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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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. – 2019-05-11
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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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.