I can't understand this in partial fractions

[arildno]

Okay, I'll post something on the Cauchy product of infinite series first:
Suppose you've got two series $S_{1}=\sum_{n=0}^{\infty}a_{n}x^{n}, S_{2}=\sum_{m=0}^{\infty}b_{m}x^{m}$
where $a_{n},b_{m}$ are coefficients for the n'th and m'th terms respectively (it is smart, and allowable, to use a different letter for the indices for the two series)

Now, suppose we wish to multiply the two series together to a single series S; how are the coefficients of S related to the coefficients of $S_{1},S_{2}$ ?

Now, let's just use a finite amount usual numbers, and see how the result OUGHT to be:
If we are to multiply the number (2+3+1) (that is, 6) with (5+7+8) (that is 20), we may do it as follows:
$(2+3+1)(5+7+8)=2*5+2*7+2*8+2*8+3*5+3*7+3*8+1*5+1*7+1*8$
The crucial point to notice that we multiply each term in the first parenthesis with each term in the other parenthesis, and then sum all the 3*3=9 terms together (3*3 terms since each parenthesis has 3 numbers).

Thus, generalizing to an infinite series, we could write our product as:
$$S=S_{1}*S_{2}=\sum_{n=0}^{\infty}a_{n}x^{n}*\sum_{m=0}^{\infty}b_{m}x^{m}=\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}a_{n}b_{m}x^{n+m}$$
where the last DOUBLE sum should be thought of as saying:
1. Fix n=0, and calculate the sum $\sum_{m=0}^{\infty}a_{0}b_{m}x^{m+0}$
that is, multiply the term $a_{0}x^{0}$ with EACH term in $S_{2}$, and sum them together

2. ADD to your result from 1. the series you get by fixing n=1, i.e, by first calculating the sum $\sum_{m=0}^{\infty}a_{1}b_{m}x^{m+1}$, and then adding all of this to the result you got from 1.
(That is, multiplying the term $a_{1}x^{1}$ with EACH term in $S_{2}$, add up the result, and add this result with that you got from 1.)
3. And so on..

Got that?

 

[arildno]

I'll assume you do, and continue:
Now, as with any sum of numbers, it shouldn't matter in which succession you add them together, right? (That is: 1+2+3=2+1+3=3+1+2 and so on)

We wish now to add up all terms in our double sum in ASCENDING powers of x.
We introduce a new index, j=n+m
Rewriting this, we have n=j-m
Now, for FIXED j, m can have any value between 0 and j.
Thus, we get the upper and lower limits on j and m:
$0\leq{j}\leq\infty,0\leq{m}\leq{j}$
and we get, by substituting j=n+m or n=j-m:
$$\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}a_{n}b_{m}x^{n+m}=\sum_{j=0}^{\infty}\sum_{m=0}^{j}a_{j-m}b_{m}x^{j}=\sum_{j=0}^{\infty}c_{j}x^{j}, c_{j}=\sum_{m=0}^{j}a_{j-m}b_{m}$$
That is, cj is the coefficient of the j'th power of x.

Example:
You were to multiply together:
$$(1+(-2x)+(-2x)^{2}+++)(1+(-2x)+(-2x)^{2}+++)$$
We can write this as the product of the series:
$$S=\sum_{n=0}^{\infty}(-2)^{n}x^{n}\sum_{m=0}^{\infty}(-2)^{m}x^{m}=\sum_{j=0}^{\infty}c_{j}x^{j}, c_{j}=\sum_{m=0}^{j}(-2)^{j-m}(-2)^{m}=\sum_{m=0}^{j}(-2)^{j}=(-2)^{j}\sum_{m=0}^{j}1=(j+1)(-2)^{j}$$

Thus, we have the series representation:
$$\frac{1}{(1+2x)^{2}}=\sum_{j=0}^{\infty}(j+1)(-2)^{j}x^{j}$$
That is, the coefficient of the j'th power of x is [itex](j+1)(-2)^{j}[/tex]
Note that this is in tune for j=4 which we already have calculated:
$$c_{4}=(4+1)(-2)^{4}=5*16=80$$