Solutions to f ' (x) = f(f(x)) ? - 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: Solutions to f ' (x) = f(f(x)) ? ( /showthread.php?tid=804) |

Solutions to f ' (x) = f(f(x)) ? - tommy1729 - 08/11/2013
We all know f ' (x) = f (x) for f(x) := C exp(x). But what is the general solution to f ' (x) = f(f(x)) ? Of course f(x) = id(0) will do. I was thinking about its fixpoints : y = f ' (y) = f(f(y)) If y is also a fixpoint of f then f '' (y) = f ' ( f(y) ) * f ' (y) = [f ' (y)] ^2 = y^2 If y is also a fixpoint of f '' (y) then y^2 = y ! Now since f ' = f(f) we get f^[3](y) * f^[2](y) = y^2 Let f ' (x) = f(f(x)) then f '' (x) = f ' (f(x)) * f ' (x). Since f ' (x) = f(f(x)) we get f '' (x) = f^[3](x) * f ' (x). Rearranging gives f '' (x)/f ' (x) = f^[3](x). Integrating gives ln( f ' (x) ) = integral f^[3](x) dx + C I feel like Im getting closer to a solution but Im stuck now. How about subtitution f^[-1](z) = x ? Anyways if we plug in f(x) = a x^b we get a b x^(b-1) = a (a x^b)^b = a a^b x^(b^2) = a^(b+1) x^(b^2) Hence we get the system a b = a^(b+1) and b-1 = b^2. b = (-1)^(1/3) or -(-1)^(2/3) ( yes can be simplified ) ab = a^(b+1) => b = a^b => a = b^(1/b) ( typical tetration ) however what about other solutions ?? regards tommy1729 RE: Solutions to f ' (x) = f(f(x)) ? - tommy1729 - 08/12/2013
Notice how the superfunction of a x^b involves the generalizations. e.g. f ' (x) = f^[2,5](x) regards tommy1729 |