Laurent expansion of 1 z 2 I need to find a Laurent expansion of 1 z 2 with centre in z 0 = 1 and P ( 1 , 2014 , 2015 )If it was 1 z , I'd rewrite the fraction like this: 1 ( z − 1 ) + 1 But with z 2 I'm not sure how to proceed...My attempt:According to the given recommendation, I've tried to do the expansion of 1 z : 1 z = 1 1 + ( z − 1 ) , a 0 = 1 , q = − ( z − 1 ) = 1 − z ∑ n = 0 ∞ ( 1 − z ) n = 1 + ( 1 − z ) + ( 1 − z ) 2 + . . .And after differentiation: 0 + 1 + 2 ( 1 − z ) + 3 ( 1 − z ) 2 + . . . = ∑ n = 0 ∞ n ( 1 − z ) n − 1

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Laurent expansion of       1          z      2      I need to find a Laurent expansion of       1          z      2       with centre in       z    0    =  1 and   P  (  1  ,  2014  ,  2015  )If it was       1    z  , I&#039;d rewrite the fraction like this:      1          (      z      −      1      )      +      1      But with       z    2   I&#039;m not sure how to proceed...My attempt:According to the given recommendation, I&#039;ve tried to do the expansion of       1    z  :      1    z    =      1          1      +      (      z      −      1      )        ,      a          0        =  1  ,  q  =  −  (  z  −  1  )  =  1  −  z      ∑          n      =      0              ∞        (  1  −  z      )          n        =  1  +  (  1  −  z  )  +  (  1  −  z      )    2    +  .  .  .And after differentiation:  0  +  1  +  2  (  1  −  z  )  +  3  (  1  −  z      )    2    +  .  .  .  =      ∑          n      =      0              ∞        n  (  1  −  z      )          n      −      1

Ashly Kaufman · Accepted Answer

Another way is:            1              z        2              =                              (          1          −                                    1              z                                )                2                    (        z        −        1                  )          2                      =                    (        1        −        z                  )          2                +        2        (        1        −        z                  )          3                +        3        (        1        −        z                  )          4                +        4        (        1        −        z                  )          5                +        …                    (        1        −        z                  )          2                      =  1  +  2  (  1  −  z  )  +  3  (  1  −  z      )    2    +  4  (  1  −  z      )    3    +  …using the product of convergent power series.

Mary Ashley · Accepted Answer

Hint: Find the laurent expansion of   1      /    z and differentiate.

Laurent expansion of 1 z 2 </msup> </mfrac> I need to find

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