simple idea ...
#1
a simple idea ... maybe considered before ?

let us approximate exp(x) with polynomials of degree n.

let us call that exp_n(x).

now we can solve for f_n(f_n(x)) = exp_n(x)

if n is large enough and then f_n(x) being a polynomial of degree < n.

( we ignore the terms of degree > n in f_n(f_n(x)) )

now if lim f_n(x) exists we got f(x) with f(f(x) = exp(x).


....

headscratch ...


regards

tommy1729
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#2
tommy1729 Wrote:headscratch ...
That doesn't work. Suppose then . Chopping off the second series gives which is false.
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#3
andydude Wrote:
tommy1729 Wrote:headscratch ...
That doesn't work. Suppose then . Chopping off the second series gives which is false.

well , thanks for your reply andy.

but i dont agree , and i assume you misunderstood what i meant.

in your example there is nothing to chop off ?

the degree of f(f(x)) = 4 just as desired.

what is ignored is more something like this

( with some imagination , example not so good )

f(x) = a + b x^4 + (x^17) / 17!

f(f(x)) = polynomial of degree 16 + ' some terms of degree > 16 '

' those terms of degree > 16 ' are caused by the x^17 part so they are dropped.

they are pretty small for some x afterall since i took x ^17 / 17!

that is important , what we ignore must be relatively small.

but exp(x) has a fast converging taylor series so *that* should probably not be a problem.

however in the OP i assumed a limit , that might be trickier.

but i believe that when taking the limit in a good way , we have chance at succeeding ...

i hope this clarifies a bit ...


regards

tommy1729
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#4
tommy1729 Wrote:i hope this clarifies a bit ...

Nope. Still head scratching.
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#5
andydude Wrote:Nope. Still head scratching.

Me too.
Most polynomials have no polynomial half iterate, so how can your suggestion work?
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