1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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]:

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

    benorin

    User Avatar
    Homework Helper

    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 [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

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

    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: Nov 25, 2005
  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

    benorin

    User Avatar
    Homework Helper

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Cauchy multiplication
  1. Cauchy Sequence (Replies: 10)

  2. Cauchy Sequence (Replies: 9)

  3. Cauchy sequence (Replies: 2)

  4. Cauchy integral (Replies: 1)

Loading...