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Math Help - Power series for tangent function

  1. #1
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    Power series for tangent function

    Derive the power series for the tangent function starting with the defining equation for Bernoulli numbers.

    \tan(z)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1} 2^{2n}(2^{2n}-1) B_{2n}}{(2n)!} z^{2n-1}
    Last edited by shawsend; October 19th 2009 at 05:40 AM.
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  2. #2
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    Thought I should post this in case anyone is interested in the derivation:

    Spoiler:

    We first consider the generating function for Bernoulli numbers ( B_n):

    \frac{z}{e^z-1}=\sum_{n=0}^{\infty}\frac{B_n z^n}{n!}

    and therefore:

    B_n=\lim_{z\to 0}\frac{d^n}{dz^n}\left(\frac{z}{e^z-1}\right)_{z=0}
    We can easily calculate the first three Bernoulli numbers and the remaining numbers can be calculated via contour integration and expressed in terms of the zeta function at the negative integers. We find:

    \{B_n\}=\{1,-1/2,1/6,0,B_4,0,B_6,0,B_8,0,\cdots\}

    where all even Bernoulli numbers are non-zero and B_{2n+1}=0 for n=1,2,3,\cdots. Now, add \frac{z}{2} to both sides of the generating expression:

    \frac{z}{e^z-1}+\frac{z}{2}=\left(\frac{z}{2}\right)+B_0+B_1 z+B_2\frac{z^2}{2}+\cdots)

    but B_1=-1/2 and the remaining odd Bernoulli numbers are zero. Thus:

    \frac{z}{e^z-1}+\frac{z}{2}=B_0+B_2\frac{z^2}{2}+\frac{B_4 z^4}{4!}+\frac{B_6 z^6}{6!}+\cdots)
    We can then write the expression as:
    \frac{z}{e^z-1}+\frac{z}{2}=\sum_{n=0}^{\infty}\frac{B_{2n}z^{2  n}}{(2n)!}

    Now note the left size can be written as:

    \frac{z}{e^z-1}+\frac{z}{2}=\frac{z}{2}\left(\frac{1+e^z}{e^z-1}\right)=\frac{z}{2}\left(\frac{e^{z/2}+e^{-z/2}}{e^{z/2}-e^{-z/2}}\right)
    so therefore:

    \frac{z}{2}\coth(\frac{z}{2})=\sum_{n=0}^{\infty}\  frac{B_{2n}z^{2n}}{(2n)!}

    Letting z=2iw then:

    iw\coth(iw)=\sum_{n=0}^{\infty}\frac{B_{2n}(2iw)^{  2n}}{(2n)!}

    which simplifies to:

    w\cot(w)=\sum_{n=0}^{\infty}\frac{(-1)^n B_{2n} 2^{2n} w^{2n}}{(2n)!}

    or:

    \cot(w)=\sum_{n=0}^{\infty}\frac{(-1)^n B_{2n} 2^{2n}w^{2n-1}}{(2n)!}

    we now use the identity:

    \cot(w)-2\cot(2w)=\tan(w) and therefore:

    \tan(w)=\sum_{n=0}^{\infty}\frac{(-1)^n B_{2n} 2^{2n} w^{2n-1}}{(2n)!}-2\sum_{n=0}^{\infty}\frac{(-)^n B_{2n}(2n)^{2n-1}w^{2n-1}}{(2n)!}

    which then simplifies to:

    \tan(w)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1} 2^{2n}\left(2^{2n}-1\right) B_{2n} w^{2n-1}}{(2n)!},\quad -\pi/2<w<\pi/2
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