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 floor functional equation tommy1729 Ultimate Fellow     Posts: 1,372 Threads: 336 Joined: Feb 2009 05/26/2011, 12:27 PM 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 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 05/27/2011, 09:06 AM (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? tommy1729 Ultimate Fellow     Posts: 1,372 Threads: 336 Joined: Feb 2009 05/27/2011, 11:59 AM 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 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 05/27/2011, 12:18 PM (This post was last modified: 05/27/2011, 12:19 PM by bo198214.) (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. tommy1729 Ultimate Fellow     Posts: 1,372 Threads: 336 Joined: Feb 2009 05/27/2011, 07:13 PM 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 Ultimate Fellow     Posts: 1,372 Threads: 336 Joined: Feb 2009 05/27/2011, 11:29 PM the easy f^(-1) ( 1 - f(z) ) does wonders bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 05/28/2011, 09:24 AM The easiest is perhaps f(x)=1/x, or f(x)=-x, if you dont need strict increase. « Next Oldest | Next Newest »

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