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edr1c

  • 2 years ago

expand the following f(z) in a Laurent series expansion valid for |z-1|>1

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

  2. edr1c
    • 2 years ago
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    oh kay i think i know where i went wrong.

  3. edr1c
    • 2 years ago
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    let u=z-1 so that z=u+1 \[\frac{ z }{ (z-1)(2-z) }=\frac{ u+1 }{ u(1-u) }=\frac{ 1 }{ u }+\frac{ 2 }{ 1-u }\]\[=\frac{ 1 }{ u }+2\sum_{n=0}^{\infty}u ^{n}\]\[=\frac{ 1 }{ z-1 }+2\sum_{n=0}^{\infty}(z-1) ^{n}\] did i make any error or its not simplified yet? the answer provided is\[=\frac{ 1 }{ z-1 }-\sum_{n=0}^{\infty}2(\frac{ 1 }{ z-1 }) ^{n+1}\]

  4. satellite73
    • 2 years ago
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    start with partial fractions i think

  5. satellite73
    • 2 years ago
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    you get \[\frac{1}{z-1}-\frac{2}{z-2}\]

  6. edr1c
    • 2 years ago
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    yep

  7. satellite73
    • 2 years ago
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    oops now i got stuck, i am not getting the right thing

  8. edr1c
    • 2 years ago
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    ouh substitution after the partial fraction works.

  9. satellite73
    • 2 years ago
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    second part is \(\frac{2}{1+(1-z)}\) right?

  10. edr1c
    • 2 years ago
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    yea

  11. satellite73
    • 2 years ago
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    so i am confused as to why this doesn't alternate, but if you got it, then i guess i have something wrong probably in my mind

  12. edr1c
    • 2 years ago
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    for the region of validity, like in this case |z-1|>1, after we let u=z-1, any series we obtain using that substitution will always be in that region? my english aint good enough lol

  13. satellite73
    • 2 years ago
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    yes, i believe so

  14. edr1c
    • 2 years ago
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    oh kay thx alot =)

  15. satellite73
    • 2 years ago
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    oh wait i am confusing myself. it is \(|z-1|>1\) not less than that is why you need to flip it

  16. edr1c
    • 2 years ago
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    so if i turn the partial fraction into \[\frac{ 1 }{ u }-\frac{ 2 }{ u }(\frac{ 1 }{ 1-\frac{ 1 }{ u } })\]should be ok right?

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