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)
Implicitly determined agm sequence
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calculus
sequences-and-series
analysis
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0I 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. – 2011-02-23
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0The 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. – 2011-02-23
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3Can't we pick $a_1$ arbitrarily? I am guessing there will always be a solution to $AGM(a, x) = b$, given $a,b \gt 0$. – 2011-02-23
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0Oh 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. – 2011-02-23
2 Answers
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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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0As at @Moron said, there are many such sequences, but I'm going to give you answer credit on elegance grounds. – 2011-02-23
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0Looking at the Wikipedia page, I strongly suspect that Moron is right, that you can always solve $AGM(x,y)=a$ for $y$ given any $x$. – 2011-02-23
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I am guessing that you have each arithmetic mean half the previous one. If so, that means that all of the geometric means are zero. That can only happen if at least one of the two starting values is zero. The other starting value can be any real number.