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 Solutions to f ' (x) = f(f(x)) ? tommy1729 Ultimate Fellow     Posts: 1,455 Threads: 350 Joined: Feb 2009 08/11/2013, 11:49 PM (This post was last modified: 08/12/2013, 12:04 AM by tommy1729.) 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^(y) * f^(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^(x) * f ' (x). Rearranging gives f '' (x)/f ' (x) = f^(x). Integrating gives ln( f ' (x) ) = integral f^(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 tommy1729 Ultimate Fellow     Posts: 1,455 Threads: 350 Joined: Feb 2009 08/12/2013, 12:10 AM Notice how the superfunction of a x^b involves the generalizations. e.g. f ' (x) = f^[2,5](x) regards tommy1729 « Next Oldest | Next Newest »

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