Multiplying and dividing macularin question

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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?
 
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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.
 
Suppose you have the two series
f(x)=\sum_{k=0}^{\infty} a_k x^k, \quad g(x)=\sum_{k=0}^{\infty} b_k x^k.
Then the product is
f(x) g(x)=\sum_{j,k=0}^{\infty} a_j b_k x^{j+k}.
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 n=j+k. At fixed n, given j \in \{0,1,\ldots n\}[/tex], k=n-j. This means that
f(x) g(x)=\sum_{n=0}^{\infty} x^n \sum_{j=0}^{n} a_j b_{n-j}.
That means to get the expansion of f g to order n you need the coefficients of the original functions to the same order, n.

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

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

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

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