Taylor series expansion of $f(x) = \frac{1}{x^2}$

Question

Expand $f$ in a Taylor’s series about a center $b = 1$, so that $f(x)=\sum_{n=0}^\infty a_n(x-1)^n $. I know a Taylor series takes the form $$\sum_{n=0}^\infty \frac{f^{n}(a)}{n!}(x-a)^n$$ but I don't know how to go from the original $f$ to that.

Answer 1

$$\sum_{n=0}^\infty \frac{f^n(a)}{n!}(x-a)^n$$

you said "let $b=1$" so here we're just using $a$ instead of $b$. So $a=b=1$. Now let's compute some derivatives....

$$f(x) = \frac{1}{x^2} = x^{-2}$$ $$f'(x) = -2x^{-3}$$ $$f''(x) = 2\cdot3x^{-4}$$ $$f^{(3)}(x) = -2\cdot3\cdot 4x^{-5} =-4!x^{-5} $$ $$f^{n}(x) = (-1)^{n} (n+1)!x^{-(n+2)}$$

$$\sum_{n=0}^\infty \frac{f^n(1)}{n!}(x-1)^n = \sum_{n=0}^\infty (-1)^n (n+1)(x-1)^n$$

so $$a_n = (-1)^n (n+1)$$

Update:

We need to find the interval of convergence for this series because it isn't necessarily true everywhere. We'll use the ratio test, $\lim_{n\to \infty}\left|\frac{a_{n+1}}{a_n}\right| = \lim_{n\to \infty} \left| \frac{(-1)^{n+1}(n+2)(x-1)^{n+1}}{(-1)^n (n+1)(x-1)^n}\right| = \lim_{n\to\infty} \left|\frac{n+2}{n+1}\right| |x-1|=|x-1|$ and from the ratio test we conclude that the series (absolutely) converges if $|x-1|<1$ but it is inconclusive if $|x-1|=1$. Let us consider if $|x-1|=1$, this is true only if $x=0$ or $x=2$. But the function isn't defined at $x=0$ so we can ignore this; so let us consider $x=2$. In this case, $\sum_{n=0}^\infty (-1)^n (n+1)$. Then the sequence $(-1)^n (n+1)$ is just $(1,-2,3,-4,5,-6,7,-8,\ldots)$ and so the partial sums are just $(1,-1,2,-2,3,-3,4,-4,\ldots)$ and so the series diverges. Therefore, the Taylor series converges when $|x-1|<1$ (i.e. when $-1<x-1<1$ that is when $0<x<2$).

Answer 2

You have the general form of a Taylor series in front of you. First, figure out what "$a$" should be for your specific question. Then, compute $f^{(n)}(a)$ for each $n$.

Answer 3

$f(x)= 1/x^2$ so $f(1)=1$

$f'(x)=-2/x^3$ so $f'(1)=-2$

$f''(1)=6/x^4$so $f''(1) =6$

Now all you have to do is plug in your series.

It is a good idea to have enough terms so one can find the pattern.

Answer 4

Here is a way to avoid the formula. Using the fact that expansions are unique, and that uniformly convergent power series can be differentiated term by term within their radius of convergence, use the geometric series to calculate

$\frac{1}{x^{2}}=-\frac{d(1/x)}{dx}=-\frac{d}{dx}\left ( \frac{1}{1-(1-x)} \right )=$

$-\frac{d}{dx}\sum^{\infty}_{n=0}(1-x)^{n}=\frac{d}{dx}\sum^{\infty}_{n=0}(-1)^{n+1}(x-1)^{n}=\sum^{\infty}_{n=1}(-1)^{n+1}n(x-1)^{n-1}=\sum^{\infty}_{n=0}(-1)^{n}(n+1)(x-1)^{n}$