A continuous function with no root
For all you crackpots who emailed me (you know who you are) with a claim
that you have found continuous roots for each continuous function, this
section should embarrass you a little. It purports to show that not all
continuous functions have continuous roots; in fact, 1) the function I
propose has NO root at all, meaning f needn't be continuous and still it
doesn't exist, and 2) the function g proposed is not complicated at all;
in fact, it is a simple polynomial. If you can prove me wrong, you will
have the enjoyment of bouncing the embarrassment back on me. (So help check
for errors please.)
Theorem: Let g(x) = 3(x3 - x)/2. I claim that there is no
f:R -> R such that f(f(x)) = g(x).
Proof: Notice first that g is odd, g(x) = 0 at the points {1, -1},
g'(x) = 0 at the points {-sqrt(3)/3 = m1, sqrt(3)/3 =
m2}, and there is a positive k such that g(k) = k and
g(-k) = -k. It is also evident that g(m1) = m2 which
means g has a 2-cycle. Recall that for a function f to exist such that
f2 = g, g must have an even number of n-classes
for even n. It suffices to show that {m1, m2} is the
only 2-cycle of g; that is, we should show that for any x
that is not one of these two numbers in the cycle, g(g(x)) = x
means only g(x) = x. We could go on to compose g with itself and
show that minus x it has only two real roots, but there is
an easier way. Notice that for all x > k, g(x) > x, and for all x < -k,
g(x) < x, outruling all numbers but (-k, k). Notice further that
f(m1, m2) = (m1, m2) and
any element x in (-k, k) - [m1, m2], iterated by g
sufficiently many times, enters [m1, m2], which
clearly rules it out from being in a cycle. Now we only have the set
[m1, m2] which means we only must check that elements
of (m1, m2) aren't in any 2-cycle. Notice that one
element must be in (m1, 0) and the other in (0, m2),
so we must only check for x < 0. Notice also that g(x) + x > 0 in
(m1, 0) and so g(x) + x < 0 in (0, m2). If x is in
(m1, 0), then g(x) > -x, and since -x is in (0, m2),
g(g(x)) < g(-x) < x, so x is not in a 2-cycle. Thus, {m1,
m2} is the only 2-cycle and g doesn't have a square root. QED.
Unfortunately, I was told at sci.math that my result was trivial. A couple of
members have posted their continuous elementary functions with no roots, and
each relies on a different principle of class theory to show that it doesn't
have a root. For example, x - 2ex + 2 has exactly one fixed point 0,
exactly one point k such that f(k) = 0, and there exist elements j such that
f(j) = k. This can't work out because a 1-cycle with arm length > 1 must square
into another 1-cycle with two elements whose image is the cycle.
Previous
Next
Back to Table of Contents
Mail me
Back to my Homepage
