question about the recursion equation: $x_{n+1}x_{n−1}=x_n^2(1−4x_n)$ - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T12:37:05Z http://mathoverflow.net/feeds/question/87463 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/87463/question-about-the-recursion-equation-x-n1x-n1x-n214x-n question about the recursion equation: $x_{n+1}x_{n−1}=x_n^2(1−4x_n)$ Jay 2012-02-03T18:15:12Z 2012-02-06T22:09:35Z <p>Hi all,</p> <p>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: $\dfrac{x_n}{x_{n+1}}$.</p> <p>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.</p> <p>Then I write the equation in the form: $f''=6f^2$, the computer provides some approach by Weierstrass elliptic function (<a href="http://www.wolframalpha.com/input/?i=d%5E2y%252Fdx%5E2%253D6y%5E2" rel="nofollow">http://www.wolframalpha.com/input/?i=d^2y%2Fdx^2%3D6y^2</a>), but it seems the Weierstrass elliptic function still has no such property as the recursion formula.</p> <p>Any method I cam apply to get the final limit of ratio, maybe without solving the general soltuions? Thanks!</p> http://mathoverflow.net/questions/87463/question-about-the-recursion-equation-x-n1x-n1x-n214x-n/87493#87493 Answer by Cristina Serpa for question about the recursion equation: $x_{n+1}x_{n−1}=x_n^2(1−4x_n)$ Cristina Serpa 2012-02-03T23:30:55Z 2012-02-03T23:30:55Z <p>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.</p> http://mathoverflow.net/questions/87463/question-about-the-recursion-equation-x-n1x-n1x-n214x-n/87523#87523 Answer by Per Alexandersson for question about the recursion equation: $x_{n+1}x_{n−1}=x_n^2(1−4x_n)$ Per Alexandersson 2012-02-04T13:09:37Z 2012-02-06T15:46:36Z <p>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.</p> <p>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 $\infty.$</p> <p><img src="http://fc09.deviantart.net/fs70/f/2012/035/7/8/data_ratio_by_paxinum-d4olp02.png" alt="alt text"></p> <p>EDIT: So here is plots for $x_0 = s$ for different values of $s$.</p> <p><a href="http://www2.math.su.se/~per/files.php?file=recursiondata_mathoverflow_87463.pdf" rel="nofollow">http://www2.math.su.se/~per/files.php?file=recursiondata_mathoverflow_87463.pdf</a></p> http://mathoverflow.net/questions/87463/question-about-the-recursion-equation-x-n1x-n1x-n214x-n/87727#87727 Answer by Barry Cipra for question about the recursion equation: $x_{n+1}x_{n−1}=x_n^2(1−4x_n)$ Barry Cipra 2012-02-06T22:09:35Z 2012-02-06T22:09:35Z <p>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.</p> <p>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</p> <p>$$r^4 = s^3(1-4s)^2(r-4s^2(1-4s))$$ and $$r^3 = s^3(1-4s)(1+4(1-4s)(r-4s^2(1-4s))$$</p> <p>If you plug this into <em>Mathematica</em> (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</p> <p>$$u^4+u^3+u^2-1=0$$ 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.</p> <p>In sum, it may be the case that you <em>usually</em> get a limit for $x_n/x_{n+1}$, but there are definitely cases where you don't.</p>