Partialbruchzerlegung Dreifache Nullstellen Beispiel Essay

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Beispiele zur Partialbruchzerlegung

Die Art der Partialbruchzerlegung wird im wesentlichen durch die Art der Nullstellen des Polynoms im Nenner bestimmt.

Einfache Nullstelle des Nenners

\(\displaystyle f(x) = \dfrac {x} {(x^2-1)} \) \(\displaystyle = \dfrac {x} {(x + 1)(x - 1)} \).

Der Nennen hat die zwei einfache Nullstellen \(\displaystyle x_0=1\) und \(\displaystyle x_1=-1\). Wir machen den Ansatz

\(\displaystyle \dfrac {A} {(x + 1)} + \dfrac {B}{(x- 1)} = \dfrac{x}{(x^2-1)} \) ,

verwendet, wobei \(\displaystyle A , B\in\R \) für die beiden zu ermittelnden reellen Konstanten stehen. Die Gleichung wird mit dem Nennerpolynom \(\displaystyle (x^2 -1) \) multipliziert:

\(\displaystyle A(x-1)+B(x+1)=x\) bzw. \(\displaystyle Ax- A + Bx + B = x \).

Wir fassen die Terme bezüglich \(\displaystyle x\) zusammen:

\(\displaystyle (A+B-1)x +(-A+B)=0\)

Und ein Koeffizientenvergleich führt zu den beiden Gleichungen

\(\displaystyle B-A = 0 \) und \(\displaystyle A+B = 1 \).

Dabei handelt es sich um ein lineares Gleichungssystem, welches wir durch Einsetzen von \(\displaystyle B=1-A \) in die erste Gleichung lösen:

\(\displaystyle 1-A-A=0\), also \(\displaystyle A=\dfrac 1 2\)

und damit ergibt sich \(\displaystyle A = B= \dfrac{1}{2} \).

Die gesuchte Partialbruchzerlegung ist also:

\(\displaystyle \dfrac {x}{x^2-1} = \dfrac{1}{2} \cdot \dfrac{1}{x + 1} + \dfrac{1}{2}\cdot \dfrac{1}{x-1} \).

Doppelte Nullstelle im Nenner

Wir suchen die Partialbruchzerlegung für \(\displaystyle \dfrac {2\, x-1}{(x-1)^2}\). Der Nenner hat hier eine doppelte Nullstelle \(\displaystyle x_0=1\). Nun führt der folgende Ansatz zum Ziel:

\(\displaystyle \dfrac {2\, x-1}{(x-1)^2} = \dfrac {A}{x-1} + \dfrac {B}{(x-1)^2} \).

Nach Multiplikation der Gleichung mit \(\displaystyle (x-1)^2\) ergibt sich:

\(\displaystyle 2\, x-1 = A (x-1) + B\) und \(\displaystyle 2\, x-1 = Ax-A + B\).

Der Koeffizientenvergleich führt zum folgenden linearen Gleichungssystem:

\(\displaystyle A = 2 \) und \(\displaystyle -A + B = -1 \)

mit der Lösung \(\displaystyle A=2 \) und \(\displaystyle B=1 \).

Damit haben wird die gesuchte Partialbruchzerlegung:

\(\displaystyle \dfrac {2x-1}{(x-1)^2} = \dfrac {2}{x-1} + \dfrac {1}{(x-1)^2}\)

Komplexe Nullstellen im Nenner

Gesucht wird die Partialbruchzerlegung von \(\displaystyle \dfrac{(1+x)^2}{x\cdot(1+x^2)}\).

Die Nullstellen lauten \(\displaystyle x_1=0\), \(\displaystyle x_2=\i\) und \(\displaystyle x_3=-\i\).

Der Ansatz für die Partialbruchzerlegung lautet:

\(\displaystyle \dfrac{(1+x)^2}{x\cdot(1+x^2)}\, = \dfrac{A}{x} + \dfrac{Bx+C}{x^2+1}\)

Wir multiplizieren die Gleichung mit dem Nenner der linken Seite und erhalten:

\(\displaystyle x^2+2x+1 = Ax^2+A+Bx^2+Cx\)

Durch Koeffizientenvergleich sehen wir sofort, dass \(\displaystyle C=2\) und \(\displaystyle A=1\) sowie \(\displaystyle A+B=1\), also \(\displaystyle B=0\).

Damit erhalten wir die Partialbruchzerlegung

\(\displaystyle \dfrac{(1+x)^2}{x\cdot(1+x^2)}\, =\dfrac{1}{x} + \dfrac{2}{1+x^2}\).

Es gibt Dinge, die den meisten Menschen unglaublich erscheinen, die nicht Mathematik studiert haben.

Archimedes

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