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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
(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.