somewhere near the beginning.

More on non-integral numbers of compositions

Filed under: Mathematics — Alex @ 10:56 am 3/14/2005

I did some thinking about the problem of taking a non-Integral number of compositions that I mentioned earlier.

What I’ve found is that it is pretty easy to define a rational number of compositions of a function, but that’s about it— there isn’t a reasonable definition that allows you to ‘add’ the number of compositions of functions because

f^{(\frac{s}{t})} \circ f^{(\frac{p}{q})} \neq f^{(\frac{p}{q})} \circ f^{(\frac{s}{t})}

in general.

There is a problem with existence and uniqueness, but I think the specific context given (f is an increasing function from [0,1] onto [0,1]) takes care of those problems— if we stipulate that if g is a rational number of compositions of f then g is increasing and maps from [0,1] onto [0,1].

Unfortunately, all this speculation doesn’t help with something as ’simple’ as finding \sin(\frac{\pi}{2}x)^{(\frac{1}{2})}.

Here’s what I have, Moore’s style:

  • Defn. A function g is said to be a q-th root of the function f only in case g^{(q)} = g \circ g \cdots g = f and the initial and final sets of g are the initial and final sets of f.
  • Prop. If f is increasing and maps from [0,1] onto [0,1], there is only one q-root of f, denoted f^{(\frac{1}{q})}, and f^{(\frac{1}{q})} is increasing.
  • Prop. \left(f^{(\frac{1}{q})}\right)^{(n)} = \left(f^{(n)}\right)^{(\frac{1}{q})} . Denote this function by f^{(\frac{n}{q})}.
  • Prop Let a_0, a_1, \ldots be a sequence of rational numbers. If for every number c>0 there is an integer N such that | a_m - a_n | < c for all m,n \geq N, then for every number d>0, there is an integer M such that \sup_{x \in [0,1]} |f^{(a_m)}(x) - f^{(a_n)}(x)| < d for all m,n \geq M.

I’m working on proving the two hardest of the above propositions.

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