# Find the Taylor expansion \color(red)\bb\text(formula)... for f(x)=1/x^2 given a=4?

## Please check my work (it has to be done using the colored parts, sorry): $f ' = \left(- 2\right) {x}^{-} 3$ $f ' ' = \left(- 2\right) \left(- 3\right) {x}^{-} 4$ $f ' ' ' = \left(- 2\right) \left(- 3\right) \left(- 4\right) {x}^{-} 5$ $f ' \left(4\right) = \left(- 2\right) \cdot {4}^{-} 3$ $f ' ' \left(4\right) = \left(- 2\right) \left(- 3\right) \cdot {4}^{-} 4$ $f ' ' ' \left(4\right) = \left(- 2\right) \left(- 3\right) \left(- 4\right) \cdot {4}^{-} 5$ \color(green)(f^n(4))=(-1)^n(n-1)!4^-n=\color(olive)(((-1)^n(n-1)!)/4^n) \color(red)(C_n=f^n(a)*1/(n!))=((-1)^n(n-1)!)/4^n*1/(n!)=((-1)^n(n-1))/4^n $\setminus \rightarrow \setminus \textcolor{red}{f \left(x\right) = \setminus {\sum}_{n = 0}^{\setminus} \infty {C}_{n} {\left(x - a\right)}^{n}} = \setminus {\sum}_{n = 0}^{\setminus} \infty \frac{{\left(- 1\right)}^{n} \left(n - 1\right)}{4} ^ n \cdot {\left(x - 2\right)}^{n}$ (Can I simplify this further?)

May 6, 2018

$\frac{1}{x} ^ 2 = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} \frac{{\left(x - 4\right)}^{n} \left(n + 1\right)}{{4}^{n + 2}}$

#### Explanation:

Let's determine the pattern for the $n t h$ derivative evaluated at $4$. We adopt the convention that the $0 t h$ derivative is just the function itself.

$f \left(x\right) = {x}^{-} 2 , f \left(4\right) = {4}^{-} 2$

$f ' \left(x\right) = - 2 {x}^{-} 3 , f ' \left(4\right) = - 2 {\left(4\right)}^{-} 3$

$f ' ' \left(x\right) = \left(- 2\right) \left(- 3\right) {x}^{-} 4 , f ' ' \left(4\right) = \left(- 2\right) \left(- 3\right) {4}^{-} 4$

$f ' ' ' \left(x\right) = \left(- 2\right) \left(- 3\right) \left(- 4\right) {x}^{-} 5 , f ' ' ' \left(4\right) = \left(- 2\right) \left(- 3\right) \left(- 4\right) {4}^{-} 5$

We can see that we start out with a positive term and then alternate between negative and positive terms, so, we have an instance of ${\left(- 1\right)}^{n} .$

The first exponent on the $4$ is $- 2$, so, since we're starting at $n = 0 ,$ we're going to have a ${4}^{-} \left(n + 2\right) = \frac{1}{4} ^ \left(n + 2\right) .$

Finally, for the factorial, the 0th derivative has a coefficient of $1$, or 1!, the first derivative has a coefficient of $- 2 ,$ or -(2!), the second derivative has a coefficient of $\left(- 2\right) \left(- 3\right) ,$ or 3!.

Knowing that the negatives are handled by the ${\left(- 1\right)}^{n} ,$ we can represent the factorials by (n+1)!

Thus,

f^((n))(4)=(-1)^n((n+1)!)/4^(n+2)

Now, using the general form of a Taylor series about $a ,$

sum_(n=0)^oof^((n))(a)(x-a)^n/(n!), we get

1/x^2=sum_(n=0)^oo(-1)^n((x-4)^n(n+1)!)/(n!4^(n+2))

We can simplify the factorials, as (n+1)! = (n+1)(n!)

1/x^2=sum_(n=0)^oo(-1)^n((x-4)^n(n+1)cancel(n!))/(cancel(n!)4^(n+2))

$\frac{1}{x} ^ 2 = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} \frac{{\left(x - 4\right)}^{n} \left(n + 1\right)}{{4}^{n + 2}}$