Recently, I got interested in the Arithmetic-Geometric mean $\mathrm{AGM}(x,y)$ because it had the surprising property that $$\int_0^{\pi/2}\frac{dt}{\sqrt{x^2\cos^2t+y^2\sin^2t}}=\frac{\pi}{2\mathrm{AGM}(x,y)}.$$ I say this is surprising because it has such a complicated definition:

If the sequences $(a_n)$ and $(g_n)$ are defined by $$\begin{align} a_{n+1}&=\tfrac12(a_n+g_n) &a_0&=x\\ g_{n+1}&=\sqrt{a_n g_n} &g_0&=y \end{align}$$ then $$\mathrm{AGM}(x,y):=\lim_{n\to\infty}a_n\ .$$

After I messed around with $\mathrm{AGM}$ and was able to prove its relation to the above elliptic integral, I asked myself the question "is there an artihmetic-harmonic mean?" The answer: yes.

The Arithmetic-Harmonic Mean:

We define the sequences $$\begin{align} a_{n+1}&=\tfrac{1}{2}(a_n+h_n) &a_0&=x\\ h_{n+1}&=\frac2{\frac1{a_n}+\frac1{h_n}} &h_0&=y \end{align}$$ and the Arithmetic-Harmonic mean is then defined as $$\mathrm{AHM}(x,y):=\lim_{n\to\infty}a_n\ .$$ Amazingly enough, we are able to find a closed-form evaluation for $\mathrm{AHM}(x,y)$ assuming $x,y>0$. We do so by noticing that $$h_{n+1}=\frac{2a_nh_n}{a_n+h_n}=\frac{a_nh_n}{a_{n+1}}$$ so that $$a_nh_n=a_{n-1}h_{n-1}=a_0h_0=xy$$ giving $$a_{n+1}=\frac12\left(a_n+\frac{xy}{a_n}\right)$$ which converges to $$\lim_{n\to\infty}a_n=\mathrm{AHM}(x,y)=\sqrt{xy}\ .$$

That being established, I wanted to know if there is a geometric-harmonic mean.

The Geometric-Harmonic Mean:

I first should define it. Let the sequences $(h_n)$ and $(g_n)$ be defined as $$\begin{align} h_{n+1}&=\frac{2}{\frac1{h_n}+\frac1{g_n}} &h_0&=x\\ g_{n+1}&=\sqrt{h_n g_n} &g_0&=y \end{align}$$ then, assuming convergence, define $$\mathrm{GHM}(x,y):=\lim_{n\to\infty}h_n\ .$$ It seems as if it will be harder to find out things about $\mathrm{GHM}$ because I cannot seem to sufficiently simplify the relationship between the two sequences as I was able to with $\mathrm{AHM}$. I feel though, that there may be a really interesting integral relationship here.

I did a little investigation of my own. One notable value of $\mathrm{AGM}$ is Gauss's Constant: $$\mathbf{g}=\mathrm{AGM}(1,\sqrt2)=\frac{(2\pi)^{3/2}}{\Gamma^2(\tfrac14)}.$$ I found $h_4$ and $g_4$ for $h_0=1$, $g_0=\sqrt{2}$ on Desmos: $$h_4\approx g_4\approx 1.18034059902$$ for which Wolfram suggests the closed form $$1.18034059902\approx \sqrt{2}\,\mathbf{g}$$ which is definitely very fishy...

So my questions: Is there some connection between $\mathrm{AGM}$ and $\mathrm{GHM}$? Is there a nice integral relationship for $\mathrm{GHM}$? Is there a closed for for $\mathrm{GHM}$?


Note that $$ a_n = \frac{1}{h_n} \, , \,b_n = \frac{1}{g_n} $$ satisfy the recurrence $$ \begin{align} a_{n+1}&=\tfrac12(a_n+b_n) &a_0&=\frac 1x\\ b_{n+1}&=\sqrt{a_n b_n} &b_0&= \frac1y \end{align} $$ so that in fact $$ \operatorname{GHM}(x, y) = \frac{1}{\operatorname{AGM}(\frac 1x, \frac 1y)} = \frac{xy}{\operatorname{AGM}(x, y)} \, . $$

  • $\begingroup$ Oh wow. so then $$\int_0^{\pi/2}\frac{dt}{\sqrt{x^{-2}\cos^2t+y^{-2}\sin^2t}}=\frac\pi2\mathrm{GHM}(x,y)$$ $\endgroup$
    – clathratus
    Jul 19 '19 at 14:40
  • $\begingroup$ @clathratus: Yes! $\endgroup$
    – Martin R
    Jul 19 '19 at 14:44
  • $\begingroup$ Really nice answer, thank you! $\endgroup$
    – clathratus
    Jul 19 '19 at 14:45
  • 3
    $\begingroup$ We can also show $\operatorname{GHM}(1,\,\sqrt{2})=\sqrt{2}/\operatorname{AGM}(1,\,\sqrt{2})$ using $\operatorname{AGM}(kx,\,ky)=k\operatorname{AGM}(x,\,y)$. $\endgroup$
    – J.G.
    Jul 19 '19 at 14:50
  • $\begingroup$ And since $\text{AGM}(x,y)=y \text{AGM}(x/y,1)$, we can restrict our attention to the function of one parameter $\text{AGM}(1,x)$ without loss of generality. $\endgroup$ Jun 11 at 17:44

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