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 TPID 8 tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 04/04/2011, 10:45 PM (04/25/2010, 10:53 AM)bo198214 Wrote: Is there an elementary real function $F$, such that $F(1+F^{-1}(x))$ is a real polynomial of degree at least 2 without real fixed points. alias TPID 8. i think there is no such F ( note F = x^(2n)^a for x^(2n) has real fixpoints ) i might have made a mistake , so i will work in steps : 1) the reason is if F is elementary and real , it must be real-analytic. ( superfunctions of polynomials cannot be " smooth but non-analytic " ) 2) since the real poly has no real fixed points , F needs to be strictly increasing. 3) if F has poles or singularities , it cannot be a superfunction of a non-linear polynomial. 4) by 1) 2) 3) F must be entire and not a polynomial. 5) if F is entire and not a polynomial it must have values f(a) = f(a + b) =/= f(a + 2b) which means it cannot be a superfunction near those points. 6) since F is entire however $F(1+F^{-1}(x))$ is always a polynomial or never => Let D be the degree of the polynomial it is suppose to be. take the D'th derivative of $F(1+F^{-1}(x))$. that should still be entire , but if $F(1+F^{-1}(x))$ is not always a polynomial , the D'th derivative sometimes is not a constant and sometimes it is. => PARADOX ! 7) so we are forced to assume F is not entire , but then where do the poles or singularities come from ? iterations of poly do not give poles or singularties and neither does solving them. ( since poly do not map finite to infinite ! ) ( see 3) ) by 7) our function F cannot have a simple branch structure. 9) but by $F(1+F^{-1}(x))$ needs to reduce. ( command simplify ) in a complicated way ! 10) the complicated way of 9) implies that the function F is not just a composition of exp , log , rational functions , sin , cos , tan hence F is not elementary. by complicated i mean that all branches or poles or singularities are parallel to the real line and F*(z) = F(z*) hence at best we F is defined on a strip. 11) since F is not definable beyond the strip with both satisfying $F(1+F^{-1}(x))$ = poly and being Coo ... F is(*) not real and elementary OR $F(1+F^{-1}(x))$ =/= real poly. (*) the only way out of that is if F is periodic with the strip. but that would mean F is still paradoxal because it has no poles or singularities or cuts in his first strip , so neither in its copies. (*) 12) keep in mind that F cannot grow faster than double exponential because it is an iteration of a polynomial ! ( too illustrate F = " about " x^(2n)^a for " about " x^(2n) ) hence the elementary compositions are limited in terms of exp !! they are also limit in terms of logs !! the number of " simplifies " for elementary functions is also very limited !! 13) combining the above it seems true that F cannot be elementary. some improvements are wanted , but i think you get the idea. tommy1729 « Next Oldest | Next Newest »

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