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Math Help - taylor series

  1. #1
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    taylor series

    derive taylor series representation for f(z)=1/1-z for |z-i|<squareroot of 2..ie..

    1/1-z=summation n=0 to infinity (z-i)^n/(1-i)^n+1
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  2. #2
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    Quote Originally Posted by durham2 View Post
    derive taylor series representation for f(z)=1/1-z for |z-i|<squareroot of 2..ie..

    1/1-z=summation n=0 to infinity (z-i)^n/(1-i)^n+1
    The easy way is to note that:

    1. \frac{1}{1 - z} = \frac{-1}{z - 1} = \frac{-1}{z - i + i - 1} = \frac{-1}{(z - i) - (1 - i)} = \frac{\frac{-1}{1 - i}}{\left( \frac{z - i}{1 - i} \right) - 1}


     = - \frac{1 + i}{2} \left[ \frac{1}{\left( \frac{z - i}{1 - i} \right) - 1} \right].


    2. \left| \frac{z - i}{1 - i}\right| < 1.


    Now recall the formula for the sum of an infinite geometric series.
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