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+--- Thread: floor functional equation (/showthread.php?tid=636)
floor functional equation - tommy1729 - 05/26/2011
let {x} = x - floor(x)
let f(x) be a nonlinear real-analytic function and satisfy for x > 0
f(f({x}/e)) = {x}/e
seems outside the books not ?
RE: floor functional equation - bo198214 - 05/27/2011
(05/26/2011, 12:27 PM)tommy1729 Wrote: let {x} = x - floor(x)
let f(x) be a nonlinear real-analytic function and satisfy for x > 0
f(f({x}/e)) = {x}/e
seems outside the books not ?
doesnt any half-iterate of f pay the bill?
RE: floor functional equation - tommy1729 - 05/27/2011
euh no.
how do you arrive at half-iterates ??
the half-iterate of a polynomial or the half-iterate of an exponential does not satisfy f(f({x}/e)) = {x}/e
we are searching for a solution to f(x) in f(f({x}/e)) = {x}/e.
not its half-iterate ?
RE: floor functional equation - bo198214 - 05/27/2011
(05/27/2011, 11:59 AM)tommy1729 Wrote: euh no.
how do you arrive at half-iterates ??
the half-iterate of a polynomial or the half-iterate of an exponential does not satisfy f(f({x}/e)) = {x}/e
we are searching for a solution to f(x) in f(f({x}/e)) = {x}/e.
not its half-iterate ?
Sorry, I meant a half-iterate of x not of f. We discussed that somewhere on the forum already. f(f(x))=x hence f(f({x}/e))={x}/e.
RE: floor functional equation - tommy1729 - 05/27/2011
yes that is true.
for those confused :
0 < x f(f({x}/e)) = {x}/e.
reduces to
0 < x < e f(f(x)) = x
in fact i noticed i made a mistake. ( when i had no computer in the neighbourhood )
if f(x) is real-analytic we get a contradiction since f(f(x)) is then also real-analytic and the equation f(f(x)) = x leads to f(f(x)) - x = 0 where f(f(x)) - x is also real-analytic.
but f(f(x)) - x = 0 for 0 < x < e
so on the interval [0,e] we simply have a constant 0 function but another function elsewhere ; this clearly is not real-analytic.
so the question reduces to finding :
non-linear Coo f(x) that satisfies for 0 < x < e => f(f(x)) = x
that should have been the OP.
RE: floor functional equation - tommy1729 - 05/27/2011
the easy f^(-1) ( 1 - f(z) ) does wonders
RE: floor functional equation - bo198214 - 05/28/2011
The easiest is perhaps f(x)=1/x, or f(x)=-x, if you dont need strict increase.