blunder ?? f(x) = f(x^2) tommy1729 Ultimate Fellow     Posts: 1,358 Threads: 330 Joined: Feb 2009 08/29/2010, 09:13 PM (This post was last modified: 08/29/2010, 09:22 PM by tommy1729.) ok here is probably a mistake. solve for NONCONSTANT f(x) f(x) = f(x^2) 1) f(x) seems not to be analytic on the unit radius , in fact constant within the unit radius because 0 is an attractor for x^2 there. 2) f(1) must diverge. however this seems a paradox with the following ' solution ' : f(x) = f(x^2) f(x) = g(ln(ln(x))) => g(ln(ln(x))) = g(ln(ln(x))+ln(2)) => g(x) = sin(x*2pi/ln(2)) => f(x) = sin(ln(ln(x))*2pi/ln(2)) but now f(x) is nonconstant and differentiable within the unit radius ?? blunder ? ok g(x) needs to be ln(2) periodic. assume g(x) must also not be defined within the unit disk. but that unit disk needs to be copied by the period. but the period is smaller than the diameter of the unit circle. thus no solution exists ? or is sin(ln(ln(x))*2pi/ln(2)) a solution afterall ? another weird thing : if we use taylor expanded anywhere we get : a_n = a_n^2 for the coefficients. this seems to imply a finite radius no matter where we expand since a_n = a_n^2 = a_n^4 = ... it seems that hence a_n = a_n^2 = a_n^3 has a radius at most ln(3) since a function cannot have 2 real periods like ln(2) and ln(3). ( since ln(2)/ln(3) is irrational ) now note that we arrive at another contradiction for f(x) = f(x^2) ; since f(x) has his copied nonvalid unit circle domains , if x satisfies that f(x) converges at x , then f(x) must also converge for x^2 imput. but that seems to rule out a lot of x's way beyond the copied unit circle domain. so f(x) = f(x^2) has no solution afterall ?? maybe if we increase the period way beyond the unit diameter. f(x) = f(x^e^4pi) has a solution ??? but it isnt sin(ln(ln(x))/2) ?? then how to find it ?? taylor ? a blunder ... maybe i need some sleep ... tommy1729 Ultimate Fellow     Posts: 1,358 Threads: 330 Joined: Feb 2009 08/30/2010, 11:40 PM the post contains some mistakes ... will correct later. meanwhile for abs(z) < 1 and <> 0 sum (n, -oo , oo) x^2^n seems a solution. « Next Oldest | Next Newest » 