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This is a long overdue follow up to another question, but it's worth asking separately.

Consider the following iterated mean:

$$a_{n+1}=\frac{a_n+b_n}{2}$$

$$b_{n+1}=a_n+b_n-\sqrt{a_n b_n}$$

$$\mu(a_0,b_0)=\lim_{n \to \infty} a_n=\lim_{n \to \infty} b_n$$

It's kind of similar to the famous arithmetic-geometric mean, however I wasn't able to find a closed form.

But then I've got the following idea. One of the ways to derive the relation between AGM and elliptic integrals is to consider the following function $f(x)=\text{agm}(1-x,1+x)$. It's even and well defined on $(-1,1)$, and (I think it was Gauss' result) we can derive the series for $1/f(x)$ and prove the famous identity.

I decided to attempt the same thing, only I forgot how Gauss derived the series, and so I tried to do it numerically, using finite differences.

Here's the result:

$$g(x)=\frac{1}{\mu(1-x,1+x)}$$

$$g(x)=1-\frac{x^2}{4}-\frac{x^4}{64}-\frac{3 x^6}{256}-\frac{161 x^8}{16384}-\frac{399 x^{10}}{65536}-\frac{5075 x^{12}}{1048576}-\frac{15817 x^{14}}{4194304}-\dots$$

The coefficients are all rational, of that I'm pretty sure, so it is likely that some closed (possibly hypergeometric like form) exists for this function.

The denominators are all powers of $2$, but I wasn't able to find any rule for the numerators.

Compare the above to the series expansion of the elliptic integral ${_2 F_1} (1/2,1/2;1;x^2)$. The denominators are exactly the same, at least for the first few terms.

It's worth noting that if we multiply by the factorials (getting back to derivatives), we obtain the following list:

$$\left\{1,-\frac{1}{2},-\frac{3}{8},-\frac{135}{16},-\frac{50715}{128},-\frac{5655825}{256},-\frac{2373958125}{1024},-\frac{673290542925}{2048}\right\}$$

We again have powers of $2$ in the denominators, and mostly multiples of $5$ in the numerators.

I can compute more with Mathematica, but I wanted to hear your suggestions on the following:

  • How do I derive the expression for the general term for a series like that (I'm pretty sure it exists and is simple enough)?

  • Maybe there's a way to directly derive the series expression from the recurrence above? If you have a reference to Gauss' AGM work I would be grateful.


Here's an illustration showing how close the above series (orange) describe the function (blue):

enter image description here


If we assume that the ratio of the successive terms is a rational function of $n$ with rational parameters, can we try some numerical search? I understand that in general, finding a pattern in a such a sequence is not possible without some additional information.

List of coefficients with sign omitted:

{1/4,1/64,3/256,161/16384,399/65536,5075/1048576,15817/4194304,3316001/1073741824,11074335/4294967296,151002287/68719476736,...}

List of ratios of the consecutive terms:

{1/16,3/4,161/192,57/92,725/912,15817/20300,3316001/4049152,11074335/13264004,151002287/177189360,...}

The search at OEIS didn't give any results.

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I'll start my further attempts at answering the questions here, because the original post is big enough.

I forgot, that @ParamanandSingh already described a method for obtaining a series like that for AGM, in this answer: https://math.stackexchange.com/a/1904059/269624. Using the same logic, I get an awkward functional equation, valid for $x\in(-1,1/3)$ (yes, $1/3$, because the derivation contains the root of $1-3x$, even though in the final expression it's $1-2x$):

$$g(x)=\frac{1}{(1-x)} g \left(2 \frac{\sqrt{x(1-2x)}}{1-x} \right)$$

Assuming the even series form:

$$g(x)=1-a_2 x^2-a_4 x^4-a_6 x^6 - \dots$$

We get:

$$(1-x)(1-a_2 x^2-a_4 x^4-a_6 x^6 - \dots)= \\ =1-2^2 a_2 \frac{x(1-2x)}{(1-x)^2}-2^4 a_4 \frac{x^2(1-2x)^2}{(1-x)^4}-2^6 a_6 \frac{x^3(1-2x)^3}{(1-x)^6} - \dots$$

This is very complicated, but let's try to expand anyway (with the help of Mathematica) and collect the powers of $x$:

$$\begin{array}( x^0: & \qquad 1=1 \\ x^1: & \qquad -1 + 4 a_2=0 \\ x^2: & \qquad -a_2 + 16 a_4=0 \\ x^3: & \qquad -3 a_2 + 64 a_6=0 \\ x^4: & \qquad -8 a_2 - 33 a_4 + 256 a_8=0 \\ x^5: & \qquad -12 a_2 - 63 a_4 - 192 a_6+1024 a_{10}=0 \end{array}$$

This gives the same values for the coefficients, however, unlike Gauss, I don't see any order in this so far.

Maybe you have any suggestions on how to improve this method?


And another way would be the hypergeometric differential equation, I will see what I can do on this road later.

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