Simple method for half iterate NOT based on a fixpoint. - Printable Version +- Tetration Forum (https://math.eretrandre.org/tetrationforum) +-- Forum: Tetration and Related Topics (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1) +--- Forum: Mathematical and General Discussion (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=3) +--- Thread: Simple method for half iterate NOT based on a fixpoint. (/showthread.php?tid=791) Simple method for half iterate NOT based on a fixpoint. - tommy1729 - 04/30/2013 Simple method for the half iterate of a taylor series. if g: R -> R x>0 => g(x) > x x,y>0 => g(x+y) > g(x) g is entire. To find f(f(x)) = g(x) Consider g(x) = g0 + g1 x + g2 x^2 + g3 x^3 + ... ( hence g(0) = g0 ) Define f(x) = f0 + f1 x + f2 x^2 + f3 x^3 + ... Now use D^n g(0) = D^n f(f(0)) = n! gn ( this follows from taylor's theorem and the chain rule ) We get g(0) = f(f(0)) = f(f0) = g0 g'(0) = f'(f(0)) f'(0) = f'(f0) f'(0) = g1 pick f0 with 0 < f0 < g0. Now pick an analytic function t(x) such that t(f^(n)(f0)) = f^(n)(0). We arrive at the set of equations : g^(1)(0) = f^(1)(f0) t(f^1(f0)) = g1 which can be solved. Now use the solutions of D^(n-1) g(0) = D^(n-1) f(f(0)) = (n-1)! g(n-1) together with t(x) to solve D^n g(0) = D^n f(f(0)) = n! gn. This is always possible because there are exactly 2 new variables in the n th equation relative to the n-1 th equation who are f^(n)(0) and f^(n)(f0) and by using t(x) we get only 1 new variable per equation. So there is no more a degree of freedom ! Doing so gives us by induction (and the lack of free parameters aka degrees of freedom) all values f^(n)(0) hence we get the values fn. As was desired. f(x) = f0 + f1 x + f2 x^2 + f3 x^3 + ... is now the unique solution based on picking f0 and t(x) assuming t(x) is picked *wisely*. By wisely it is meant that 1) All solution fn are real 2) There are NO branches too choose when we solve for the new variable in each step from equation n-1 to equation n. Note that if for all n : gn > 0 (*) => fn > 0. ( * = COND 1 ) If that is the case we can repeat the procedure to find g^[1/4](x) or equivalently f^[1/2]. HOWEVER it is not certain if the SAME t(x) can be used again. Notice a similar method will probably work finding g^[1/3](x). (with a t1(x) AND t2(x) to remove 2 degrees of freedom) Also note that COND 1 easily gives a method for g^[1/2^m] for integer m. Notice that for a real z : 0