Hey guys,

a polynomial pendant of moving from base e over eta to sqrt(2)

may be moving from x^2+1 to x^2 to x^2-1,

i.e. from no fixpoint to one fixpoint to two fixpoints on the real axis.

Now for polynomials there is a technique to iterate them, that is not applicable to exp(x), which I will describe here applied to the example f(x)=x^2+1.

It is the iteration at infinity.

To see whats happening there one moves the fixpoint at infinity to 0, by conjugating with 1/x:

can be developed into a powerseries at 0, knowing that:

has a so called super-attracting fixpoint at 0, which means that . In this case one can solve the Böttcher equation:

(where 2 is the first power with non-zero coefficient in the powerseries of ).

Iterating can then be done similarly to the Schröder iteration:

.

Again similar to the Schröder case, we have an alternative expression:

If we even roll back our conjugation with we get:

.

Numerically this also looks very convincing:

The following is the half-iterate h of x^2+1 accompanied by the identity function and x^2+1 itself. Computed with n=9.

And the verification h(h(x))-(x^2+1)

This may lead into a new way of computing fractional iterates of exp, because we just approximate exp(x) with polynomials and approximate the half-iterate of exp with the half-iterate of these polynomials.

a polynomial pendant of moving from base e over eta to sqrt(2)

may be moving from x^2+1 to x^2 to x^2-1,

i.e. from no fixpoint to one fixpoint to two fixpoints on the real axis.

Now for polynomials there is a technique to iterate them, that is not applicable to exp(x), which I will describe here applied to the example f(x)=x^2+1.

It is the iteration at infinity.

To see whats happening there one moves the fixpoint at infinity to 0, by conjugating with 1/x:

can be developed into a powerseries at 0, knowing that:

has a so called super-attracting fixpoint at 0, which means that . In this case one can solve the Böttcher equation:

(where 2 is the first power with non-zero coefficient in the powerseries of ).

Iterating can then be done similarly to the Schröder iteration:

.

Again similar to the Schröder case, we have an alternative expression:

If we even roll back our conjugation with we get:

.

Numerically this also looks very convincing:

The following is the half-iterate h of x^2+1 accompanied by the identity function and x^2+1 itself. Computed with n=9.

And the verification h(h(x))-(x^2+1)

This may lead into a new way of computing fractional iterates of exp, because we just approximate exp(x) with polynomials and approximate the half-iterate of exp with the half-iterate of these polynomials.