Taylor series for $\frac{x^2 - 4x + 4}{2x^2 - 8x + 5}$ at $x = 2$

Question

To find the Taylor series of $f(x) = \dfrac{x^2 - 4x + 4}{2x^2 - 8x + 5}$ at $x = 2$, I can either:

• Differentiate up to $n$th degree and determine the series by observation;
• Reduce $f(x)$ to a commonly-known Taylor series expansion and infer from there.

I chose the latter option, but I am not sure if my approach is correct. Reproduced below:

$f(x) = \dfrac{x^2 - 4x + 4}{2x^2 - 8x + 5} = \dfrac{3 + 2x^2 - 8x + 5}{2(2x^2 - 8x + 5)}$

Let $g = 2x^2 - 8x + 5$, then

$f(x) = \dfrac{3 + g}{2g} = \dfrac{1}{2} - 3{\dfrac{1}{1 - (1 + 2g)}}$

From the identity $\dfrac{1}{1 - x} = \sum^{\infty}_{n = 0} x^n$,

$\therefore f(x) = \dfrac{1}{2} - 3 \sum^{\infty}_{n = 0}{(4x^2 - 16x +11)^n}$

But I don't think I am correct because Wolfram Alpha's expansion is:

Answer 1

HINT

Let

$$\dfrac{x^2 - 4x + 4}{2x^2 - 8x + 5}=\dfrac{(x-2)^2}{2(x-2)^2-3}$$

Answer 2

Here is another way to expand your function: we can observe that $$\frac{x^2 - 4x + 4}{2x^2 - 8x + 5} = \frac{(x-2)^2}{2(x-2)^2-3}$$

What is the Taylor expansion of $\dfrac{x^2}{2x^2-3}$?