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We know that $\pi$ can be computed by Arithmetic Geometric mean using Gauss-Legendre procedure which does provide fastest convergence rate as well with a guarantee of $2^n$ bits of $\pi$ at $n$th iteration.

This leads to the question what classes of continued fractions can be given as Arithmetic geometric procedure?

1. Does $C(x) = x + \frac{1^{2}}{2x + \frac{3^{2}}{2x + \frac{5^{2}}{2x + \frac{7^{2}}{2x + \cdots}}}}$ have a rapid Arithmetic geometric procedure at $x\in\Bbb Z_{\neq1}$ (note $C(1)=\frac{4}{\pi}$)?

2. Which algebraic numbers that have continued fraction expansion, admit rapid Arithmetic geometric procedure?

3. Since $\pi$ has rapid Arithmetic geometric procedure we have $\zeta(2n)$ having rapid Arithmetic geometric procedure. Does $\zeta(2n+1)$ admit rapid Arithmetic geometric procedure?

4. Supposing the fastest approximation for a number is through a rapid Arithmetic geometric procedure, is that number transcendental? Is there any connection to transcendentality through existence of rapid Arithmetic geometric procedure?

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  • $\begingroup$ With regard to 2., for all algebraic numbers, Newton iteration has even lower computational complexity than AGM (faster by a $\log(n)$ factor). I don't know about AGM iteration for $\zeta(2n+1)$, but there are other algorithms that are almost as fast (slower by a $\log^c(n)$ factor). $\endgroup$
    – Fredrik Johansson
    Commented Sep 20, 2015 at 23:00
  • $\begingroup$ @FredrikJohansson So could you elaborate the AGM technique for algebraic numbers? $\endgroup$
    – Turbo
    Commented Sep 21, 2015 at 1:06
  • $\begingroup$ I don't believe this is related to numerical analysis. $\endgroup$
    – David Ketcheson
    Commented Sep 21, 2015 at 3:10
  • $\begingroup$ @FredrikJohansson what is the AGM technique for algebraic numbers? $\endgroup$
    – Turbo
    Commented Feb 24, 2017 at 2:38
  • $\begingroup$ @Turbo I meant that Newton iteration for computing algebraic numbers is faster than AGM iteration itself, regardless of whether the AGM can be used to compute algebraic numbers. $\endgroup$
    – Fredrik Johansson
    Commented Feb 24, 2017 at 11:37

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