I have a function(adsbygoogle = window.adsbygoogle || []).push({});

[tex]2-z^2-2\cos z,[/tex]

which has a zero at [itex]z=0[/itex].

I have determined the Maclaurin series for [itex]f[/itex]:

[tex]\sum_{j=2}^\infty(-1)^{j-1}\frac{2z^{2j}}{(2j)!},[/tex]

and now I have to determine the coefficients [itex]a_{-j},~\forall j>0,[/itex] in the Laurent series for a function [itex]h[/itex], which is defined as [itex]h(z)=1/f(z)[/itex].

For this purpose I want to use Cauchy multiplication, where I multiply the series for f by the series for h, because I know that the result is the series for 1, i.e. a 1 with all the following terms equal zero. However, when applying this, I do not get any useful result, because the positive exponents in the series for f dominate the negative exponents in the series for h, and hence I am unable to determine the coefficients [itex]a_{-j}[/itex] of the [itex]z^{-j}[/itex] terms in the series for h.

Could anyone give me a hint to a solving procedure, please?

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Cauchy multiplication

**Physics Forums | Science Articles, Homework Help, Discussion**