# What is the Taylor series expansion of f(x) = 1/x^2 at a=1?

Jun 20, 2015

The Taylor series expansion, in general, is written as:

sum_(n=0)^(oo) f^n(a)/(n!)(x-a)^n

So, we will have to take $n$ derivatives of $\frac{1}{x} ^ 2$. $n = 3$ is the bare minimum in my opinion if you want to see a significant chunk of a pattern, but let's just stop at $n = 4$; this derivative isn't too bad, I guess. You just get pretty large numbers past $n = 4$.

${f}^{0} \left(x\right) = \textcolor{b l u e}{f \left(x\right)} = {x}^{-} 2 \textcolor{b l u e}{= \frac{1}{x} ^ 2}$

$\textcolor{b l u e}{f ' \left(x\right)} = - 2 {x}^{-} 3 \textcolor{b l u e}{= - \frac{2}{x} ^ 3}$

$\textcolor{b l u e}{f ' ' \left(x\right)} = 6 {x}^{-} 4 \textcolor{b l u e}{= \frac{6}{x} ^ 4}$

$\textcolor{b l u e}{f ' ' ' \left(x\right)} = - 24 {x}^{-} 5 \textcolor{b l u e}{= - \frac{24}{x} ^ 5}$

$\textcolor{b l u e}{f ' ' ' ' \left(x\right)} = 120 {x}^{-} 6 = \textcolor{b l u e}{\frac{120}{x} ^ 6}$

Now, let's plug them in. Just remember:

• $x \to a$, except for the term ${\left(x - a\right)}^{n}$, which remains as ${\left(x - a\right)}^{n}$.
• $n$ does vary from $0$ to your choice of $n$. We chose ${n}_{e n d} = 4$.
• $a$ is a constant we choose. We chose $a = 1$ for simplicity, and because it's close to $x = 0$ (if we let $x \to a$ as $a \to 0$, then $x \to 0$, and our Taylor series becomes more accurate, usually). $a$ does not vary.

Generally, the sum is written out to be:
= f^0(1)/(0!)(x-1)^0 + (f'(1))/(1!)(x-1)^1 + (f''(1))/(2!)(x-1)^2 + (f'''(1))/(3!)(x-1)^3 + (f''''(1))/(4!)(x-1)^4 + ...

Now plug in your newly-acquired derivatives:
= (1/(1^2))/(1!)(1) + (-2/1^3)/(1!)(x-1) + (6/1^4)/(2!)(x-1)^2 + (-24/1^5)/(3!)(x-1)^3 + (120/1^6)/(4!)(x-1)^4 + ...

Simplify:
$= 1 + \left(- 2\right) \left(x - 1\right) + \frac{6}{2} {\left(x - 1\right)}^{2} + \frac{- 24}{6} {\left(x - 1\right)}^{3} + \frac{120}{24} {\left(x - 1\right)}^{4} + \ldots$

And simplify some more:
$= \textcolor{b l u e}{1 - 2 \left(x - 1\right) + 3 {\left(x - 1\right)}^{2} - 4 {\left(x - 1\right)}^{3} + 5 {\left(x - 1\right)}^{4} + \ldots}$