Chapter 9 - Power Series - Review Exercises - Page 705: 36

\[\begin{align} & \left. a \right)-2\left( x+\frac{\pi }{2} \right)+\frac{8}{3!}{{\left( x+\frac{\pi }{2} \right)}^{3}}-\frac{32}{5!}{{\left( x+\frac{\pi }{2} \right)}^{5}} \\ & \left. b \right){{\sum\limits_{k=0}^{\infty }{\frac{{{\left( -1 \right)}^{k+1}}{{\left( 2 \right)}^{2k+1}}}{\left( 2k+1 \right)!}\left( x+\frac{\pi }{2} \right)}}^{2k+1}} \\ \end{align}\]

\[\begin{align} & f\left( x \right)=\sin 2x,\text{ }a=-\frac{\pi }{2} \\ & \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{ } \\ & =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{ }\left( \mathbf{1} \right) \\ & \text{First}\text{, we calculate some derivatives and their value at }x=-\frac{\pi }{2} \\ & f\left( x \right)=\sin 2x\to f\left( -\frac{\pi }{2} \right)=\sin 2\left( -\frac{\pi }{2} \right)=0 \\ & f'\left( x \right)=2\cos 2x\to f'\left( -\frac{\pi }{2} \right)=2\cos 2\left( -\frac{\pi }{2} \right)=-2 \\ & f''\left( x \right)=-4\sin 2x\to f''\left( -\frac{\pi }{2} \right)=-4\sin 2\left( -\frac{\pi }{2} \right)=0 \\ & f'''\left( x \right)=-8\cos 2x\to f'''\left( -\frac{\pi }{2} \right)=-8\cos 2\left( -\frac{\pi }{2} \right)=8 \\ & {{f}^{\left( 4 \right)}}\left( x \right)=16\sin 2x\to {{f}^{\left( 4 \right)}}\left( -\frac{\pi }{2} \right)=16\sin 2\left( -\frac{\pi }{2} \right)=0 \\ & {{f}^{\left( 5 \right)}}\left( x \right)=32\cos 2x\to {{f}^{\left( 5 \right)}}\left( -\frac{\pi }{2} \right)=32\cos 2\left( -\frac{\pi }{2} \right)=-32 \\ & \text{Then}\text{,} \\ & f\left( -\frac{\pi }{2} \right)=0,\text{ }f'\left( -\frac{\pi }{2} \right)=2,\text{ }f''\left( -\frac{\pi }{2} \right)=0,\text{ }f'''\left( -\frac{\pi }{2} \right)=-8 \\ & {{f}^{\left( 4 \right)}}\left( -\frac{\pi }{2} \right)=0,\text{ }{{f}^{\left( 5 \right)}}\left( -\frac{\pi }{2} \right)=-32 \\ & \\ & \left. a \right)\text{ Substituting the previous result into the formula }\left( \mathbf{1} \right). \\ & \left( \text{Using the nonzero coefficients} \right) \\ & =-\frac{2}{1!}{{\left( x+\frac{\pi }{2} \right)}^{1}}+\frac{8}{3!}{{\left( x+\frac{\pi }{2} \right)}^{3}}-\frac{32}{5!}{{\left( x+\frac{\pi }{2} \right)}^{5}} \\ & \text{Simplifying} \\ & =-2\left( x+\frac{\pi }{2} \right)+\frac{8}{3!}{{\left( x+\frac{\pi }{2} \right)}^{3}}-\frac{32}{5!}{{\left( x+\frac{\pi }{2} \right)}^{5}} \\ & \\ & \left. b \right)\text{ Rewrite the power series} \\ & ={{\left( -1 \right)}^{0+1}}\frac{2}{1!}{{\left( x+\frac{\pi }{2} \right)}^{1}}+{{\left( -1 \right)}^{1+1}}\frac{{{2}^{3}}}{3!}{{\left( x+\frac{\pi }{2} \right)}^{3}}+{{\left( -1 \right)}^{2+1}}\frac{{{2}^{5}}}{5!}{{\left( x+\frac{\pi }{2} \right)}^{5}} \\ & \text{Using summation notation}\text{, we obtain} \\ & {{\sum\limits_{k=0}^{\infty }{\frac{{{\left( -1 \right)}^{k+1}}{{\left( 2 \right)}^{2k+1}}}{\left( 2k+1 \right)!}\left( x+\frac{\pi }{2} \right)}}^{2k+1}} \\ \end{align}\]