Chapter 9 - Power Series - 9.3 Taylor Series - 9.3 Exercises - Page 695: 21

\[\begin{align} & \left. a \right)1-\frac{1}{2}{{\left( x-\frac{\pi }{2} \right)}^{2}}+\frac{1}{4!}{{\left( x-\frac{\pi }{2} \right)}^{4}}-\frac{1}{6!}{{\left( x-\frac{\pi }{2} \right)}^{6}} \\ & \left. b \right)\sum\limits_{k=0}^{\infty }{\frac{{{\left( -1 \right)}^{k}}}{\left( 2k \right)!}{{\left( x-\frac{\pi }{2} \right)}^{2k}}} \\ \end{align}\]

\[\begin{align} & f\left( x \right)=\sin x,\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{ }\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=\frac{\pi }{2} \\ & f\left( x \right)=\sin x\to f\left( \frac{\pi }{2} \right)=\sin \left( \frac{\pi }{2} \right)=1 \\ & f'\left( x \right)=\cos x\to f'\left( \frac{\pi }{2} \right)=\cos \left( \frac{\pi }{2} \right)=0 \\ & f''\left( x \right)=-\sin x\to f''\left( \frac{\pi }{2} \right)=-\sin \left( \frac{\pi }{2} \right)=-1 \\ & f'''\left( x \right)=-\cos x\to f'''\left( \frac{\pi }{2} \right)=-\cos x\left( \frac{\pi }{2} \right)=0 \\ & {{f}^{\left( 4 \right)}}\left( x \right)=\sin x\to {{f}^{\left( 4 \right)}}\left( \frac{\pi }{2} \right)=\sin \left( \frac{\pi }{2} \right)=1 \\ & {{f}^{\left( 5 \right)}}\left( x \right)=\cos x\to {{f}^{\left( 5 \right)}}\left( \frac{\pi }{2} \right)=\cos \left( \frac{\pi }{2} \right)=0 \\ & {{f}^{\left( 6 \right)}}\left( x \right)=-\sin x\to {{f}^{\left( 6 \right)}}\left( \frac{\pi }{2} \right)=-\sin \left( \frac{\pi }{2} \right)=-1 \\ & \\ & \left. a \right)\text{ Substituting the previous result into the formula }\left( \mathbf{1} \right). \\ & \left( \text{Using the nonzero coefficients} \right) \\ & =1+\frac{\left( -1 \right)}{2!}{{\left( x-\frac{\pi }{2} \right)}^{2}}+\frac{\left( 1 \right)}{4!}{{\left( x-\frac{\pi }{2} \right)}^{4}}+\frac{\left( -1 \right)}{6!}{{\left( x-\frac{\pi }{2} \right)}^{6}} \\ & =1-\frac{1}{2}{{\left( x-\frac{\pi }{2} \right)}^{2}}+\frac{1}{4!}{{\left( x-\frac{\pi }{2} \right)}^{4}}-\frac{1}{6!}{{\left( x-\frac{\pi }{2} \right)}^{6}} \\ & \\ & \left. b \right)\text{ Rewrite the power series} \\ & =\frac{{{\left( -1 \right)}^{0}}}{\left( 2\cdot 0 \right)!}{{\left( x-\frac{\pi }{2} \right)}^{\left( 2\cdot 0 \right)}}+\frac{{{\left( -1 \right)}^{1}}}{\left( 2\cdot 1 \right)!}{{\left( x-\frac{\pi }{2} \right)}^{\left( 2\cdot 1 \right)}}+\frac{{{\left( -1 \right)}^{2}}}{\left( 2\cdot 2 \right)!}{{\left( x-\frac{\pi }{2} \right)}^{\left( 2\cdot 2 \right)}} \\ & +\frac{{{\left( -1 \right)}^{3}}}{\left( 2\cdot 3 \right)!}{{\left( x-\frac{\pi }{2} \right)}^{\left( 2\cdot 3 \right)}} \\ & \text{Using summation notation}\text{, we obtain} \\ & \sum\limits_{k=0}^{\infty }{\frac{{{\left( -1 \right)}^{k}}}{\left( 2k \right)!}{{\left( x-\frac{\pi }{2} \right)}^{2k}}} \\ \end{align}\]