# How do you write the taylor series about 0 for the function (1+x)^3?

##### 1 Answer
Aug 2, 2015

The Taylor series computes as $1 + 3 x + 3 {x}^{2} + {x}^{3}$ as expected...

#### Explanation:

Let $f \left(x\right) = {\left(1 + x\right)}^{3}$

Then:

$f ' \left(x\right) = 3 {\left(1 + x\right)}^{2}$

$f ' ' \left(x\right) = 6 \left(1 + x\right)$

$f ' ' ' \left(x\right) = 6$

So $f \left(0\right) = 1$, $f ' \left(0\right) = 3$, $f ' ' \left(0\right) = 6$ and $f ' ' ' \left(0\right) = 6$

So the Taylor series for $f \left(x\right)$ about $0$ gives us:

f(x) = f(0)/(0!) + (f'(0))/(1!)x + (f''(0))/(2!)x^2 + (f'''(0))/(3!)x^3

$= \frac{1}{1} + \frac{3}{1} x + \frac{6}{2} {x}^{2} + \frac{6}{6} {x}^{3}$

$= 1 + 3 x + 3 {x}^{2} + {x}^{3}$