^{2}\notag\\ &\,\,\,\,\,\,\,\,= 1 + \left(\frac{1}{2}\right)^{3}(2kk')^{2} + \left(\frac{1\cdot 3}{2\cdot 4}\right)^{3}(2kk')^{4} + \left(\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\right)^{3}(2kk')^{6} + \cdots\notag\end{align} and this is valid for $ 0 \leq k \leq 1/\sqrt{2}$.
This is a fundamental result which is used by Ramanujan to derive certain series for $ 1/\pi$. With this we complete the background material in the theory of hypergeometric series which is needed to understand Ramanujan's Modular Equations and Approximations to $ \pi$.
In the next post we will define a modular equation and also derive such equations as done by Jacobi in his Fundamenta Nova.
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If $ a + b + (1/2)$ is neither zero nor a negative integer and if $ |x| < 1$ and $ |4x(1 - x)| < 1$, then $$F\left(a, b; a + b + \frac{1}{2}; 4x(1 - x)\right) = F\left(2a, 2b; a + b + \frac{1}{2}; x\right)$$
A curious thing to observe here is that the term on the left does not change if we replace $ x$ by $ (1 - x)$, but the right term does change its value. If we restrict ourselves to real positive values of $ x$ then we need to understand that in the right term the symbol $ x$ should actually be replaced by minimum of $ x$ and $ 1 - x$ or we can say that the identity as it is holds only when $ 0 \leq x \leq 1/2$.
Again to establish the identity we start with function $ y = F(a, b; a + b + 1/2; z)$ which satisfies the differential equation $$z(1 - z)\frac{d^{2}y}{dz^{2}} + \left(a + b + \frac{1}{2} - (a + b + 1)z\right)\frac{dy}{dz} - aby = 0$$ Putting $ z = 4x(1 - x)$ we get $$\frac{dy}{dz} = \dfrac{\dfrac{dy}{dx}}{\dfrac{dz}{dx}} = \frac{1}{4(1 - 2x)}\frac{dy}{dx}$$ and \begin{align} \frac{d^{2}y}{dz^{2}} &= \dfrac{\dfrac{d}{dx}\left(\dfrac{1}{4(1 - 2x)}\dfrac{dy}{dx}\right)}{\dfrac{dz}{dx}}\notag\\ &= \frac{1}{16(1 - 2x)}\left(\frac{1}{1 - 2x}\frac{d^{2}y}{dx^{2}} + \frac{2}{(1 - 2x)^{2}}\frac{dy}{dx}\right)\notag\\ &= \frac{1}{16(1 - 2x)^{2}}\left(\frac{d^{2}y}{dx^{2}} + \frac{2}{1 - 2x}\frac{dy}{dx}\right)\notag\end{align} and after some symbolic manipulations we arrive at the following differential equation $$x(1 - x)\frac{d^{2}y}{dx^{2}} + \left(a + b + \frac{1}{2} - (2a + 2b + 1)x\right)\frac{dy}{dx} - 4aby = 0$$ This is clearly satisfied by $ y = F(2a, 2b; a + b + 1/2; x)$ and hence the identity follows. On putting $ a = b = 1/4, x = k^{2}$ we see that $ 4x(1 - x) = 4k^{2}(1 - k^{2}) = (2kk')^{2}$ and therefore we obtain the formula $$K = \frac{\pi}{2}F\left(\frac{1}{4},\frac{1}{4}; 1; (2kk')^{2}\right)$$ or in simpler and expanded form $$\frac{2K}{\pi} = 1 + \left(\frac{1}{4}\right)^{2}(2kk')^{2} + \left(\frac{1\cdot 5}{4\cdot 8}\right)^{2}(2kk')^{4} + \left(\frac{1\cdot 5\cdot 9}{4\cdot 8\cdot 12}\right)^{2}(2kk')^{6} + \cdots$$ and this is valid for $ 0 \leq k \leq 1/\sqrt{2}$.
Clausen's Formula
It turns out we can square the above identity on both sides and get another series which is like the hypergeometric series but slightly more complex. In order to proceed further we introduce the generalized hypergeometric series $ {}_{3}F_{2}$ defined by $${}_{3}F_{2}(a, b, c; d, e; z) = \sum_{n = 0}^{\infty}\frac{(a)_{n}(b)_{n}(c)_{n}}{(d)_{n}(e)_{n}}\frac{z^{n}}{n!}$$ The differential equation satisfied by $ _{3}F_{2}$ is obtained in the similar way as done for the hypergeometric function and we have $$\{\Theta(\Theta + d - 1)(\Theta + e - 1) - z(\Theta + a)(\Theta + b)(\Theta + c)\}\,{}_{3}F_{2} = 0$$ This needs to be simplified by a laborious symbolic manipulation as follows: \begin{align}&\Theta(\Theta + d - 1)(\Theta + e - 1)y = \Theta(\Theta + d - 1)(zy' + (e - 1)y)\notag\\ &\,\,\,\,= \Theta\{z(zy' + (e - 1)y)' + (d - 1)(zy' + (e - 1)y)\}\notag\\ &\,\,\,\,= \Theta\{z(zy'' + ey') + z(d - 1)y' + (d - 1)(e - 1)y\}\notag\\ &\,\,\,\,=\Theta\{z^{2}y'' + (d + e - 1)zy' + (d - 1)(e - 1)y\}\notag\\ &\,\,\,\,= z\{z^{2}y'' + (d + e - 1)zy' + (d - 1)(e - 1)y\}'\notag\\ &\,\,\,\,= z\{z^{2}y''' + 2zy'' + (d + e - 1)zy'' + (d + e - 1)y' + (d - 1)(e - 1)y'\}\notag\\ &\,\,\,\,= z\{z^{2}y''' + (d + e + 1)zy'' + (d + e - 1 + de - d - e + 1)y'\}\notag\\ &\,\,\,\,= z\{z^{2}y''' + (d + e + 1)zy'' + dey'\}\notag\end{align} and \begin{align}&z(\Theta + a)(\Theta + b)(\Theta + c)y = z(\Theta + a)(\Theta + b)(zy' + cy)\notag\\ &\,\,\,\,= z(\Theta + a)\{z(zy' + cy)' + b(zy' + cy)\}\notag\\ &\,\,\,\,= z(\Theta + a)\{z(zy'' + y' + cy') + bzy' + bcy\}\notag\\ &\,\,\,\,= z(\Theta + a)\{z^{2}y'' + (b + c + 1)zy' + bcy\}\notag\\ &\,\,\,\,= z\{z(z^{2}y'' + (b + c + 1)zy' + bcy)' + a(z^{2}y'' + (b + c + 1)zy' + bcy)\}\notag\\ &\,\,\,\,= z\{z(z^{2}y''' + 2zy'' + (b + c + 1)zy'' + (b + c + 1)y' + bcy')\notag\\ &\,\,\,\,\,\,\,\,\,\,\,\,+ a(z^{2}y'' + (b + c + 1)zy' + bcy)\}\notag\\ &\,\,\,\,= z\{z^{3}y''' + (a + b + c + 3)z^{2}y'' + (b + c + 1 + bc + a(b + c + 1))zy' + abcy\}\notag\end{align} Finally the differential equation turns out to be \begin{align}&z^{2}(z - 1)y''' + \{(a + b + c + 3)z - (d + e + 1)\}zy''\notag\\ &\,\,\,\,\,\,\,\,+\{((a + 1)(b + c + 1) + bc)z - de\}y' + abcy = 0\notag\end{align} Using this differential equation Thomas Clausen in 1828 found a way to express the $ _{3}F_{2}$ series as a square of the usual hypergeometric function and this lead to some sufficient conditions under which a generalized hypergeometric series could be guaranteed to be positive. His result is famously called the Clausen's Formula which is as follows: $$\left(F\left(a, b; a + b + \frac{1}{2}; z\right)\right)^{2} = \,{}_{3}F_{2}\left(2a, 2b, a + b; 2a + 2b, a + b + \frac{1}{2}; z\right)$$ The conditions for validity turn out to be $ |z| < 1$ and both $ a + b + 1/2, 2a + 2b$ should not be zero or a negative integer.The equation satisfied by the series on the right side is \begin{align}&z^{2}(z - 1)y''' - 3z\left(a + b + \frac{1}{2} - (a + b + 1)z\right)y''\notag\\ &\,\,\,\,+[\{2(a^{2} + b^{2} + 4ab) + 3(a + b) + 1\}z - (a + b)(2a + 2b + 1)]y'\notag\\ &\,\,\,\, + 4ab(a + b)y = 0\notag\end{align} On the other hand the function $ v = F(a, b; a + b + 1/2; z)$ satisfies the differential equation $$z(z - 1)v'' + \{(a + b + 1)z - a - b - 1/2\}v' + abv = 0$$ or $$(z^{3} - z^{2})v'' + \{(a + b + 1)z^{2} - (a + b + 1/2)z\}v' + abzv = 0\tag{1}$$ Differentiating with respect to $ z$ we get \begin{align}&(z^{3} - z^{2})v''' + \{(a + b + 4)z^{2} - (a + b + 5/2)z\}v''\notag\\ &\,\,\,\,+\{(2a + 2b + ab + 2)z - (a + b + 1/2)\}v' + abv = 0\tag{2}\end{align} Let's put $ y = v^{2}$ so that $$ y' = 2vv', y'' = 2v'^{2} + 2vv'', y''' = 6v'v'' + 2vv'''$$ Now mulplying $(1)$ by $ 6v'$ and $(2)$ by $ 2v$ and adding the equations we get \begin{align}&(z^{3} - z^{2})(6v'v'' + 2vv''') + 6\{(a + b + 1)z^{2} - (a + b + 1/2)z\}v'^{2}\notag\\ &\,\,\,\, + 6abzvv' + 2\{(a + b + 4)z^{2} - (a + b + 5/2)z\}vv''\notag\\ &\,\,\,\, + 2\{(2a + 2b + ab + 2)z - (a + b + 1/2)\}vv' + 2abv^{2} = 0\notag\end{align} or \begin{align}& (z^{3} - z^{2})(6v'v'' + 2vv''') + 3z\{(a + b + 1)z - (a + b + 1/2)\}(2v'^{2} + 2vv'')\notag\\ &\,\,\,\,+ 2z(z - 1)(-2a - 2b + 1)vv'' + \{2z(a + b + 2ab + 1) - (a + b + 1/2)\}(2vv')\notag\\ &\,\,\,\, + 2abv^{2} = 0\notag\end{align} or \begin{align}&(z^{3} - z^{2})(6v'v'' + 2vv''') + 3z\{(a + b + 1)z - (a + b + 1/2)\}(2v'^{2} + 2vv'')\notag\\ &\,\,\,\, + 2(-2a - 2b + 1)[z(z - 1)vv'' + \{(a + b + 1)z - (a + b + 1/2)\}vv' + abv^{2}]\notag\\ &\,\,\,\, + 2(2a + 2b - 1)[\{(a + b + 1)z - (a + b + 1/2)\}vv' + abv^{2}]\notag\\ &\,\,\,\, + \{2z(a + b + 2ab + 1) - (a + b + 1/2)\}(2vv') + 2abv^{2} = 0\notag\end{align} or \begin{align}&(z^{3} - z^{2})(6v'v'' + 2vv''') + 3z\{(a + b + 1)z - (a + b + 1/2)\}(2v'^{2} + 2vv'')\notag\\ &\,\,\,\, + 2(a + b - 1/2)\{(a + b + 1)z - (a + b + 1/2)\}(2vv')\notag\\ &\,\,\,\, + \{2z(a + b + 2ab + 1) - (a + b + 1/2)\}(2vv') + 4ab(a + b)v^{2}= 0\notag\end{align} and finally \begin{align}&(z^{3} - z^{2})(6v'v'' + 2vv''') + 3z\{(a + b + 1)z - (a + b + 1/2)\}(2v'^{2} + 2vv'')\notag\\ &\,\,\,\, + [\{2(a^{2} + b^{2} + 4ab) + 3(a + b) + 1\}z - (a + b)(2a + 2b + 1)](2vv')\notag\\ &\,\,\,\, + 4ab(a + b)v^{2}= 0\notag\end{align} Thus $ y = v^{2} = (F(a, b; a + b + 1/2; z))^{2}$ satisfies the differential equation \begin{align}&z^{2}(z - 1)y''' - 3z\left(a + b + \frac{1}{2} - (a + b + 1)z\right)y''\notag\\ &\,\,\,\, + [\{2(a^{2} + b^{2} + 4ab) + 3(a + b) + 1\}z - (a + b)(2a + 2b + 1)]y'\notag\\ &\,\,\,\, + 4ab(a + b)y = 0\notag\end{align} This establishes the Clausen's Formula. Immediate application of this formula in the context of elliptic integrals is obtained by putting $ a = b = 1/4, z = (2kk')^{2}$ as follows: $$\left(\frac{2K}{\pi}\right)^{2} = 1 + \left(\frac{1}{2}\right)^{3}(2kk')^{2} + \left(\frac{1\cdot 3}{2\cdot 4}\right)^{3}(2kk')^{4} + \left(\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\right)^{3}(2kk')^{6} + \cdots$$ or in the more striking form as \begin{align}&\left(1 + \left(\frac{1}{2}\right)^{2}k^{2} + \left(\frac{1\cdot 3}{2\cdot 4}\right)^{2}k^{4} + \left(\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\right)^{2}k^{6} + \cdots\right)^{2}\notag\\ &\,\,\,\,\,\,\,\,= 1 + \left(\frac{1}{2}\right)^{3}(2kk')^{2} + \left(\frac{1\cdot 3}{2\cdot 4}\right)^{3}(2kk')^{4} + \left(\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\right)^{3}(2kk')^{6} + \cdots\notag\end{align} and this is valid for $ 0 \leq k \leq 1/\sqrt{2}$.
This is a fundamental result which is used by Ramanujan to derive certain series for $ 1/\pi$. With this we complete the background material in the theory of hypergeometric series which is needed to understand Ramanujan's Modular Equations and Approximations to $ \pi$.
In the next post we will define a modular equation and also derive such equations as done by Jacobi in his Fundamenta Nova.
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