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Write $f(x)=\frac{1}{x^2-4x+3}$ as a Taylor series centered at $a=2$ and determine the interval of convergence.

I'm having problems with constructing the Taylor series. So far, I've gotten

$$f(2)=-1$$ $$f'(2)=0$$ $$f''(2)=-2$$ $$f'''(2)=0$$ $$f''''(2)=-7584$$

but I can't seem to find the pattern. Am I missing something here?

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  • $\begingroup$ By the way, $f''''(2)=-24$. $\endgroup$ – Claude Leibovici Mar 15 '17 at 10:08
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It is known (being a geometric series) that $$\frac{1}{1-u}=1+u+u^2+u^3+\dots\qquad |u|<1$$ and so by replacing $u$ by $u^2$ we get $$\frac{1}{1-u^2}=1+u^2+u^4+u^6+\dots\qquad |u|<1$$

We have$$\begin{align} \frac{1}{x^2-4x+3}&=\frac{1}{(3-x)(1-x)}\\ &=-\frac{1}{2}\bigg[\frac{1}{3-x}-\frac{1}{1-x}\bigg]\\ &=-\frac{1}{2}\bigg[\frac{1}{1-(x-2)}+\frac{1}{1+(x-2)}\bigg]\\ &=-\frac{1}{2}\cdot\frac{2}{[1-(x-2)^2]}\\ &=- \frac{1}{1-(x-2)^2}\qquad \text{write $x-2=u$ to get}\\ &=-\frac{1}{1-u^2}\\ &=-\bigg[1+u^2+u^4+u^6 +\dots\bigg]\\ &=-1-u^2-u^4-u^6-\dots\\ &=-1-(x-2)^2-(x-2)^4-(x-2)^6-\dots \end{align}$$ Since we must have $|x-2|<1$, we get $1<x<3$.

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Hint

Start using $x=y+2$. This gives $x^2-4x+3=y^2-1$. So $$f(x)=\frac{1}{x^2-4x+3}=\frac{1}{y^2-1}=-\frac{1}{2 (1-y)}-\frac{1}{2 (1+y)}$$ Now, use well known expansions around $y=0$.

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