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I need help with the following calculus problem:

Use completing the square and the geometric series to get the Taylor expansion about ${x=2}$ of ${\frac{1}{x^{2}+4x+3}}$

So far I have the following: By completing the square. It can be shown that

${\frac{1}{x^{2}+4x+3} = \frac{1}{(x+2)^{2}-1}}$

We want to use the geometric series, so we recognize that
${\frac{1}{(x+2)^{2}-1} = -1\times\frac{1}{1-(x+2)^{2}}}$

Using what we know about the geometric series, it is seen that
${\frac{1}{1-(x+2)^2} = \sum\limits_{k=0}^\infty (x+2)^{2k}}$ and our Taylor series about ${x = -2}$ is ${-\sum\limits_{0}^{\infty}(x+2)^{2k}}$

I am stuck, however on figuring out how to use this expansion about ${x=-2}$ to find the expansion about ${x=2}$. How is this done, without calculating the derivative at ${x=2}$ of the original function and trying to find a pattern for the summation?

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  • $\begingroup$ The Taylor expansion that you have given is not centered around $x=0$. That would have the have the form $\sum a_nx^n$. The Taylor series that you have given is centered around $-2$. $\endgroup$
    – Tim Raczkowski
    Jan 12, 2015 at 22:56
  • $\begingroup$ Is this because of the restriction on the geometric series that abs(x) < 1. So in my case abs( (x+2)^2 ) < 1 and therefore -3 < x <-1 is my domain of convergence? $\endgroup$
    – James Bender
    Jan 12, 2015 at 23:09
  • $\begingroup$ The Taylor expansion about $x=x_0$ has the form $\sum a_n(x-x_0)^n$. Since, the Taylor series you gave is in terms of powers of $(x+2)$, or $(x-(-2))$, $x_0=-2$. $\endgroup$
    – Tim Raczkowski
    Jan 12, 2015 at 23:19
  • $\begingroup$ Maybe I'm going about it the wrong way, but do you know how you'd find this series about x=2? Using only completing the square and the geometric series. $\endgroup$
    – James Bender
    Jan 12, 2015 at 23:27
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    $\begingroup$ For what it's worth, it seems extremely likely to me that the original problem has a type and should either read $x^2 - 4x + 3$ in the denominator or ask for the series to be centered around $-2$. If we assume that the problem is correctly written, then Jack's answer is definitely the way to go: partial fractions and geometric series. At least that's how I see this problem from the perspective of the skills and understanding it attempts to reflect. $\endgroup$
    – davidlowryduda ♦
    Jan 13, 2015 at 0:21

1 Answer 1

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$$\frac{1}{x^2+4x+3}=\frac{1}{(x+1)(x+3)}=\frac{1}{2}\left(\frac{1}{x+1}-\frac{1}{x+3}\right)=\frac{1}{2}\left(\frac{1}{(x-2)+3}-\frac{1}{(x-2)+5}\right)$$ Now just use: $$\frac{1}{z+3}=\frac{\frac{1}{3}}{1+\frac{z}{3}}=\sum_{n\geq 0}\frac{(-1)^n}{3^{n+1}}z^n $$ to get: $$\frac{1}{x^2+4x+3}=\frac{1}{2}\sum_{n\geq 0}\left(\frac{1}{3^{n+1}}-\frac{1}{5^{n+1}}\right)(-1)^n (x-2)^n.$$

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