09/23/2010, 11:20 PM

f(x) = f(g(x))

its a simple equation.

and i talked about it before. as did e.g. gottfried.

in particular for a certain given g(x) the equation is solvable for f(x) by using the super of g(x) and a 1-periodic function...

but that is not the end of the story.

what if f(x) is given and we want to find g(x) ?

well for starters g(x) might not be unique.

if f(x) = f(exp(x)) then we must also have f(x) = f(x + 2pi i)

the pattern is clear

f(x) = f(g(x))

=> g(x) = g(g1(x)) => f(x) = f(g1(x))

and we can now replace g(x) with g1(x) , g1(x) with g2(x) and repeat ...

until we arrive at a moebius function.... if we ever ...

i guess we could call g_n(x) the invariants of f(x).

and i guess one could say : hey , this is just the construction of a riemann surface in disguise.

but those are just names !

another question is the following :

if we go up the chain , from g_n+1 to g_n , indefinitely , then what f(x) satisfies that ?

thus : f(x) = f(g_-n(x)) for all n. => f(x) = ??

if g_(-)n is not cyclic and f(x) is coo do we have a uniqueness condition on f(x) ?

is f(x) then a fractal or a constant ? ( i assume if we require f(x) to be coo and g_-n is dense ( non-repelling point ) then f(x) must be constant )

also related : is lim n-> oo g_(-)n(x) convergent or divergent ?

is brouwers fixed point theorem related ?

i believe lim n-> oo g_-n(x) is divergent because

g(x) = g(g(x)) is the associated equation , and this has no solution.

we do not accept g_(-)n(x) = x of course ...

and finally of course the following

f(x) = f(a(x)) = f(b(x))

which belong more too the "functional equation category" perhaps ...

that equation is trickier than it seems ; at first sight it seems impossible since we have sin(super(a)/2pi) = sin(super(b)/2pi) ( each side as solution )

but then again g_2 and g_3 e.g. satisfy it.

i assume lim n -> oo f(x) = f(a_g_(-)n(x)) = f(b_g_(-)n(x))

has no solution.

quite a brainstorm ....

tommy1729

its a simple equation.

and i talked about it before. as did e.g. gottfried.

in particular for a certain given g(x) the equation is solvable for f(x) by using the super of g(x) and a 1-periodic function...

but that is not the end of the story.

what if f(x) is given and we want to find g(x) ?

well for starters g(x) might not be unique.

if f(x) = f(exp(x)) then we must also have f(x) = f(x + 2pi i)

the pattern is clear

f(x) = f(g(x))

=> g(x) = g(g1(x)) => f(x) = f(g1(x))

and we can now replace g(x) with g1(x) , g1(x) with g2(x) and repeat ...

until we arrive at a moebius function.... if we ever ...

i guess we could call g_n(x) the invariants of f(x).

and i guess one could say : hey , this is just the construction of a riemann surface in disguise.

but those are just names !

another question is the following :

if we go up the chain , from g_n+1 to g_n , indefinitely , then what f(x) satisfies that ?

thus : f(x) = f(g_-n(x)) for all n. => f(x) = ??

if g_(-)n is not cyclic and f(x) is coo do we have a uniqueness condition on f(x) ?

is f(x) then a fractal or a constant ? ( i assume if we require f(x) to be coo and g_-n is dense ( non-repelling point ) then f(x) must be constant )

also related : is lim n-> oo g_(-)n(x) convergent or divergent ?

is brouwers fixed point theorem related ?

i believe lim n-> oo g_-n(x) is divergent because

g(x) = g(g(x)) is the associated equation , and this has no solution.

we do not accept g_(-)n(x) = x of course ...

and finally of course the following

f(x) = f(a(x)) = f(b(x))

which belong more too the "functional equation category" perhaps ...

that equation is trickier than it seems ; at first sight it seems impossible since we have sin(super(a)/2pi) = sin(super(b)/2pi) ( each side as solution )

but then again g_2 and g_3 e.g. satisfy it.

i assume lim n -> oo f(x) = f(a_g_(-)n(x)) = f(b_g_(-)n(x))

has no solution.

quite a brainstorm ....

tommy1729