# Laurent Series of rational function about $z=0$

Find the Laurent expansion for $$\frac{\exp(\frac 1{z^2})}{z-1}$$ about $z=0$. I know that $\exp( \frac 1{z^2}) = \sum_{n=0}^\infty \frac{z^{-2n}}{n!}$ and $\frac 1{z-1}=-\sum_{n=0}^\infty z^n$

-
And do you know how to multiply power series: en.wikipedia.org/wiki/Power_series#Multiplication_and_division ? – Dirk Mar 20 '12 at 8:07
I was thinking of that but I was having difficult time simplifying it and finding a direct formula for the coefficients. Does it matter that the power series we are multiplying have different powers of z because in the given example they both have (x-c)^n – caligurl11 Mar 20 '12 at 13:42

Write $$\exp(1/z^2) = \sum_{n=-\infty}^\infty a_n z^n \qquad\text{and}\qquad \frac{1}{z-1} = \sum_{n=-\infty}^\infty b_n z^n,$$ i.e. $a_{-2n} = 1/n!$ and $a_k = 0$ otherwise, and $b_n = -1$ for $n \ge 0$ and $b_n = 0$ for $n < 0$. Multiply the two series: $$\left(\sum_{n=-\infty}^\infty a_n z^n\right)\left(\sum_{n=-\infty}^\infty b_n z^n\right) = \left(\sum_{n=-\infty}^\infty c_n z^n\right)$$ and start identifying the $c_n$:s. We get $$c_0 = a_0 b_0 + a_{-2}b_2 + a_{-4}b_4 + \cdots = -\left( \frac1{0!} + \frac{1}{1!} + \frac1{2!} + \cdots \right) = -e,$$ $$c_1 = a_0 b_1 + a_{-2}b_3 + a_{-4}b_5 + \cdots = -\left( \frac1{0!} + \frac{1}{1!} + \frac1{2!} + \cdots \right) = -e,$$ etc. In other words, $c_n = -e$ for all $n \ge 0$. For the negative indices you get $$c_{-1} = a_{-2} b_1 + a_{-4}b_3 + a_{-6}b_5 + \cdots = -\left( \frac{1}{1!} + \frac1{2!} + \frac1{3!}+ \cdots \right) = -(e-1),$$ $$c_{-2} = a_{-2} b_0 + a_{-4}b_2 + a_{-6}b_4 + \cdots = -\left( \frac{1}{1!} + \frac1{2!} + \frac1{3!}+ \cdots \right) = -(e-1),$$ $$c_{-3} = a_{-4} b_1 + a_{-6}b_3 + a_{-8}b_5 + \cdots = -\left( \frac{1}{2!} + \frac1{3!} + \frac1{4!}+ \cdots \right) = -(e-1-1),$$ $$c_{-4} = a_{-4} b_0 + a_{-6}b_2 + a_{-8}b_4 + \cdots = -\left( \frac{1}{2!} + \frac1{3!} + \frac1{4!}+ \cdots \right) = -(e-1-1).$$

By now, the pattern should be obvious, if $n = -2k$, $$c_{-2k} = c_{-2k+1} = - \left(e - 1 - \frac1{1!} - \frac1{2!} - \cdots \frac1{k!}\right) = -e + \sum_{j=0}^k \frac{1}{j!}.$$

-
perfect answer! thanks a lot. – d13 May 6 '13 at 13:09

I have not read the previous anwer in detail. The short cut way is to notice the relationship that $$f^{n}=\frac{n!}{2\pi i}\oint \frac{f(z)dz}{(z-c)^{n+1}}$$

Notice that for Laurent series we have $$a_{n}=f^{n}/n!=\frac{1}{2\pi i}\oint \frac{f(z)dz}{(z-c)^{n+1}}$$

Thus the Laurent series coefficient for $f=\frac{e^{z^{-2}}}{z-1}$ is $$a_{n}=\frac{1}{2\pi i}\oint \frac{e^{z^{-2}}dz}{z^{n+1}(z-1)}$$

Consider the Cauchy integral formula which says $$f(c)=\frac{1}{2\pi i}\oint \frac{f(z)dz}{z-c}$$

Thus let $$g=\frac{e^{z^{-2}}}{z^{n+1}}, a_{n}=g(1)=e$$ should suffice. I might be wrong somewhere though.

-