[math-fun] Interesting ovoids
This discussion of ovals reminds me of a problem I was working on a while ago, which I gave up since I was not getting anywhere. The problem starts with the remarkable observation that if we take a(n+1) = (a(n)+1)/a(n-1), the resulting sequence is periodic with period five. (I saw this in one of Martin Gardner's columns in Scientific American; I don't know what issue at this point. He puts it into the mouth of the fictional Dr. Matrix, but I think he gave a reference for it later. It probably also appeared in one of his books.) (This excludes cases where a zero appears in the sequence. It is sufficient to take a(1) and a(2) both positive.) I asked, what happens if the 1 in that equation is replaced by some other constant, c: a(n+1) = (a(n)+c)/a(n-1). The case c=0 gives us period 6: starting with x, y, the sequence continues y/x, 1/x, 1/y, x/y, and then loops. In general, the sequence exhibits quasi-periodic behavior. It goes up and down regularly, with the average distance between peaks being some real number. This quasi-period is between 4 and 5 for c>1, and between 5 and 6 for 0 < c < 1; however in general the quasi-period depends on the initial values as well as on c. (Of course, if the quasi-period is rational, the sequence is actually periodic, with period equal to the numerator of the quasi-period.) Note that such a sequence has a fixed point u_c where u_c = (1+sqrt(4c+1))/2. (So in particular, u_1 is the golden ratio, phi, aka tau.) When you graph a(n) vs. a(n+1), with suitable parameters the result is an ovoid. If a(1), a(2) is close to u_c,u_c you get an oval, approaching a circle as you approach u_c,u_c. When the starting values are far from this point, the graph approximates an isoceles right triangle, with the right angle near the origin. The obvious next step is to find a non-trivial function f_c which satisfies the functional equation f_c(x,y) = f_c(y,(y+c)/x). The function must obviously have a minimum or maximum at u_c,u_c. Then the ovals should be the curves defined by f_c(x,y) = k. This is where I got stuck - I wasn't able to find such an f_c. Maybe somebody with better analytic skills than I have will be able to. Regardless, have a look at some of these curves. Franklin T. Adams-Watters ________________________________________________________________________ Check Out the new free AIM(R) Mail -- 2 GB of storage and industry-leading spam and email virus protection.
On 2007 Feb. 22, <franktaw@netscape.net> wrote:
This discussion of ovals reminds me of a problem I was working on a while ago, which I gave up since I was not getting anywhere.
The problem starts with the remarkable observation that if we take a(n+1) = (a(n)+1)/a(n-1), the resulting sequence is periodic with period five. (I saw this in one of Martin Gardner's columns in Scientific American; I don't know what issue at this point. He puts it into the mouth of the fictional Dr. Matrix, but I think he gave a reference for it later. It probably also appeared in one of his books.) (This excludes cases where a zero appears in the sequence. It is sufficient to take a(1) and a(2) both positive.)
I'm not sure where I saw it first, but my guess is Math. Mag. 64:2 (1991 Apr.) p. 133, partial solution of Problem 1343 (proposed by Ron Graham the previous year) The a. part of that problem involved the recursion above. Another reference is Chapter 5 of _Global Behavior of Nonlinear Difference Equations of Higher Order with Applications_, V.L. Kocic and G. Ladas (Kluwer, 1993).
I asked, what happens if the 1 in that equation is replaced by some other constant, c: a(n+1) = (a(n)+c)/a(n-1). The case c=0 gives us period 6: starting with x, y, the sequence continues y/x, 1/x, 1/y, x/y, and then loops.
In general, the sequence exhibits quasi-periodic behavior. It goes up and down regularly, with the average distance between peaks being some real number. This quasi-period is between 4 and 5 for c>1, and between 5 and 6 for 0 < c < 1; however in general the quasi-period depends on the initial values as well as on c.
True. Here's a nice example from my old notes: With a(1) = 1 and a(2) = 2 and c = (-4 +/- Sqrt(13))/3, we get period 7. I am not sure of the correct concise terminology when the sequence itself is not periodic. (Quasi- or pseudo- or almost- periodic?) But what I can say conjecturally about these (and some related) recursions is that the sequence is merely a restriction to domain Z+ of a continuous function which is (_precisely_) periodic. David W. Cantrell
Earlier today, I wrote:
On 2007 Feb. 22, <franktaw@netscape.net> wrote:
This discussion of ovals reminds me of a problem I was working on a while ago, which I gave up since I was not getting anywhere.
The problem starts with the remarkable observation that if we take a(n+1) = (a(n)+1)/a(n-1), the resulting sequence is periodic with period five. (I saw this in one of Martin Gardner's columns in Scientific American; I don't know what issue at this point. He puts it into the mouth of the fictional Dr. Matrix, but I think he gave a reference for it later. It probably also appeared in one of his books.) (This excludes cases where a zero appears in the sequence. It is sufficient to take a(1) and a(2) both positive.)
I'm not sure where I saw it first, but my guess is Math. Mag. 64:2 (1991 Apr.) p. 133, partial solution of Problem 1343 (proposed by Ron Graham the previous year) The a. part of that problem involved the recursion above.
Another reference is Chapter 5 of _Global Behavior of Nonlinear Difference Equations of Higher Order with Applications_, V.L. Kocic and G. Ladas (Kluwer, 1993).
I asked, what happens if the 1 in that equation is replaced by some other constant, c: a(n+1) = (a(n)+c)/a(n-1). The case c=0 gives us period 6: starting with x, y, the sequence continues y/x, 1/x, 1/y, x/y, and then loops.
In general, the sequence exhibits quasi-periodic behavior. It goes up and down regularly, with the average distance between peaks being some real number. This quasi-period is between 4 and 5 for c>1, and between 5 and 6 for 0 < c < 1; however in general the quasi-period depends on the initial values as well as on c.
True. Here's a nice example from my old notes:
With a(1) = 1 and a(2) = 2 and c = (-4 +/- Sqrt(13))/3, we get period 7.
I am not sure of the correct concise terminology when the sequence itself is not periodic. (Quasi- or pseudo- or almost- periodic?) But what I can say conjecturally about these (and some related) recursions is that the sequence is merely a restriction to domain Z+ of a continuous function which is (_precisely_) periodic.
Here's an example. Suppose a(1) = 1 and a(2) = 2 and c = 1/2. The sequence itself is not periodic. The continuous function, of which the sequence is a restriction, has a period of roughly 5.248; a graph of one cycle is shown at <http://img453.imageshack.us/img453/744/recursioncu0.gif>. Its maximum and minimum are roughly 2.549 and 0.741, resp. David W. Cantrell
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