• 0 Vote(s) - 0 Average
• 1
• 2
• 3
• 4
• 5
 Solutions to f ' (x) = f(f(x)) ? tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 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^[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 tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 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 »

 Possibly Related Threads... Thread Author Replies Views Last Post Constructing real tetration solutions Daniel 4 6,146 12/24/2019, 12:10 AM Last Post: sheldonison Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) Gottfried 91 156,562 03/03/2011, 03:16 PM Last Post: Gottfried Infinite towers & solutions to Lambert W-function brangelito 1 6,234 06/16/2010, 02:50 PM Last Post: bo198214 Comparing the Known Tetration Solutions bo198214 23 42,553 08/29/2007, 06:15 PM Last Post: jaydfox

Users browsing this thread: 1 Guest(s)