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Thread: 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:

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

    I cannot normally find the Maclaurin expansion since $\displaystyle \cot(0)$ 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; Apr 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:

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

    I cannot normally find the Maclaurin expansion since $\displaystyle \cot(0)$ is undefined.
    Right, there is no Maclaurin expansion for $\displaystyle \cot(x)$ for just that reason: the above series expansion for $\displaystyle \cot(x)$ 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 $\displaystyle x\cdot\cot(x)$ 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" $\displaystyle \frac{z}{e^z-1}=\sum_{n=0}^\infty \frac{B_n}{n!}z^n$ of $\displaystyle B_n$, then first derives the Maclaurin expansion for $\displaystyle z\cdot \mathrm{coth}(z)$, and finally gets the Maclaurin expansion for $\displaystyle z\cdot\cot(h)$ via the identity $\displaystyle z\cdot\cot(z)=\mathrm{i}z\cdot\mathrm{coth}(\mathr m{i}z)$.
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