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Suppose I have a sequence of real numbers such that AGM $(a_{n},a_{n+1}) = 1/2^n$. Would that sequence be unique and how would I go about finding the individual $a_{n}$? (AGM is the arithmetic geometric mean)

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I am not sure what you mean by uniqueness here. Since if I take $a$ and $b$ to get the AGM then every pair in the process i.e $(a_1,b_1)$, $(a_2,b_2)$, $(a_3,b_3)$, and so on will have the same AGM right? So in some sense uniqueness would mean that if I start with two distinct numbers $a$ and $b$, then I will never be able to get $\frac{1}{2^n}$ as my AGM (i.e. it forces both the numbers to be $\frac{1}{2^n}$). Is that your claim? Correct me if my interpretation is wrong. –  user17762 Feb 23 '11 at 0:28
    
The only constraint on the $a_{n}$ here is that AGM$(a_{n},a_{n+1})=1/2^{n}$. I think you're confusing the $a$s and $b$s in the def of the AGM with the sequence of $a_{n}$ that I'm using here. –  deoxygerbe Feb 23 '11 at 0:32
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Can't we pick $a_1$ arbitrarily? I am guessing there will always be a solution to $AGM(a, x) = b$, given $a,b \gt 0$. –  Aryabhata Feb 23 '11 at 0:37
    
Oh Ok. I get it. Moron's comment clarified it further. You have one degree of freedom in choosing one of the $a_i$'s and then as Moron says if there is a solution to $AGM(a,x) = b$, then the sequence is determined. –  user17762 Feb 23 '11 at 0:46

1 Answer 1

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One series that works is generated by noting that $AGM(x,\frac{x}{2})=\frac {3 \pi x}{8K(\frac{1}{9})}$ where $K(x)$ is the complete elliptic integral of the first kind. So we can take $a_n=\frac{8K(\frac{1}{9})}{3\pi 2^n}$ Numerically, $K(\frac{1}{9})\approx 1.617386$ This yields $AGM(x,\frac{x}{2})\approx 0.7283955 x$ and $a_n\approx \frac{1.37288}{2^n}$. I suspect you could apply a dither, raising the odd $n$ and lowering the even $n$, but haven't checked.

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As at @Moron said, there are many such sequences, but I'm going to give you answer credit on elegance grounds. –  deoxygerbe Feb 23 '11 at 1:18
    
Looking at the Wikipedia page, I strongly suspect that Moron is right, that you can always solve $AGM(x,y)=a$ for $y$ given any $x$. –  Ross Millikan Feb 23 '11 at 1:37

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