# How do you find the taylor series of 1 1x^2?

Aug 9, 2015

The Taylor series basically hands us back $11 {x}^{2}$ ...

#### Explanation:

The Taylor series for $f \left(x\right)$ about $0$ is:

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

For $f \left(x\right) = 11 {x}^{2}$, we have

$f ' \left(x\right) = 22 x , f ' ' \left(x\right) = 22 , f ' ' ' \left(x\right) = 0 , \ldots$

So

$f \left(0\right) = 0$, $f ' \left(0\right) = 0$, $f ' ' \left(0\right) = 22$, $f ' ' ' \left(0\right) = 0$, ...

and the Taylor series is:

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

=0+0/(1!)x+22/(2!)x^2+0/(3!)x^3+...

$= 11 {x}^{2}$