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Math Help - Expansion of 1/f(z)

  1. #1
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    Expansion of 1/f(z)

    Hello,
    If I have an expansion of a certain function- f(z) = \sum_{n = 0}^\infty a_n z^n, how do I derive the expantion of 1/f(z) ?
    Namely, I am looking for a series b_m such that:

    \sum_{m = -\infty}^\infty b_m z^m = \frac{1}{\sum_{n = 0}^\infty a_n z^n}

    And I'm looking to express it as a function of the given series of  a_n

    Thank you
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  2. #2
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    I don't know if this would work, but since f(z) is a complex function, you should be able to write it in terms of its real and imaginary parts, and from there be able to find \overline{f(z)}.

    You should know that \frac{1}{f(z)} = \frac{\overline{f(z)}}{f(z)\overline{f(z)}} = \frac{\overline{f(z)}}{\left|f(z)\right|^2}
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  3. #3
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by dudyu View Post
    Hello,
    If I have an expansion of a certain function- f(z) = \sum_{n = 0}^\infty a_n z^n, how do I derive the expantion of 1/f(z) ?
    Namely, I am looking for a series b_m such that:

    \sum_{m = -\infty}^\infty b_m z^m = \frac{1}{\sum_{n = 0}^\infty a_n z^n}



    And I'm looking to express it as a function of the given series of  a_n

    Thank you
    If , then is...

    (1)

    ... and the b coefficients are found as follows...



    (2)

    Kind regards

    \chi \sigma
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  4. #4
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    Quote Originally Posted by chisigma View Post
    If , then is...

    (1)

    ... and the b coefficients are found as follows...



    (2)

    Kind regards

    \chi \sigma
    Thank you!
    and if a_0 is equal to zero, do I simply start from a_1 ?
    Namely-
    b_n = - \frac{1}{a_1} \sum_{k = 2}^n a_k b_{n-k}
    ?
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  5. #5
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by dudyu View Post
    Thank you!
    and if a_0 is equal to zero, do I simply start from a_1 ?
    Namely-
    b_n = - \frac{1}{a_1} \sum_{k = 2}^n a_k b_{n-k}
    ?
    If and You can write...

    (1)

    (2)

    ... and then compute the b coefficients imposing that...

    (3)

    Kind regards

    \chi \sigma
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