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:
\( g(x)=\frac{1}{(\left(\frac{1}{x}\right)^2+1}=\frac{x^2}{x^2+1} \)
\( g(x) \) can be developed into a powerseries at 0, knowing that:
\( \frac{1}{x+1}=1-x+x^2-x^3\pm\dots \)
\( g(x)=x^2-x^4+x^6\mp \dots \)
\( g(x) \) has a so called super-attracting fixpoint at 0, which means that \( g'(x)=0 \). In this case one can solve the Böttcher equation:
\( \beta(g(x))=\beta(x)^2 \) (where 2 is the first power with non-zero coefficient in the powerseries of \( g(x) \)).
Iterating \( g(x) \) can then be done similarly to the Schröder iteration:
\( g^t(x)=\beta^{-1}(\beta(x)^{2^t}) \).
Again similar to the Schröder case, we have an alternative expression:
\( g^t(x)=\lim_{n\to\infty} g^{-n}\left(g^n(x)^{2^t}\right) \)
If we even roll back our conjugation with \( 1/x \) we get:
\( f^t(x)=\lim_{n\to\infty} f^{-n}\left(f^n(x)^{2^t}\right) \).
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 \( \sum_{n=0}^N \frac{x^n}{n!} \) 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:
\( g(x)=\frac{1}{(\left(\frac{1}{x}\right)^2+1}=\frac{x^2}{x^2+1} \)
\( g(x) \) can be developed into a powerseries at 0, knowing that:
\( \frac{1}{x+1}=1-x+x^2-x^3\pm\dots \)
\( g(x)=x^2-x^4+x^6\mp \dots \)
\( g(x) \) has a so called super-attracting fixpoint at 0, which means that \( g'(x)=0 \). In this case one can solve the Böttcher equation:
\( \beta(g(x))=\beta(x)^2 \) (where 2 is the first power with non-zero coefficient in the powerseries of \( g(x) \)).
Iterating \( g(x) \) can then be done similarly to the Schröder iteration:
\( g^t(x)=\beta^{-1}(\beta(x)^{2^t}) \).
Again similar to the Schröder case, we have an alternative expression:
\( g^t(x)=\lim_{n\to\infty} g^{-n}\left(g^n(x)^{2^t}\right) \)
If we even roll back our conjugation with \( 1/x \) we get:
\( f^t(x)=\lim_{n\to\infty} f^{-n}\left(f^n(x)^{2^t}\right) \).
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 \( \sum_{n=0}^N \frac{x^n}{n!} \) and approximate the half-iterate of exp with the half-iterate of these polynomials.