Cardanische Formel für x^3 = p x + q im casus irreducibilis, in dem es genau drei reelle Nullstellen gibt.
Author: Prof. Dr. Dörte Haftendorn
Date: 24.05.2011

Cardanische Formeln

x^3=p*x+q

x=\sqrt[3]{\frac{q}{2}+\sqrt{R}}+\sqrt[3]{\frac{q}{2}-\sqrt{R}}=\sqrt[3]{a}+\sqrt[3]{b}=a3+b3=x_1

R=(\frac{q}{2})^{2}-(\frac{p}{3})^{3}

R<0, Gauß-Ebene

Variiere p und q
mit den 
Pfeiltasten!

ebenso x_2=a3_2+b3_2  und x_3=a3_3 + b3_3

Formeln 2

f

α/3

Schritt 2

Schritt 3

Schritt 4

Lösungen

a = \frac{q}{2} + i* \sqrt{-R}

b = \frac{q}{2} - i* \sqrt{-R}

f(x)=x^3-p x - q

Ha 2011

Formeln 1

Schritt 1

p = 7 (slider from -5 to 7, step 0.1)
q = 5 (slider from -5 to 5, step 0.1)
R: (q / 2.0)^(2.0) - (p / 3.0)^(3.0)
A: (q / 2.0, sqrt(-(R)))
B: (x(A), -y(A))
O = Intersect(xAxis, yAxis)
E = (1, 0)
a = Segment(A, O)
b = Segment(O, B)
α = Angle(E, O, A)
B3_1: b^(1.0 / 3.0) * (cos(-(α) / 3.0), sin(-(α) / 3.0))
A3_1: a^(1.0 / 3.0) * (cos(α / 3.0), sin(α / 3.0))
k3 = Circle(O, A3_1)
k = Circle(O, A)
A3_2: a^(1.0 / 3.0) * (cos(α / 3.0 + 120.0°), sin(α / 3.0 + 120.0°))
A3_3: a^(1.0 / 3.0) * (cos(α / 3.0 + 240.0°), sin(α / 3.0 + 240.0°))
B3_3: b^(1.0 / 3.0) * (cos(-(α) / 3.0 + 120.0°), sin(-(α) / 3.0 + 120.0°))
B3_2: b^(1.0 / 3.0) * (cos(-(α) / 3.0 + 240.0°), sin(-(α) / 3.0 + 240.0°))
c = Segment(B3_3, A3_3)
d = Segment(A3_2, B3_2)
e = Segment(A3_1, B3_1)
f(x) = x^(3.0) - p * x - q
g = false
h = true
X1 = Mirror(O, e)
X2 = Mirror(O, c)
X3 = Mirror(O, d)
i = Segment(O, A3_1)
β = Angle(E, O, A3_1)
j = Segment(B3_3, O)
l = Segment(A3_2, O)
m = Segment(B3_2, O)
n = Segment(A3_3, O)
r = Segment(O, B3_1)
o = false
s = false
t = false
u = false
v = false
w = true
Text2: "x_1 = " + (x(X1)) + "
x_2 = " + (x(X2)) + "
x_3 = " + (x(X3))