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: 14


$$T_{2}(x) =x-\frac{1}{2} x ^{2}$$ $$T_{3}(x) =x-\frac{1}{2 } x^{2}+\frac{1}{3 } x^{3}$$

Work Step by Step

Given$$f(x)=\ln (x+1), \quad a=0$$ Since \begin{aligned} f(x) &=\ln (x+1)& & f(0)=0 \\ f^{\prime}(x) &=\frac{1 }{x+1} & & f(0)=1 \\ f^{\prime \prime}(x) &=\frac{-1 }{(x+1)^2} & f(0) &=-1 \\ f^{\prime \prime \prime}(x) &=\frac{2 }{(x+1)^3} & f(0) &=2 \end{aligned} Then \begin{aligned} T_{2}(x) &=f(a)+f^{\prime}(a)(x-a)+\frac{f^{\prime \prime}(a)}{2 !}(x-a)^{2}\\ &=0+1(x-0)+\frac{-1}{2 !}(x-0)^{2} \\ &=x-\frac{1}{2} x ^{2} \end{aligned} and \begin{aligned} T_{3}(x) &=f(a)+f^{\prime}(a)(x-a)+\frac{f^{\prime \prime}(a)}{2 !}(x-a)^{2}+\frac{f^{\prime \prime \prime}(a)}{3 !}(x-a)^{3} \\ &=0+1(x-0)+\frac{-1}{2 !}(x-0)^{2}+\frac{2}{3 !}(x-0)^{3} \\ &=x-\frac{1}{2 } x^{2}+\frac{1}{3 } x^{3} \end{aligned}
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