08/05/2022, 12:34 AM
(This post was last modified: 08/05/2022, 12:54 AM by bo198214.

*Edit Reason: wording*)
And now to the culmination of two-fixed points regular iteration: parabolic plosion!

We have quite a similar situation as with the exponentials where the base transits from \(b<\eta\) to \(b>\eta\).

But this time with the fractional linear function:

$$ f(z) = \frac{c+z}{1+z} $$

The fixed points are \(\pm\sqrt{c}\).

By varying c from positive to negative the two fixed points on the real axis unite at \(c=0\) being a parabolic fixed point and then turn into conjugate complex fixed points.

Everything is very similar here to moving \(b\) from \(<\eta\) to \(>\eta\) for exponentials \(b^z\),

EXCEPT that this is the perfect world!

Everything we had wished for the exponentials like:

* iteration at the lower fixed point is the same function as iteration at the upper fixed point

* iterating at one of the complex fixed points would give real values on the real axis

* crescent iteration is regular iteration

* iteration at the parabolic fixed point is analytic there

* \(b\mapsto \exp_b^{\circ t}(z) \) is analytic at \(\eta\)

is true here! (Maybe a bit too perfect world like the Witnesses of Jahwe would imagine!)

So we again have an explicit formula for the iteration of \(f\), however it is a bit more difficult too derive and I will leave it to you to check this formula (for \(f^{\circ s+t} = f^{\circ s}\circ f^{\circ t}\) or even derive it yourself.)

$$ f^{\circ t}(z) = \frac{q_t(\sqrt{c})c + z}{1+q_t(\sqrt{c})z}, q_t(s)=\frac{(1+s)^t-(1-s)^t}{s((1+s)^t+(1-s)^t)}$$

interesting to find out that \(q_t(\sqrt{c})\) is always real for real \(c,t\), particularly for \(c<0\) where \(s=\sqrt{c}\) is imaginary. This comes from the series expansion

$$ (1+z)^t = \sum_{n=0}^\infty \binom{t}{n} z^n $$

Then

\begin{align}

(1+z)^t - (1-z)^t &= 2\sum_{n=0}^\infty \binom{t}{2n+1} z^{2n+1}\\

(1+z)^t + (1-z)^t &= 2\sum_{n=0}^\infty \binom{t}{2n} z^{2n}\\

(1+ix)^t - (1-ix)^t &= 2i\sum_{n=0}^\infty \binom{t}{2n+1} (-1)^n x^{2n+1}\\

(1+ix)^t + (1-ix)^t &= 2\sum_{n=0}^\infty \binom{t}{2n} (-1)^n x^{2n}

\end{align}

All in all \(q_t(s)\) is that's why real valued for real and imaginary \(s\).

Also something that is not immediately visible: \(t\mapsto q_t(s)\) is a periodic function for imaginary \(s\). I have to admit that I don't see how to prove this directly, anyone any ideas?

Here a picture of \(t\mapsto q_t(\sqrt{c})\) for \(c=-0.5\). It reminds me quite of tan:

On the other hand

this is something we know already from regular iteration: it is periodic in t with period \(\frac{2\pi i}{\log(d)}\) where d is the derivative at the fixed point it is developed on.

We counter check:

\begin{align}

f'(z) &= -\frac{c + z}{(z + 1)^2} + \frac{1}{z + 1} \\

f'(\sqrt{c}) &= -\frac{c+\sqrt{c}}{(\sqrt{c}+1)^2} + \frac{1}{\sqrt{c}+1} = \frac{1-c}{(1+\sqrt{c})^2}

\end{align}

That \(\frac{2\pi i}{\log(f'(\sqrt{c}))}\) is even real is already difficult to see, we need the power series development of log:

$$ \log(1+z) = \sum_{n=1}^\infty (-1)^{n-1} \frac{x^n}{n} $$

Then we use this with our above expression:

$$ \log\left(\frac{1-c}{(1+\sqrt{c})^2}\right) = \log(1-c) - 2\log(1+\sqrt{c}) $$

We show that the expression on the right side is purely imaginary.

To make it clearer that we are talking about \(c<0\), I substitute \( c = (ix)^2\):

\begin{align}

\log(1+x^2) - 2\log(1+ix) &= \sum_{n=1}^\infty (-1)^{n-1}\frac{ x^{2n}}{n} - 2\sum_{n=1}^\infty (-1)^{2n-1} \frac{(ix)^{2n}}{2n} - 2 \sum_{n=0}^\infty (-1)^{2n} \frac{(ix)^{2n+1}}{2n+1}\\

&= \sum_{n=1}^\infty (-1)^{n-1}\frac{ x^{2n}}{n} - \sum_{n=1}^\infty (-1) (-1)^n\frac{(x)^{2n}}{n} - 2i \sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{2n+1}\\

&= - 2i \sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{2n+1}

\end{align}

How can such a simple statement about an expression being real/imaginary, be so tedious to show!

Ok, but from this follows that the period

$$ \frac{2\pi i}{\log(1-c) - 2\log(1+\sqrt{c})} $$

is real for \(c<0\) and that it is the same period (though negative) at the other fixed point \(-\sqrt{c}\).

Also interesting to check the limit for \(s=0\)! This is the parabolic case from my previous posts:

$$ \lim_{s\to 0} q_t(s) = t $$

because

\begin{align}

q_t(z)=\frac{2\sum_{n=0}^\infty \binom{t}{2n+1} z^{2n+1}}{2z\sum_{n=0}^\infty \binom{t}{2n} z^{2n}} =

\frac{z(\binom{t}{1} + z r_1(t))}{z(1+z r_2(t))} = \frac{t + z r_1(t)}{1+z r_2(t)}

\end{align}

From this representation even follows that \(q_t(s)\) is analytic at \(s=0\)! And so is our original function:

$$ f_s^{\circ t}(z) = \frac{q_t(\sqrt{c})c + z}{1+q_t(\sqrt{c})z} = \frac{q_t(s)s^2 + z}{1+q_t(s)z} $$

analytic in \(s\) at \(0\).

And here the perfect world from regular iteration of linear fractional functions as motion picture:

We have quite a similar situation as with the exponentials where the base transits from \(b<\eta\) to \(b>\eta\).

But this time with the fractional linear function:

$$ f(z) = \frac{c+z}{1+z} $$

The fixed points are \(\pm\sqrt{c}\).

By varying c from positive to negative the two fixed points on the real axis unite at \(c=0\) being a parabolic fixed point and then turn into conjugate complex fixed points.

Everything is very similar here to moving \(b\) from \(<\eta\) to \(>\eta\) for exponentials \(b^z\),

EXCEPT that this is the perfect world!

Everything we had wished for the exponentials like:

* iteration at the lower fixed point is the same function as iteration at the upper fixed point

* iterating at one of the complex fixed points would give real values on the real axis

* crescent iteration is regular iteration

* iteration at the parabolic fixed point is analytic there

* \(b\mapsto \exp_b^{\circ t}(z) \) is analytic at \(\eta\)

is true here! (Maybe a bit too perfect world like the Witnesses of Jahwe would imagine!)

So we again have an explicit formula for the iteration of \(f\), however it is a bit more difficult too derive and I will leave it to you to check this formula (for \(f^{\circ s+t} = f^{\circ s}\circ f^{\circ t}\) or even derive it yourself.)

$$ f^{\circ t}(z) = \frac{q_t(\sqrt{c})c + z}{1+q_t(\sqrt{c})z}, q_t(s)=\frac{(1+s)^t-(1-s)^t}{s((1+s)^t+(1-s)^t)}$$

interesting to find out that \(q_t(\sqrt{c})\) is always real for real \(c,t\), particularly for \(c<0\) where \(s=\sqrt{c}\) is imaginary. This comes from the series expansion

$$ (1+z)^t = \sum_{n=0}^\infty \binom{t}{n} z^n $$

Then

\begin{align}

(1+z)^t - (1-z)^t &= 2\sum_{n=0}^\infty \binom{t}{2n+1} z^{2n+1}\\

(1+z)^t + (1-z)^t &= 2\sum_{n=0}^\infty \binom{t}{2n} z^{2n}\\

(1+ix)^t - (1-ix)^t &= 2i\sum_{n=0}^\infty \binom{t}{2n+1} (-1)^n x^{2n+1}\\

(1+ix)^t + (1-ix)^t &= 2\sum_{n=0}^\infty \binom{t}{2n} (-1)^n x^{2n}

\end{align}

All in all \(q_t(s)\) is that's why real valued for real and imaginary \(s\).

Also something that is not immediately visible: \(t\mapsto q_t(s)\) is a periodic function for imaginary \(s\). I have to admit that I don't see how to prove this directly, anyone any ideas?

Here a picture of \(t\mapsto q_t(\sqrt{c})\) for \(c=-0.5\). It reminds me quite of tan:

On the other hand

this is something we know already from regular iteration: it is periodic in t with period \(\frac{2\pi i}{\log(d)}\) where d is the derivative at the fixed point it is developed on.

We counter check:

\begin{align}

f'(z) &= -\frac{c + z}{(z + 1)^2} + \frac{1}{z + 1} \\

f'(\sqrt{c}) &= -\frac{c+\sqrt{c}}{(\sqrt{c}+1)^2} + \frac{1}{\sqrt{c}+1} = \frac{1-c}{(1+\sqrt{c})^2}

\end{align}

That \(\frac{2\pi i}{\log(f'(\sqrt{c}))}\) is even real is already difficult to see, we need the power series development of log:

$$ \log(1+z) = \sum_{n=1}^\infty (-1)^{n-1} \frac{x^n}{n} $$

Then we use this with our above expression:

$$ \log\left(\frac{1-c}{(1+\sqrt{c})^2}\right) = \log(1-c) - 2\log(1+\sqrt{c}) $$

We show that the expression on the right side is purely imaginary.

To make it clearer that we are talking about \(c<0\), I substitute \( c = (ix)^2\):

\begin{align}

\log(1+x^2) - 2\log(1+ix) &= \sum_{n=1}^\infty (-1)^{n-1}\frac{ x^{2n}}{n} - 2\sum_{n=1}^\infty (-1)^{2n-1} \frac{(ix)^{2n}}{2n} - 2 \sum_{n=0}^\infty (-1)^{2n} \frac{(ix)^{2n+1}}{2n+1}\\

&= \sum_{n=1}^\infty (-1)^{n-1}\frac{ x^{2n}}{n} - \sum_{n=1}^\infty (-1) (-1)^n\frac{(x)^{2n}}{n} - 2i \sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{2n+1}\\

&= - 2i \sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{2n+1}

\end{align}

How can such a simple statement about an expression being real/imaginary, be so tedious to show!

Ok, but from this follows that the period

$$ \frac{2\pi i}{\log(1-c) - 2\log(1+\sqrt{c})} $$

is real for \(c<0\) and that it is the same period (though negative) at the other fixed point \(-\sqrt{c}\).

Also interesting to check the limit for \(s=0\)! This is the parabolic case from my previous posts:

$$ \lim_{s\to 0} q_t(s) = t $$

because

\begin{align}

q_t(z)=\frac{2\sum_{n=0}^\infty \binom{t}{2n+1} z^{2n+1}}{2z\sum_{n=0}^\infty \binom{t}{2n} z^{2n}} =

\frac{z(\binom{t}{1} + z r_1(t))}{z(1+z r_2(t))} = \frac{t + z r_1(t)}{1+z r_2(t)}

\end{align}

From this representation even follows that \(q_t(s)\) is analytic at \(s=0\)! And so is our original function:

$$ f_s^{\circ t}(z) = \frac{q_t(\sqrt{c})c + z}{1+q_t(\sqrt{c})z} = \frac{q_t(s)s^2 + z}{1+q_t(s)z} $$

analytic in \(s\) at \(0\).

And here the perfect world from regular iteration of linear fractional functions as motion picture: