Multiplying and dividing macularin question

[freshman2013]

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?

 

[HallsofIvy]

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.

 

[vanhees71]

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 $0<|x|<r_{<}$.