# Prove the AGM identity using only Hypergeometric series

Arithmetic Geometric Mean can be represented by a Hypergeometric function:

$$\text{agm}(1,p)=\frac{1}{{_2F_1} \left(\frac{1}{2},\frac{1}{2};1;1-p^2 \right)}$$

$$0<p \leq 1$$

One of the main properties of the AGM is the following identity:

$$\text{agm}(1,p)=\frac{1+p}{2}\text{agm} \left(1,\frac{2\sqrt{p}}{1+p} \right)$$

This allows the infinite product representation of the AGM.

I wanted to know if it's possible to prove this identity by directly using the Hypergeometric series.

For the Hypergeometric function the identity will take the following form:

$${_2F_1} \left(\frac{1}{2},\frac{1}{2};1;1-p^2 \right)=\frac{2}{1+p} {_2F_1} \left(\frac{1}{2},\frac{1}{2};1;\frac{(1-p)^2}{(1+p)^2} \right)$$

In the series form it will be:

$$\sum_{k=0}^\infty \frac{1}{k!^2} \left(\frac{1}{2}\right)^2_k (1-p^2)^k=\frac{2}{1+p} \sum_{k=0}^\infty \frac{1}{k!^2} \left(\frac{1}{2}\right)^2_k \frac{(1-p)^{2k}}{(1+p)^{2k}}$$

I think the following substitution will simplify things:

$$1-p=2x$$

$$\sum_{k=0}^\infty \frac{1}{k!^2} \left(\frac{1}{2}\right)^2_k 2^{2k}x^k(1-x)^k=\frac{1}{1-x} \sum_{k=0}^\infty \frac{1}{k!^2} {\left(\frac{1}{2}\right)^2_k} \frac{x^{2k}}{(1-x)^{2k}}$$

I haven't been able to prove this identity from the series.

Comparing terms in this form is useless, since the partial sums of the series are not equal (the second series converges much faster).

The only idea I have is to use the uniqueness of the power series, which requires expanding everything, so there are only powers of $x$ left.

We have:

$$(1-x)^k=\sum_{l=0}^k (-1)^l \left(\begin{matrix} k \\ l \end{matrix} \right) x^l$$

$$\frac{1}{(1-x)^{2k+1}}=(2k)! \sum_{n=0}^\infty (-1)^{n+2k}~ (n+2k)_{2k} ~x^n$$

Here $(n+2k)_{2k}$ actually means falling factorial, not rising factorial, like above. $(n+2k)_{2k}=(n+2k)(n+2k-1)(n+2k-2) \cdots$. I don't know what other notation to use in this case.

Now I'm stuck. I don't know how to get the single power series for $x$ on each side so we can compare them.

• Just out of curiosity, may I ask how came this fascination for means ? I am always interested by your posts. – Claude Leibovici Aug 26 '16 at 8:03
• @ClaudeLeibovici, I don't know how my brain works. I switch topics of interest fast. It's just that the topic of means is so rich and has so many applications that I can't stop thinking about it. Especially, since I joined Math.SE – Yuriy S Aug 26 '16 at 8:07
• While you are on topic of means I would want you to explore the agm concept for complex numbers instead of just for real numbers. See this very illuminating paper by David A. Cox e-periodica.ch/cntmng?pid=ens-001:1984:30::87 – Paramanand Singh Aug 26 '16 at 9:11
• @ParamanandSingh, if you want to be suitably impressed, there is also a concept of p-adic agm – Yuriy S Aug 26 '16 at 9:20
• I have given another approach which uses the differential equation for the hypergeometric function. This proof is also from Gauss. – Paramanand Singh Aug 26 '16 at 10:51

## 2 Answers

Use the function $$f(x) = \frac{1}{\operatorname{agm}(1 + x, 1 - x)}\tag{1}$$ which has the property that $$f(x) = \frac{1}{1 + x}\cdot f\left(\frac{2\sqrt{x}}{1 + x}\right)\tag{2}$$ and use the series expansion $$f(x) = 1 + a_{1}x^{2} + a_{2}x^{4} + \cdots\tag{3}$$ (note that the function $f$ is even) and find coefficients $a_{n}$ by using series expansion $(3)$ in functional equation $(2)$. This is how Gauss derived the formula $$f(x) = {}_{2}F_{1}\left(\frac{1}{2}, \frac{1}{2};1; x^{2}\right)\tag{4}$$ See this post of mine for details. Note also that $$f(x) = \frac{1}{\operatorname{agm}(1, \sqrt{1 - x^{2}})}\tag{5}$$ and hence on putting $1 - x^{2} = p^{2}$ we get $$\operatorname{agm}(1, p) = \dfrac{1}{{}_{2}F_{1}\left(\dfrac{1}{2}, \dfrac{1}{2};1; 1 - p^{2}\right)}\tag{6}$$

BTW the evaluation of coefficients $a_{n}$ for general $n$ is difficult but Gauss solved it completely. See this paper for more details of the calculation of $a_{n}$.

Another route to the relation between hypergeometric series is to use the differential equation satisfied by the function $y(x) = {}_{2}F_{1}(a, b; c; x)$. The differential equation satisfied by $y$ is given as $$x(1 - x)y'' + \{c - (a + b + 1)x\}y' - aby = 0\tag{7}$$ Let's just write $F$ in place of ${}_{2}F_{1}$ and then we see that $y = F(a, b; 2b; x)$ satisfies the equation $$x(1 - x)y'' + \{2b - (a + b + 1)x\}y' - aby = 0$$ Putting $x = 4z/(1 + z)^{2}$ we can see that that $$\frac{dF}{dx} = \frac{(1 + z)^{3}}{4(1 - z)}\cdot\frac{dF}{dz}$$ and $$\frac{d^{2}F}{dx^{2}} = \frac{(1 + z)^{5}}{16(1 - z)^{3}}\left((1 - z^{2})\frac{d^{2}F}{dz^{2}} + (4 - 2z)\frac{dF}{dz}\right)$$ and $$x(1 - x) = \frac{4z(1 - z)^{2}}{(1 + z)^{4}}$$ and thus after some algebraic manipulation we get $$z(1 - z)(1 + z)^{2}\frac{d^{2}F}{dz^{2}} + 2(1 + z)(b - 2az + bz^{2} - z^{2})\frac{dF}{dz} - 4ab(1 - z)F = 0$$ and the function $F(a, b; 2b; 4z/(1 + z)^{2})$ satisfies this equation. If we put $F = (1 + z)^{2a}G$ we get $$z(1 - z^{2})\frac{d^{2}G}{dz^{2}} + 2\{b - (2a - b + 1)z^{2}\}\frac{dG}{dz} - 2az(1 + 2a - 2b)G = 0$$ It is easily seen that $G(-z)$ also satisfies this equation and hence $G$ is an even function and we can put $z^{2} = t$ to get $$t(1 - t)\frac{d^{2}G}{dt^{2}} + \left(b + \frac{1}{2} - \left(2a - b + \frac{3}{2}\right)t\right)\frac{dG}{dt} - a\left(a - b + \frac{1}{2}\right)G = 0\tag{8}$$ Comparing this with equation $(7)$ we see that solution $G$ is given by $$G = F(a, a - b + 1/2; b + 1/2; t) = F(a, a - b + 1/2; b + 1/2; z^{2})\tag{9}$$ However the way we reached equation $(8)$ shows us that its solution is given by $$G = (1 + z)^{-2a}F(a, b; 2b; 4z/(1 + z)^{2})\tag{10}$$ Both the solutions $(9)$ and $(10)$ are analytic in neighborhood of $0$ and they are equal to $1$ at $z = 0$ and therefore they are equal and we get the quadratic transformation of Gauss $$F\left(a, b; 2b; \frac{4z}{(1 + z)^{2}}\right) = (1 + z)^{2a}F\left(a, a - b + \frac{1}{2}; b + \frac{1}{2}; z^{2}\right)\tag{11}$$ Putting $a = 1/2, b = 1/2$ we get $$F\left(\frac{1}{2}, \frac{1}{2}; 1; \frac{4z}{(1 + z)^{2}}\right) = (1 + z)F\left(\frac{1}{2}, \frac{1}{2}; 1; z^{2}\right)\tag{12}$$ and putting $z = (1 - p)/(1 + p)$ so that $p = (1 - z)/(1 + z)$ we get $$F\left(\frac{1}{2}, \frac{1}{2}; 1; 1 - p^{2}\right) = \frac{2}{1 + p}F\left(\frac{1}{2}, \frac{1}{2}; 1; \frac{(1 - p)^{2}}{(1 + p)^{2}}\right)\tag{13}$$ which is the result in question. Using similar technique we can prove another quadratic transformation $$F\left(a, b; a + b + \frac{1}{2}; 4z(1 - z)\right) = F\left(2a, 2b; a + b + \frac{1}{2}; z\right)\tag{14}$$ See this post and the next one for more details.

• Thank you for the reference. I didn't know Gauss have done it like this. This means that I can potentially develop powers series for every iterated mean I want – Yuriy S Aug 26 '16 at 8:04
• @YuriyS: you need to get the right functional equation for the corresponding series so that the coefficients can be easily evaluated. – Paramanand Singh Aug 26 '16 at 8:05
• @ParamanandSingh, that's still more clear to me, than 'random' integral substitutions (which most people use to demonstrate the integral form) or general properties of elliptic curves (which I know nothing about, and what to do with) – Yuriy S Aug 26 '16 at 8:12
• @YuriyS: fully agree with you. The paper from Stacy Langton (linked in the answer) was the reason for which I started blogging. I was so inspired by the proof of Gauss connecting AGM with hypergeometric functions and elliptic integrals that I wrote an entire series of blog posts on that. From then on the journey into these topics (including theta functions) has never ceased to amaze me. – Paramanand Singh Aug 26 '16 at 8:16
• Excellent work. I hope you don't mind that I wait a day or two before accepting the answer – Yuriy S Aug 26 '16 at 11:25

I think it is faster to go through an elliptic integral. Given $a,b\in\mathbb{R}^+$ we may define $$E(a,b)=\int_{0}^{+\infty}\frac{dx}{\sqrt{(a^2+x^2)(b^2+x^2)}} \tag{1}$$ and notice that by Lagrange's identity $(a^2+x^2)(b^2+x^2)=(ax+bx)^2+(ab-x^2)^2$ and a suitable change of variable we have $$E(a,b)=E\left(\frac{a+b}{2},\sqrt{ab}\right) \tag{2}$$ hence: $$E(a,b)=E(AGM(a,b),AGM(a,b))=\frac{1}{AGM(a,b)}\int_{0}^{+\infty}\frac{dx}{x^2+1}\tag{3}$$ and $$AGM(a,b) = \frac{\pi}{2 E(a,b)}\tag{4}$$ so our claim is equivalent to proving that $$\frac{2}{\pi}\int_{0}^{+\infty}\frac{dx}{\sqrt{(1+x^2)(p^2+x^2)}} = \phantom{}_2 F_1\left(\frac{1}{2},\frac{1}{2};1;1-p^2\right) \tag{5}$$ but by setting $u=\frac{1}{1+x^2}$, that simply follows from Euler's integral representation for $\phantom{}_2 F_1$.

• @YuriyS: but still, Landen and Euler's identities for hypergeometric functions are essentially given by change of variables / symmetry arguments. The only difference in "style" is that you may perform them in a integral or in a differential equation: also Paramanand set $x=\frac{1-p}{1+p}$ at some point. – Jack D'Aurizio Aug 26 '16 at 14:18
• So your request may sound like "how to exploit a symmetry without exploiting a symmetry?" :D – Jack D'Aurizio Aug 26 '16 at 14:20
• @ Jack D'Aurizio, I'm trying to find a general method for linking Hypergeometric functions to some iterated algorithm. – Yuriy S Aug 26 '16 at 14:52
• @YuriyS: that is the way (Carson's way): if an integral is invariant under some substitution, it is related with an iterated mean. In the opposite direction, if you want to represent an iterated mean as an integral, you have to look at integrals that are invariant under some substitution. Given the hypergeometric differential equation, the change of variables of interest are given by the Sturm-Liouville theory and Fuchs' theorem (en.wikipedia.org/wiki/Fuchs%27_theorem) since the involved differential equations exhibit high symmetry. – Jack D'Aurizio Aug 26 '16 at 15:01
• @JackD'Aurizio: Thanks for the link on Fuchs' theorem. I was not aware of general theorem for dealing with differential equations having symmetry. Both my proofs in the answer were small building blocks towards a systematic exposition of the theory of modular equations. +1 for the approach based on integral transformation which may look slightly indirect, but is pretty standard in elliptic integrals. – Paramanand Singh Aug 26 '16 at 19:21