# Use Lagrange Remainder to estimate the error term

The question asked us to find an upper bound for the absolute error in estimating $$\sqrt{x}$$ for $$3\leq x \leq 5$$ with the quadratic $$0.75+0.375x-0.015625x^2$$. A close look reveals tha the quadratic is the second Taylor polynomial of $$\sqrt{x}$$ at $$x=0$$. So I worked to find the error term as given by the Lagrange remainder:

$$R_3(x)=\frac{f'''(c)}{3!}x^3=\frac{3}{8}c^{-\frac{5}{2}}\frac{x^3}{3!}$$

for some c between 0 and x. Now since $$3\leq x \leq 5$$, $$0\leq c \leq 5$$. Then I was stuck. Wouldn't the error be really big when $$c$$ is close to $$0$$?

The answer given in the book is $$\frac{3^{-\frac{5}{2}}}{2^4}$$. Sorry for asking stupid questions. I would really appreciate any help!

• There is no Taylor series of $\sqrt{x}$ at $x=0$ – gammatester Oct 6 '18 at 12:34

## 1 Answer

That is the taylor series at point $$x_0=4$$.

For $$3 \le x \le 5$$, we have $$|x-x_0| \le 1$$.

We have $$3\le c \le 5$$

$$5^{-\frac52} \le c^{-\frac52} \le 3^{-\frac52}$$

Hence an upper bound for the error is $$\frac{3}{8}\cdot 3^{-\frac52}\cdot\frac{1}{3!}=\frac{3^{-\frac52}}{16}$$

• Oops just realised the stupid mistake that I made. Thank you for your answer! – M. W Oct 6 '18 at 12:51