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The arithmetic–geometric mean of 2 values $a_0$,$b_0$, is the value to which the arithmetic and geometric values converge, being $$a_n=\frac{a_{n-1}+b_{n-1}}{2} \text{ and } b_n=\sqrt{a_{n-1} .b_{n-1}}$$ with $$\lim_{n \to \infty} a_n = \lim_{n \to \infty} b_n = AGM(a_0,b_0)$$

The arithmetic and the geometric values have a simple geometrical interpretation

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if $a_0$ is the segment FE, and $b_0$ is the segment EG, then the radius CD is the arithmetic mean, and the segment ED is the geometric mean.

Is possible to draw the arithmetic–geometric mean on that circle (or there is another geometrical interpretation maybe not involving circles)?

Glorfindel
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zugatevej
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1 Answers1

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The angular average of the distance between the perimeter of an ellipse and its centre:

\begin{align*} \langle r \rangle_{\theta} &= \frac{1}{2\pi} \int_{0}^{2\pi} r \, d\theta \\ &= \frac{1}{2\pi} \int_{0}^{2\pi} \frac{a b}{\sqrt{a^2\sin^2 \theta+b^2\cos^2 \theta}} \, d\theta \\ &= \frac{2ab}{\pi} \int_{0}^{\pi/2} \frac{d\theta}{\sqrt{a^2\sin^2 \theta+b^2\cos^2 \theta}} \\ &= \frac{ab}{\operatorname{agm}(a,b)} \end{align*}

Ng Chung Tak
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  • That's an useful answer, but would be more useful if were possible to construct the AGM geometrically. (Note 1: your a,b are not my $a_0, b_0$). (Note 2: In the circle, it reduces to $AGM(r,r)=\frac {r^2} {r}$) – zugatevej Mar 02 '17 at 20:03
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    @zugatevej: If it were possible to construct the AGM in a finite process, it would seem that it would not follow the limit-driven definition given in the OP. – Brian Tung Mar 02 '17 at 23:00
  • @Brian Tung: a and b are known from the start. You are saying that cannot be constructed. A limit does not guarantee that the limit value can not be constructed with finite steps. Also, the construction is not limited to compass and rule. – zugatevej Mar 03 '17 at 10:04
  • @zugatevej: That's not my intent. I mean to say that whatever construction there is would not follow the derivation given above. It would not be pedagogically direct, even though it might be proven that it's correct. – Brian Tung Mar 03 '17 at 16:30