Cardanische Formel für x^3 = p x + q  im Fall R>0, in dem es genau eine reelle Lösung gibt.
Author: Prof. Dr. Dörte Haftendorn
Date: 24.05.2011

Cardanische Formeln, R>0

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!

aber x_2 und x_3   sind echt-komplexe Zahlen 

Formeln 2

f

α/3

Schritt 2

Schritt 3

Lösungen

a = \frac{q}{2} +  \sqrt{R}=x(A)

b = \frac{q}{2} -  \sqrt{R}=x(B)

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

Ha 2011

Formeln 1

Schritt 1

Schritt 4

p = -3 (slider from -10 to 0, step 0.01)
q = -2 (slider from -10 to 7, step 0.1)
R: (q / 2)^(2) - (p / 3)^(3)
A: (q / 2 + sqrt(R), 0)
B: (q / 2 - sqrt(R), 0)
O = Intersect(xAxis, yAxis)
E = (1, 0)
a: x(A)
b: x(B)
α = Angle(E, O, A)
B3: b^(1 / 3) * (1, 0)
A3: a^(1 / 3) * (1, 0)
k3 = Circle(O, A3)
k = Circle(O, A)
A3_1: a^(1 / 3) * (cos(α / 3 + 120°), sin(α / 3 + 120°))
A3_2: a^(1 / 3) * (cos(α / 3 + 240°), sin(α / 3 + 240°))
B3_2: b^(1 / 3) * (cos((-α) / 3 + 120°), sin((-α) / 3 + 120°))
B3_1: b^(1 / 3) * (cos((-α) / 3 + 240°), sin((-α) / 3 + 240°))
c = Segment(B3_2, A3_2)
d = Segment(A3_1, B3_1)
e = Segment(A3, B3)
g = false
h = true
X1: (x(A3) + x(B3), 0)
i = Segment(O, A3)
β = Angle(E, O, A3)
j = Segment(B3_2, O)
l = Segment(A3_1, O)
m = Segment(B3_1, O)
n = Segment(A3_2, O)
r = Segment(O, B3)
o = false
s = false
u = true
v = false
w = true
u_3 = Vector(O, A3_1)
v_3 = Vector(O, B3_1)
u_2 = Vector(O, A3_2)
v_2 = Vector(O, B3_2)
x_3: v_3 + u_3
x_2: u_2 + v_2
Text2: "x_1 = " + (x(X1)) + "
x_2 = " + x_2 + "
x_3 = " + x_3
t = 0.1 (slider from -5 to 5, step 0.1)
f(x) = t * (x^(3) - p * x - q)
Text7: "=" * x(A)
Text8: "=" * x(B)