Iteration of fractional linear maps - Revision history

Bo198214: /* Iterates */

2013-05-01T15:56:40Z

Iterates

2013-05-01T15:55:27Z

Iterates: fixed sign error

2013-05-01T15:44:28Z

Iterates

2013-05-01T15:23:56Z

Iterates: finished

2013-05-01T14:47:17Z

Iterates: intermediate

2013-05-01T14:31:55Z

Schröder coordinate at the fixpoints: save state

2013-05-01T13:35:30Z

Fixpoints

2013-05-01T13:34:10Z

Fixpoints

2013-05-01T13:33:16Z

Fixpoints: elegance

2013-05-01T13:32:26Z

Fixpoints: fixed typo

@@ Line 50: / Line 50: @@
 $s_t := (1-q)^t - (1+q)^t$
-$f^t(z) = \frac{(-d s_t + a_2 s_{t+1})z + s_t b}{s_t c z - d s_t + a_2 (1-q^2) s_{t-1}} = \frac{(-d+a_2 \frac{s_{t+1}}{s_t})z + b}{cz-d+a_2(1-q^2)\frac{s_{t-1}}{s_t}}$
+$f^t(z) = \frac{(-d s_t + a_2 s_{t+1})z + s_t b}{s_t c z - d s_t + a_2 (1-q^2) s_{t-1}} = \frac{(-d+a_2 \frac{s_{t+1}}{s_t})z + b}{cz-d+a_2\frac{(1-q^2)s_{t-1}}{s_t}}$
 We see indeed that this is independent of the sign of $q$, i.e. regardless at which fixpoint the iterate is developed, the result is the same.

@@ Line 45: / Line 45: @@
 $\sigma_1^{-1}(w) = \frac{L_2w-L_1}{w-1}$
-$f_1^t(z) = \frac{L_2K^t\frac{z-L_1}{z-L_2} - L_1}{K^t\frac{z-L_1}{z-L_2}-1} = \frac{(L_2K^t-L_1)z -L_2K^tL_1 + L_2L_1}{(K^t-1)z-K^tL_1+L_2} =
+$f_1^t(z) = \frac{L_2K^t\frac{z-L_1}{z-L_2} - L_1}{K^t\frac{z-L_1}{z-L_2}-1} = \frac{(L_2K^t-L_1)z -L_2K^tL_1 + L_2L_1}{(K^t-1)z+K^tL_1-L_2} =
-\frac{(L_2K^t-L_1)z + (K^t-1)\frac{b}{c}}{(K^t-1)z - K^tL_1 + L_2} = \frac{(L_2 (1-q)^t-L_1 (1+q)^t)z + ((1-q)^t-(1+q)^t)\frac{b}{c}}{((1-q)^t-(1+q)^t)z - (1-q)^tL_1 + (1+q)^tL_2}$
+\frac{(L_2K^t-L_1)z + (K^t-1)\frac{b}{c}}{(K^t-1)z + K^tL_1 - L_2} = \frac{(L_2 (1-q)^t-L_1 (1+q)^t)z + ((1-q)^t-(1+q)^t)\frac{b}{c}}{((1-q)^t-(1+q)^t)z + (1-q)^tL_1 - (1+q)^tL_2}$
 $s_t := (1-q)^t - (1+q)^t$
-$f^t(z) = \frac{(-d s_t + a_2 s_{t+1})z + s_t b}{s_t c z + d s_t - a_2 (1-q^2) s_{t-1}} = \frac{(-d+a_2 \frac{s_{t+1}}{s_t})z + b}{cz+d-a_2(1-q^2)\frac{s_{t-1}}{s_t}}$
+$f^t(z) = \frac{(-d s_t + a_2 s_{t+1})z + s_t b}{s_t c z - d s_t + a_2 (1-q^2) s_{t-1}} = \frac{(-d+a_2 \frac{s_{t+1}}{s_t})z + b}{cz-d+a_2(1-q^2)\frac{s_{t-1}}{s_t}}$
 We see indeed that this is independent of the sign of $q$, i.e. regardless at which fixpoint the iterate is developed, the result is the same.

@@ Line 41: / Line 41: @@
 The iterates $f^t$ of $f$ can then be given by:
-$f^t(z) = \sigma_k^{-1}(f'(L_k)^t \sigma_k(z))$
+$f_k^t(z) = \sigma_k^{-1}(f'(L_k)^t \sigma_k(z))$
 $\sigma_1^{-1}(w) = \frac{L_2w-L_1}{w-1}$
-$f^t(z) = \frac{L_2K^t\frac{z-L_1}{z-L_2} - L_1}{K^t\frac{z-L_1}{z-L_2}-1} = \frac{(L_2K^t-L_1)z -L_2K^tL_1 + L_2L_1}{(K^t-1)z-K^tL_1+L_2} =
+$f_1^t(z) = \frac{L_2K^t\frac{z-L_1}{z-L_2} - L_1}{K^t\frac{z-L_1}{z-L_2}-1} = \frac{(L_2K^t-L_1)z -L_2K^tL_1 + L_2L_1}{(K^t-1)z-K^tL_1+L_2} =
 \frac{(L_2K^t-L_1)z + (K^t-1)\frac{b}{c}}{(K^t-1)z - K^tL_1 + L_2} = \frac{(L_2 (1-q)^t-L_1 (1+q)^t)z + ((1-q)^t-(1+q)^t)\frac{b}{c}}{((1-q)^t-(1+q)^t)z - (1-q)^tL_1 + (1+q)^tL_2}$

← Older revision		Revision as of 15:23, 1 May 2013
Line 47:		Line 47:
	$f^t(z) = \frac{L_2K^t\frac{z-L_1}{z-L_2} - L_1}{K^t\frac{z-L_1}{z-L_2}-1} = \frac{(L_2K^t-L_1)z -L_2K^tL_1 + L_2L_1}{(K^t-1)z-K^tL_1+L_2} =		$f^t(z) = \frac{L_2K^t\frac{z-L_1}{z-L_2} - L_1}{K^t\frac{z-L_1}{z-L_2}-1} = \frac{(L_2K^t-L_1)z -L_2K^tL_1 + L_2L_1}{(K^t-1)z-K^tL_1+L_2} =
	\frac{(L_2K^t-L_1)z + (K^t-1)\frac{b}{c}}{(K^t-1)z - K^tL_1 + L_2} = \frac{(L_2 (1-q)^t-L_1 (1+q)^t)z + ((1-q)^t-(1+q)^t)\frac{b}{c}}{((1-q)^t-(1+q)^t)z - (1-q)^tL_1 + (1+q)^tL_2}$		\frac{(L_2K^t-L_1)z + (K^t-1)\frac{b}{c}}{(K^t-1)z - K^tL_1 + L_2} = \frac{(L_2 (1-q)^t-L_1 (1+q)^t)z + ((1-q)^t-(1+q)^t)\frac{b}{c}}{((1-q)^t-(1+q)^t)z - (1-q)^tL_1 + (1+q)^tL_2}$
		+
		+	$s_t := (1-q)^t - (1+q)^t$
		+
		+	$f^t(z) = \frac{(-d s_t + a_2 s_{t+1})z + s_t b}{s_t c z + d s_t - a_2 (1-q^2) s_{t-1}} = \frac{(-d+a_2 \frac{s_{t+1}}{s_t})z + b}{cz+d-a_2(1-q^2)\frac{s_{t-1}}{s_t}}$
		+
		+	We see indeed that this is independent of the sign of $q$, i.e. regardless at which fixpoint the iterate is developed, the result is the same.

← Older revision		Revision as of 14:47, 1 May 2013
Line 45:		Line 45:
	$\sigma_1^{-1}(w) = \frac{L_2w-L_1}{w-1}$		$\sigma_1^{-1}(w) = \frac{L_2w-L_1}{w-1}$

−	$f^t(z) = \frac{L_2K^t\frac{z-L_1}{z-L_2} - L_1}{K^t\frac{z-L_1}{z-L_2}-1} = \frac{(L_2K^t-L_1)z -L_2K^tL_1 + L_2L_1}{(K^t-1)z-K^tL_1+L_2} = $	+	$f^t(z) = \frac{L_2K^t\frac{z-L_1}{z-L_2} - L_1}{K^t\frac{z-L_1}{z-L_2}-1} = \frac{(L_2K^t-L_1)z -L_2K^tL_1 + L_2L_1}{(K^t-1)z-K^tL_1+L_2} =
		+	\frac{(L_2K^t-L_1)z + (K^t-1)\frac{b}{c}}{(K^t-1)z - K^tL_1 + L_2} = \frac{(L_2 (1-q)^t-L_1 (1+q)^t)z + ((1-q)^t-(1+q)^t)\frac{b}{c}}{((1-q)^t-(1+q)^t)z - (1-q)^tL_1 + (1+q)^tL_2}$

@@ Line 22: / Line 22: @@
 == [[Schröder coordinate]] at the fixpoints ==
-$\sigma_k(f(z))=f'(L_k) \sigma(z)$, $\sigma_k$ is unique up to a multiplicative constant.
+Let $\sigma_k(f(z))=f'(L_k) \sigma(z)$, $\sigma_k$ is unique up to a multiplicative constant.
-$\sigma_1\left(\frac{az+b}{cz+d}\right) = \frac{(a-cL_2)z + b-dL_2}{(a-cL_1)z + b -dL_2} = \frac{a_2(1-q)z+}{a_2(1+q)z + }$.
+Then $\sigma_1(z) = \frac{z-L_1}{z-L_2}$, $\sigma_2(z)=\frac{z-L_2}{z-L_1}$.

@@ Line 6: / Line 6: @@
 $L_{1,2}=-\frac{d-a}{2c} \pm \sqrt{\left(\frac{d-a}{2c}\right)^2 + \frac{b}{c}}$, or for later use:
-$cL_{1,2}=a_2-d \pm a_2\underbrace{\sqrt{1-\frac{ad-bc}{a_2^2}}}_{=:q}$, where $a_2=\frac{a+d}{2}$.
+$cL_{1,2}=-d + a_2 \pm a_2\underbrace{\sqrt{1-\frac{ad-bc}{a_2^2}}}_{=:q}$, where $a_2=\frac{a+d}{2}$.
 == Derivative at the fixpoints ==