# Asymptotic behavior of recurrence $x_{n+1}=\mbox{Stdev}(x_1,\dots,x_n)$

Here $$x_1>0$$ is the initial condition and $$x_{n+1}$$ is defined by

$$x_{n+1}=\Big[\frac{1}{n}\sum_{k=1}^n x_k^2 -\frac{1}{n^2}\Big(\sum_{k=1}^n x_k\Big)^2 \Big]^{1/2}.$$ Clearly, $$x_n=\lambda_n \cdot x_1$$ where $$\lambda_1,\lambda_2,\dots$$ is a sequence of positive real numbers not depending on the initial condition $$x_1$$. For instance, $$\lambda_1=1,\lambda_2=0$$. It seems, based on empirical evidence, that $$\lambda_n\sim \gamma \cdot n^{-\alpha}$$ for some constants $$\gamma\approx 0.54$$ and $$\alpha\approx 0.35$$.

Substituting $$x_k$$ by its asymptotic expansion in the first formula defining the sequence $$(x_n)$$, one gets the following, after trivial computations and rearrangements: $$\alpha^2=(1-2\alpha)(1-\alpha)^2$$. Let $$\tau=(13+3\sqrt{33})^{1/3}$$. Then we have, according to Mathematica:

$$\alpha=\frac{1}{3}\Big(2+\frac{2^{5/3}}{\tau}-\frac{\tau}{2^{2/3}}\Big)\approx 0.3522011.$$

Is this correct? Also, I could not find how to determine the value of $$\gamma$$. Could you express $$\gamma$$ as a solution of some equation (as I did for $$\alpha$$) or using a series or anything else?

• Put $\lambda_n=\gamma\,n^{-\alpha}\,(1+\mu_n)$, then empirically $\mu_n=O(1/n)$; this implies $$0=\zeta(\alpha)+\sum_{n=1}^\infty n^{-\alpha}\mu_n=\zeta(2\alpha)+\sum_{n=1}^\infty n^{-2\alpha}(2\mu_n+\mu_n^2)$$ (even under weaker assumptions), and $\mu_n=\beta/n+o(1/n)$ results in $$\beta=\frac\alpha2-\frac{1-3\alpha^2}7,$$ so that $n^\alpha(1-\beta/n)\lambda_n$ is a better estimate for $\gamma$ than just $n^\alpha\lambda_n$. Commented Aug 4, 2022 at 8:55
• Further, it seems that $\mu_n-\beta/n=O(n^{-\rho})$ with $\rho=(1-\alpha)(3-2\alpha)$. Commented Aug 4, 2022 at 9:22