04/04/2011, 10:45 PM
(04/25/2010, 10:53 AM)bo198214 Wrote: [ -> ]Is there an elementary real function, such that
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
Let D be the degree of the polynomial it is suppose to be.
take the D'th derivative of
that should still be entire , but if
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
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 is(*) not real and elementary OR
(*) 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