Taylor expansion for $\frac{1}{x^2 + 2x + 2}$ I'm trying to find the Taylor expansion of the function:
$$ f(x) = \frac{1}{x^2 + 2x + 2} $$
about the point $ x = 0 $. I have worked out the terms up to the fourth derivative, which was very tedious. I found:
$$ f(x) = \frac{1}{2} - \frac{1}{2} x + \frac{1}{4} x^2 + 0 x^3 - \frac{1}{8} x^4 + O \left( x^5 \right) $$
I notice powers of two in the denominator, but I'm not sure of the pattern (and to calculate the next term to confirm would entail another tedious product rule).
Any ideas? Thanks!
 A: Herein, we present an approach using partial fraction expansion.  Proceeding, we write
$$\begin{align}
\frac{1}{x^2+2x+2}&=\frac{1}{i2}\left(\frac{1}{x+1-i}-\frac{1}{x+1+i}\right)\\\\
&=\text{Im}\left(\frac{1}{x+1-i}\right)\\\\
&=\text{Im}\left(\frac{1}{1-i}\frac{1}{1+\frac{x}{1-i}}\right)\\\\
&=\sum_{n=0}^\infty \frac{(-1)^n\sin((n+1)\pi/4)}{2^{(n+1)/2}}\,x^n\\\\
&=\sum_{n=1}^\infty \frac{(-1)^{n-1}\sin(n\pi/4)}{2^{n/2}}\,x^{n-1}\\\\
&=\frac12-\frac x2 +\frac{x^2}{4}+\cdots
\end{align}$$ 
A: Note that
$$f(x)=\frac{1}{x^2 + 2x + 2}=\frac{1/2}{1+(x+x^2/2)}\\=\frac{1}{2}\left(1-(x+x^2/2)+(x+x^2/2)^2-((x+x^2/2)^3+o(x^3)\right).$$
In general, for any positive integer $n$,
$$2f(x)=1+\sum_{k=1}^n x^k(1+x/2)^k+o(x^n)=
\sum_{k=1}^n (-1)^k\sum_{j=0}^{n-k} \frac{1}{2^j}\binom{k}{j}x^{j+k}+o(x^n)\\
=1+\sum_{k=1}^n \frac{x^m}{2^m}\left(\sum_{k=1}^m(-2)^k\binom{k}{m-k}\right)+o(x^n).$$
P.S. Note that $a_m$, the coefficient of $x^m$, is related to the sequence A009116. It follows immediately from its generating function that it satisfies the linear sequence 
$$a_{m+1}=-\left(a_m+\frac{a_{m-1}}{2}\right).$$
A: You can write
$$
f(x) = \frac{1}{2}\cdot\frac{1}{1+(x+\frac{x^2}{2})}.
$$
Setting $\stackrel{\rm def}{=} x+\frac{x^2}{2}\xrightarrow[x\to0]{}0$, you can use the Taylor expansion of $\frac{1}{1+u} = \sum_{n=0}^{\infty} (-1)^n u^n$ around $0$ to get
$$\begin{align}
f(x) &= \frac{1}{2}\left(1-u+u^2-u^3+u^4+o(u^4)\right)\\
&= \frac{1}{2}\left(1-(x+\frac{x^2}{2})+(x+\frac{x^2}{2})^2-(x+\frac{x^2}{2})^3+(x+\frac{x^2}{2})^4+o(x^4)\right)\\
&= \frac{1}{2}\left(1-x-\frac{x^2}{2}+(x^2+x^3+\frac{x^4}{4})-(x^3+\frac{3}{2}x^4)+x^4+o(x^4)\right)
\end{align}$$
(where when we expand we neglect all terms $x^k$ for $k>4$), which gives
$$\begin{align}
f(x) &= \frac{1}{2}\left(1-x+\frac{x^2}{2}-\frac{1}{4}x^4+o(x^4)\right)
\end{align}$$
i.e., the result you found. Note that with this technique, going to order say $x^9$ is much simpler than by differentiating repeatedly 9 times.
A: $$
x^2 + 2x + 2 = (x+1)^2 + 1
$$
and, with  $\ s = x+1 \ $ you have  
$$
\frac 1{x^2 + 2x + 2} = \frac 1{(x+1)^2 + 1} = \frac 1{s^2 + 1} = \text{D} ( \arctan s ).
$$
Now, just derive the Taylor expansion of  $\ \arctan s \ $ and subsitute $\ s = x +1 $
A: The pattern is 
$$
\frac{1}{x^2+2x+2}=\frac12+\sum_{k\in \Bbb{Z}^+ | k-3\not\in \Bbb{Z}} (-1)^\left\lfloor \frac34 k\right\rfloor2^{-\left\lfloor \frac{k}2+1\right\rfloor}
$$
where $\lfloor s \rfloor$ means the greatest integer not exceeding $s$, and the sum on $k$ skips numbers of the form $4n+3$.
Thus
$$
\frac12 - \frac{x}2 +\frac{x^2}{4} - \frac{x^4}{8} + \frac{x^5}{8} - \frac{x^6}{16}  + \frac{x^8}{32} -\frac{x^9}{32} + \frac{x^{10}}{64} \cdots
$$
A: $$  f(x) = \frac{1}{x^2 + 2x + 2} = a_0 + a_1 x + a_2 x^2 + \cdots  $$
\begin{align*}
&& (x^2 + 2x + 2)f(x) &= 1 \quad ({}+ 0 x + 0x^2 + 0x^3 + \cdots) \\
    &\implies& 2a_0 &= 1  \\
    && 2a_1 + 2a_0 &= 0  \\
    && 2a_2 + 2a_1 + a_0 &= 0  \\
    && 2a_3 + 2a_2 + a_1 &= 0  \\
    && 2a_4 + 2a_3 + a_2 &= 0  \\
    && 2a_5 + 2a_4 + a_3 &= 0  \\
    &&& \vdots  \\
 &\implies& a_0 &= \frac{1}{2}  \\
    && a_1 &= \frac{-1}{2}  \\
    && a_2 &= \frac{1}{4}  \\
    && a_3 &= \frac{1}{2}\left(\frac{2}{4} + \frac{-1}{2}\right) = 0  \\
    &&& \vdots  \\
    && a_{k \geq 3} &= \frac{1}{2}\left(2 \cdot a_{k-1} + a_{k-2}\right)  \\
    &&& \vdots  \\
\end{align*}
A: HINT
I would say - simply divide the number $1$ manually expression $(2+2x+x^2)$:
More attention and gradually comes out:  
$1:(2+2x+x^2) = 1/2 - x/2 + x^2/4 - x^4/8 + x^5/8 - x^6/16 + \cdots$
$\Rightarrow f(x)=1/2 \cdot(1 - x + x^2/2 - x^4/4 + x^5/4 - x^6/8 + \cdots)$
A: There is a much simpler way involving no contrived derivatives, binomial expansions, etc., that unfortunately does not easily generalize to any other quadratic reciprocal.
If we notice that $x^2+2x+2$ occurs in the factorization of a certain binomial:
$$ x^8 - 16 = \left( x^4 - 4 \right) \left( x^4 + 4 \right) = \left( x^4 - 4 \right) \left( x^2 - 2x + 2 \right) \left( x^2 + 2x + 2 \right) = \left( x^6 - 2x^5 + 2x^4 - 4x^2 + 8x - 8 \right) \left( x^2 + 2x + 2 \right) $$
Rearranging:
$$ \frac{1}{x^2 + 2x + 2} = \frac{x^6 - 2x^5 + 2x^4 - 4x^2 + 8x - 8}{x^8 - 16} $$
We can bring the right-hand side to the form of a simple geometric series:
$$ \frac{1}{x^2 + 2x + 2} = \frac{\frac{8}{16} - \frac{8}{16} x + \frac{4}{16} x^2 - \frac{2}{16} x^4 + \frac{2}{16} x^5 - \frac{1}{16} x^6}{\frac{16}{16} - \frac{x^8}{16}} = \frac{\frac{1}{2} - \frac{1}{2} x + \frac{1}{4} x^2 - \frac{1}{8} x^4 + \frac{1}{8} x^5 - \frac{1}{16} x^6}{1 - \frac{x^8}{16}} = \left( \frac{1}{2} - \frac{1}{2} x + \frac{1}{4} x^2 - \frac{1}{8} x^4 + \frac{1}{8} x^5 - \frac{1}{16} x^6 \right) \left( 1 + \frac{x^8}{16} + \frac{x^{16}}{256} + \cdots \right) $$
Clearly, there is no overlap between the terms when multiplying out, so if we write only the coefficients, we see that they group together in sets of eight:
$$ \frac{1}{x^2 + 2x + 2} = \left[ \frac{1}{2}, -\frac{1}{2}, \frac{1}{4}, 0, -\frac{1}{8}, \frac{1}{8}, -\frac{1}{16}, 0, \frac{1}{16}, -\frac{1}{32}, \frac{1}{32}, 0, -\frac{1}{64}, \frac{1}{64}, -\frac{1}{128}, 0, \cdots \right] $$
in an elegant pattern!
