Laurent Series and Number of Terms

**mathmaniac234** · October 27th 2010, 03:27 PM

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!

**Opalg** · October 28th 2010, 10:22 AM

Originally Posted by mathmaniac234

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).

**chisigma** · October 28th 2010, 11:56 AM

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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