find taylor series to fourth term I'm wondering if there is faster method than just calculating derivatives with finding taylor series up to 4 term of function $\displaystyle f(x)=\frac{(1+x^4)}{(1+2x)^3(1-2x)^2}$  
 A: The Taylor expansion at $0$ of $(1+x)^\alpha$, $\alpha\in\mathbb{R}$ is
$$\sum_{k=0}^{+\infty}\left(\begin{array}$\alpha\\k\end{array}\right)x^k$$
Where $$\left(\begin{array}$\alpha\\k\end{array}\right):=\frac{\alpha(\alpha-1)\cdots(\alpha-k+1)}{k!};\\\left(\begin{array}$\alpha\\0\end{array}\right):=1$$
So, your function being $f(x)=(1+x^4)(1+(2x))^{-3}(1+(-2x))^{-2}$ you may also try a symbolic computation thourgh products and compositions.
Also, it might help noting that $f(x)=(1+x^4)(1+2x)^{-1}(1-4x^2)^{-2}$
I do not know whether it actually saves time, though.
A: The idea is to use the formula
$$
\frac{1}{(1-x)^t} = \sum_{n=0}^\infty \binom{n+t-1}{n} x^n,
$$
which can be proved by induction on $t$.
Using this we get
$$
\frac{1+x^4}{(1+2x)^3(1-2x)^2} = \\
(1+O(x^4))(1 - 6x + 24x^2 - 80x^3+O(x^4))(1 + 4x + 12x^2 + 32x^3 + O(x^4)) = \\
1 + (-6+4)x + (24-6\cdot 4+12)x^2 + (-80 + 24\cdot 4 - 6\cdot 12 + 32)x^3 +
O(x^4) = \\
1 - 2x + 12x^2 - 24x^3 + O(x^4).
$$
A: Since the Taylor series of $g(x)=\frac{x^4}{(1+2x)^3(1-2x)^2}$ in a neighbourhood of the origin is given by $x^4+o(x^4)$, it is enough to compute the Taylor series of:
$$\begin{eqnarray*} h(x)&=&\frac{1}{(1+2x)^3(1-2x)^2}\\&=&\left(1-6 x+24 x^2-80 x^3+240 x^4+o(x^4)\right)\left(1+4 x+12 x^2+32 x^3+80 x^4+o(x^4)\right)\\&=&1 - 2 x + 12 x^2 - 24 x^3 + 96 x^4+o(x^4)\tag{1}\end{eqnarray*}$$
giving:
$$ \frac{1+x^4}{(1+2x)^3(1-2x)^2}=1 - 2 x + 12 x^2 - 24 x^3 + 9\color{red}{7} x^4+o(x^4).\tag{2}$$
In $(1)$ we just used the Taylor series for $\frac{1}{(1-z)^m}$, given by:
$$\frac{1}{(1-z)^m}=\sum_{n\geq 0}\binom{n+m-1}{n}\, z^n\tag{3}$$
and took the Cauchy product of two such series. A different approach is given by:
$$ \frac{1}{(1-4x^2)^2}=1+8 x^2+48 x^4+o(x^4),\tag{4} $$
that just leads to $(1)$ since:
$$ \frac{1}{1+2x}=1 - 2 x + 4 x^2 - 8 x^3 + 16 x^4+o(x^4).\tag{5} $$
