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Homework Help: Cauchy multiplication

  1. Nov 24, 2005 #1
    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]:

    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?
  2. jcsd
  3. Nov 25, 2005 #2


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    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 [Broken].

    A quick version is:

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

    [tex]\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}[/tex] also converges absolutely.

    Alternately, look here, under the heading A Variant.

    In particular, we know that

    [tex]f(z)=2-z^2-2\cos z,[/tex]

    which is an even function and which has the Maclaurin series


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

    [tex]h(z)=\sum_{j=-\infty}^{\infty} a_{j}z^{j}[/tex]

    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

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

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

    [tex]h(z)=\sum_{k=-2}^{\infty} a_{2k}z^{2k}[/tex]

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

    [tex]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) [/tex]

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

    [tex]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) [/tex]
    [tex]=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 ,[/tex]

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

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

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

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

    plug-in the know value of [itex]a_{-4}=-\frac{1}{2}[/itex], see if you can determine the rest of the sequence from the. Good luck.
    Last edited by a moderator: May 2, 2017
  4. Nov 27, 2005 #3
    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 [itex]\frac{-2a_{-4}}{4!}=1\Leftrightarrow a_{-4}=-12[/itex]. And then, using the recursion formula (the very last equation of your post), we get [itex]a_{-2}=-\frac{2}{5}[/itex].

    These two coefficients are the only non-zero coefficients of negative powers of z in the Laurent series.
  5. Nov 27, 2005 #4


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    My bad, your right. Thank you.
  6. Nov 27, 2005 #5
    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.
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