16
$\begingroup$

I'm interested in the following set of nonlinear difference equations:

$$x_{n+1} = \frac{c + x_n}{x_{n-1}},\; x_1 = x_0 = 1 \qquad \textrm{for } c > 0$$

For $c=1$ the sequence is periodic with period 5 and range {1,2,3}. For other values of $c$ the sequence appears to be oscillatory. For small $c$, the supremum seems to be slightly more than $1+2c$, while for larger $c$ it seems to approach $3c$.

Is there a simple formula for the bounds of the sequence in this general case? What about the range? Are the sequences periodic? Are they dense within the bounds?

c = 1.1

$\endgroup$
  • 2
    $\begingroup$ Solving nonlinear difference equations of higher order is usually difficult if not impossible. I do not want to discourage you as this one might very well have an elegant solution, but it is just something to keep in mind. Generlly we try to make substitutions (like e.g. $y_n = x_n/x_{n-1}$) in order to transform them into a linear difference equation, but I could not find a way of doing so right away. $\endgroup$ – flawr Dec 21 '15 at 11:19
  • $\begingroup$ Thanks. I'm certainly not expecting a full solution. I'm mainly wondering about the bounds (especially the supremum) and whether or not the sequences are periodic for $c \neq 1$. $\endgroup$ – Uri Zarfaty Dec 21 '15 at 11:25
  • $\begingroup$ it does look like for each c there is some energy being conserved $\endgroup$ – mercio Nov 10 '17 at 11:12
7
+100
$\begingroup$

From experimentation with complex orbits, it looks like the points travel on a curve which is symmetric in $x,y$ and has degree $2$ in $x$. So the curves have equations of the form $a_0 + a_1(x+y) + a_2xy + a_3(x+y)^2 + a_4xy(x+y) + a_5x^2y^2 = 0$.

It is very unlikely that $a_5$ is nonzero because this would give lots of genus $3$ curves (assuming they're not all singular), with a transformation that gives infinitely many rational points on them except when the orbits are periodic, and that can only happen for a countable number of curves. So it would contradict Faltings's theorem. So first I assume $a_5=0$.

Then, since the map $(x,y) \mapsto ((c+y)/x,y)$ preserves them, for a given $y$, the two corresponding $x$ solutions multiply to $c+y$, which means the equation is of the form $b_2(y)x^2 + b_1(y)x + b_0(y)$ where $b_0(y)=(c+y)b_2(y)$, and so $(a_0+a_1y+a_3y^2) = (c+y)(a_3+a_4y) = ca_3+(ca_4+a_3)y+a_4y^2$.

Therefore $a_3=a_4, a_1=(c+1)a_4$ and $a_0=ca_4$, which gives $a_4(c + (c+1)(x+y) + (x+y)^2+xy(x+y)) + a_2xy = 0$,

which means $\frac{c+(c+1)(x+y)+(x+y)^2+xy(x+y)}{xy} = -a_2/a_4$ is constant.


So, the transformation $(x,y) \mapsto (y,(c+y)/x)$ conserves the quantity $\frac{c(1+x+y) + (x+y)(1+x)(1+y))}{xy}$ whenever $xy(c+y) \neq 0$

For almost all fixed values of $c$ and $k$, the curve $\frac{c(1+x+y) + (x+y)(1+x)(1+y))}{xy} = k$ should be an elliptic curve, of which this map is an automorphism. (so most likely a translation of infinite order)


If you start from $(1,1)$, then the curve you obain has the equation $c(1+x+y)+(x+y)(1+x)(1+y) = (3c+8)xy$, or $f(x,y) = c+(c+1)x+(c+1)y+xx-(3c+6)xy+yy+xxy+xyy = 0$.

To get the highest point, you need to intersect it with the curve $\frac{df}{dx} = 0$, so
$c+1+2x-(3c+6)y+2xy+y^2=0$

You get $(2+2y)x = (3c+6)y-(c+1+y^2)$.

Plugging this into the original equation gives you a degree $5$ equation on $y$. $y= -1$ will always be a root because the curve has an horizontal asymptote there, leaving $4$ other real roots, among which you want the second and third largest.


If (like me) you're interested in the dynamics and orbits shapes of this map in the other parts of the plane,

when $k = -1$ the equation factors into $(x+y+1)(c+xy+x+y)$, so a line and an hyperbola, both genus $0$ curves. The map switches one with the other, and doing it twice is doing an automorphism of any of those two curves. Dynamics of automorphisms of genus $0$ curves are well understood. If $c > 3/4$ the two fixpoints are real, so a point gets attracted into a $2$-cycle. If $c < 3/4$ then they are complex, so the map acts more like a "rotation" of the curves. If $c = 3/4$ we get a double (real) fixpoint, so it's more like a "translation".

when $k = -c$ the equation factors into $(x+1)(y+1)(c+x+y)$, so three lines. The line $y=-1$ is sent onto the line $x=-1$ then to $x+y=-c$ then to $y=-1$ again. In this case, the discriminant is $(c-2)²-3$, so the behaviours change at $c= 2\pm \sqrt 3$

when $k = \infty$ the equation becomes $xy=0$ (or $xyz = 0$ if you're working in the projective plane), so three lines. The line $y=0$ is sent to the line $x=0$ then to the blowup of a point at infinity, then to the line at infinity $(z=0)$ then to another blowup then back to $y=0$. This long trip sends $(x,0)$ to $(x/c,0)$. Thus we get an attractive $5$-cycle most of the time.

There are another two values of $k$ (the roots of $8-k-13c+10ck-ck^2+16c^2$) where the curve is singular (there is a node where the small loops disappear), and so the smooth locus of the curve has genus $0$ that is fixed by the transformation.

For any other values of $k$, the curve is an elliptic curve (with one or two real components), and the map is some infinite order translation in the generic case.

If you compute the $j$ invariants, those special values of $k$ correspond to poles, and their order is the number of genus $0$ curves the transformation cycles through. The map $k \mapsto j(C_k)$ is then a rational function of degree $5+3+2+1+1 = 12$.

All the curves go through the points $(-1,0),(-1,c),(0,-1),(0,-c)$, have common vertical and horizontal asymptotes (so they all go to the same point on the two blowups I mentioned before), plus they each have their own oblique asymptote, so that's another common point on the line at infinity.

If you blow up the projective plane at those $4$ finite points, those $3$ points at infinity, then blowup again at the two common asymptotes, the original map should maybe become a nice automorphism of a smooth elliptic surface


As for periodicity, I expect the set of $c$ where this happens is dense (and those values of $c$ are all algebraic). One would need to investigate how the $j$-invariant and the translation varies while $k$ and $c$ varies.

$\endgroup$
  • 1
    $\begingroup$ +1 there is no need to worry about the lines $x = 0, y = 0, y = -c$, If $x,y > 0$, then $y, \frac{c+y}{x} > 0$. the sequence $(x_n,x_{n+1})$ will never reaches those lines. How did you figure out the invariant, the best I can get is in the limit $c \to \infty$, the invariant takes the form $\frac{x}{\alpha} + \frac{\alpha}{x} + \frac{y}{\alpha} + \frac{\alpha}{y}$ where $\alpha = \frac{1+\sqrt{1+4c}}{2}$. $\endgroup$ – achille hui Nov 10 '17 at 21:50
  • 2
    $\begingroup$ first I plotted a complex orbit to get reassurance that $f(x,y)$ was symmetric and had degree $2$ in $x$ and $y$ separately. It might have been $3$ and still be an elliptic surface, but I think anything higher was impossible due to Faltings theorem. Then I checked that determinants made with $[1;x+y;xy;(x+y)^2;(x+y)xy;(xy)^2]$ vectors were all $0$ to be sure that there indeed was an equation to be found. Finally I used that for a given $y$ the two $x$ solutions satisfied $x_1x_2 = c+y$ and tinkered a bit. $\endgroup$ – mercio Nov 10 '17 at 23:37
  • $\begingroup$ I get the part " $f(x,y)$ had degree $2$ in $x$ and $y$". The involvement of Faltings theorem is beyond me. In any event, thanks for the info. $\endgroup$ – achille hui Nov 10 '17 at 23:51
  • 1
    $\begingroup$ @mercio Wow, great job finding an invariant! (+1) $\endgroup$ – Frpzzd Nov 11 '17 at 0:16
  • $\begingroup$ Could you please explicate how you derived the conserved quantity? Thank you. $\endgroup$ – Hans Nov 16 '17 at 18:36
4
$\begingroup$

Computational evidence suggest the following conjectures.

For $1\ne c>0$ a sequence $\{z_n\}=\{(x_n,x_{n+1})\}$ is a dense set of a smooth curve which bounds a convex shape. When $c$ tends to infinity this shape tends to an isosceles right angled triangle with sides of order $c$, when $c$ tends to zero the shape tends to an ellipse, see the graphs.

$c=2^{-2}, 2^{-1},\dots, 2^7$ (I remark that the diameter of the shape is not a monotone function of $c$):

enter image description here

$c=2^{8}, \dots, 2^{11}$:

enter image description here

$c=2^{-3}, \dots, 2^{-22}$:

enter image description here

Everybody feel free to find the equations of the curves and to prove the conjectures, I'll try to do this too. In particular, if these equations are polynomial then their coefficients can be found by solving the respective system of linear equations.

$\endgroup$
2
$\begingroup$

This is an partial answer. Please consider this as a supplement to merico's excellent answer.

Introduce auxillary sequences $(y_n), (z_n), (w_n)$ by $y_n = x_{n+1}, z_n = x_{n+2}, w_n = x_{n+3}$.
Rewrite the defining recurrence relation of $(x_n)$ to a more symmetric form: $$x_{n+1} = \frac{c+x_n}{x_{n-1}}\quad \iff\quad x_{n+1}x_{n-1} = c + x_n$$ In terms of the auxillary sequences, this becomes $$x_n z_n = y_n + c\quad\text{ and }\quad y_n w_n = z_n + c.$$ As a result, $$ (1+x_n)z_n = z_n + z_n x_n = z_n + (y_n+c) = y_n + (z_n + c) = y_n + y_nw_n = y_n(1+w_n) $$ Multiply both side by $\frac{(1+y_n)(1+z_n)}{y_nz_n}$, we find $$\frac{(1+x_n)(1+y_n)(1+z_n)}{y_n} = \frac{(1+y_n)(1+z_n)(1+w_n)}{z_n} = \frac{(1+x_{n+1})(1+y_{n+1})(1+z_{n+1})}{y_{n+1}}$$ This means following expression $$\frac{(1+x_n)(1+y_n)(1+z_n)}{y_n} = \frac{(1+x_n)(1+y_n)(x_n+y_n+c)}{x_ny_n}$$ is independent of $n$ and they are invariant of the system. Up to an additive offset $c$, the second form is the invariant found by merico.

Since $x_1 = y_1 = 1, z_1 = c+1$, the sequence of points $(x_n, y_n, z_n)$ lies on intersection of following two surfaces in $\mathbb{R}^3$: $$(1+x)(1+y)(1+z) = 4(c+2)y\quad\text{ and } xz = y + c$$

Parametrize the curves within above interesection by some parameter $s$ and taking logarithm derivatives, we obtain $$\frac{x'}{1+x} + \frac{y'}{1+y} + \frac{z'}{1+z} = \frac{y'}{y}\quad\text{ and }\quad \frac{x'}{1+x} + \frac{z'}{1+z} = \frac{y'}{y+c} $$ At those $s$ where where $y$ is extremal, $y' = 0$.

Above equations implies $x = z$ and hence $x^2 = y + c$. Furthermore, $$(1+x)^2(1+y) = 4(c+2)y \iff (1+2x+y+c)(1+y) = 4(c+2)y\\ {\large\Downarrow}\\ x = -\frac{y^2 - 3(c+2)y + (c+1)}{2(y+1)} $$ Substitute this back into $x^2 = y + c$, we find the extremal values of $y$ are roots of a quartic polynomial.

$$(y^2 - 3(c+2)y + (c+1))^2 - 4(y+1)^2(y+c) = 0 \tag{*1}$$

For $c > 0$, this polynomial has $4$ distinct real roots: $0 \le \lambda_1(c) < \lambda_2(c) < \lambda_3(c) < \lambda_4(c)$.

Start from $x_1 = x_2 = 1$, it is easy to see all $x_n, y_n$ generated are positive. This means the extremal values of $y$ also need to satisfy: $$y^2 - 3(c+2) + (c+1) \le 0 \iff \frac32(c+2) - \Delta(c) \le y \le \frac32(c+2) + \Delta(c)\tag{*2} $$ where $\Delta(c) = \frac12\sqrt{9c^2+32(c+1)}$.

If one make an implicit plot of the roots of $(*1)$. One find in general,

$$\lambda_1(c) < \frac32(c+2) - \Delta(c) < \lambda_2(c) < \lambda_3(c) < \frac32(c+2) + \Delta(c) < \lambda_4(c)$$

From this, we can deduce in general, $x_{n+1} = y_n$ are bounded between the two middle two roots $\lambda_2(c), \lambda_3(c)$ of equation $(*1)$.

$\endgroup$
  • $\begingroup$ Could you please explain further what leads you to think of the steps from "As a result" on to derive the conserved quantity? $\endgroup$ – Hans Nov 16 '17 at 18:35
1
$\begingroup$

You will have noticed that, for $c=1$, sequences are globally periodic i.e. are periodic for (almost any) initial values. I have been studying these recurrence relations from that point of view. For example, putting $c=0$ gives globally periodic sequences of period 6 (however, you are excluding 0). I have been using the invariant $\frac{(x+1)(y+1)(x+y+c)}{xy}$, which differs by a constant from the one noted above - it is possibly easier to use in some contexts in that it is completely factorised into linear factors.

$\endgroup$
  • 1
    $\begingroup$ How do you derive the invariant? $\endgroup$ – Hans Nov 16 '17 at 18:36
  • $\begingroup$ I am not sure this answer will help. However, I have been studying more general recurrence relations with multilinear defining relationships. In terms of finding an invariant I only managed it for some simple forms and only because someone had told me that there were results of this type. (They may even have told me the sort of invariants to look for.) I think they said the idea was due to Gerry Ladas. Stan $\endgroup$ – S. Dolan Nov 17 '17 at 8:19
  • $\begingroup$ OK. But can you tell me how you find the current specific invariant in your answer? Also can you provide some references for finding invariants of a finite difference in general? Does it have to do with group theory? $\endgroup$ – Hans Nov 17 '17 at 19:34
0
$\begingroup$

Suppose we have variables $t_0,t_1,t_2,b_2$ and define quantities $$ b_3 = (b_2^2t_1^2-t_2^2)/t_0^2,\; b_4 = (2b_2^2t_1^4-2t_1^2t_2^2-b_2^2t_0^3t_2)/(t_0^3t_1^2),\; p_1 = b_2^2,\; p_2=-b_3,\; p_3=-b_4. $$ Now define the three sequences $\;w_n, t_n,x_n\;$ with the initial values and recursion equations $ w_1 = 1,\quad w_2 = b_2,\quad w_3 = b_3,\quad w_4 = b_4b_2,\quad w_n = (p_1 w_{n-1}w_{n-3} + p_2 w_{n_2}^2)/w_{n-4},$ $ t_3 = p_2t_1 + p_1t_0t_2/t_1,\quad t_n = (p_1 t_{n-1}t_{n-3} + p_2 t_{n_2}^2)/t_{n-4},\quad x_n := t_{n+2}t_{n-3}/(t_nt_{n-1}).$

Now if $\;c=-w_6/(w_3b_2),\;$ then $\;x_{n+1}x_{n-1}=b_4x_n+c.$ By choosing $b_2$ as a square root, we get $\;b_4=1\;$ the original difference equation. Note that $\;b_3^3 = (x_{n-1}+b_4)(x_n+b_4)(x_{n+1}+b_4)/x_n\;$ and $\;b_3^3 = (x_n+b_4)(x_{n+1}+b_4)(b_4(x_n+x_{n+1})+c)/(x_nx_{n+1})\;$ is easily shown to be invariant.

The $w_n$ is an elliptic divisibility sequence, associated with multiples of a point on an elliptic curve, Jacobi theta functions, and Weierstrass sigma functions, and $t_n$ is one of 3 companion sequences.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.