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

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

edr1c Group TitleBest ResponseYou've already chosen the best response.0
oh kay i think i know where i went wrong.
 one year 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\[=\frac{ 1 }{ z1 }\sum_{n=0}^{\infty}2(\frac{ 1 }{ z1 }) ^{n+1}\]
 one year ago

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

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

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

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

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

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

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