ChapterZero » More on non-integral numbers of compositions

More on non-integral numbers of compositions

Mathematics — Alex @ 10:56 am

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 ( is an increasing function from [0,1] onto [0,1] ) takes care of those problems— if we stipulate that if is a rational number of compositions of then 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 is said to be a -th root of the function only in case $g^{(q)} = g \circ g \cdots g = f$ and the initial and final sets of are the initial and final sets of .
Prop. If is increasing and maps from onto , there is only one -root of , 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 there is an integer such that for all $m,n \geq N$ , then for every number , there is an integer such that $\sup_{x \in [0,1]} |f^{(a_m)}(x) - f^{(a_n)}(x)| < d$ for all $m,n \geq M$ .