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Math Help - Power series for 1/cosh(z)

  1. #1
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    Power series for 1/cosh(z)

    How would I go about computing the series expansion of 1/cosh(z), or at least the first few terms? I know the series of cosh(z), but trying to manipulate that, or trying to form it using the e^z definition of cosh, always gives me something complicated and looking very far away from a term-by-term definition.

    Any thoughts?
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  2. #2
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    Krizalid's Avatar
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    Don't know if you're working with complex numbers, but in the real case we have \frac{1}{\cosh x}=\frac{2}{e^{x}+e^{-x}}=\frac{2e^{x}}{1+e^{2x}}=2e^{x}\sum\limits_{j=0  }^{\infty }{\left( -e^{2x} \right)^{j}}.
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  3. #3
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    Quote Originally Posted by Krizalid View Post
    Don't know if you're working with complex numbers, but in the real case we have \frac{1}{\cosh x}=\frac{2}{e^{x}+e^{-x}}=\frac{2e^{x}}{1+e^{2x}}=2e^{x}\sum\limits_{j=0  }^{\infty }{\left( -e^{2x} \right)^{j}}.
    Sorry, I should have been clearer. I'm working over the complex numbers.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Krizalid View Post
    Don't know if you're working with complex numbers, but in the real case we have \frac{1}{\cosh x}=\frac{2}{e^{x}+e^{-x}}=\frac{2e^{x}}{1+e^{2x}}=2e^{x}\sum\limits_{j=0  }^{\infty }{\left( -e^{2x} \right)^{j}}.
    I was reprimanded before for giving a series that wasn't a power series. But, I think for all intents and purposes this is much better than some \sum_{n\in\mathbb{N}}a_nx^n where a_n is craaaaaaaaaaazy!
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  5. #5
    MHF Contributor chisigma's Avatar
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    Let suppose that is...

    f(z)= \frac{1}{\cosh z}= a_{0} + a_{1}\cdot z + a_{2}\cdot z^{2} + a_{3}\cdot z^{3} + a_{4}\cdot z^{4} + \dots (1)

    ... and that we want to compute the a_{n} till to n=4. Because is...

    \cosh z = 1 + \frac{z^{2}}{2!} + \frac{z^{4}}{4!} + \dots (2)

    ... from (1) we derive that is...

     (1 + \frac{z^{2}}{2!} + \frac{z^{4}}{4!} + \dots)\cdot (a_{0} + a_{1}\cdot z + a_{2}\cdot z^{2} + a_{3}\cdot z^{3} + a_{4}\cdot z^{4} + \dots)=1 (3)

    ... so that is...

    a_{0}=1

    a_{1}=0

    a_{2}=-\frac{a_{0}}{2!}= -\frac{1}{2}

    a_{3}=0

    a_{4}=-\frac{a_{2}}{2!} - \frac{a_{0}}{4!} = -\frac{13}{24} (4)

    Kind regards

    \chi \sigma
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