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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!

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  • $\begingroup$ There is no Taylor series of $\sqrt{x}$ at $x=0$ $\endgroup$ – gammatester Oct 6 '18 at 12:34
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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 $$\frac{3}{8}\cdot 3^{-\frac52}\cdot\frac{1}{3!}=\frac{3^{-\frac52}}{16}$$

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

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