03/07/2008, 10:56 AM
Gottfried Wrote:Code:G0= |G1=
1 . . . . . | 1 . . . . .
0 3/2 . . . . | 0 3/2 . . . .
0 1/4 9/4 . . . | 0 -1/4 9/4 . . .
0 0 3/4 27/8 . . | 0 0 -3/4 27/8 . .
0 0 1/16 27/16 81/16 . | 0 0 1/16 -27/16 81/16 .
0 0 0 9/32 27/8 243/32 | 0 0 0 9/32 -27/8 243/32
But Gottfried, you didnt use the translation! For applying the regular iteration you have to *move* the fixed point to zero. The function graph is just translated so that the fixed point is situated at 0. No stretching or whatever is allowed. Here that would be
\( g_0(z)=(z-\frac{1}{4})^2+ (z-\frac{1}{4})-\frac{1}{16} + \frac{1}{4}=z^2-2\frac{1}{4}z+\frac{1}{16}+z-\frac{1}{4}-\frac{1}{16}+\frac{1}{4}=z^2+\frac{1}{2}z \)
\( g_1(z)=(z+\frac{1}{4})^2+(z+\frac{1}{4})-\frac{1}{16} - \frac{1}{4} = z^2+2\frac{1}{4}z+\frac{1}{16}+z+\frac{1}{4}-\frac{1}{16}-\frac{1}{4}=z^2+\frac{3}{2}z \)
This is because regular half iteration is characterized as *the* half iteration that has at the fixed point \( a \) the slope \( f'(a)^{1/2} \). If you change the slope at the fixed point by the transformation, and you do by your transformation
\( g_1'(0)=(-\frac{1}{4}z^2+\frac{3}{2}z)'|_{z=0}=\frac{3}{2}\neq\frac{1}{2}=(2z+1)|_{z=-\frac{1}{4}}=f'(-\frac{1}{4}) \),
it is no more *regular* half iteration.