## Calculus with Applications (10th Edition)

$$\approx - 0.030459202$$
\eqalign{ & \ln 0.97 \cr & {\text{Using the result of Taylor polynomial of degree }}4{\text{ at }}x = 0{\text{ found in the exercise 11}} \cr & f\left( x \right) = \ln \left( {1 - x} \right){\text{ we found that}} \cr & {P_4}\left( x \right) = - x - \frac{1}{2}{x^2} - \frac{1}{3}{x^3} - \frac{1}{4}{x^4} \cr & {\text{To approximate }}\ln 0.97,{\text{ we must evaluate }}\ln \left( {1 - 0.03} \right),{\text{ then taking }}x = 0.03 \cr & {P_4}\left( {0.03} \right) = - \left( {0.03} \right) - \frac{1}{2}{\left( {0.03} \right)^2} - \frac{1}{3}{\left( {0.03} \right)^3} - \frac{1}{4}{\left( {0.03} \right)^4} \cr & {P_4}\left( {0.03} \right) = - 0.03 - 0.00045 - 0.000009 - 0.000000202 \cr & {P_4}\left( {0.03} \right) = - 0.030459202 \cr & {\text{Thus}}{\text{, }}\ln 0.97 \approx - 0.030459202 \cr}