# Cauchy multiplication

1. Nov 24, 2005

### 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?

2. Nov 25, 2005

### benorin

Laurent Series via Cauchy Product

Rudin gives the formula for the Cauchy product of series as it is presented in this link.

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.

Last edited: Nov 25, 2005
3. Nov 27, 2005

### sigmund

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.

4. Nov 27, 2005