Growth of the sequence given by $x_0=1, x_{n+1}=\sqrt{x_n^2+\frac{1}{x_n^2}}$

Question

Find the growth of the sequence given by $x_0=1, x_{n+1}=\sqrt{x_n^2+\frac{1}{x_n^2}}$ (In other words, how fast does the function grow?)

First off, I tried to show that it's an increasing sequence and has no fixed upper bound (limit). For the sake of contradiction, assume that the upper limit is $L$ and then, from the fact that $x_{n+1}=\sqrt{x_n^2+\frac{1}{x_n^2}}$, we will get $L^4=L^4+1$ leading to a contradiction. Also, $\frac{x_{n+1}}{x_n}=\sqrt{1+\frac{1}{x_n^4}} >1$. This implies that the function is increasing without having any fixed upper limit, i.e., $\displaystyle\lim_{n\to \infty} x_n = \infty$.

Is there any better way to show that (instead of what's done above?)

Now, it boils down to finding how fast the sequence grows.

Also, can we find a closed form expression of the $n^\text {th}$ term using generating functions? After that, everything would be easy I hope.

Perhaps it would be simpler to define $y_n = x_n ^ 2$, with the recursive formula $y _ {n + 1} = y_ n + \frac 1 {y_n}$ — idok, Mar 03 '18 at 19:12
@idok Apparently, yes. But how do you explicitly write $y_n$. It seems some form of continued fractions actually — Mathejunior, Mar 03 '18 at 19:14
Virtual duplicate of Showing a recursion sequence isn't bounded $a_{n+1}=a_n+\frac 1 {a_n}$. — dxiv, Mar 03 '18 at 19:15
@dxiv Wow, that's awesome and has some really cool proofs. Btw, still, how do we get a closed form? — Mathejunior, Mar 03 '18 at 19:18
@Mathbg Don't think you'll find a closed form, but you can still get the asymptotic behavior. See several related questions (and answers) here. — dxiv, Mar 03 '18 at 19:24

robjohn · Answer 1 · 2018-03-04T01:42:49.150

3

Since $$ x_{n+1}^2=x_n^2+\frac1{x_n^2}\tag1 $$ we have $$ x_{n+1}^4-x_n^4=2+\frac1{x_n^4}\tag2 $$ Since $x_n\ge1$, we get $$ 2n\le x_n^4-x_0^4\le3n\tag3 $$ Combining $(2)$ and $(3)$ allows us to improve the estimate $$ 2n\le x_n^4-x_0^4\le2n+1+\frac12\log(2n-1)\tag4 $$ Thus, $$ x_n\sim(2n)^{1/4}+\frac{2^{1/4}}{16}\frac{\log(n)}{n^{3/4}}\tag5 $$

edited Mar 04 '18 at 01:42

answered Mar 03 '18 at 22:01

robjohn

345,667

Somos · Accepted Answer · 2022-08-06T12:21:46.997

The original question asked about $\;x_0=1,\; x_{n+1}=\sqrt{x_n^2+1/{x_n^2}}.\;$ For technical reasons use the sequence $\;a_n:=1/x_n^2\;$ instead. The recursion is now $\;a_0=1,\; a_{n+1}=1/(a_n+1/a_n).\;$ Let us define $x:=1/n,\; y:=\log(x)\;$ and the sequence of power series in $x$ and $y$ as $$b_n:=\sqrt{x/2}\Bigg(1 + \Big(c+y\frac18\Big)x + \Big(\frac{1+8c+48c^2}{32} + y\frac{1+12c}{32} +y^2\frac{3}{128}\Big)x^2 + \dots\Bigg)$$ satisfies the same recursion as $\;a_n\;$ while $c=-0.43078559...$ satisfies the initial value $a_0=1$.

The series expansion of $b_n$ implies that $\;x_n \approx (2n)^{1/4}\;$ for large values of $\;n\;$ for any $x_0>0.$ For example, $x_{5000} = 10.001495167\dots.$

score 1 · Answer 3 · answered Nov 12 '21 at 13:41

Here's a heuristic approach to the asymptotic behaviour using calculus.

Let $y(n) = \sqrt{x(n)}$, and replace the resulting recursion $y(n+1) = y(n)+\frac{1}{y(n)}$ by this ODE for $z(n) \simeq y(n)$

$$z'(n) = \frac{1}{z(n)}, z(0)=1\tag{1}$$

whose solution is $z(n) = \sqrt{1+2n}$. Hence $x(n)= \sqrt{z(n)} = (2n+1)^{1/4} \to (2n)^{1/4}$.

Growth of the sequence given by $x_0=1, x_{n+1}=\sqrt{x_n^2+\frac{1}{x_n^2}}$

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