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Math Help - complex power series

  1. #1
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    complex power series

    the problem is to find the power series about the origin of the given function:

    \frac{{1 + z}}<br />
{{1 - z}},{\text{ }}\left| z \right| < 1

    this is my work...

    \frac{{1 + z}}<br />
{{1 - z}} = \left( {1 + z} \right)\left( {\frac{1}<br />
{{1 - z}}} \right) = \left( {1 + z} \right)\left( {\sum\limits_{n = 0}^\infty  {z^n } } \right) = \sum\limits_{n = 0}^\infty  {z^n }  + \sum\limits_{n = 0}^\infty  {z^{n + 1} }  = \sum\limits_{n = 0}^\infty  {\left( {z^n  + z^{n + 1} } \right)}  = 1 + 2\sum\limits_{n = 0}^\infty  {z^{n + 1} }


    my question is if this is the form required or is there is some way to get the 1 back inside the sum? or would it be better to leave it as <br />
\sum\limits_{n = 0}^\infty  {\left( {z^n  + z^{n + 1} } \right)} ?
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by xifentoozlerix View Post
    the problem is to find the power series about the origin of the given function:

    \frac{{1 + z}}<br />
{{1 - z}},{\text{ }}\left| z \right| < 1

    this is my work...

    \frac{{1 + z}}<br />
{{1 - z}} = \left( {1 + z} \right)\left( {\frac{1}<br />
{{1 - z}}} \right) = \left( {1 + z} \right)\left( {\sum\limits_{n = 0}^\infty  {z^n } } \right) = \sum\limits_{n = 0}^\infty  {z^n }  + \sum\limits_{n = 0}^\infty  {z^{n + 1} }  = \sum\limits_{n = 0}^\infty  {\left( {z^n  + z^{n + 1} } \right)}  = 1 + 2\sum\limits_{n = 0}^\infty  {z^{n + 1} }


    my question is if this is the form required or is there is some way to get the 1 back inside the sum? or would it be better to leave it as <br />
\sum\limits_{n = 0}^\infty  {\left( {z^n  + z^{n + 1} } \right)} ?
    It's only my opinion. Maybe your teacher views it a different way, maybe you prefer it a certain way...

    It's better not to leave z^{n+1} inside the series, but rather z^n. More beautiful, since the indices are n !


    =\sum_{n=0}^\infty z^n+\sum_{n=0}^\infty z^{n+1}

    Change the power in the second one.

    =\sum_{n=0}^\infty z^n+\sum_{n={\color{red}1}}^\infty z^n

    =1+2 \sum_{n={\color{red}1}}^\infty z^n

    or =\left(2 \sum_{n=0} z^n \right)-1
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