# Tetration Forum

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(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