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Mar 27, 1992, 8:57:43 AM3/27/92

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David Petrie wrote:

] Here's a somewhat heuristic theoretical explanation of what's going on,

] atleast around the point (1/4,0).

]

] When we iterate the equation x := x^2 + 1/4+epsilon, starting at x = 0,

] x increases slowly to 1/2, and after it passes 1/2, it zooms off rapidly

] to infinity. So the interesting behavior is when x = 1/2. Let x = y+1/2.

] Then our equation reads y := y^2 + y + epsilon, or using subscripts

] y_(n+1) = (y_n)^2 + y_n + epsilon.

]

] The y_n 's are increasing smoothly and slowly, atleast near y = 0, so it

] is reasonable to consider y to be a function of the continuous variable n,

] and y_(n+1) - y_n is very close to y'(n) (the derivative of y).

]

] So our equation now reads y'(n) = y^2 + epsilon. This has the solution

] y = a*tan(a*n+c) where a = Sqrt(epsilon). The intitial point and end point

] of our iteration correspond to consecutive poles of the tangent function,

] giving a*n = pi, where a = Sqrt(expsilon) and n is the number of iterations

] to leave the set, exactly what Mr. Boll has found.

]

] A similar method seems to work around the point (-3/4,0), but I haven't

] completed the analysis (it seems to require a second degree diffeq).

]

] I did a little more experimentation, and found something really neat at

] the points (-1.25,epsilon). There, n*epsilon/pi (n = #iterations) jumps

] around chaotically, but it is always very close to an integer or a half

] integer for epsilon very small. Wierd!

]

]

] David Petry

]

] Here's a somewhat heuristic theoretical explanation of what's going on,

] atleast around the point (1/4,0).

]

] When we iterate the equation x := x^2 + 1/4+epsilon, starting at x = 0,

] x increases slowly to 1/2, and after it passes 1/2, it zooms off rapidly

] to infinity. So the interesting behavior is when x = 1/2. Let x = y+1/2.

] Then our equation reads y := y^2 + y + epsilon, or using subscripts

] y_(n+1) = (y_n)^2 + y_n + epsilon.

]

] The y_n 's are increasing smoothly and slowly, atleast near y = 0, so it

] is reasonable to consider y to be a function of the continuous variable n,

] and y_(n+1) - y_n is very close to y'(n) (the derivative of y).

]

] So our equation now reads y'(n) = y^2 + epsilon. This has the solution

] y = a*tan(a*n+c) where a = Sqrt(epsilon). The intitial point and end point

] of our iteration correspond to consecutive poles of the tangent function,

] giving a*n = pi, where a = Sqrt(expsilon) and n is the number of iterations

] to leave the set, exactly what Mr. Boll has found.

]

] A similar method seems to work around the point (-3/4,0), but I haven't

] completed the analysis (it seems to require a second degree diffeq).

]

] I did a little more experimentation, and found something really neat at

] the points (-1.25,epsilon). There, n*epsilon/pi (n = #iterations) jumps

] around chaotically, but it is always very close to an integer or a half

] integer for epsilon very small. Wierd!

]

]

] David Petry

]

A reference for this approach is the Guckenheimer & Holmes book,

"Nonlinear Oscillations ...", section 6.8. I agree, it is "heuristic".

The point -3/4 can be dealt with similarly. The function f(z) = z^2-3/4

has fixed point -1/2 with eigenvalue -1. So we use f(f(z)) to get

eigenvalue +1. Transform coordinates z = -1/2 + y so that we are

interested in what happens for y near 0. Consider the

transformations z^2 - 3/4 + i*r, for small, positive, r. After some

algebra, we get the differential equation:

dy 4 3 2 2

---- = y - 2 y + 2 I r y - 2 I r y - r

dt

(Here I = sqrt(-1).) The time the trajectory spends near 0 (say between

-1 and 1) is the integral

1

/

| 1

| ----------------------------------- dy

| 4 3 2 2

/ y - 2 y + 2 I r y - 2 I r y - r

-1

Its asymptotics are now shown in the articles called

"Asymptotics of an integral" in sci.math.symbolic. We conclude that

if n is the number of steps required for divergence, then

n is asymptotic to Pi/(2*r). Actually, this is the number of steps

for the composition f(f(z)), so the number of steps for f(z) itself

is double this, Pi/r. So: Mr. Boll's asymptotic behavior is vindicated.

This is a heuristic argument, so there could be some work done to make it

rigorous...

Now, is -5/4 the next spot to study? Or how about -2 ?

--

Gerald A. Edgar Internet: ed...@mps.ohio-state.edu

Department of Mathematics Bitnet: EDGAR@OHSTPY

The Ohio State University telephone: 614-292-0395 (Office)

Columbus, OH 43210 -292-4975 (Math. Dept.) -292-1479 (Dept. Fax)

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