- #1
sigmund
- 23
- 0
I have a function
[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?
[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?