f( f(x) ) = exp(x) solved ! ! !

[tommy1729]

you do realise sqrt is not an entire function !?

thus not a counterexample.

So what, then consider \( f_n(x)=\exp(-x^{2n}) \).
Do you guess what it converges to?

ok but , does the limit f_n ( f_n (x) ) converge to a strictly rising entire function that maps R to a subset of R that has no real fixpoints ? ( disregarding removable singularities )

my f oo does : exp(x) ( again disregarding removable singularities ( x = 0 ) )

high regards

tommy1729

keep in mind that we know :

if g(x) is a monotonic function defined on an interval I, then g(x) is differentiable almost everywhere on I , i.e. the set of numbers x in I such that g(x) is not differentiable in x has at most Lebesgue measure zero.

if f(f(x)) = g(x) then f(x) is also a monotonic function.

( we may replace g and f with 'my' f_n and g_n )

high regards

tommy1729