UWIS, Mathematik 1, Lösung Serie 10

Thomas Kuster 18. Januar 2004

1 Differentialgleichungen

1.1

x\frac{d}{dx}(1 + t) = x^{2}
(1 + t)dx = x^{2}dt
∫ 1 + t dx = ∫ x^{2} dt
−\frac{1}{x} + c_{1} = ln (1 + t) + c_{2} c = c_{2} − c_{1}
x = \frac{−1}{ln (1 + t) + c}

1.2

ẋ = −2x + e^{3t}

homogene Lösung:

ẋ_{h} = −2x_{h}
x_{h} = c_{1}e^{−2t}

Ansatz C(t)c_{1}e^{−2t} (auch x = c_{1}e^{−2t} + Ae^{3t} möglich):

Ċ(t)c_{1}e^{−2t} + C(t)c_{1}(−2)e^{−2t} = −2C(t)c_{1}e^{−2t} + e^{3t}
Ċ(t)c_{1}e^{−2t} = e^{3t}
Ċ(t) = \frac{e^{3t}}{c_{1}e^{−2t}} c_{2} = \frac{1}{c_{1}}
Ċ(t) = c_{2}e^{3t}e^{2t}
Ċ(t) = c_{2}e^{5t}
C(t) = \frac{c_{2}}{5} e^{5t}

Lösung:

x = \frac{c_{2}}{5} e^{5t}c_{1}e^{−2t} c = c_{1}c_{2}
x = \frac{c}{5}e^{3t}

1.3

ẋ + x sin (t) = sin (t)
ẋ = −x sin (t) + sin (t)
ẋ = sin (t)(1 − x)
\frac{ẋ}{(1 − x)} = sin (t)
\frac{dx}{(1 − x)dt} = sin (t)
\frac{1}{(1 − x)}dx = sin (t)dt
∫ \frac{1}{(1 − x)} dx = ∫ sin (t) dt
ln (1 − t) = − cos (t) + c_{1}
1 − x = e^{− cos (t)+c_{1}}
x = 1 − e^{− cos (t)+c_{1}}
x = 1 − ce^{− cos (t)}

2 Newtonverfahren

Die Tangente der Funktion f an der Stelle x_{0} (nullte Näherung der Nullstelle der Funktion f) wird mit der x-Achse geschnitten, dieser Schnittpunkt ergibt den neuen Punkt x_{1} (erste Näherung der Nullstelle der Funktion f).

f(x_{n}) + f′(x_{n})(x_{n+1} − x_{n}) = 0
f(x_{n}) + f′(x_{n})x_{n+1} − f′(x_{n})x_{n} = 0
f(x_{n}) − f′(x_{n})x_{n} = −f′(x_{n})x_{n+1}
\frac{f′(x_{n})x_{n} − f(x_{n})}{f′(x_{n})} = x_{n+1}

f(x) = tan x − x
f′(x) = 1 + tan^{2}x − 1 = tan^{2}x

\frac{(tan^{2}x_{n}) x_{n} − (tan x_{n} − x_{n})}{tan^{2}x} = x_{n+1}
\frac{(tan^{2}4.5) 4.5 − (tan 4.5 − 4.5)}{tan^{2}4.5} = 4.4936
\frac{(tan^{2}4.49) 4.49 − (tan 4.49 − 4.49)}{tan^{2}4.49} = 4.4934

3 Mac Laurinsche Reihen

3.1

3.1.1

f(t) = (1 + t)^{\frac{1}{2}}
f′(t) = \frac{1}{2}(1 + t)^{− \frac{1}{2}}
f′′(t) = −\frac{1}{4}(1 + t)^{− \frac{3}{2}}
f′′′(t) = \frac{3}{8}(1 + t)^{− \frac{5}{2}}
f^{(4)}(t) = −\frac{15}{16}(1 + t)^{− \frac{7}{2}}
T (t) =(1 + 0)^{\frac{1}{2}} + \frac{1}{2}(1 + 0)^{− \frac{1}{2}} (t − 0) − \frac{1}{4}(1 + 0)^{− \frac{3}{2}} (t − 0)^{2} \frac{1}{2}

+ \frac{3}{8}(1 + 0)^{− \frac{5}{2}} (t − 0)^{3} \frac{1}{6} − \frac{15}{16}(1 + 0)^{− \frac{7}{2}} (t − 0)^{4} \frac{1}{24}

T (t) =1 + \frac{1}{2}t − \frac{1}{8}t^{2} + \frac{1}{16}t^{3} − \frac{5}{128}t^{4}

3.1.2 eine Ableitung mehr als in 3.1.1 und g^{(n)}(t) = f^{(n+1)}(t)

g(t) = (1 + t)^{− \frac{1}{2}}
g′(t) = −\frac{1}{2}(1 + t)^{− \frac{3}{2}}
g′′(t) = \frac{3}{4}(1 + t)^{− \frac{5}{2}}
g′′′(t) = −\frac{15}{8}(1 + t)^{− \frac{7}{2}}
g^{4}(t) = \frac{105}{16}(1 + t)^{− \frac{9}{2}}
T (t) = 1 − \frac{1}{2}t + \frac{3}{8}t^{2} − \frac{5}{16}t^{3} + \frac{35}{128}t^{4}

UWIS, Mathematik 1, Lösung Serie 10 4 3.1.3

f(t) = ln (\frac{1 + t}{1 − t}) = ln (1 + t) − ln (1 − t)
f′(t) = \frac{1}{1 + t} + \frac{1}{1 − t}
f′′(t) = −\frac{1}{(1 + t)^{2}} + \frac{1}{(1 − t)^{2}}
f′′′(t) = \frac{2}{(1 + t)^{3}} + \frac{2}{(1 − t)^{3}}
f^{(4)}(t) = −\frac{6}{(1 + t)^{4}} + \frac{6}{(1 − t)^{4}}
f^{(5)}(t) = \frac{24}{(1 + t)^{5}} + \frac{24}{(1 − t)^{5}}
T (t) = 2t + \frac{2}{3}t^{3} + \frac{2}{5}t^{5}

3.1.4

f(t) = tan (t)
f′(t) = 1 + tan^{2} (t)
f′′(t) = 2 tan (t) (1 + tan^{2} (t)) = 2 tan (t) + 2 tan^{3} (t)
f′′′(t) = 2 + 2 tan^{2} (t) + 6 tan^{2} (t) (1 + tan^{2} (t))
= 2 + 8 tan^{2} (t) + 6 tan^{4} (t)
f^{(4)}(t) = 16 tan (t) (1 + tan^{2} (t)) + 24 tan^{3} (t) (1 + tan^{2} (t))
= 16 tan (t) + 40 tan^{3} (t) + 24 tan^{5} (t)
f^{(5)}(t) = 16 (1 + tan^{2} (t)) + 120 tan^{2} (t) (1 + tan^{2} (t)) +
120 tan^{4} (t) (1 + tan^{2} (t))
= 16 + 136 tan^{2} (t) + 240 tan^{4} (t) + 120 tan^{6} (t)

3.2

\frac{1 + t}{1 − t} = 2
1 + t = 2 − 2t
3t = 1
t = \frac{1}{3}

T (\frac{1}{3}) = 2\frac{1}{3} + \frac{2}{3} (\frac{1}{3})^{3} + \frac{2}{5} (\frac{1}{3})^{5} = \frac{2}{3} + \frac{2}{81} + 21215
T (\frac{1}{3}) = \frac{842}{1215} = 0.69300

Zum Vergleich ln(2) = 0.69315 der Fehler ist sicher kleiner t^{7} < 0.00046.

4 Taylorpolynome verschiedenen Grades

> series(sin(t),t=0,1);
                         O(t)
> series(sin(t),t=0,3);
                              3
                       t + O(t )
> series(sin(t),t=0,5);
                           3      5
                  t - 1/6 t  + O(t )
> series(sin(t),t=0,7);
                      3          5      7
             t - 1/6 t  + 1/120 t  + O(t )

allgemein:

sin t = ∑_{k=0}^{∞} (−1)^{k} \frac{t^{2k+1}}{(2k + 1)!}

5 komplexe Zahlen

5.1

5.2

5.2.1

z = \frac{5 − 5i}{−1 + 2i} = \frac{(5 − 5i)(−1 − 2i)}{(−1 + 2i)(−1 − 2i)} = \frac{−5 − 10i + 5i + 10i^{2}}{1 − 4i^{2}} = \frac{−15 − 5i}{5}
z = −3 − i

5.2.2

z = \frac{1}{(1 + 2i)^{3}} = \frac{(1 − 2i)^{3}}{((1 + 2i)(1 − 2i))^{3}} = \frac{(−3 − 4i)(1 − 2i)}{(1 − 4i^{2})^{3}} = \frac{−11 + 2i}{5^{3}}
z = \frac{−11}{125} + \frac{2}{125}i