I am trying to find the Taylor series of the following expression about $x = 1$:
\begin{align} \frac{4x-5}{2x^2-4x+5} \end{align} My steps are as follows:
\begin{align} 2x^2-4x+5 = 2(x-1)^2+3 \quad (\text{by completing the square}) \end{align} \begin{align} \therefore \quad\frac{4x-5}{2x^2-4x+5} &= \frac{4x-5}{2(x-1)^2+3} \\ \\ &= \frac{4(x-1)-1}{2(x-1)^2+3} \\ \\ &= \frac{4(x-1)-1}{3}\cdot\frac{1}{1+\frac{2}{3}(x-1)^2} \\ \\ &= \frac{4(x-1)-1}{3}\cdot\sum_{n=0}^{\infty}(-1)^{n}\left(\frac{2^n}{3^{n}}(x-1)^{2n}\right) \\ \\ &= (4(x-1)-1)\cdot\sum_{n=0}^{\infty}(-1)^{n}\left(\frac{2^n}{3^{n+1}}(x-1)^{2n}\right) \\ \\ &= \sum_{n=0}^{\infty}(-1)^{n}\left(\frac{2^{n+2}}{3^{n+1}}(x-1)^{2n+1}\right) - \sum_{n=0}^{\infty}(-1)^{n}\left(\frac{2^{n}}{3^{n+1}}(x-1)^{2n}\right) \end{align}
I am pretty sure my answer is wrong even though I think that my steps are correct. I am thinking that the Taylor series should turn out as a single summation term. But I can't see how to simplify this any further. Is anyone able to advise whether my method and answer is correct?
Let
$$f(x)=\frac{4x-5}{2x^2-4x+5}$$
and
$$g(x)=f(x+1)$$.
we expand $g(x)$ arround $x=0$.
$$g(x)=\frac{4x-1}{2x^2+3}$$
$$=\frac{4x-1}{3}\frac{1}{1+\frac{2}{3}x^2}$$
$$=\frac{4x-1}{3}\sum_{n=0}^{+\infty}(-\frac{2x^2}{3})^n$$.
finally
$f(x)=g(x-1)$.