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In this paper, the authors make the passing remark "simple analysis reveals that" the coupled sequence

$\mu_k = \theta_k \left(\theta_{k-1}^{-1} - 1\right)$

$\theta_{k+1} = \frac{\sqrt{\theta_k^4 + 4\theta_k^2} - \theta_k^2}{2}$

is such that $\mu_k$ asymptotically equals $1 - \frac{3}{k} + \mathcal{O}(1/k^2)$, but I have been unable to prove this. I was wondering if anyone could point me in the right direction as to how I would show something like this?

The Wind-Up Bird
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1 Answers1

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This question is very similar to MSE question 2861768 and several other questions. The idea from these questions is to use the Stolz theorem.

For simplicity, define $\,f(x) := (\sqrt{4x^2+x^4}-x^2)/2.\,$ Define a sequence $\,a_{n+1} = f(a_n)\,$ where $\,a_1>0.\,$ Next, define $$ b_n := 1/a_n, \quad\text{and} \quad e(x) := \frac12 + x\sqrt{1+\frac1{4x^2}} \tag{1}$$ where $\, b_{n+1} = e(b_n)\,$ and $\,\lim_{ x\to \infty} e(x)-x = \frac12.\,$ By the Stolz theorem we get $\, b_n = \frac{n}2 + O(1)\,$ and thus, $\, a_n = \frac2{n} + O(n^{-2}).$

The inverse function of $\,f(x)\,$ is $\,f^{(-1)}(x) = x/\sqrt{1-x},\,$ and we get

$$ \mu_n := a_n\!\left(\!\frac1{a_{n-1}} \!-\! 1 \!\right)\!=\! a_n\!\left(\!\frac{\sqrt{1\!-\!a_n}}{a_n} \!-\! 1\!\right) \!=\! 1 \!-\! \frac32 a_n \!+\! O(a_n^2) \tag{2}$$ but since $\,a_n = 2/n + O(n^{-2})\,$ now $\, \mu_n = 1 - 3/n + O(n^{-2}) \,$ which is what we wanted to prove.

ADVANCED results (only if you are interested):

Define $\,x := 1/n, \, y := \ln(x).\,$ For some constant $\,c\,$ only depending on $\,a_1,\,$ for example, if $\,a_1=1\,$ then $\,c \approx -1.2929418036,\,$ we get an expansion of $\,a_n = g(1/n) = h(2n-c-\ln(2))\,$ where $$ g(x) \approx 2\,x + (c+y)\,x^2 +((c+c^2) + (1+2c)y + y^2)\,x^3/2 + O(x^4) \tag{3}$$ and $$ h(x) \approx 4 / (x + \ln(x + \ln(x + \frac16 + \ln(x + \frac{29}{72} + \ln(x + \frac{437}{640} + \dots))))). \tag{4}$$

The expansion $(3)$ implies that $$ a_{n+1}(1/a_n - 1) = 1 - 3\,x + (-1-3c-3y)\,x^2/2 + O(x^3). \tag{5}$$

Notice that $\,\frac1{n+1}=\frac{x}{x+1}\,$ and thus the recursion for $\,a_n\,$ is equivalent to $\,g(\frac{x}{1+x}) = f(g(x))\,$ and from which the coefficients of the series expansion of $\,g(x)\,$ can be solved for term by term.

MSE question 3215 has an answer to the question of the convergence rate of $\,a_n\,$ for general $\,f(x).\,$

Somos
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  • Could you make this a bit more clear? Why did you define $x=1/n$, when $1/n$ does not appear in the iteration? Furthermore, where does the $log$ come from, and what is the intuition for incorporating a $log$? What is the constant $c$, and where did you get this expansion for $a_n$? – The Wind-Up Bird Feb 07 '19 at 01:47
  • Please read the accepted answer and my answer in MSE question 2861768 "Calculate $\lim_{n \rightarrow \infty } \frac{n (1- na_n)}{\log n} $" – Somos Feb 07 '19 at 02:22
  • A reference to an answer to the general question is given in MSE question 3215 "Convergence of $\sqrt{n}x_n$ wehre $x_{n+1}=\sin(x_n)$". – Somos Feb 07 '19 at 13:08