Re: Can we find the function?

David W. Cantrell <DWCantrell@xxxxxxxxxxx> writes:

David W. Cantrell <DWCantrell@xxxxxxxxxxx> wrote:
Robert Israel <israel@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote:
evan <evan.e.thomas@xxxxxxxxx> writes:
[snip]
How about f(x) = 1/x if x != 0,
0 if x = 0
f(f(x)) = x, certainly a polynomial function.

... but also a monomial.

Right. But, as preface to my example:

If f(f(x)) were allowed to be a monomial, I would use simply f(x) = 1/x.
[Note that I'm taking the codomain of f to be R*, the one-point extension
of R, so that we have f(0) = oo and f(oo) = 0.] Then f(f(x)) = x.

--------------------------------

Now for the desired example:

f(x) = 2/(x + 1)^2 - 1 yields f(f(x)) = (x + 1)^4/2 - 1

Note that again I am taking the codomain to be R*. Thus, f(-1) = oo and
f(oo) = -1.

Perhaps I should make a comment for anyone who thinks that using R* is
somehow "cheating". If we want to have a _real_ non-polynomial function
f(x) such that f(f(x)) is a polynomial with more than one term, then just
modify my function above slightly to give

| { -1 if x = -1,
| f(x) = {
| { 2/(x + 1)^2 - 1 otherwise.

Then we still have f(f(x)) = (x + 1)^4/2 - 1.

David

Or somewhat more generally consider

f(x) = b (x-a)^(-n) + a for x <> a,
f(a) = a

where n is a positive integer.

Then f(f(x)) = b^(1-n) (x-a)^(n^2) + a.

Similarly,

f(x) = b |x-a|^sqrt(n) + a

where n is an even positive integer that isn't a square.

Then f(f(x)) = b |b|^sqrt(n) (x-a)^n + a.
--
Robert Israel israel@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
.

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