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edr1c
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expand the following f(z) in a Laurent series expansion valid for z1>1
 2 years ago
 2 years ago
edr1c Group Title
expand the following f(z) in a Laurent series expansion valid for z1>1
 2 years ago
 2 years ago

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edr1c Group TitleBest ResponseYou've already chosen the best response.0
\[f(z)=\frac{ z }{ (z1)(2z) }\]
 2 years ago

edr1c Group TitleBest ResponseYou've already chosen the best response.0
oh kay i think i know where i went wrong.
 2 years ago

edr1c Group TitleBest ResponseYou've already chosen the best response.0
let u=z1 so that z=u+1 \[\frac{ z }{ (z1)(2z) }=\frac{ u+1 }{ u(1u) }=\frac{ 1 }{ u }+\frac{ 2 }{ 1u }\]\[=\frac{ 1 }{ u }+2\sum_{n=0}^{\infty}u ^{n}\]\[=\frac{ 1 }{ z1 }+2\sum_{n=0}^{\infty}(z1) ^{n}\] did i make any error or its not simplified yet? the answer provided is\[=\frac{ 1 }{ z1 }\sum_{n=0}^{\infty}2(\frac{ 1 }{ z1 }) ^{n+1}\]
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
start with partial fractions i think
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
you get \[\frac{1}{z1}\frac{2}{z2}\]
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
oops now i got stuck, i am not getting the right thing
 2 years ago

edr1c Group TitleBest ResponseYou've already chosen the best response.0
ouh substitution after the partial fraction works.
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
second part is \(\frac{2}{1+(1z)}\) right?
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
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
 2 years ago

edr1c Group TitleBest ResponseYou've already chosen the best response.0
for the region of validity, like in this case z1>1, after we let u=z1, any series we obtain using that substitution will always be in that region? my english aint good enough lol
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
yes, i believe so
 2 years ago

edr1c Group TitleBest ResponseYou've already chosen the best response.0
oh kay thx alot =)
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
oh wait i am confusing myself. it is \(z1>1\) not less than that is why you need to flip it
 2 years ago

edr1c Group TitleBest ResponseYou've already chosen the best response.0
so if i turn the partial fraction into \[\frac{ 1 }{ u }\frac{ 2 }{ u }(\frac{ 1 }{ 1\frac{ 1 }{ u } })\]should be ok right?
 2 years ago
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