## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 11 - Infinite Series - 11.7 Taylor Series - Exercises - Page 588: 29

#### Answer

Taylors Series: $\dfrac{1}{x}=\Sigma_{n = 0}^{\infty} (-1)^n(x-1)^n$ with convergence interval $(0,2)$.

#### Work Step by Step

Taylors Series: $T(x)=\Sigma_{n = 0}^{\infty} \dfrac{f^n (c)}{n!}(x-c)^n$ If we consider $c=0$, then $T(x)$ is also known as Maclaurin Series. We have the Maclaurin series for $f(x)$ as follows: $f(x)=f(0)+f^{\prime}(0) x+\dfrac{f^{\prime \prime}(0) x^2}{2!}+\dfrac{f^{\prime \prime \prime}(0) x^3}{3!}+\dfrac{f^{\prime \prime \prime \prime}(0) x^4}{4!}+...$ From Table $2$, we notice $f(x)=\dfrac{1}{x+1}=\Sigma_{n = 0}^{\infty}(-1)^nx^n=1-x+x^2-x^3+.....$; converges for $|x| \lt 1$ Now, Taylors Series becomes: $f(x)=\dfrac{1}{1+(x-1)}=\Sigma_{n = 0}^{\infty} (-1)^n(x-1)^n$; converges for $|x-1| \lt 1$ or, $0 \lt x\lt 2$ So, we get Taylors Series: $\dfrac{1}{x}=\Sigma_{n = 0}^{\infty} (-1)^n(x-1)^n$ with convergence interval $(0,2)$.

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