# Laurent Series and Number of Terms

• Oct 27th 2010, 03:27 PM
mathmaniac234
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?

• Oct 28th 2010, 10:22 AM
Opalg
Quote:

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 $\displaystyle \sinh z = z + \frac1{3!}z^3 + \frac1{5!}z^5 + \frac1{7!}z^7 + \ldots = z\bigl(1 + \frac16z^2 + \frac1{120}z^4 + \frac1{5040}z^6 + \ldots\bigr)$. Therefore $\displaystyle \frac1{\sinh z} = \frac1z\bigl(1 + \frac16z^2 + \frac1{120}z^4 + \frac1{5040}z^6 + \ldots\bigr)^{-1}$. Suppose that the inverse of that last expression in parentheses is $\displaystyle a_0+a_2z^2 + a_4z^4+a_6z^6+\ldots$ (assuming, as is fairly obvious from the series in parentheses, that its inverse must also consist only of even powers of z). Then

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

Now compare coefficients of the constant terms on both sides to see that $\displaystyle a_0=1$. Then compare successively the coefficients of $\displaystyle z^2,\ z^4$ and $\displaystyle z^6$ to find $\displaystyle a_2,\ a_4,\ a_6$. After dividing by z, you have the coefficients of $\displaystyle z^{-1},\ z,\ z^3$ and $\displaystyle z^5$ in the Laurent series for $\displaystyle 1/\sinh z$. The numbers get fairly unwieldy, so check your results with the answer given here for csch(x).
• Oct 28th 2010, 11:56 AM
chisigma
The calculation is a little 'more pleasant' if we start considering the function...

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

... where...

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

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

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

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

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

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

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

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

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

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

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

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

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

The (5) permit us to write...

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

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$