# Chapter 12 - Sequences and Series - 12.3 Taylor Polynomials at 0 - 12.3 Exercises - Page 631: 30

$$\approx 1.996239408$$

#### Work Step by Step

\eqalign{ & \root 4 \of {15.88} \cr & {\text{Using the result of Taylor polynomial of degree }}4{\text{ at }}x = 0{\text{ found in the exercise 10}} \cr & f\left( x \right) = \root 4 \of {x + 16} {\text{ we found that}} \cr & {P_4}\left( x \right) = 2 + \frac{1}{{32}}x - \frac{3}{{4096}}{x^2} + \frac{7}{{262,144}}{x^3} - \frac{{77}}{{67,108,864}}{x^4} \cr & {\text{To approximate }}\root 4 \of {15.88} ,{\text{ we must evaluate }}\sqrt { - 0.12 + 16} ,{\text{ then taking }}x = - 0.12 \cr & {P_4}\left( { - 0.12} \right) = 2 + \frac{1}{{32}}\left( { - 0.12} \right) - \frac{3}{{4096}}{\left( { - 0.12} \right)^2} + \frac{7}{{262,144}}{\left( { - 0.12} \right)^3} - \frac{{77}}{{67,108,864}}{\left( { - 0.12} \right)^4} \cr & {P_4}\left( { - 0.12} \right) = 2 - 0.00375 - 0.000010546 - 0.000000046 - 0.000000237 \times {10^{ - 3}} \cr & {P_4}\left( { - 0.12} \right) = 1.996239408 \cr & {\text{Thus}}{\text{, }}\root 4 \of {15.88} \approx 1.996239408 \cr}

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