# What is the coefficient of x^2 in the Taylor series for (1+x)^-2 about a=0?

Aug 31, 2015

If you were to want the entire Taylor series, I would start by taking the necessary number of derivatives to make it to the ${x}^{2}$ term (i.e. the $n = 2$ term). That means up to and including the 2nd derivative.

$\textcolor{g r e e n}{{f}^{\left(0\right)} \left(x\right)} = f \left(x\right) = {\left(1 + x\right)}^{- 2} \textcolor{g r e e n}{= \frac{1}{1 + x} ^ 2}$

$\textcolor{g r e e n}{f ' \left(x\right)} = - 2 {\left(1 + x\right)}^{- 3} = \textcolor{g r e e n}{- \frac{2}{1 + x} ^ 3}$

$\textcolor{g r e e n}{f ' ' \left(x\right)} = 6 {\left(1 + x\right)}^{- 4} = \textcolor{g r e e n}{\frac{6}{1 + x} ^ 4}$

The general formula for the Taylor series is:

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

For $N = 2$ (otherwise known as "truncating at the second order term") and centered around $a = 0$, this is:

= f^((0))(0)/(0!)(x-0)^0 + (f'(0))/(1!)(x-0)^1 + (f''(0))/(2!)(x-0)^2

= (1/(1+(0))^2)/(0!)*1 + (-2/(1+(0))^3)/(1!)x + (6/(1+(0))^4)/(2!)x^2

$= \left(\frac{1}{1} ^ 2\right) + \left(- \frac{2}{1} ^ 3\right) x + \left(\frac{6}{2 \cdot {\left(1\right)}^{4}}\right) {x}^{2}$

$= 1 - \left(\frac{2}{1}\right) x + \left(\frac{6}{2}\right) {x}^{2}$

$= \textcolor{b l u e}{1 - 2 x + 3 {x}^{2}}$

So your coefficient is $3$. Now, you could have also done only the 2nd derivative, but this shows you the pattern for the entire Taylor series. You could write out more terms:

$= 1 - 2 x + 3 {x}^{2} - 4 {x}^{3} + 5 {x}^{4} - \ldots$

Note that for the series, $a$ does not vary, but $n$ does. Also, ${\left(x - a\right)}^{n}$ is not affected by $x \to a$.