Tommy these are far apart from proofs. I looked up your profile to see your math education. You wrote "professor" ... What do you think this question is for? To discriminate people with not much math education? Its the opposite; this point is for being prepared at which level you can expect and give math explanations to that person. And I expect that you correct your wrong declaration. The level we are currently talking is high school maths. Nothing wrong about it, but it lacks the punch of real proofs, the certainty that the findings are solid, etc.
To only give an example how shallow reasoning may fail:
Wrong.
You can have sequences \( f_n \) of analytic functions that converge to a function that is not analytic, not even continuous.
And curiously your example \( g_n(x)=\exp(x)-\exp(-nx^2) \) is such an example. It converges to the function that is \( \exp(x) \) for \( x\neq 0 \) and that is 0 at 0. This function is not continuous at 0.
Or consider the sequence \( f_n(x)=\sqrt{\frac{1}{n}+x^2} \) it converges to \( f(x)=|x| \). Which is surely not differentiable at 0, while each \( f_n \) is analytic.
And your "proofs" of 1) and 3) do not contain any usuable conclusion that would answer these questions.
For you its a problem for others its a luck. You know you can apply regular iteration whenever a function has fixed point...
To only give an example how shallow reasoning may fail:
tommy1729 Wrote:2) since they converge , f is analytic.
because both sequences f_n and g_n are continu and analytic.
...
thus for all finite n , f_n is analytic.
thus only candidate for not being analytic is f_oo.
but that is absurd.
since lim n-> oo f_oo(x) - f_n(x) = 0.
Wrong.
You can have sequences \( f_n \) of analytic functions that converge to a function that is not analytic, not even continuous.
And curiously your example \( g_n(x)=\exp(x)-\exp(-nx^2) \) is such an example. It converges to the function that is \( \exp(x) \) for \( x\neq 0 \) and that is 0 at 0. This function is not continuous at 0.
Or consider the sequence \( f_n(x)=\sqrt{\frac{1}{n}+x^2} \) it converges to \( f(x)=|x| \). Which is surely not differentiable at 0, while each \( f_n \) is analytic.
And your "proofs" of 1) and 3) do not contain any usuable conclusion that would answer these questions.
Quote:one of the problems encountered in f(f(x)) = a^x
is that i might have an annoying zero : a^x = x. for a real x.
For you its a problem for others its a luck. You know you can apply regular iteration whenever a function has fixed point...