# Thread: Error in taylor polynomial

1. ## Error in taylor polynomial

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

This is what I've done so far:

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

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

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

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

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

This is what I've done so far:

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

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

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

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

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:

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

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

So now find an upper bound for |R_n|

CB