Answer
$$ \approx 1.01488$$
Work Step by Step
$$\eqalign{
& \sqrt {1.03} \cr
& {\text{Using the result of Taylor polynomial of degree }}4{\text{ at }}x = 0{\text{ found }} \cr
& {\text{in the exercise 19 above }}\left( {page{\text{ 662}}} \right){\text{. for }}f\left( x \right) = \sqrt {x + 1} {\text{ }} \cr
& {\text{we found that}} \cr
& {P_4}\left( x \right) = 1 + \frac{1}{2}x - \frac{1}{8}{x^2} + \frac{1}{{16}}{x^3} - \frac{5}{{128}}{x^4} \cr
& {\text{To approximate }}\sqrt {1.03} ,{\text{ we must evaluate }}\sqrt {1 + 0.03} ,{\text{ }} \cr
& {\text{then taking }}x = 0.03 \cr
& {P_4}\left( {0.03} \right) = 1 + \frac{1}{2}\left( {0.03} \right) - \frac{1}{8}{\left( {0.03} \right)^2} + \frac{1}{{16}}{\left( {0.03} \right)^3} - \frac{5}{{128}}{\left( {0.03} \right)^4} \cr
& {P_4}\left( {0.03} \right) = 1 + 0.015 - 0.0001125 + 0.000001687 - 0.00003164 \times {10^{ - 3}} \cr
& {P_4}\left( {0.02} \right) = 1.01488 \cr
& {\text{Thus}}{\text{, }}\sqrt {1.03} \approx 1.01488 \cr} $$
