Rudolf Rabenstein edited untitled.tex  almost 9 years ago

Commit id: 9f7d02422d03d464ef94391577bd583a211062b4

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\begin{align}  \frac{1}{z(z^2-\frac14)} &= \frac{1}{z(z-\frac12)(z+\frac12)} = \frac{2}{z-\frac12} + \frac{2}{z+\frac12} -\frac{4}{z}  \\  \frac{1}{z^2-\frac14} &= \frac{z}{(z-\frac12)(z+\frac12)} = \frac12\frac{1}{z-\frac12} - \frac12\frac{1}{z+\frac12} \qquad\text{wie vorher} \qquad\text{(wie vorher)}  \end{align}  Einsetzen in~\eqref{eq:20} gibt  \begin{equation}