| Let P(x) be any n-th degree complex polynomial in x. Let S(a) be the series
of complex numbers s[0], s[1], s[2], ... where s[0] = a and s[k+1] = P(s[k]).
For what values of a does S(a) converge?
Observations:
S(a) converges if P(a) = a , i.e. a is root of P(x) - x.
S(a) converges if S(b) converges and P(a) = b , i.e. a is a root
of P(x) - b.
So to generate a (potentially) infinite set of complex numbers for which S(a)
converges:
Find the roots of P(x) - x.
For each distinct value a[i] you already have, find the
roots of P(x) - a[i] and add them to the list.
Repeat as necessary.
In the original problem P(x) = x^2 - 2.
The roots of x^2 - 2 - x are -1 and +2.
The roots of x^2 - 2 + 1 are +&-1.
The roots of x^2 - 2 - 2 are +&-2.
The roots of x^2 - 2 - 1 are +&-sqrt(3).
The root of x^2 - 2 + 2 is 0.
The roots of x^2 - 2 - +&-sqrt(3) are +&-sqrt(2 -&+ sqrt(3)).
The roots of x^2 - 2 - 0 is +&-sqrt(2).
etc, etc and so forth.
In general, if S(a) converges then S( +&-sqrt(2 - a) ) converges.
Some other observations on the original problem:
Since x^2 - 2 is symmetric, if S(a) converges then S(-a) converges.
If |a| > 2 then a^2 - 2 > |a| and hence S(a) diverges.
Non-convergent cyclic behavior is possible:
Let P(a) = b and P(b) = a. This will occur if a is a root
of P(P(x)) - x. Then S(a) will be cyclic.
k
Let a be a root of P (x) - x. Then S(a) will be cyclic.
Let b be such that S(b) is (ultimately) cyclic and let a be
k
a root of P (x) - b. Then S(a) is ultimately cyclic.
In the original problem:
S(-1/2+sqrt(5)/2) = -1/2+sqrt(5)/2, -1/2-sqrt(5)/2, -1/2+sqrt(5)/2, ...
I'll have to think some more about necessary conditions for convergence.
Mike.
|
| Define f(n) by the following (where k is a positive integer):
n
Sum f(i) = k
i\n
Prove that:
n \ f(n)
( a\b means a evenly divides b )
2 3 4
Thus, f(1) = k, f(2) = k - f(1), f(3) = k - f(1), f(4) = k - f(2) - f(1),
5 6 7
f(5) = k - f(1), f(6) = k - f(3) - f(2) - f(1), f(7) = k - f(1), et cetera.
|
| It should be noted that the sequence doesn't really "converge". That is,
although a[i] = 2 implies a[i+1] = 2, if a[i] = 2+d (for a small d) then
a[i+1] = 2 + 4d + d^2 is actually further away from 2. The same is true
of any n-th degree polynomial P(x) if n > 1 and the n-th order coefficient
is >= 1.
Let's call (2) and (-1) 1-cycles, and (-1/2+sqrt(5)/2, -1/2-sqrt(5)/2) a
2-cycle.
There are n-cycles for all n > 0. This can be seen by noting the behaviour of
the family of curves x[i] = 2^x[i-1]-2, drawn as x[0] vs x[i], and noting that
the curve intersects the line x[i] = x[0] exactly 2^i times. Thus, there are
2 values that repeat every iteration (2 and -1), 4 values that repeat every 2
iterations (2, -1, -1/2+sqrt(5)/2, and -1/2-sqrt(5)/2), 8 values that repeat
every 3 iterations, et cetera.
Note that given any value a such that |a| <= 2 and any non-zero d, there are
an infinite number of values in the range |x-a| < |d| that are part of some
repeating cycles.
There are also an infinite number of values in |x-a| < |d| that never cycle,
or fall into a cycle. Since the sequence doesn't really "converge", it must
either fall into a cycle or never fall into a cycle -- it can't converge to a
cycle. Clearly, the value pi-3 never falls into a cycle; since that would
imply that pi is the root of a polynomial with rational coefficients.
- Gilbert
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