Recent questions tagged taylor-series

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GATE IN 2021 | Question: 26
$f\left ( Z \right )=\left ( Z-1 \right )^{-1}-1+\left ( Z-1 \right )-\left ( Z-1 \right )^{2}+ \cdots$ is the series expansion of $\frac{-1}{Z\left ( Z-1 \right )}$ for $\left | Z-1 \right |< 1$ $\frac{1}{Z\left ( Z-1 \right )}$ for $\left | Z-1 \right |< 1$ $\frac{1}{\left ( Z-1 \right )^{2}}$ for $\left | Z-1 \right |< 1$ $\frac{-1}{\left ( Z-1 \right )}$ for $\left | Z-1 \right |< 1$

$f\left ( Z \right )=\left ( Z-1 \right )^{-1}-1+\left ( Z-1 \right )-\left ( Z-1 \right )^{2}+ \cdots$ is the series expansion of$\frac{-1}{Z\left ( Z-1 \right )}$ for $...

Arjun

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Arjun asked Feb 19, 2021

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GATE2016-5
In the neighborhood of $z=1$, the function $f(z)$ has a power series expansion of the form $f(z)$ = $1$ + $(1-z)$ + $(1-z)^2+ \ldots$ Then $f(z)$ is $\frac{1}{z}$ $\frac{-1}{z-2}$ $\frac{z-1}{z+}$ $\frac{1}{2z-1}$

In the neighborhood of $z=1$, the function $f(z)$ has a power series expansion of the form $f(z)$ = $1$ + $(1-z)$ + $(1-z)^2+ \ldots$Then $f(z)$ is$\frac{1}{z}$$\frac{-1}...

Milicevic3306

7.9k points

Milicevic3306 asked Mar 26, 2018

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