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Thread: Laurent Series and Number of Terms

  1. #1
    Feb 2010

    Laurent Series and Number of Terms

    Hello there,

    How would I find the laurent series about 0 of 1/sinh(z) up to and including the z^5 term?

    Thank you in advance!
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  2. #2
    MHF Contributor
    Opalg's Avatar
    Aug 2007
    Leeds, UK
    Quote Originally Posted by mathmaniac234 View Post
    How would I find the Laurent series about 0 of 1/sinh(z) up to and including the z^5 term?
    This is not a particularly pleasant calculation. You know (presumably) that the power series for sinh(z) is \sinh z = z + \frac1{3!}z^3 + \frac1{5!}z^5 + \frac1{7!}z^7 + \ldots = z\bigl(1 + \frac16z^2 + \frac1{120}z^4 + \frac1{5040}z^6 + \ldots\bigr). Therefore \frac1{\sinh z} = \frac1z\bigl(1 + \frac16z^2 + \frac1{120}z^4 + \frac1{5040}z^6 + \ldots\bigr)^{-1}. Suppose that the inverse of that last expression in parentheses is a_0+a_2z^2 + a_4z^4+a_6z^6+\ldots (assuming, as is fairly obvious from the series in parentheses, that its inverse must also consist only of even powers of z). Then

    \bigl(1 + \frac16z^2 + \frac1{120}z^4 + \frac1{5040}z^6 + \ldots\bigr)\bigl(a_0+a_2z^2 + a_4z^4+a_6z^6+\ldots\bigr) = 1.

    Now compare coefficients of the constant terms on both sides to see that a_0=1. Then compare successively the coefficients of z^2,\ z^4 and z^6 to find a_2,\ a_4,\ a_6. After dividing by z, you have the coefficients of z^{-1},\ z,\ z^3 and z^5 in the Laurent series for 1/\sinh z. The numbers get fairly unwieldy, so check your results with the answer given here for csch(x).
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  3. #3
    MHF Contributor chisigma's Avatar
    Mar 2009
    near Piacenza (Italy)
    The calculation is a little 'more pleasant' if we start considering the function...

    \displaystyle \varphi (z)= \frac{\sinh z}{z} = \sum_{k=0}^{\infty} b_{k}\ z^{k} (1)

    ... where...

    \displaystyle b_{k}=\left\{\begin{array}{ll} \frac{1}{(2k+1)!} ,\,\, k\ even \\{}\\0 ,\,\, k\ odd\end{array}\right. (2)

    Now we consider \frac{1}{\varphi(z)}, that is analitic in z=0 so that we can write...

    \displaystyle \frac{1}{\varphi (z)} = \frac{z}{\sinh z} = \sum_{n=0}^{\infty} a_{n}\ z^{n} (3)

    The a_{n} can be computed considering that is...

    \displaystyle \sum_{n=0}^{\infty} a_{n}\ z^{n}\ \sum_{k=0}^{\infty} b_{k}\ z^{k} = \sum_{n=0}^{\infty} z^{n}\ \sum_{k=0}^{n} a_{k}\ b_{n-k} = 1 (4)

    ... so that is [details of the calculation are omitted] ...

    \displaystyle a_{0}\ b_{0} = 1 \implies a_{0}=1

    \displaystyle a_{0}\ b_{1} + a_{1}\ b_{0} = 0 \implies a_{1}=0

    \displaystyle a_{0}\ b_{2} + a_{1}\ b_{1} + a_{2}\ b_{0} =0 \implies a_{2}= -\frac{1}{6}

    \displaystyle a_{0}\ b_{3} + a_{1}\ b_{2} + a_{2}\ b_{1} + a_{3}\ b_{0} = 0 \implies a_{3}=0

    \displaystyle a_{0}\ b_{4} + a_{1}\ b_{3} + a_{2}\ b_{2} + a_{3}\ b_{1} + a_{4}\ b_{0} = 0 \implies a_{4} = \frac{7}{360}

    \displaystyle a_{0}\ b_{5} + a_{1}\ b_{4} + a_{2}\ b_{3} + a_{3}\ b_{2} + a_{4}\ b_{1} + a_{5}\ b_{0} = 0 \implies a_{5}=0

    \displaystyle a_{0}\ b_{6} + a_{1}\ b_{5} + a_{2}\ b_{4} + a_{3}\ b_{3} + a_{4}\ b_{2} + a_{5}\ b_{1} + a_{6}\ b_{0} = 0 \implies a_{6} = -\frac{31}{15120} (5)

    The (5) permit us to write...

    \displaystyle \frac{1}{\varphi(z)} = \frac{z}{\sinh z} = 1 -\frac{1}{6}\ z^{2} + \frac{7}{360}\ z^{4} - \frac{31}{15120}\ z^{6} + ... (6)

    ... and from (6) we derive...

    \displaystyle \frac{1}{z\ \varphi(z)} = \frac{1}{\sinh z} = \frac{1}{z} -\frac{1}{6}\ z + \frac{7}{360}\ z^{3} - \frac{31}{15120}\ z^{5} + ... (7)

    Kind regards

    \chi \sigma
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