Fractional iteration of x^2+1 at infinity and fractional iteration of exp

[bo198214]

Anyway, I think this is not analytic at 0. The iterates of g so formed have a branch point at 0, and also a complementary one at infinity (note that if there is a BP at 0, there must be one at inf, since "circling about inf" is equivalent to circling about 0). The conjugate simply exchanges these two branch points. This would explain how it can approach \( |x| \) as \( t \rightarrow 0 \).

Well, the behaviour is comparable to that of \( x^{2^t} \), i.e. x taken to a non-integer number has a branchpoint at 0,oo. There is anyway the well-known proposition that the fractional iteration of exp can not be entire.
Even [1] shows that that there is no solution of f(f(x))=ax^2+bx+c in the complex plane.

[1] Rice, R. E., Schweizer, B., & Sklar, A. (1980). When is \(f(f(z))=az^2+bz+c\)? Am. Math. Mon., 87, 252–263.

[Catullus]

exp(x) + x has a " true " fixpoint at oo.

Exp(∞) may be a larger infinity.

[JmsNxn]

exp(x) + x has a " true " fixpoint at oo.

Exp(∞) may be a larger infinity.

Hmmmmm, you'd have to qualify that using some kind of framework. No idea what that would be. You could use something like Hardy spaces, and refer to \(1/\exp(\infty)\) in the right half plane as smaller than \(1/\infty\) in the right half plane. But then, you'd have to qualify how you mean this. Typically we're not referring to growth hierarchies. And they don't apply to Tommy's comment.

[Catullus]

exp(x) + x has a " true " fixpoint at oo.

Exp(∞) may be a larger infinity.

Hmmmmm, you'd have to qualify that using some kind of framework. No idea what that would be. You could use something like Hardy spaces, and refer to \(1/\exp(\infty)\) in the right half plane as smaller than \(1/\infty\) in the right half plane. But then, you'd have to qualify how you mean this. Typically we're not referring to growth hierarchies. And they don't apply to Tommy's comment.

Exp(\( \aleph_0 \)) = ℶ1.

[JmsNxn]

exp(x) + x has a " true " fixpoint at oo.

Exp(∞) may be a larger infinity.

Hmmmmm, you'd have to qualify that using some kind of framework. No idea what that would be. You could use something like Hardy spaces, and refer to \(1/\exp(\infty)\) in the right half plane as smaller than \(1/\infty\) in the right half plane. But then, you'd have to qualify how you mean this. Typically we're not referring to growth hierarchies. And they don't apply to Tommy's comment.

Exp() = beth 1.

\(\aleph_0 \neq \infty\)

[Catullus]

exp(x) + x has a " true " fixpoint at oo.

Exp(∞) may be a larger infinity.

Hmmmmm, you'd have to qualify that using some kind of framework. No idea what that would be. You could use something like Hardy spaces, and refer to \(1/\exp(\infty)\) in the right half plane as smaller than \(1/\infty\) in the right half plane. But then, you'd have to qualify how you mean this. Typically we're not referring to growth hierarchies. And they don't apply to Tommy's comment.

Exp() = beth 1.

\(\aleph_0 \neq \infty\)

Aleph zero is infinite.

[JmsNxn]

Exp(∞) may be a larger infinity.

Hmmmmm, you'd have to qualify that using some kind of framework. No idea what that would be. You could use something like Hardy spaces, and refer to \(1/\exp(\infty)\) in the right half plane as smaller than \(1/\infty\) in the right half plane. But then, you'd have to qualify how you mean this. Typically we're not referring to growth hierarchies. And they don't apply to Tommy's comment.

Exp() = beth 1.

\(\aleph_0 \neq \infty\)

Aleph zero is infinite.

Yes, but not in the sense of complex analysis limits. The point \(\infty\) on the Riemann sphere, is not the same thing as Aleph zero.

[tommy1729]

on the real line i make two notions of fixpoints at + oo :

those that touch the identity axis ( in the limit ).

and those that do not.

what that means is another thing.

and complex analysis is also another thing.