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 selfreference equation tommy1729 Ultimate Fellow Posts: 1,471 Threads: 352 Joined: Feb 2009 08/13/2011, 04:02 PM a(b(x)) = solve(b(x)*x) b(c(x)) = solve(c(x)*x) c(a(x)) = solve(a(x)*x) it seems closely related to iterations , recursions and continued fractions but its still somewhat puzzling ... there are many equivalent equations , but i dont know if they are helpfull. perhaps taking derivates on both sides will help ? regards tommy1729 BenStandeven Junior Fellow Posts: 27 Threads: 3 Joined: Apr 2009 08/19/2011, 08:11 PM (08/13/2011, 04:02 PM)tommy1729 Wrote: a(b(x)) = solve(b(x)*x) b(c(x)) = solve(c(x)*x) c(a(x)) = solve(a(x)*x) it seems closely related to iterations , recursions and continued fractions but its still somewhat puzzling ... there are many equivalent equations , but i dont know if they are helpfull. perhaps taking derivates on both sides will help ? regards tommy1729 Other than the obvious solution, a(0) = b(0) = c(0) = 0, we can simplify to: a(b(x)) = solve(b(x)) b(c(x)) = solve(c(x)) c(a(x)) = solve(a(x)) by dividing the solvand equations by x. Then we get b(a(b(x))) = 0 = b(a(0)) c(b(c(x))) = 0 = c(b(0)) a(c(a(x))) = 0 = a(c(0)) Using the original equations again: a(b(x)) = a(0) b(c(x)) = b(0) c(a(x)) = c(0) These three quantities can then be any numbers at all, apparently. « Next Oldest | Next Newest »

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