# Thread: Error in taylor polynomial

1. ## Error in taylor polynomial

Suppose that $f(x)=\sqrt{1+x}$ and let $p_2$ denote the second Taylor polynomial for $f$ about $0$. If $x\in [0,1]$ then show that the absolute error in the approximation $f(x)\approx p_2(x)$ does not exceed $\frac{1}{16}$

This is what I've done so far:

$f(x)=\sqrt{1+x}\Rightarrow f(0)=1$

$f'(x)=\frac{1}{2}(1+x)^{-\frac{1}{2}}\Rightarrow f'(0)=\frac{1}{2}$

$f''(x)=-\frac{1}{4}(1+x)^{-\frac{3}{2}}\Rightarrow f''(0)=-\frac{1}{4}$

$p_2(x)=1+\frac{1}{2}x-\frac{1}{8}x^2$

What should I do next? I'm thinking of using the Lagrange formula, but not too sure how I would use it

2. Originally Posted by acevipa
Suppose that $f(x)=\sqrt{1+x}$ and let $p_2$ denote the second Taylor polynomial for $f$ about $0$. If $x\in [0,1]$ then show that the absolute error in the approximation $f(x)\approx p_2(x)$ does not exceed $\frac{1}{16}$

This is what I've done so far:

$f(x)=\sqrt{1+x}\Rightarrow f(0)=1$

$f'(x)=\frac{1}{2}(1+x)^{-\frac{1}{2}}\Rightarrow f'(0)=\frac{1}{2}$

$f''(x)=-\frac{1}{4}(1+x)^{-\frac{3}{2}}\Rightarrow f''(0)=-\frac{1}{4}$

$p_2(x)=1+\frac{1}{2}x-\frac{1}{8}x^2$

What should I do next? I'm thinking of using the Lagrange formula, but not too sure how I would use it
The Lagrange form of the remainder is:

$R_N=\dfrac{f^{(3)}(\xi)x^3}{3!}$

for some $\xi$ in $[0,1]$.

So now find an upper bound for |R_n|

CB