In [1]:

%display latex

Ableitungen¶

In [2]:

x = var('x')
f = x^2 * exp(2*x+3)
f

Out[2]:

$\newcommand{\Bold}[1]{\mathbf{#1}}x^{2} e^{\left(2 \, x + 3\right)}$

In [3]:

f.diff(x)

Out[3]:

$\newcommand{\Bold}[1]{\mathbf{#1}}2 \, x^{2} e^{\left(2 \, x + 3\right)} + 2 \, x e^{\left(2 \, x + 3\right)}$

In [4]:

f.diff(x).collect(exp(2*x+3))

Out[4]:

$\newcommand{\Bold}[1]{\mathbf{#1}}2 \, {\left(x^{2} + x\right)} e^{\left(2 \, x + 3\right)}$

In [5]:

f.diff(x, x).collect(exp(2*x+3))

Out[5]:

$\newcommand{\Bold}[1]{\mathbf{#1}}2 \, {\left(2 \, x^{2} + 4 \, x + 1\right)} e^{\left(2 \, x + 3\right)}$

In [6]:

f.diff(x).factor()

Out[6]:

$\newcommand{\Bold}[1]{\mathbf{#1}}2 \, {\left(x + 1\right)} x e^{\left(2 \, x + 3\right)}$

In [7]:

df = f.diff(x)

Integration¶

In [8]:

ff = df.integral(x)
ff

Out[8]:

$\newcommand{\Bold}[1]{\mathbf{#1}}\frac{1}{2} \, {\left(2 \, x^{2} e^{3} - 2 \, x e^{3} + e^{3}\right)} e^{\left(2 \, x\right)} + \frac{1}{2} \, {\left(2 \, x e^{3} - e^{3}\right)} e^{\left(2 \, x\right)}$

In [9]:

ff.factor()

Out[9]:

$\newcommand{\Bold}[1]{\mathbf{#1}}x^{2} e^{\left(2 \, x + 3\right)}$

Matrizen¶

In [10]:

M = Matrix([[1,2,3], [4,5,6], [7,8,9]])
M

Out[10]:

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrr} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{array}\right)$

In [11]:

M.kernel()

Out[11]:

$\newcommand{\Bold}[1]{\mathbf{#1}}\mathrm{RowSpan}_{\Bold{Z}}\left(\begin{array}{rrr} 1 & -2 & 1 \end{array}\right)$

In [12]:

M * vector([1,-2,1])

Out[12]:

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(0,\,0,\,0\right)$

Gewöhnliche Differentialgleichungen¶

In [13]:

phi = function('\phi')
y = phi(x)
y

Out[13]:

$\newcommand{\Bold}[1]{\mathbf{#1}}{\rm \phi}\left(x\right)$

In [14]:

dgl = y.diff(x,x) + 4*y == 0
dgl

Out[14]:

$\newcommand{\Bold}[1]{\mathbf{#1}}4 \, {\rm \phi}\left(x\right) + \frac{\partial^{2}}{(\partial x)^{2}}{\rm \phi}\left(x\right) = 0$

In [15]:

lsg = desolve(dgl, y)
lsg

Out[15]:

$\newcommand{\Bold}[1]{\mathbf{#1}}K_{2} \cos\left(2 \, x\right) + K_{1} \sin\left(2 \, x\right)$

In [16]:

lsg.variables()

Out[16]:

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(K_{1}, K_{2}, x\right)$

In [17]:

K1, K2, dummy = lsg.variables()
y1 = lsg.subs({K1: 1, K2: 0})
y2 = lsg.subs({K1: 0, K2: 1})
y1, y2

Out[17]:

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\sin\left(2 \, x\right), \cos\left(2 \, x\right)\right)$

In [18]:

pl1 = plot(y1, -5, 5)
pl2 = plot(y2, -5, 5, color='red')
pl1 + pl2

Out[18]:

In [19]:

dgl = y.diff(x) == sqrt(x*y)
dgl

Out[19]:

$\newcommand{\Bold}[1]{\mathbf{#1}}\frac{\partial}{\partial x}{\rm \phi}\left(x\right) = \sqrt{x {\rm \phi}\left(x\right)}$

In [20]:

lsg = desolve(dgl, y)
lsg

Out[20]:

$\newcommand{\Bold}[1]{\mathbf{#1}}-\frac{2 \, {\left(\sqrt{x \phi\left(x\right)} x - 3 \, \phi\left(x\right)\right)}}{3 \, \sqrt{\phi\left(x\right)}} = C$

In [21]:

lsg2 = solve(lsg, y)[0]
lsg2

Out[21]:

$\newcommand{\Bold}[1]{\mathbf{#1}}\phi\left(x\right) = \frac{1}{3} \, \sqrt{x \phi\left(x\right)} x + \frac{1}{2} \, C \sqrt{\phi\left(x\right)}$

In [22]:

glg2 = (lsg2^2).expand().canonicalize_radical()
glg2

Out[22]:

$\newcommand{\Bold}[1]{\mathbf{#1}}\phi\left(x\right)^{2} = \frac{1}{3} \, C x^{\frac{3}{2}} \phi\left(x\right) + \frac{1}{36} \, {\left(4 \, x^{3} + 9 \, C^{2}\right)} \phi\left(x\right)$

In [23]:

lsg3 = solve(glg2, y)
lsg3

Out[23]:

$\newcommand{\Bold}[1]{\mathbf{#1}}\left[\phi\left(x\right) = \frac{1}{9} \, x^{3} + \frac{1}{3} \, C x^{\frac{3}{2}} + \frac{1}{4} \, C^{2}, \phi\left(x\right) = 0\right]$

In [24]:

g = lsg3[0].rhs()
g.variables()

Out[24]:

$\newcommand{\Bold}[1]{\mathbf{#1}}\left(C, x\right)$

In [25]:

C = g.variables()[0]

In [26]:

pl1 = plot(g.subs({C:0}), 0, 2, thickness=2)
pl2 = plot(g.subs({C:-1}), 0, 2, color='red', thickness=2)

In [27]:

pl1 + pl2

Out[27]:

In [28]:

pl2a = plot(g.subs({C:-1}), 0, (3/2)^(2/3), color='red', thickness=2, linestyle="--")
pl2b = plot(g.subs({C:-1}), (3/2)^(2/3), 2, color='red', thickness=2)
pl3 = plot(g.subs({C:1}), 0, 1.4, color='green', thickness=2)
pl1 + pl2a + pl2b + pl3

Out[28]: