floor functional equation

[tommy1729]

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 ?

[bo198214]

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?

[tommy1729]

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 ?

[bo198214]

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.

[tommy1729]

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.

[tommy1729]

the easy f^(-1) ( 1 - f(z) ) does wonders

[bo198214]

The easiest is perhaps f(x)=1/x, or f(x)=-x, if you dont need strict increase.