Iteration für den Turm von Hanoi
Author: Prof. Dr. Dörte Haftendorn

Ziehe n, die Anzahl der Scheiben. Darunter steht als letztes die Zahl den nötigen Umlegungen.

exponentieller Zerfall

Ha Aug. 08

y=x

exponentieller  Zerfall
Faktor q mit 0<q<1

q = 0.7 (slider from 0 to 10, step 0.1)
r = 20
k = 2
f(x) = (q * x)
c = 0.20000000000000004
n = 8 (slider from 1 to 15, step 1)
A: (16, 0)
x0: x(A)
L1 = IterationList(f, x0, n)
L3 = Sequence(Segment[(Element[L1, i], Element[L1, i + 1]), (Element[L1, i + 1], Element[L1, i + 1])], i, 1, n - 1)
L4 = Sequence(Segment[(Element[L1, i + 1], Element[L1, i + 1]), (Element[L1, i + 1], Element[L1, i + 2])], i, 1, n - 1)
y_A = Segment((x0, 0), (x0, f(x0)))
ntes = Element(L1, n + 1)
a = Segment((Element[L1, 2], 0), (Element[L1, 2], Element[L1, 2]))
Liste1 = Sequence(Segment[(0, Element[L1, i + 1]), (Element[L1, i], Element[L1, i + 1])], i, 1, n - 1)
Text1: "f(x) = " + f
L2 = Sequence(Element[L1, i], i, 1, n)
Text4: L2 + ""
pli = Sequence((i, Element[L2, i + 1]), i, 1, n)
yli = Sequence(Segment[(0, Element[L2, j + 1]), (20, Element[L2, j + 1])], j, 1, n - 1)
h(x) = (x(A) * q^(x))
g(x) = x
i = OrthogonalLine(Element[pli, 1], yAxis)
j = OrthogonalLine((3, Element[L1, 3]), yAxis)
l = OrthogonalLine((5, Element[L1, 5]), yAxis)
m = OrthogonalLine((4, Element[L1, 4]), yAxis)
p = OrthogonalLine((6, Element[L1, 6]), yAxis)
s = OrthogonalLine((7, Element[L1, 7]), yAxis)
t = OrthogonalLine((8, Element[L1, 8]), yAxis)
B: (0, 16)