Melanie Siebenhofer, Jänner 2018

Integrale über alle Bereiche

Baum 1

In [1]:
var('s0, s1, s2, s3, c')
Out[1]:
(s0, s1, s2, s3, c)

Fall 1: $2 \leq c$

\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*}

In [2]:
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)

\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*}

In [3]:
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:

\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*}

In [4]:
i3 = d.integrate(s3, 0, 1 - s1 - s2).integrate(s2, 1/2 - 1/2*s1, 1 - s1).integrate(s1, 1/2, 1)

Fall 4:

\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*}

In [5]:
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:

\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*}

In [6]:
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:

\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*}

In [7]:
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:

\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*}

In [8]:
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)

\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*}

In [9]:
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:

\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*}

In [10]:
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)

\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*}

In [11]:
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)

\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*}

In [12]:
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:

\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*}

In [13]:
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)

\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*}

In [14]:
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:

\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*}

In [15]:
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)
In [16]:
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

In [17]:
P = (s0, 1 - s0 - s2 - s3, s2, s3)
D = jacobian( P, (s0, s2, s3) )
d = sqrt( (D.transpose() * D).det() ) #verallgemeinerte Funktionaldeterminante

\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*}

In [18]:
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)

\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*}

In [19]:
i16 = d.integrate(s3, 1/2 - 1/2*s2, 1 - s0 - s2).integrate(s2, s0, 1 - 2*s0).integrate(s0, 0, 1/3)
In [20]:
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

In [21]:
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:

\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*}

In [22]:
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)

\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*}

In [23]:
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:

\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*}

In [24]:
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)

\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*}

In [25]:
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:

\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*}

In [26]:
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)

\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*}

In [27]:
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:

\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*}

In [28]:
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)

\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*}

In [29]:
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:

\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*}

In [30]:
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)
In [31]:
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:

\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*}

In [32]:
P = (s0, s1, s2, 1 - s0 - s1 - s2)
D = jacobian( P, (s0, s1, s2) )
d = sqrt( (D.transpose() * D).det() ) #verallgemeinerte Funktionaldeterminante
In [33]:
n = d.integrate(s2, 0, 1 - s1 - s0).integrate(s1, 0, 1 - s0).integrate(s0, 0, 1)
n
Out[33]:
1/3
In [34]:
def F(c):
    return 1/n*(b1(c) + b23(c) + b45(c))
In [35]:
plot(F, (c,-1,3))
Out[35]:

$\rightarrow$ Atome in 1 und 2

In [37]:
F(3) #teste, ob Funktionswert = 1
Out[37]:
1
In [48]:
F1(c) = 1/n * 2*(i20 + i15)
F1
Out[48]:
c |--> 2/3*c^3 - 2*c^2 + 2*c - 2/3
In [49]:
F2(c) = 1/n * 2*(i15 + i17 + i19 + i23)
F2
Out[49]:
c |--> 2/3*c^3 - 2*c^2 + 2*c - 2/3
In [50]:
F3(c) = 1/n * 2*(i15 + i17 + i19 + i21 + i24 + i25)
F3
Out[50]:
c |--> -2/9*c^3 + 2*c^2 - 4*c + 7/3
In [51]:
F4(c) = 1/n * (i1 + i2 + i3 + i4 + i5 + i6 + i7 + i8 + i9 + i10 + i11 + i12 + i13 + i14 + 2*(i16 + i18 + i22) )
F4
Out[51]:
c |--> 1

Verteilungsfunktion

\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

In [52]:
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
In [53]:
plot(f, (c, -1, 3))
Out[53]: