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Math Help - Maclaurin/ Taylor Series for Cotangent

  1. #1
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    Maclaurin/ Taylor Series for Cotangent

    I was wondering how the Maclaurin series expansion for cotangent is derived using the Bernoulli numbers.

    It is given as:

    \cot(x)=\sum^{\infty}_{n=0}\frac{(-1)^n2^{2n}B_{2n}x^{2n-1}}{(2n)!}

    I cannot normally find the Maclaurin expansion since \cot(0) is undefined.

    Could I derive this cotangent representation from the series representation of sine and cosine? Or tangent? If not, how would I?

    Thank you very much.
    Last edited by Anthonny; April 26th 2010 at 08:23 PM.
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by Anthonny View Post
    I was wondering how the Maclaurin series expansion for cotangent is derived using the Bernoulli numbers.

    It is given as:

    \cot(x)=\sum^{\infty}_{n=0}\frac{(-1)^n2^{2n}B_{2n}x^{2n-1}}{(2n)!}

    I cannot normally find the Maclaurin expansion since \cot(0) is undefined.
    Right, there is no Maclaurin expansion for \cot(x) for just that reason: the above series expansion for \cot(x) is a so called Laurent-series, not a Maclaurin series.

    Could I derive this cotangent representation from the series representation of sine and cosine? Or tangent? If not, how would I?
    For one, you could determine the Maclaurin expansion for x\cdot\cot(x) the usual way.
    For another: much depends on what definition of the Bernoully numbers you are using. The derivation that I know of uses the "definition" \frac{z}{e^z-1}=\sum_{n=0}^\infty \frac{B_n}{n!}z^n of B_n, then first derives the Maclaurin expansion for z\cdot \mathrm{coth}(z), and finally gets the Maclaurin expansion for z\cdot\cot(h) via the identity z\cdot\cot(z)=\mathrm{i}z\cdot\mathrm{coth}(\mathr  m{i}z).
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