Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 9 - Further Applications of the Integral and Taylor Polynomials - 9.4 Taylor Polynomials - Exercises - Page 492: 1


$$T_2=x,\ \ T_3= x-\frac{1}{6} x^{3}$$

Work Step by Step

Since \begin{array}{ll} {f(x)=\sin x,} & {f(0)=0} \\ {f^{\prime}(x)=\cos x,} & {f^{\prime}(0)=1} \\ {f^{\prime \prime}(x)=-\sin x,} & {f^{\prime \prime}(0)=0} \\ {f^{\prime \prime \prime}(x)=-\cos x,} & {f^{\prime \prime \prime}(0)=-1} \end{array} Then \begin{aligned} T_{2}(x) &=f(a)+\frac{f^{\prime}(a)}{1 !}(x-a)+\frac{f^{\prime \prime}(a)}{2 !}(x-a)^{2} \\ &=0+(x-0)+(0)(x-0)^{2} \\ &=x \end{aligned} and \begin{aligned} T_{3}(x) &=f(a)+\frac{f^{\prime}(a)}{1 !}(x-a)+\frac{f^{\prime \prime}(a)}{2 !}(x-a)^{2}+\frac{f^{\prime \prime \prime}(a)}{3 !}(x-a)^{3} \\ &=0+(0)(x-0)+(x-0)^{2}-\frac{1}{6}(x-0)^{3} \\ &=x-\frac{1}{6} x^{3} \end{aligned}
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