<s>[INST] Finding n terms of a series by multiplying two known series The exact question: 

Find the first 4 non-zero terms of the series for $$e^{-x}\cos(x)$$ by
  multiplying two known series. Hint: use pointers on the 2nd series.

I know $$e^{x} = \sum_{k=0}^{\infty} \frac{x^{k}}{k!} = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + ... $$
and
$$\cos(x) = \sum_{k=0}^{\infty} \frac{x^{2k}}{2k!} = 1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} + \frac{x^{6}}{6!} + ...  $$
so,
for $$e^{-x}\cos(x)$$ we just plug in the -x for x and get
$$e^{-x} = \sum_{k=0}^{\infty} \frac{(-1)^kx^{k}}{k!} = 1 - x + \frac{x^{2}}{2!} - \frac{x^{3}}{3!} + ... $$
cos(x) stays the same, but...what do we do from there? He gave us the answer, which I think is (not sure because the prof just wrote it on the board and didn't really say if that was the "final answer"): 
$$1 - x + \frac{x^{3}}{6} - \frac{x^{4}}{6} + ... $$
Can anyone explain to me what's going on here? What are pointers even? I went to the tutors at school and no one knows what that is and they told me I would just have to FOIL out the first four terms each series, but...is that even right? They didn't want to do it because it would take too long, but is that how this is supposed to be done?
 [/INST] $(1 - x + \frac 12 x^2 - \frac 16 x^3 + \frac 1{24} x^4)(1 - \frac 12x^2 + \frac 1{24} x^4)$
We can ignore the higher powered terms.  When we multiply two terms together the power of $x$ equals sum of the powers of $x$ in the two factors.
To get the $x^0$ coefficient then, it will be the product of the two constant terms. $a_0 = 1$
To get the $x^1$ coefficient, we multiply the constant term from each series by the $x^1$ terms from the other.
$a_1 = 1\cdot -1 + 1\cdot 0 = -1$ 
The $x^2$ terms, now it starts to get a little bit more complicated.
We multiply the each constant term by the others $x^2$ term and then multiply the two $x^1$ terms (if we had two $x^1$ terms)
$a_2 = \frac 12 - \frac 12 = 0$
$a_3 = -\frac 16 + \frac 12 = \frac 13$
$a_4 = \frac 1{24} + \frac {1}{24} - \frac 14 = -\frac 16$
or perhaps this is easier.
$\sum_\limits{n=0}^{\infty} a_n x^n \sum_\limits{m=0}^{\infty} b_m x^m = \sum_\limits{n=0}^{\infty} \sum_\limits{i=0}^n a_ib_{n-i} x^n$ 
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