Pi and the Mandelbrot set
Gerald Edgar
] 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)