Here is the task: using standard expansions, expand $f(x) = x \sqrt{4-x}$ to Maclaurin's series.
I calculated derivatives up to $f^{(5)}(x)$, and got some results.
Fortunately, in Maclaurin's expansion I need to calcalate $f^{(n)}(0)$, which simplifies the task a little. Here is what I've got:
$d(n) = 1 \times 3 \times 5 \times \ldots \times (2n-1) = \frac{(2n)!}{2^nn!}$ (a function for calculating product of first n odd numbers)
$r(n)= \begin{cases} (2(n - 2) - 1) \times (r(n-1) + d(n-3)),&\text{if $n > 3$;}\\ 3,&\text{if $n = 3$.} \end{cases} $
$r(n)$ is calculated iteratively or recursively, and it frightens me. However, factorial function is iterative/recursive too, so my $r(n)$ is not worse, but I still somewhat displeased by it. What about derivatives,
$f^{(n)}(0)= \begin{cases} 0,&\text{if $n = 0$;}\\ 2,&\text{if $n = 1$;}\\ -1/2,&\text{if $n = 2$;}\\ -r(n) \times 2^{n-2},&\text{if $n \geq 3$.} \end{cases} $ ($f^{(0)}(x) = f(x)$, if this is confusing)
And it seems to work. I just calculated derivatives and noticed how next derivative is produced from previous. How the series is written down is obvious. But I didn't use any "standard expansions". Could this task be done easier and could the $f^{(n)}(0)$ formula be more beautiful?