Find the Laurent expansion for $f(z)=\frac{\exp{(1/z^2)}}{z-1}$ about $z=0$.

I was able to determine the series for each of the factors. We have $$e^{1/z^2}=1+\frac{1}{z^2}+\frac{1}{2!z^4}+\frac{1}{3!z^6}+\cdots=\sum_{n=0}^{\infty}\frac{1}{n!z^{2n}}.$$

Also, the series $$\frac{1}{z-1}=-\frac{1}{1-z}=-\sum_{n=0}^{\infty}z^n.$$

Now all I can think of to do is multiply these two series together to get the result, but I dread attempting that again (for like the fourth time). Perhaps I am not seeing an easier way? Any help would be appreciated.


1 Answer 1



Start replacing $z$ by $\frac 1x$. So $$f(x)=\frac{x }{1-x}\,e^{x^2}$$ Now, using Taylor series, we then have $$f(x)=x \Big(\sum_{n=0}^{\infty}x^{n}\Big)\Big(\sum_{m=0}^{\infty}\frac{x^{2m}}{m!}\Big)=\sum_{i=0}^{\infty}{a_i}{x^i}$$ Now, may be, you could identify coefficient $a_k$ for an identical power $k$ in both sides.

If you start, you will find $$x+x^2+2 x^3+2 x^4+\frac{5 x^5}{2}+\frac{5 x^6}{2}+\cdots=a_0+a_1x+a_2x^2+a_3x^3+a^4x^4+a_5x^5+\cdots$$ where appears some nice patterns (I let you finding the general expression).

  • $\begingroup$ @giorgiomugnaini. Thanks but I knew that when I answered. Isn't funny to see how beautiful is this Laurent series ? Cheers :-) $\endgroup$ Apr 9, 2015 at 11:22
  • $\begingroup$ I appreciate your answer, but I must admit that I do not understand it. Could you explain what you mean, please? $\endgroup$ Apr 9, 2015 at 18:25
  • $\begingroup$ @ClaudeLeibovici For $0<z<1$, i.e. $x>1$, I do not think your series converges. If it did then it would be positive while $\frac{\exp{(1/z^2)}}{z-1}$ and $\frac{x }{1-x}\,e^{x^2}$ look negative $\endgroup$
    – Henry
    Nov 18, 2020 at 22:10

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