Recall the problem I had earlier of trying to define a non-integral number of compositions of a function. Specifically, if
is an increasing function that maps onto, what meaning can we assign to
, (the composition of
with itself
times, where
)?
I presented some ideas I had come up with earlier this week, and was surprised to see that they were nowhere near the mark.
A little background on the problem: Dr. J. said one of his friends had posed it to him, and he couldn’t figure it out, even over a period of years. Everytime that friend saw him, he would ask him if he had figured it out yet, so eventually Dr. J. asked someone for a hint. Apparently, the hint was enough of an aid for him to figure the problem out.
He passed the hint along to us, really more of a definition— define
to be
where
is increasing.
You can check that this works out to give you the properties that you would expect:
,
(the identity), etc. But how do you find the particular
,
, given
? And are they unique?
It took me a while to get over the outlandishness of this definition, but I can see that it is plausible–
is an increasing function from
, and
is an increasing function to
, so their composition determines some increasing function from
onto
. Only, is
in the range of this function?
Apparently the theory this is concerned with is that of ‘differentiable flows’, and is useful in engineering.
There is one complication, but it’s pretty simple to overcome. The problem is this, as the definition lies,
cannot be just any increasing function. Look at the definition for a while, and see if you can figure out why not (it has to do with
).
The problem is, by the definition,
and
is increasing, so if
,
, while if
,
. So if
crosses the identity function, or even touches it (then
), the definition doesn’t make sense. But this can be overcome by noticing that if you partition
into a maximal partition
where
are the fixed points of
and call
the restriction of
to
, then each
maps from its domain back onto its domain (because of
is increasing). So if you let
and
, then
, so the original definition applies, and
and we can define
so that
Once again, there is a problem here at the fixed points
. I don’t think it would be cheating too much to just stipulate that
.