It seems that I'm [censored] blind in searching the power series expansion of $$f(x):=\frac{2x-2}{x^2-2x+4}$$ in $x=0$.
I've tried a lot, e.g., partiell fraction decomposition, or regarding $f(x)=\left(\log((x+1)^2+3)\right)'$ -- without success.
I' sure that I'm overseeing a tiny little missing link; dear colleagues, please give me a hint.
The given $f(x)$ is given by \begin{align} f(x) = \frac{2(x-1)}{x^2 - 2x + 4} \end{align} and can be seen as, where $a = 1 + \sqrt{3} i$ and $b = 1 - \sqrt{3} i$, such that $ab=4$, \begin{align} f(x) = \frac{-2 \, (1 - x)}{ab \, \left(1 - \frac{x}{a}\right) \left( 1 - \frac{x}{b}\right)} \end{align} for which \begin{align} \ln f(x) &= \ln\left(- \frac{4}{ab}\right) + \ln(1 - x) - \ln\left(1 - \frac{x}{a}\right) - \ln\left(1 - \frac{x}{b}\right) \\ &= \pi i - \ln 2 + \ln(1 - x) - \ln\left(1 - \frac{x}{a}\right) - \ln\left(1 - \frac{x}{b}\right) \\ &= \pi i - \ln 2 - \sum_{n=1}^{\infty} \left( 1 - \frac{1}{a^{n}} - \frac{1}{b^{n}} \right) \, \frac{x^{n}}{n} \\ &= \pi i - \ln 2 + \sum_{n=1}^{\infty} \left( \frac{a^{n} + b^{n}}{4^{n}} - 1 \right) \, \frac{x^{n}}{n} \\ &= \pi i - \ln 2 + \sum_{n=1}^{\infty} \frac{a^{n} + b^{n} - 4^{n} }{n} \, \left(\frac{x}{4}\right)^{n} \\ &= \pi i - \ln 2 + \sum_{n=1}^{\infty} \frac{ 2 \, \cos\left(\frac{n \pi }{3}\right) - 2^{n} }{n} \, \left(\frac{x}{2}\right)^{n}. \end{align} Now differentiating both sides leads to \begin{align} \frac{f'}{f} &= \sum_{n=1}^{\infty} \left( 2 \, \cos\left(\frac{n \pi }{3}\right) - 2^{n} \right) \, \left(\frac{x}{2}\right)^{n} \\ &= \frac{1}{2} \, \sum_{n=1}^{\infty} \left(2 \, \cos\left(\frac{n \pi}{3}\right) - 2^{n} \right) \, \left(\frac{x}{2}\right)^{n-1} \\ &= \sum_{n=0}^{\infty} \left( \cos\left(\frac{(n+1) \pi}{3}\right) - 2^{n} \right) \, \left(\frac{x}{2}\right)^{n}. \end{align} Expanding the first few terms provides \begin{align} - \frac{f'}{f} = \frac{1}{2} + \frac{5 \, x}{4} + \frac{5 \, x^{2}}{4} + \frac{17 \, x^{3}}{16} + \frac{31 \, x^{4}}{32} + \mathcal{O}(x^{5}) \end{align}
Given
$$ f(x) = \frac{ 2 x - 2 }{ x^2 - 2 x + 4 }. \tag 1 $$
Assuming you want to expand $f(x)$.
Let $$ \phi_\pm = 1 \pm \mathtt{i} \sqrt{3}. \tag 2 $$ We can write (1) as $$ f(x) = \frac{ 1 }{ x - \phi_+ } + \frac{ 1 }{ x - \phi_- }. \tag 3 $$
Whence $$ f(x) = - \sum_{k=0}^\infty \left( \frac{1}{{\phi_+}^{k+1}} + \frac{1}{{\phi_-}^{k+1}} \right) x^k. \tag 4 $$
Note that $$ \phi_\pm = 1 \pm \mathtt{i} \sqrt{3} = 2 {\exp}\big( \pm \mathtt{i} \pi / 3 \big). \tag 5 $$
Putting (5) in (4) yields
$$ \bbox[16px,border:2px solid #800000] { \frac{ 2 x - 2 }{ x^2 - 2 x + 4 } = - \sum_{k=0}^\infty {\cos}\big( [k+1] \pi / 3 \big) \left( \frac{x}{2} \right)^k. } \tag {I} $$
Simple expansion yields
$$ \bbox[16px,border:2px solid #800000] { \frac{ 2 x - 2 }{ x^2 - 2 x + 4 } = - \frac{1}{2} + \frac{x}{4} + \frac{x^2}{4} + \frac{x^3}{16} - \frac{x^4}{32} - \frac{x^5}{32} - \frac{x^6}{128} + \cdots } \tag{II} $$
If you want to expand $$ \frac{f'(x)}{f(x)}, \tag 8 $$
then use the simple relation $$ \frac{f'(x)}{f(x)} = \frac{g''(x)}{g'(x)} - \frac{g'(x)}{g(x)}. \tag 9 $$
Note that $$ \frac{g''(x)}{g'(x)} = \frac{2x}{2x-2} = - \sum_{k=0}^\infty x^k \tag {10}. $$
Combination with (II) yields
$$ \bbox[16px,border:2px solid #800000] { \frac{f'(x)}{f(x)} = + \sum_{k=0}^\infty \left( 2^{-k} {\cos}\big( [k+1] \pi / 3 \big) - 1 \right) x^k. } \tag {III} $$
Simple expansion yields
$$ \bbox[16px,border:2px solid #800000] { \frac{f'(x)}{f(x)} = - \frac{1}{2} - \frac{5x}{4} - \frac{5x^2}{4} - \frac{17x^3}{16} - \frac{31x^4}{32} - \frac{31x^5}{32} - \frac{127x^6}{128} \cdots } \tag{IV} $$
You could try noting $$\frac{1}{(x-2)^{2}}=\frac{1}{2}\frac{d}{dx}\left(\frac{1}{1-(x/2)}\right)$$ Then $$f(x)=2(x-1)\frac{1}{2}\frac{d}{dx}\left(\frac{1}{1-(x/2)}\right)$$ lends itself to a power-series expansion immediately...
Given any function $f$, if we restrict to where $f$ is nonzero (so that either $\log f$ or $\log -f$ exists), we can write $\frac{f'}{f}=\frac{d}{dt}\log |f|$. If $f=c \prod (x-\alpha_i)^{k_i}$ is a rational function, $\log f = \log c + \sum k_i \log (x-\alpha_i)$ and so $$\frac{f'}{f}=\frac{d}{dt}\log |f| = \sum k_i \frac{1}{x-\alpha}.$$
Now, use the fact that $\frac{1}{1-x}=\sum x^i$, and the factorization that $(2x-2)/(x^2−2x+4)= 2(x-1)(x-1+\sqrt{3})^{-1}(x-1-\sqrt{3})^{-1}$.
Since $$ x^2-2x+4=(x-1)^2+3=(x-2u)(x-2\bar{u}), $$ with $$ u=e^{i\frac\pi3}, $$ we have \begin{eqnarray} f(x)&=&\frac{2x-2}{(x-2u)(x-2\bar{u})}=\frac{1}{x-2u}+\frac{1}{x-2\bar{u}}=-\frac{\bar{u}}{2}\cdot\frac{1}{1-\frac{\bar{u}}{2}x}-\frac{u}{2}\cdot\frac{1}{1-\frac{u}{2}x}\\ &=&-\frac{\bar{u}}{2}\sum_{n=0}^\infty\left(\frac{\bar{u}}{2}\right)^nx^n-\frac{u}{2}\sum_{n=0}^\infty\left(\frac{u}{2}\right)^nx^n =-\sum_{n=0}^\infty\left(\frac{\bar{u}}{2}\right)^{n+1}x^n-\sum_{n=0}^\infty\left(\frac{u}{2}\right)^{n+1}x^n\\ &=&-\sum_{n=0}^\infty\frac{u^{n+1}+\bar{u}^{n+1}}{2^{n+1}}x^n=-\sum_{n=0}^\infty\frac{1}{2^n}\cos\left(\frac{n+1}{3}\pi\right)x^n. \end{eqnarray} Also \begin{eqnarray} \frac{f'(x)}{f(x)}&=&\left[\ln |f(x)|\right]'=[\ln2+\ln|x-1|-\ln|x^2-2x+4|]'\\ &=&\frac{1}{x-1}-\frac{2x-2}{x^2-2x+4}=-\frac{1}{1-x}-f(x)\\ &=&-\sum_{n=0}^\infty x^n-\sum_{n=0}^\infty\frac{1}{2^n}\cos\left(\frac{n+1}{3}\pi\right)x^n\\ &=&-\sum_{n=0}^\infty\left[1+\frac{1}{2^n}\cos\left(\frac{n+1}{3}\pi\right)\right]x^n. \end{eqnarray}
censored. Can't you just differentiate again and again and try to spot a pattern between the successive derivatives? – Khallil Jul 25 '15 at 17:47