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edr1c

  • 3 years ago

laurent series expansion?

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  1. edr1c
    • 3 years ago
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    \[f(z)=\frac{ 1 }{ z(z-1)^{2} }\]

  2. edr1c
    • 3 years ago
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    expanding about the point z=1, let u=z-1 then z=u+1 \[f(z)=\frac{ 1 }{ (u+1)u ^{2} }=\frac{ 1 }{ u ^{2} }\frac{ 1 }{ 1-(-u) }\]\[f(z)=\frac{ 1 }{ u ^{2} }\sum_{n=0}^{\infty}(-1)^{n}(u)^{n}\]\[f(z)=\frac{ 1 }{ (z-1) ^{2} }\sum_{n=0}^{\infty}(-1)^{n}(z-1)^{n}\]\[f(z)=\sum_{n=0}^{\infty}\frac{ (-1)^{n} }{ (z-1)^{n-1} }\]my steps correct for z=1?

  3. edr1c
    • 3 years ago
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    (z-1)^n-2 for the denominator*

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