Chapter 9 - Power Series - Review Exercises - Page 705: 35

\[\begin{align} & \left. a \right)4x-\frac{64}{3}{{x}^{3}}+\frac{1024}{5}{{x}^{5}} \\ & \left. b \right)\sum\limits_{k=0}^{\infty }{\frac{{{\left( -1 \right)}^{k}}{{\left( 4x \right)}^{2k+1}}}{2k+1}} \\ \end{align}\]

\[\begin{align} & f\left( x \right)={{\tan }^{-1}}4x,\text{ }a=0 \\ & \text{Using the definition of Taylor series for a function}\left( page\,685 \right) \\ & \sum\limits_{k=0}^{\infty }{\frac{{{f}^{\left( k \right)}}\left( a \right)}{k!}{{\left( x-a \right)}^{k}}}\text{ }\left( \mathbf{1} \right) \\ & =f\left( a \right)+f'\left( a \right)\left( x-a \right)+\frac{f''\left( a \right)}{2!}{{\left( x-a \right)}^{2}}+\cdots \\ & \text{First}\text{, we calculate some derivatives and their value at }x=0 \\ & f\left( x \right)={{\tan }^{-1}}4x\to f\left( 0 \right)={{\tan }^{-1}}\left( 0 \right)=0 \\ & f'\left( x \right)=\frac{4}{1+16{{x}^{2}}}\to f'\left( 0 \right)=4 \\ & f''\left( x \right)=4{{\left( 1+16{{x}^{2}} \right)}^{-2}}\left( 32x \right)=\frac{-128x}{{{\left( 1+16{{x}^{2}} \right)}^{2}}}\to f''\left( 0 \right)=0 \\ & \text{Using a CAS to find }f'''\left( x \right),\text{ }{{f}^{\left( 4 \right)}}\left( x \right)\text{ and }{{f}^{\left( 5 \right)}}\left( x \right) \\ & f'''\left( x \right)=\frac{128\left( 48{{x}^{2}}-1 \right)}{{{\left( 1+16{{x}^{2}} \right)}^{3}}}\to f'''\left( 0 \right)=-128 \\ & {{f}^{\left( 4 \right)}}\left( x \right)=-\frac{24576x\left( 16{{x}^{2}}-1 \right)}{{{\left( 1+16{{x}^{2}} \right)}^{4}}}\to {{f}^{\left( 4 \right)}}\left( 0 \right)=0 \\ & {{f}^{\left( 5 \right)}}\left( x \right)=\frac{24576x\left( 1280{{x}^{4}}-160{{x}^{2}}+1 \right)}{{{\left( 1+16{{x}^{2}} \right)}^{5}}}\to {{f}^{\left( 5 \right)}}\left( 0 \right)=24576 \\ & \text{Then}\text{,} \\ & f\left( 0 \right)=0 \\ & f'\left( 0 \right)=4 \\ & f''\left( 0 \right)=0 \\ & f'''\left( 0 \right)=-128 \\ & {{f}^{\left( 4 \right)}}\left( 0 \right)=0 \\ & {{f}^{\left( 5 \right)}}\left( 0 \right)=24576 \\ & \\ & \left. a \right)\text{ Substituting the previous result into the formula }\left( \mathbf{1} \right). \\ & \left( \text{Using the nonzero coefficients} \right) \\ & =4x-\frac{128}{3!}{{x}^{3}}+\frac{24576}{5!}{{x}^{5}} \\ & \text{Simplifying} \\ & =4x-\frac{64}{3}{{x}^{3}}+\frac{1024}{5}{{x}^{5}} \\ & \\ & \left. b \right)\text{ Rewrite the power series} \\ & =\frac{{{\left( 4 \right)}^{1}}}{1}x-\frac{{{\left( 4 \right)}^{3}}}{3}{{x}^{3}}+\frac{{{\left( 4 \right)}^{5}}}{5}{{x}^{5}} \\ & \text{Using summation notation}\text{, we obtain} \\ & \sum\limits_{k=0}^{\infty }{\frac{{{\left( -1 \right)}^{k}}{{\left( 4x \right)}^{2k+1}}}{2k+1}} \\ \end{align}\]