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Multiplying and dividing macularin question

  1. Nov 15, 2013 #1
    When you multiply or divide macularin series to find, let's say, the first 3 nonzero terms of the polynomial, how many terms of each series should be included in the multipication/division?
  2. jcsd
  3. Nov 15, 2013 #2


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    All of them if you want an exact value.

    If you are looking for an approximation, then you can truncate numerator and/or denominator. But how many terms you need to keep depends upon what accuracy you want.
  4. Nov 15, 2013 #3


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    Suppose you have the two series
    [tex]f(x)=\sum_{k=0}^{\infty} a_k x^k, \quad g(x)=\sum_{k=0}^{\infty} b_k x^k.[/tex]
    Then the product is
    [tex]f(x) g(x)=\sum_{j,k=0}^{\infty} a_j b_k x^{j+k}.[/tex]
    Now within the common range of convergence you are allowed to interchange the terms as you like. So we can reorder the zeros in powers [itex]n=j+k[/itex]. At fixed [itex]n[/itex], given [itex]j \in \{0,1,\ldots n\}[/itex][/tex], [itex]k=n-j[/itex]. This means that
    [tex]f(x) g(x)=\sum_{n=0}^{\infty} x^n \sum_{j=0}^{n} a_j b_{n-j}.[/tex]
    That means to get the expansion of [itex]f g[/itex] to order [itex]n[/itex] you need the coefficients of the original functions to the same order, [itex]n[/itex].

    For division you set
    [tex]\frac{f(x)}{g(x)}=\sum_{k=0}^{\infty} c_k x^k.[/tex]
    Then you have
    [tex]f(x)=\sum_{n=0}^{\infty} a_n x^n=g(x) \sum_{k=0}^{\infty} c_k x^k=\sum_{j=0}^{\infty} b_j x^k \sum_{k=0}^{\infty} c_k x^k=\sum_{n=0}^{\infty} x^n \sum_{j=0}^{n} b_j c_{n-j}.[/tex]
    Comparing coefficients you get the set of equations
    [tex]a_n=\sum_{j=0}^{n} b_j c_{n-j},[/tex]
    from which you can recursively determine the [itex]c_k[/itex].

    Starting from [itex]n=0[/itex] you get
    [tex]a_0=b_0 c_0 \; \Rightarrow c_0=\frac{a_0}{b_0},[/tex]
    then for [itex]n=1[/itex]
    [tex]a_1=b_0 c_1+b_1 c_0 \; \Rightarrow c_1=\frac{a_1-b_1 c_0}{b_0}[/tex]
    or generally
    [tex]a_n=\sum_{j=0}^{n} b_j c_{n-j} \; \Rightarrow \; c_n=\frac{a_n-\sum_{j=1}^{n} b_j c_{n-j}}{b_0}.[/tex]
    As you see, you must have [itex]b_0 \neq 0[/itex] in order to have a well-defined power series for the fraction [itex]f/g[/itex]. That's clear, because if [itex]b_0=0[/itex] then [itex]g(0)=0[/itex], and the function [itex]f/g[/itex] has a singularity at [itex]x=0[/itex].

    If the power series of [itex]g[/itex] starts at the power [itex]n_0[/itex], i.e., if [itex]c_k=0[/itex] for [itex]0 \leq k<n_0[/itex] and [itex]c_{n_0} \neq 0[/itex], then you can write
    [tex]g(x)=x^{n_0} \sum_{k=0}^{\infty} \tilde{b}_{k} x^k=x^{n_0} \tilde{g}(x), \quad \tilde{b}_k=b_{n_0+k}[/tex]
    and use the above considerations for
    [tex]\frac{f(x)}{\tilde{g}(x)}=\sum_{k=0}^{\infty} \tilde{c}_k x^k.[/tex]
    Then you get recursively the [itex]\tilde{c}_n[/itex] as explained above and then you finally find
    [tex]\frac{f(x)}{g(x)}=\frac{1}{x^{n_0}} \sum_{k=0}^{\infty} \tilde{c}_k x^k.[/tex]
    That means that in this case [itex]f/g[/itex] has a pole of order [itex]n_0[/itex]. The convergence radius of the corresponding Laurent series is still given by the smaller of the convergence radii [itex]r_{<}[/itex] of the power series expansion of [itex]f[/itex] and [itex]g[/itex] around 0, i.e., the Laurent series is convergent for [itex]0<|x|<r_{<}[/itex].
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