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Kernel: SageMath (stable)
Melanie Siebenhofer, Jänner 2018

Integrale über alle Bereiche

Baum 1

var('s0, s1, s2, s3, c')
(s0, s1, s2, s3, c)

Fall 1: 2≤c2 \leq c

2≤cs0=1−s1−s2−s30≤s1≤131−2⋅s1≤s2≤1−s10≤s3≤1−s1−s2\begin{align*} 2 \leq c \quad &\\ s_0 = & 1 - s_1 - s_2 - s_3\\ 0 \leq s_1 \leq & \frac{1}{3}\\ 1 - 2 \cdot s_1 \leq s_2 \leq & 1 - s_1\\ 0 \leq s_3 \leq & 1 - s_1 - s_2\\ \end{align*}
P = (1 - s1 - s2 - s3, s1, s2, s3)
D = jacobian( P, (s1, s2, s3) )
d = sqrt( (D.transpose() * D).det() ) #verallgemeinerte Funktionaldeterminante
i1 = d.integrate(s3, 0, 1 - s1 - s2).integrate(s2, 1 - 2*s1, 1 - s1).integrate(s1, 0, 1/3)
2≤cs0=1−s1−s2−s313≤s1≤1212−12⋅s1≤s2≤1−s10≤s3≤1−s1−s2\begin{align*} 2 \leq c \quad &\\ s_0 = & 1 - s_1 - s_2 - s_3\\ \frac{1}{3} \leq s_1 \leq & \frac{1}{2}\\ \frac{1}{2} - \frac{1}{2} \cdot s_1 \leq s_2 \leq & 1 - s_1\\ 0 \leq s_3 \leq & 1 - s_1 - s_2\\ \end{align*}
i2 = d.integrate(s3, 0, 1 - s1 - s2).integrate(s2, 1/2 - 1/2*s1, 1 - s1).integrate(s1, 1/3, 1/2)

Fall 2:

2≤cs0=1−s1−s2−s312≤s1≤112−12s1≤s2≤1−s10≤s3≤1−s1−s2\begin{align*} 2 \leq c \quad &\\ s_0 = & 1 - s_1 - s_2 - s_3\\ \frac{1}{2} \leq s_1 \leq & 1 \\ \frac{1}{2} - \frac{1}{2}s_1 \leq s_2 \leq & 1 - s_1\\ 0 \leq s_3 \leq & 1 - s_1 - s_2\\ \end{align*}
i3 = d.integrate(s3, 0, 1 - s1 - s2).integrate(s2, 1/2 - 1/2*s1, 1 - s1).integrate(s1, 1/2, 1)

Fall 4:

2≤cs0=1−s1−s2−s30≤s1≤1312−12s1≤s2≤1−2s10≤s3≤12−12s2\begin{align*} 2 \leq c \quad &\\ s_0 = & 1 - s_1 - s_2 - s_3\\ 0 \leq s_1 \leq & \frac{1}{3}\\ \frac{1}{2} - \frac{1}{2}s_1 \leq s_2 \leq & 1 - 2 s_1\\ 0 \leq s_3 \leq & \frac{1}{2} - \frac{1}{2} s_2 \\ \end{align*}
i4 = d.integrate(s3, 0, 1/2 - 1/2*s2).integrate(s2, 1/2 - 1/2*s1, 1 - 2*s1).integrate(s1, 0, 1/3)

Fall 5:

2≤cs0=1−s1−s2−s313≤s1≤121−2s1≤s2≤12−12s112−12s1−s2≤s3≤1−s1−s2\begin{align*} 2 \leq c \quad &\\ s_0 = & 1 - s_1 - s_2 - s_3\\ \frac{1}{3} \leq s_1 \leq & \frac{1}{2}\\ 1 - 2s_1 \leq s_2 \leq & \frac{1}{2} - \frac{1}{2} s_1 \\ \frac{1}{2} - \frac{1}{2}s_1 - s_2 \leq s_3 \leq & 1 - s_1 - s_2\\ \end{align*}
i5 = d.integrate(s3, 1/2 - 1/2*s1 - s2, 1 - s1 - s2).integrate(s2, 1 - 2*s1, 1/2 - 1/2*s1).integrate(s1, 1/3, 1/2)

Fall 6:

2≤cs0=1−s1−s2−s312≤s1≤10≤s2≤12−12s112−12s1−s2≤s3≤1−s1−s2\begin{align*} 2 \leq c \quad &\\ s_0 = & 1 - s_1 - s_2 - s_3\\ \frac{1}{2} \leq s_1 \leq & 1\\ 0 \leq s_2 \leq & \frac{1}{2} - \frac{1}{2} s_1 \\ \frac{1}{2} - \frac{1}{2}s_1 - s_2 \leq s_3 \leq & 1 - s_1 - s_2\\ \end{align*}
i6 = d.integrate(s3, 1/2 - 1/2*s1 - s2, 1 - s1 - s2).integrate(s2, 0, 1/2 - 1/2*s1).integrate(s1, 1/2, 1)

Fall 7:

2≤cs0=1−s1−s2−s313≤s1≤120≤s2≤13−23s112−12s1−s2≤s3≤s1+s2\begin{align*} 2 \leq c \quad &\\ s_0 = & 1 - s_1 - s_2 - s_3\\ \frac{1}{3} \leq s_1 \leq & \frac{1}{2}\\ 0 \leq s_2 \leq & \frac{1}{3} - \frac{2}{3} s_1 \quad\\ \frac{1}{2} - \frac{1}{2}s_1 - s_2 \leq s_3 \leq & s_1 + s_2\\ \end{align*}
i7 = d.integrate(s3, 1/2 - 1/2*s1 - s2, s1 + s2).integrate(s2, 0, 1/3 - 2/3*s1).integrate(s1, 1/3, 1/2)
2≤cs0=1−s1−s2−s315≤s1≤1312−32s1≤s2≤13−23s112−12s1−s2≤s3≤s1+s2\begin{align*} 2 \leq c \quad &\\ s_0 = & 1 - s_1 - s_2 - s_3\\ \frac{1}{5} \leq s_1 \leq & \frac{1}{3}\\ \frac{1}{2} - \frac{3}{2} s_1 \leq s_2 \leq & \frac{1}{3} - \frac{2}{3} s_1 \quad\\ \frac{1}{2} - \frac{1}{2}s_1 - s_2 \leq s_3 \leq & s_1 + s_2\\ \end{align*}
i8 = d.integrate(s3, 1/2 - 1/2*s1 - s2, s1 + s2).integrate(s2, 1/2 - 3/2*s1, 1/3 - 2/3*s1).integrate(s1, 1/5, 1/3)

Fall 8:

2≤cs0=1−s1−s2−s313≤s1≤1213−23s1≤s2≤1−2s112−12s1−s2≤s3≤12−12s2\begin{align*} 2 \leq c \quad &\\ s_0 = & 1 - s_1 - s_2 - s_3\\ \frac{1}{3} \leq s_1 \leq & \frac{1}{2}\\ \frac{1}{3} - \frac{2}{3} s_1 \leq s_2 \leq & 1 - 2 s_1\\ \frac{1}{2} - \frac{1}{2} s_1 - s_2 \leq s_3 \leq & \frac{1}{2} - \frac{1}{2} s_2\\ \end{align*}
i9 = d.integrate(s3, 1/2 - 1/2*s1 - s2, 1/2 - 1/2*s2).integrate(s2, 1/3 - 2/3*s1, 1 - 2*s1).integrate(s1, 1/3, 1/2)
2≤cs0=1−s1−s2−s30≤s1≤1512−32s1≤s2≤12−12s112−12s1−s2≤s3≤12−12s2\begin{align*} 2 \leq c \quad &\\ s_0 = & 1 - s_1 - s_2 - s_3\\ 0 \leq s_1 \leq & \frac{1}{5}\\ \frac{1}{2} - \frac{3}{2} s_1 \leq s_2 \leq & \frac{1}{2} - \frac{1}{2} s_1\\ \frac{1}{2} - \frac{1}{2} s_1 - s_2 \leq s_3 \leq & \frac{1}{2} - \frac{1}{2} s_2\\ \end{align*}
i10 = d.integrate(s3, 1/2 - 1/2*s1 - s2, 1/2 - 1/2*s2).integrate(s2, 1/2 - 3/2*s1, 1/2 - 1/2*s1).integrate(s1, 0, 1/5)
2≤cs0=1−s1−s2−s315≤s1≤1313−23s1≤s2≤12−12s112−12s1−s2≤s3≤12−12s2\begin{align*} 2 \leq c \quad &\\ s_0 = & 1 - s_1 - s_2 - s_3\\ \frac{1}{5} \leq s_1 \leq & \frac{1}{3}\\ \frac{1}{3} - \frac{2}{3} s_1 \leq s_2 \leq & \frac{1}{2} - \frac{1}{2} s_1\\ \frac{1}{2} - \frac{1}{2} s_1 - s_2 \leq s_3 \leq & \frac{1}{2} - \frac{1}{2} s_2\\ \end{align*}
i11 = d.integrate(s3, 1/2 - 1/2*s1 - s2, 1/2 - 1/2*s2).integrate(s2, 1/3 - 2/3*s1, 1/2 - 1/2*s1).integrate(s1, 1/5, 1/3)

Fall 11:

2≤cs0=1−s1−s2−s315≤s1≤1313−s1≤s2≤12−32s11−2s1−2s2≤s3≤s1+s2\begin{align*} 2 \leq c \quad &\\ s_0 = & 1 - s_1 - s_2 - s_3\\ \frac{1}{5} \leq s_1 \leq & \frac{1}{3}\\ \frac{1}{3} - s_1 \leq s_2 \leq & \frac{1}{2} - \frac{3}{2} s_1\\ 1 - 2s_1 - 2 s_2 \leq s_3 \leq & s_1 + s_2\\ \end{align*}
i12 = d.integrate(s3, 1 - 2*s1 - 2*s2, s1 + s2).integrate(s2, 1/3 - s1, 1/2 - 3/2*s1).integrate(s1, 1/5, 1/3)
2≤cs0=1−s1−s2−s30≤s1≤1513−s1≤s2≤13−23s11−2s1−2s2≤s3≤s1+s2\begin{align*} 2 \leq c \quad &\\ s_0 = & 1 - s_1 - s_2 - s_3\\ 0 \leq s_1 \leq & \frac{1}{5}\\ \frac{1}{3} - s_1 \leq s_2 \leq & \frac{1}{3} - \frac{2}{3} s_1\\ 1 - 2s_1 - 2 s_2 \leq s_3 \leq & s_1 + s_2\\ \end{align*}
i13 = d.integrate(s3, 1 - 2*s1 - 2*s2, s1 + s2).integrate(s2, 1/3 - s1, 1/3 - 2/3*s1).integrate(s1, 0, 1/5)

Fall 12:

2≤cs0=1−s1−s2−s30≤s1≤1513−23s1≤s2≤12−32s11−2s1−2s2≤s3≤12−12s2\begin{align*} 2 \leq c \quad &\\ s_0 = & 1 - s_1 - s_2 - s_3\\ 0 \leq s_1 \leq & \frac{1}{5} \\ \frac{1}{3} - \frac{2}{3} s_1 \leq s_2 \leq & \frac{1}{2} - \frac{3}{2} s_1\\ 1 - 2s_1 - 2 s_2 \leq s_3 \leq & \frac{1}{2} - \frac{1}{2} s_2\\ \end{align*}
i14 = d.integrate(s3, 1 - 2*s1 - 2*s2, 1/2 - 1/2*s2).integrate(s2, 1/3 - 2/3*s1, 1/2 - 3/2*s1).integrate(s1, 0, 1/5)

def b1(c):
    if c >= 2:
        return i1 + i2 + i3 + i4 + i5 + i6 + i7 + i8 + i9 + i10 + i11 + i12 + i13 + i14
    else:
        return 0

Baum 2 + 3

P = (s0, 1 - s0 - s2 - s3, s2, s3)
D = jacobian( P, (s0, s2, s3) )
d = sqrt( (D.transpose() * D).det() ) #verallgemeinerte Funktionaldeterminante
s1=1−s0−s2−s31≤c≤20≤s0≤13c−13s0≤s2≤c−2s0−132−12c−12s2≤s3≤1−s0−s2\begin{align*} s_1 = & 1 - s_0 - s_2 - s_3 \\ 1 \leq c \leq & 2\\ 0 \leq s_0 \leq & \frac{1}{3}c - \frac{1}{3} \\ s_0 \leq s_2 \leq& c - 2s_0 -1 \\ \frac{3}{2} - \frac{1}{2}c - \frac{1}{2}s_2 \leq s_3 \leq &1- s_0 - s_2 \\ \end{align*}
i15 = d.integrate(s3, 3/2 - 1/2*c - 1/2*s2, 1 - s0 - s2).integrate(s2, s0, c - 2*s0 - 1).integrate(s0, 0, 1/3*c - 1/3)
s1=1−s0−s2−s32≤c0≤s0≤13s0≤s2≤1−2⋅s012−12⋅s2≤s3≤1−s0−s2\begin{align*} s_1 = & 1 - s_0 - s_2 - s_3 \\ 2 \leq c \quad & \\ 0 \leq s_0 \leq & \frac{1}{3} \\ s_0 \leq s_2 \leq& 1 - 2 \cdot s_0 \\ \frac{1}{2} - \frac{1}{2} \cdot s_2 \leq s_3 \leq &1- s_0 - s_2 \\ \end{align*}
i16 = d.integrate(s3, 1/2 - 1/2*s2, 1 - s0 - s2).integrate(s2, s0, 1 - 2*s0).integrate(s0, 0, 1/3)

def b23(c):
    if 1 <= c and c < 2:
        return 2 * i5(c=c)
    if c >= 2:
        return 2 * i16
    else:
        return 0

Baum 4 + 5

P = (s0, s1, 1 - s0 - s1 - s3, s3)
D = jacobian( P, (s0, s1, s3) )
d = sqrt( (D.transpose() * D).det() ) #verallgemeinerte Funktionaldeterminante

Fall 1.1:

s2=1−s0−s1−s343≤c≤20≤s1≤12c−23s1≤s3≤1332−12c−12s3≤s0≤1−s1−s3\begin{align*} s_2 = 1 - s_0 - s_1 - s_3 \\ \frac{4}{3} \leq c \leq & 2\\ 0 \leq s_1 \leq & \frac{1}{2}c - \frac{2}{3} \\ s_1 \leq s_3 \leq & \frac{1}{3}\\ \frac{3}{2} - \frac{1}{2}c - \frac{1}{2}s_3\leq s_0 \leq & 1 - s_1 - s_3 \\ \end{align*}
i17 = d.integrate(s0, 3/2 - 1/2*c - 1/2*s3, 1 - s1 - s3).integrate(s3, s1, 1/3).integrate(s1, 0, 1/2*c - 2/3)
s2=1−s0−s1−s32≤c0≤s1≤13s1≤s3≤1312−12⋅s3≤s0≤1−s1−s3\begin{align*} s_2 = 1 - s_0 - s_1 - s_3 \\ 2 \leq c \quad &\\ 0 \leq s_1 \leq & \frac{1}{3} \\ s_1 \leq s_3 \leq & \frac{1}{3}\\ \frac{1}{2} - \frac{1}{2} \cdot s_3 \leq s_0 \leq & 1 - s_1 - s_3 \\ \end{align*}
i18 = d.integrate(s0, 1/2 - 1/2*s3, 1 - s1 - s3).integrate(s3, s1, 1/3).integrate(s1, 0, 1/3)

Fall 1.2:

s2=1−s0−s1−s343≤c≤212c−23≤s1≤13c−13s1≤s3≤c−2s1−132−12c−12s3≤s0≤1−s1−s3\begin{align*} s_2 = 1 - s_0 - s_1 - s_3 \\ \frac{4}{3} \leq c \leq & 2\\ \frac{1}{2}c - \frac{2}{3} \leq s_1 \leq & \frac{1}{3}c - \frac{1}{3} \\ s_1 \leq s_3 \leq & c - 2s_1 - 1\\ \frac{3}{2} - \frac{1}{2}c - \frac{1}{2}s_3 \leq s_0 \leq & 1 - s_1 - s_3 \\ \end{align*}
i19 = d.integrate(s0, 3/2 - 1/2*c - 1/2*s3, 1 - s1 - s3).integrate(s3, s1, c - 2*s1 - 1).integrate(s1, 1/2*c - 2/3, 1/3*c - 1/3)
s2=1−s0−s1−s31≤c≤430≤s1≤13c−13s1≤s3≤c−2s1−132−12c−12s3≤s0≤1−s1−s3\begin{align*} s_2 = 1 - s_0 - s_1 - s_3 \\ 1 \leq c \leq & \frac{4}{3}\\ 0 \leq s_1 \leq & \frac{1}{3}c - \frac{1}{3} \\ s_1 \leq s_3 \leq & c - 2s_1 - 1\\ \frac{3}{2} - \frac{1}{2}c - \frac{1}{2}s_3 \leq s_0 \leq & 1 - s_1 - s_3 \\ \end{align*}
i20 = d.integrate(s0, 3/2 - 1/2*c - 1/2*s3, 1 - s1 - s3).integrate(s3, s1, c - 2*s1 - 1).integrate(s1, 0, 1/3*c - 1/3)

Fall 2.1:

s2=1−s0−s1−s332≤c≤20≤s1≤23c−11−13c≤s3≤12−12s1s3≤s0≤1−s1−s3\begin{align*} s_2 = 1 - s_0 - s_1 - s_3 \\ \frac{3}{2} \leq c \leq & 2\\ 0 \leq s_1 \leq & \frac{2}{3}c - 1 \\ 1 - \frac{1}{3}c \leq s_3 \leq &\frac{1}{2} - \frac{1}{2}s_1\\ s_3 \leq s_0 \leq & 1 - s_1 - s_3 \\ \end{align*}
i21 = d.integrate(s0, s3, 1 - s1 - s3).integrate(s3, 1 - 1/3*c, 1/2 - 1/2*s1).integrate(s1, 0, 2/3*c - 1)
s2=1−s0−s1−s32≤c0≤s1≤1313≤s3≤12−12s1s3≤s0≤1−s1−s3\begin{align*} s_2 = 1 - s_0 - s_1 - s_3 \\ 2 \leq c \quad & \\ 0 \leq s_1 \leq & \frac{1}{3} \\ \frac{1}{3} \leq s_3 \leq &\frac{1}{2} - \frac{1}{2}s_1\\ s_3 \leq s_0 \leq & 1 - s_1 - s_3 \\ \end{align*}
i22 = d.integrate(s0, s3, 1 - s1 - s3).integrate(s3, 1/3, 1/2 - 1/2*s1).integrate(s1, 0, 1/3)

Fall 2.2.1:

s2=1−s0−s1−s343≤c≤320≤s1≤12c−2313≤s3≤c−2s1−132−12c−12s3≤s0≤1−s1−s3\begin{align*} s_2 = 1 - s_0 - s_1 - s_3 \\ \frac{4}{3} \leq c \leq & \frac{3}{2}\\ 0 \leq s_1 \leq & \frac{1}{2}c - \frac{2}{3} \\ \frac{1}{3} \leq s_3 \leq & c - 2s_1 - 1 \\ \frac{3}{2} - \frac{1}{2}c - \frac{1}{2}s_3 \leq s_0 \leq & 1 - s_1 - s_3 \\ \end{align*}
i23 = d.integrate(s0, 3/2 - 1/2*c - 1/2*s3, 1 - s1 - s3).integrate(s3, 1/3, c - 2*s1 - 1).integrate(s1, 0, 1/2*c - 2/3)
s2=1−s0−s1−s332≤c≤223c−1≤s1≤12c−2313≤s3≤c−2s1−132−12c−12s3≤s0≤1−s1−s3\begin{align*} s_2 = 1 - s_0 - s_1 - s_3 \\ \frac{3}{2} \leq c \leq & 2\\ \frac{2}{3}c - 1 \leq s_1 \leq & \frac{1}{2}c - \frac{2}{3} \\ \frac{1}{3} \leq s_3 \leq & c - 2s_1 - 1 \\ \frac{3}{2} - \frac{1}{2}c - \frac{1}{2}s_3 \leq s_0 \leq & 1 - s_1 - s_3 \\ \end{align*}
i24 = d.integrate(s0, 3/2 - 1/2*c - 1/2*s3, 1 - s1 - s3).integrate(s3, 1/3, c - 2*s1 - 1).integrate(s1, 2/3*c - 1, 1/2*c - 2/3)

Fall 2.2.2:

s2=1−s0−s1−s332≤c≤20≤s1≤23c−113≤s3≤1−13c32−12c−12s3≤s0≤1−s1−s3\begin{align*} s_2 = 1 - s_0 - s_1 - s_3 \\ \frac{3}{2} \leq c \leq & 2 \\ 0 \leq s_1 \leq & \frac{2}{3}c - 1 \\ \frac{1}{3} \leq s_3 \leq & 1 - \frac{1}{3}c \\ \frac{3}{2} - \frac{1}{2}c - \frac{1}{2}s_3 \leq s_0 \leq & 1 - s_1 - s_3 \\ \end{align*}
i25 = d.integrate(s0, 3/2 - 1/2*c - 1/2*s3, 1 - s1 - s3).integrate(s3, 1/3, 1 - 1/3*c).integrate(s1, 0, 2/3*c - 1)

def b45(c):
    if 1 <= c and c < 4/3:
        return 2*i20(c=c)
    if 4/3 <= c and c < 3/2:
        return 2*(i17(c=c) + i19(c=c) + i23(c=c))
    if 3/2 <= c and c < 2:
        return 2*(i17(c=c) + i19(c=c) + i21(c=c) + i24(c=c) + i25(c=c))
    if c >= 2:
        return 2*(i18 + i22)
    else:
        return 0

Normierung

um Verteilungsfunktion zu erhalten

Berechne dazu das Integral über den gesamten Bereich:

s0+s1+s2+s3=10≤s0≤10≤s1≤10≤s2≤10≤s3≤1\begin{align*} s_0 + s_1 + s_2 + s_3 = & 1 \\ 0 \leq s_0 \leq & 1 \\ 0 \leq s_1 \leq & 1 \\ 0 \leq s_2 \leq & 1 \\ 0 \leq s_3 \leq & 1 \\ \end{align*}
P = (s0, s1, s2, 1 - s0 - s1 - s2)
D = jacobian( P, (s0, s1, s2) )
d = sqrt( (D.transpose() * D).det() ) #verallgemeinerte Funktionaldeterminante

n = d.integrate(s2, 0, 1 - s1 - s0).integrate(s1, 0, 1 - s0).integrate(s0, 0, 1)
n
1/3
def F(c):
    return 1/n*(b1(c) + b23(c) + b45(c))

plot(F, (c,-1,3))
Image in a Jupyter notebook

→\rightarrow Atome in 1 und 2

F(3) #teste, ob Funktionswert = 1
1
F1(c) = 1/n * 2*(i20 + i15)
F1
c |--> 2/3*c^3 - 2*c^2 + 2*c - 2/3
F2(c) = 1/n * 2*(i15 + i17 + i19 + i23)
F2
c |--> 2/3*c^3 - 2*c^2 + 2*c - 2/3
F3(c) = 1/n * 2*(i15 + i17 + i19 + i21 + i24 + i25)
F3
c |--> -2/9*c^3 + 2*c^2 - 4*c + 7/3
F4(c) = 1/n * (i1 + i2 + i3 + i4 + i5 + i6 + i7 + i8 + i9 + i10 + i11 + i12 + i13 + i14 + 2*(i16 + i18 + i22) )
F4
c |--> 1

Verteilungsfunktion

F(x)={0fu¨r x<1,23x3−2x2+2x−23fu¨r 1≤x<32,−29x3+2x2−4x+73fu¨r 32≤x<2,1sonst\begin{equation} F(x) = \begin{cases} 0 & \text{für } x < 1, \\ \frac{2}{3} x^3 - 2x^2 + 2x - \frac{2}{3} & \text{für } 1 \leq x < \frac{3}{2}, \\ -\frac{2}{9}x^3 + 2x^2 - 4x + \frac{7}{3} & \text{für } \frac{3}{2} \leq x < 2, \\ 1 & \text{sonst} \end{cases} \end{equation}

Dichtefunktion

def f(x):
    if 1 <= x and x < 4/3:
        return (diff(F1(c)))(c=x)
    if 4/3 <= x and x < 3/2:
        return (diff( F2(c)))(c=x)
    if 3/2 <= x and x < 2:
        return (diff(F3(c)))(c=x)
    if c >= x:
        return (diff(F4(c)))(c=x)
    else:
        return 0

plot(f, (c, -1, 3))
Image in a Jupyter notebook