Using Cauchy Multiplication to Find Coefficients in Laurent Series for 1/f(z)

[sigmund]

I have a function

2-z^2-2\cos z,
which has a zero at z=0.

I have determined the Maclaurin series for f:

\sum_{j=2}^\infty(-1)^{j-1}\frac{2z^{2j}}{(2j)!},
and now I have to determine the coefficients a_{-j},~\forall j>0, in the Laurent series for a function h, which is defined as h(z)=1/f(z).
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 a_{-j} of the z^{-j} terms in the series for h.

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

 

[benorin]

Laurent Series via Cauchy Product

Rudin gives the formula for the Cauchy product of series as it is presented in this http://mwt.e-technik.uni-ulm.de/world/lehre/basic_mathematics/di/node14.php3 .

A quick version is:

Suppose \sum_{n=0}^{\infty} a_n and \sum_{n=0}^{\infty} b_n converge absolutely. Then

\left( \sum_{n=0}^{\infty} a_n\right) \left( \sum_{n=0}^{\infty} b_n\right) = \sum_{n=0}^{\infty} \sum_{k=0}^{n} a_{k}b_{n-k} also converges absolutely.

Alternately, look here, under the heading A Variant.

In particular, we know that

f(z)=2-z^2-2\cos z,

which is an even function and which has the Maclaurin series

f(z)=\sum_{j=2}^{\infty}(-1)^{j-1}\frac{2z^{2j}}{(2j)!}

We also know that f(z)h(z)=1; so suppose that the Laurent series for h is given by

h(z)=\sum_{j=-\infty}^{\infty} a_{j}z^{j}

What do we already know about said Laurent series? Well, if f(z) has a zero of order m at z=0, then h(z) has a pole of order m at z=0. Since

f^{(n)}(0)=0\mbox{ for } n=0,1,2,3,\mbox{ but }f^{(4)}(0)=-2,

we have m=4, and hence a_{-j}=0, \forall j>4. We know h(z) is an even function (since the same is true of f(z)), and therefore a_{2k+1}=0, \forall k\in\mathbb{Z}. Our refined guess at the Laurent series for h(z) is

h(z)=\sum_{k=-2}^{\infty} a_{2k}z^{2k}

Far enough, on with the so-called nitty-gritty:

f(z)h(z)= \left( \sum_{j=2}^{\infty}(-1)^{j-1}\frac{2z^{2j}}{(2j)!}\right) \left( \sum_{k=-2}^{\infty} a_{2k}z^{2k}\right)

re-index the sums to start at zero so that the Cauchy product is "nice"...

f(z)h(z)= \left( 2\sum_{j=0}^{\infty}(-1)^{j+1}\frac{z^{2j+4}}{(2j+4)!}\right) \left( \sum_{k=0}^{\infty} a_{2k-4}z^{2k-4}\right)
=2\sum_{n=0}^{\infty}\sum_{k=0}^{n} a_{2n-2k-4} z^{2n-2k-4} (-1)^{k+1} \frac{z^{2k+4}}{(2k+4)!} = 2\sum_{n=0}^{\infty} z^{2n}\sum_{k=0}^{n}(-1)^{k+1} \frac{a_{2n-2k-4}}{(2k+4)!} =1 ,

and hence, equating coefficients of like powers of z on both sides, we have:

the only surviving term on the right corresponds to n=0, so that k=0 also, yielding \frac{-2a_{-4}}{1!}=1\Rightarrow a_{-4}=-\frac{1}{2};

all other other powers of z on the right have coefficients of zero so that

2\sum_{k=0}^{n}(-1)^{k+1} \frac{a_{2n-2k-4}}{(2k+4)!} =0,\forall n\geq 1

plug-in the know value of a_{-4}=-\frac{1}{2}, see if you can determine the rest of the sequence from the. Good luck.

 

[sigmund]

(...)
and hence, equating coefficients of like powers of z on both sides, we have:
the only surviving term on the right corresponds to n=0, so that k=0 also, yielding \frac{-2a_{-4}}{1!}=1\Rightarrow a_{-4}=-\frac{1}{2};
(...)

Actually, I cannot follow you here. The sides of which equation are you talking of?

If you set n=0 and k=0 in the last equation (the result of the Cauchy multiplication), you should get \frac{-2a_{-4}}{4!}=1\Leftrightarrow a_{-4}=-12. And then, using the recursion formula (the very last equation of your post), we get a_{-2}=-\frac{2}{5}.

These two coefficients are the only non-zero coefficients of negative powers of z in the Laurent series.

 

[benorin]

My bad, your right. Thank you.

 

[sigmund]

(...)
the only surviving term on the right corresponds to n=0, so that k=0 also,
(...)

Well, I think I have got it now. The series for 1 does only consist of a constant term, hence we set n=0. Then k=0 also, and we determine the coefficient a_{-4}, using the Cauchy product. All the other coefficients in the series for 1 are zero. Thus we get the recursion formula (stated in your post) from which we can determine the other coeffients in the Laurent series for h.