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Given two recurrence relations
$x_{n+1}^2 = x_n^2 + \frac{c_1^2}{x_n^2}$ and $y_{n+1}^2 = y_n^2 + \frac{c_2^2}{y_n^2}$ for $c_1, c_2\in \mathbb{R}^+$ and $x_1=y_1=c_3\in\mathbb{R}^+$.

We want to study the convergence properties of $r_n := \frac{y_n^2 c_1}{x_n^2 c_2}$ as $n\rightarrow \infty$.

I strongly suspect that $r_n$ converges to 1 as $n\rightarrow \infty$ but am unable to prove it yet. I was also able to show that 1 is its the only fixed point and an attractor in the sense that $r_n > 1 \Rightarrow r_{n+1} < r_n$ and $r_n < 1 \Rightarrow r_{n+1} > r_n$. I was then able to prove that the $r_n$ does converge, as $r_n$ is strictly monotonous for $n\geq 2$. I would like to show that $r_n$ cannot asymptotically near another limit than 1 or find a counter-example for it.

amWhy
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Blackliquid
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    Why do you need squares ? Your first equation should be $x_{n+1}=x_n+c_1^2/x_n$, – Jean Marie Dec 04 '23 at 23:18
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    You can substitute $\tilde{x}_n=x_n/c_1$ and $\tilde{y}_n=y_n/c_2$ taking into account the comment by @JeanMarie The recurrence relations satisfied by $\tilde{x}_n$ and $\tilde{y}_n$ will coincide besides the initial conditions. – Ryszard Szwarc Dec 04 '23 at 23:38

2 Answers2

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Hint : If, as I have advised, your first recurrence relationship is

$$x_{n+1}=x_n+(c_1^2)/x_n,\tag{1}$$

setting $x_n:=c_1 e^{X_n}$, relationship (1) becomes :

$$c_1 e^{X_{n+1}}=c_1 (e^{X_n}+e^{-X_n})=2 c_1 \cosh(X_n),\tag{2}$$

Simplifying by $c_1$ and taking the logarithm on both sides, we get :

$$X_{n+1}=\underbrace{\ln(2 \cosh(X_n))}_{f(X_n)}$$

which is a recurrence relationship independent from $c_1$.

If we define, in the same way :

$$y_{n+1}=y_n+(c_2^2)/y_n,\tag{1}$$

Same operation ; we set $y_n:=c_2 e^{Y_n}$ with the same final relationship :

$$Y_{n+1}=\underbrace{\ln(2 \cosh(Y_n))}_{f(Y_n)}$$

Can you take it from here ?

enter image description here

In blue, the graphical representation of function $f$.

Jean Marie
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  • Intuitively, the derivative of cosh(x), tanh(x) saturates for $x\rightarrow\infty$ so i can see how the fraction of fixed point iterations converge, but I dont know how to show it. – Blackliquid Dec 07 '23 at 10:01
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As suggested by @Jean Marie we will study the equations $$x_{n+1}=x_n+{c_1^2\over x_n},\quad y_{n+1}=y_n+{c_2^2\over y_n}$$ Let $u_n={x_n\over c_1}$ and $v_n={y_n\over c_2}.$ Then $u_n$ and $v_n$ satisfy the same recurrence relation: namely $u_{n+1}=u_n+u_n^{-1}.$ Raising to the second power gives $$u_{n+1}^2=u_n^2+2+{1\over u_n^2},\quad {\rm which \ implies}\\ u_n^2=u_1^2+2(n-1)+{1\over u_1^2}+\ldots +{1\over u_{n-1}^2} \quad (*)$$ As $u_j^2\ge 2(j-1)$ we get $$2(n-1)\le u_n^2\le 2(n-1)+u_1^2+{1\over u_1^2}+{1\over 2}+{1\over 4}+\ldots + {1\over 2(n-2)}$$ Finally for $a=u_1^2+u_1^{-2}$ we obtain $$(2n-2)^{1/2}\le u_n\le (2n-2+\log [(n-1)^{1/2}] +a)^{1/2} \quad (**)$$ The estimate implies $$\lim {u_n\over \sqrt{2n}}=1$$ The same is valid for $v_n,$ hence the ratio $u_n/v_n$ tends to $1.$

Remark 1 The reason $\sqrt{2n}$ shows up is associated with the continuous version of the equation $x_{n+1}-x_n=x^{-1}_n,$ i.e. with $${dx\over dt}={1\over x}$$ whose solution are of the form $$x=\sqrt{2t+c}$$

Remark 2 The lower estimate in $(**)$ can be improved. We can use the second inequality in $(**)$ for $(*),$ applied to $k=2,3,\ldots, n-1,$ to get $$u_n^2\ge 2(n-1)+a +\sum_{k=2}^{n-1}{1\over 2(k-1)+a+\log [(k-1)^{1/2}]}$$ Next $$\sum_{k=1}^{n-2}{1\over 2k+a+[\log k^{1/2}]} \ge \int\limits_1^{n-1}{dx\over 2x+a+\log [x^{1/2}]}\\ =\int\limits_{e^a}^{n-1}{dx\over 2x+\log x}\ge \int\limits_{e^a}^{n-1}{dx\over 3x}=\log[(n-1)^{1/3}]-a/3$$

Ryszard Szwarc
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