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.

Ha Aug. 08

y=x

Begrenztes  Wachstum

d = 4 (slider from -5 to 5, step 0.05)
m = 0.7 (slider from 0 to 10, step 0.1)
r: 8
k = 6.4
f(x) = (1 / 2 * (x + r / x^(k - 1)))
c = 0.20000000000000004
n = 14 (slider from 1 to 15, step 1)
A: (4, 0)
x0: x(A)
L1 = IterationList(f, x0, n)
g(x) = x
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)
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)
B = Point(yAxis)
i = OrthogonalLine(B, yAxis)
j = OrthogonalLine(Element[pli, 1], yAxis)
l = OrthogonalLine(Element[pli, 2], yAxis)
p = OrthogonalLine(Element[pli, 3], yAxis)
q = OrthogonalLine(Element[pli, 4], yAxis)
s = OrthogonalLine(Element[pli, 5], yAxis)
T1: "Heronfolge r=" + r + ", k=" + k + " x_0=" + x0 + ""
C = Point(pli)
D = Point(l)
E = Point(p)
F = Point(pli)
G = Point(s)
H = Point(pli)
I = Point(s)
J = Point(pli)
K = Point(pli)
L = Point(pli)
M = Point(l)
N = Point(pli)
O = Point(pli)
t = PolyLine(C, D, E, F, G, H, I, J, K, L, M, N, O)
Wurzel: r^(1 / k)
h: y = Wurzel
Text1: "f(x) = " + (LaTeX[f]) + ""