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 493: 21

Answer

\begin{aligned} T_{n}(x) =e+e(x-1)+\frac{e}{2}(x-1)^{2}+\frac{e}{6}(x-1)^{3}+\cdots+\frac{e}{n !}(x-1)^{n} \end{aligned}

Work Step by Step

Since \begin{array}{ll} {f (x)=e^{x},} & {f (1)=e} \\ {f^{\prime}(x)=e^{x},} & {f^{\prime}(1)=e} \\ {f^{\prime \prime}(x)=e^{x},} & {f^{\prime \prime}(1)=e} \\ {f^{\prime \prime \prime}(x)=e^{x},} & {f^{\prime \prime \prime}(1)=e} \\ {\vdots} & {\vdots} \\ {f^{(n)}(x)=e^{x}} & {f^{(n)}(1)=e}\end{array} Then \begin{aligned} T_{n}(x) &=f(1)+\frac{f^{\prime}(1)}{1 !}(x-1)+\frac{f^{\prime \prime}(1)}{2 !}(x-1)^{2}+\cdots+(x-1)^{n} \frac{f^{(n)}(1)}{n !} \\ &=e+e(x-1)+\frac{e}{2}(x-1)^{2}+\frac{e}{6}(x-1)^{3}+\cdots+\frac{e}{n !}(x-1)^{n} \end{aligned}
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