1. ## Taylor series

Find the Taylor series for:
$$f(x) = \tfrac{1}{x^2}, a=1$$ Center at a = 1.

I've written down up to the 4th derivative and its value at 1, but I always have trouble actually getting the taylor series notation.

$$\sum \tfrac{(-1)^n(n(n+1))}{n!}(x-1)^n$$ is what I got but I know it's not right...thanks for any help

2. ## Re: Taylor series

Assume you can write $\displaystyle x^{-2} = c_0 + c_1(x - 1) + c_2(x - 1)^2 + c_3(x - 1)^3 + \dots$

Let $\displaystyle x = 1$ and we find $\displaystyle c_0 = 1$.

Differentiate both sides

$\displaystyle -2x^{-3} = c_1 + 2c_2(x - 1) + 3c_3(x - 1)^2 + 4c_4(x - 1)^3 + \dots$

Let $\displaystyle x = 1$ to find $\displaystyle c_1 = -2$.

Differentiate both sides

$\displaystyle 6x^{-4} = 2c_2 + 6c_3(x - 1) + 12c_4(x - 1)^2 + 20c_5(x - 1)^3 + \dots$

Let $\displaystyle x = 1$ to find $\displaystyle c_2 = 3$

Differentiate both sides

$\displaystyle -24x^{-5} = 6c_3 + 24c_4(x - 1) + 60c_5(x - 1)^2 + \dots$

Let $\displaystyle x = 1$ to find $\displaystyle c_3 = -4$.

I expect you're now seeing a pattern...

$\displaystyle \frac{1}{x^2} = 1 - 2(x - 1) + 3(x - 1)^2 - 4(x - 1)^3 + \dots = \sum_{n = 0}^{\infty}(-1)^n(n + 1)(x - 1)^n$

3. ## Re: Taylor series

i always do these problems by calculating the derivatives first.

$f(1) = 1$
$f'(1) = f'(x)|_{x=1} = \frac{-2}{x^3}|_{x=1} = -2$
$f''(1) = f''(x)|_{x=1} = \frac{6}{x^4}|_{x=1} = 6$

a quick inductive proof (which you should supply) shows that:

$f^{(n)}(x) = (n+1)!(-1)^{n+1}x^{-(n+2)}$

so $f^{(n)}(1) = (n+1)!(-1)^{n+1}$

therefore, the Taylor series centered at a = 1, is:

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

4. ## Re: Taylor series

Originally Posted by Intrusion
Find the Taylor series for:
$$f(x) = \tfrac{1}{x^2}, a=1$$ Center at a = 1.

I've written down up to the 4th derivative and its value at 1, but I always have trouble actually getting the taylor series notation.

$$\sum \tfrac{(-1)^n(n(n+1))}{n!}(x-1)^n$$ is what I got but I know it's not right...thanks for any help
A fast way is setting $x=1+\xi$ so that You obtain...

$f(x)= \frac{1}{(1+\xi)^{2}}= - \frac{d}{d \xi} \frac{1}{1+\xi}=$

$- \frac{d}{d \xi} \sum_{n=0}^{\infty} (-1)^{n} \xi^{n}= \sum_{n=0}^{\infty} (-1)^{n-1} n\ \xi^{n-1} = \sum_{n=0}^{\infty} (-1)^{n-1} n\ (x-1)^{n-1}$ (1)

Kind regards

$\chi$ $\sigma$