Question

Hi all,

I am trying to slove the recursion equation: $x_{n+1}x_{n−1}=x_n^2(1−4x_n)$ in the form of $x_n=x_n(x_1,x_2)$ or $x_n=x_n(c_1,c_2)$, and finally get the limit of the ratio: $\dfrac{x_n}{x_{n+1}}$.

I tried the way of setting: $x_n=f(n)$, and use the 1st order taylor expansion of $f(n+1)$ and $f(n−1)$, then get the first differential equation: $(f')2=4f^3$, the general solution is: $f(x)=\dfrac{4}{(2x−c_1)^2}$, but it does not fit the original recursion equation.

Then I write the equation in the form: $f''=6f^2$, the computer provides some approach by Weierstrass elliptic function (http://www.wolframalpha.com/input/?i=d^2y%2Fdx^2%3D6y^2), but it seems the Weierstrass elliptic function still has no such property as the recursion formula.

Any method I cam apply to get the final limit of ratio, maybe without solving the general soltuions? Thanks!

Answer 1

I decided to compute the ratio $x_{30}/x_{29}$ for various start values $x_0 = x_1 = s$ For $s>0.42$, the computations overflows for me, so I could not compute that part.

The image shows the ratio on the y axis, and start value on the x axis. The images are essentially identical for $x_{31}/x_{30}$, so it is motivated to take 30 as an approximation of $\infty.$

EDIT: So here is plots for $x_0 = s$ for different values of $s$.

http://www2.math.su.se/~per/files.php?file=recursiondata_mathoverflow_87463.pdf

Answer 2

Not really the answer you're looking for, but possibly helpful: The fact that the sequence $0,0,1/4,0,0,1/4,0,0,1/4,...$ satisfies the recursion equation offers a glimpse into what's making things hard here. At the very least, there are starting pairs $(x_1,x_2)$ close to $(0,0)$ that stay close to this three-peat for an arbitrarily long time before they (presumably) diverge and do something else.

More generally, one can look for a three-peat of the form $r,s,t,r,s,t,...$ and a derive a pair of algebraic equations relating $r$ and $s$. If I've done the algebra correctly, one gets

$$r^4 = s^3(1-4s)^2(r-4s^2(1-4s))$$ and $$r^3 = s^3(1-4s)(1+4(1-4s)(r-4s^2(1-4s))$$

If you plug this into Mathematica (which my friend Paul Zorn graciously did for me) you get a raft of real and imaginary roots. The general solution is unpleasant to behold, but it simplifies considerably if you set $r=s=(1+u)/4$. Ignoring the "trivial" case $r=s=0$, this boils down to

$$u^4+u^3+u^2-1=0$$ which has $u=0.682327803828...$ for a root, corresponding to $r=s=0.420581951$ (with $t=-0.286974759$), explaining the cut-off Paxinum ran up against in his graph.

In sum, it may be the case that you usually get a limit for $x_n/x_{n+1}$, but there are definitely cases where you don't.

Answer 3

If your goal is get the limit of the ratio there's no need to explicitly solve the recursion equation. When $n$ is enough big we have $\frac{x_{n+1}}{x_{n+2}}=\frac{x_{n}}{x_{n+1}}=\frac{x_{n-1}}{x_{n}}$. Then $\frac{x_{n+1}}{x_{n+2}}=\frac{\frac{x_{n}^{2}}{x_{n-1}}\left(1-4x_{n}\right)}{\frac{x_{n+1}^{2}}{x_{n}}\left(1-4x_{n+1}\right)}$ is equivalent to $\left(\frac{x_{n}}{x_{n+1}}\right)^{2}=\left(\frac{x_{n}}{x_{n+1}}\right)^{2}\frac{1-4x_{n}}{1-4x_{n+1}}$, and so $1=\frac{1-4x_{n}}{1-4x_{n+1}}$, which gives $x_{n+1}=x_{n}$. Then the limit of the ratio $\frac{x_{n}}{x_{n+1}}$ is 1.