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

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

edr1c
 2 years ago
Best ResponseYou've already chosen the best response.0let 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\[=\frac{ 1 }{ z1 }\sum_{n=0}^{\infty}2(\frac{ 1 }{ z1 }) ^{n+1}\]

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

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

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

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

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

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1so 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

edr1c
 2 years ago
Best ResponseYou've already chosen the best response.0for 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

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

edr1c
 2 years ago
Best ResponseYou've already chosen the best response.0so 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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