Chapter 9 - Power Series - 9.3 Taylor Series - 9.3 Exercises - Page 694: 9

\[\begin{align} & \left. a \right)1-x+\frac{1}{2!}{{x}^{2}}-\frac{1}{3!}{{x}^{3}} \\ & \left. b \right)\sum\limits_{k=0}^{\infty }{\frac{{{\left( -1 \right)}^{k}}}{k!}{{\left( x \right)}^{k}}} \\ & \left. c \right)\left( -\infty ,\infty \right) \\ \end{align}\]

\[\begin{align} & f\left( x \right)={{e}^{-x}} \\ & \text{Using the definition of Maclaurin series for }f\text{ }\left( \mathbf{1} \right) \\ & \sum\limits_{n=0}^{\infty }{\frac{{{f}^{\left( k \right)}}\left( 0 \right)}{k!}}{{x}^{k}}=f\left( 0 \right)+f'\left( 0 \right)x+\frac{f''\left( 0 \right)}{2!}{{x}^{2}}+\cdots +\frac{{{f}^{\left( n \right)}}\left( 0 \right)}{n!}{{x}^{k}}+\cdots \\ & \\ & \left. a \right)\text{ We need to calculate the first four terms}\text{, so} \\ & f\left( x \right)={{e}^{-x}} \\ & f'\left( x \right)=-{{e}^{-x}} \\ & f''\left( x \right)={{e}^{-x}} \\ & f'''\left( x \right)=-{{e}^{-x}} \\ & \text{Evaluating for }x=0 \\ & f\left( 0 \right)={{e}^{-0}}=1 \\ & f'\left( 0 \right)=-{{e}^{-0}}=-1 \\ & f''\left( 0 \right)={{e}^{-0}}=1 \\ & f'''\left( 0 \right)={{e}^{-0}}=-1 \\ & \text{Substituting the previous result into the formula }\left( \mathbf{1} \right) \\ & =1+\left( -1 \right)x+\frac{1}{2!}{{x}^{2}}+\frac{\left( -1 \right)}{3!}{{x}^{2}} \\ & \text{Therefore},\text{ the first four nonzero terms of the Maclaurin series } \\ & \text{of the function }f\left( x \right)={{e}^{-x}}\text{ are} \\ & =1-x+\frac{1}{2!}{{x}^{2}}-\frac{1}{3!}{{x}^{3}} \\ & \\ & \left. b \right)\text{ Rewrite the power series} \\ & =1-x+\frac{1}{2!}{{x}^{2}}-\frac{1}{3!}{{x}^{3}}\cdots \\ & =\frac{{{\left( -1 \right)}^{0}}}{0!}{{\left( x \right)}^{0}}+\frac{{{\left( -1 \right)}^{1}}}{1!}{{\left( x \right)}^{1}}+\frac{{{\left( -1 \right)}^{2}}}{2!}{{\left( x \right)}^{2}}-\frac{{{\left( -1 \right)}^{3}}}{3!}{{\left( x \right)}^{3}}\cdots \\ & \text{Using he summation notation we can write} \\ & =\sum\limits_{k=0}^{\infty }{\frac{{{\left( -1 \right)}^{k}}}{k!}{{\left( x \right)}^{k}}} \\ & \\ & \left. c \right)\text{Using the ratio test to determine the interval of convergence} \\ & \sum\limits_{k=0}^{\infty }{\frac{{{\left( -1 \right)}^{k}}}{k!}{{\left( x \right)}^{k}}}\to {{a}_{n}}=\frac{{{\left( -1 \right)}^{k}}}{k!}{{\left( x \right)}^{k}} \\ & \text{By the ratio test }\underset{k\to \infty }{\mathop{\lim }}\,\left| \frac{{{a}_{k+1}}}{{{a}_{k}}} \right| \\ & =\underset{k\to \infty }{\mathop{\lim }}\,\frac{\left| \frac{{{\left( -1 \right)}^{k+1}}{{x}^{k+1}}}{\left( k+1 \right)!} \right|}{\left| \frac{{{\left( -1 \right)}^{k}}{{x}^{k}}}{k!} \right|} \\ & =\underset{k\to \infty }{\mathop{\lim }}\,\left| \frac{{{\left( -1 \right)}^{k+1}}{{x}^{k+1}}}{\left( k+1 \right)!}\cdot \frac{k!}{{{\left( -1 \right)}^{k}}{{x}^{k}}} \right| \\ & \text{Simplifying} \\ & =\underset{k\to \infty }{\mathop{\lim }}\,\left| \frac{{{\left( -1 \right)}^{k}}\left( -1 \right){{x}^{k}}\cdot x}{\left( k+1 \right)k!}\cdot \frac{k!}{{{\left( -1 \right)}^{k}}{{x}^{k}}} \right| \\ & =\underset{k\to \infty }{\mathop{\lim }}\,\left| \frac{-x}{k+1} \right| \\ & =\left| -x \right|\underset{k\to \infty }{\mathop{\lim }}\,\left| \frac{1}{k+1} \right| \\ & \text{Evaluating the limit} \\ & \underset{k\to \infty }{\mathop{\lim }}\,\left| \frac{{{a}_{k+1}}}{{{a}_{k}}} \right|=0 \\ & \text{Since the limit is less than 1 for all }x, \\ & \text{the Maclaurin series converges for all real numbers}\text{, then} \\ & \text{The interval of convergence is }\left( -\infty ,\infty \right) \\ \end{align}\]