Chapter 9 - Power Series - Review Exercises - Page 705: 37

\[\begin{align} & \left. a \right)\frac{1}{0!}{{x}^{0}}+\frac{9}{2!}{{x}^{2}}+\frac{81}{4!}{{x}^{4}} \\ & \left. b \right)\sum\limits_{k=0}^{\infty }{\frac{{{\left( 3x \right)}^{2k}}}{\left( 2k \right)!}} \\ \end{align}\]

\[\begin{align} & f\left( x \right)=\cosh 3x,\text{ }a=0 \\ & \text{Using the definition of Taylor series for a function}\left( page\,685 \right) \\ & \sum\limits_{k=0}^{\infty }{\frac{{{f}^{\left( k \right)}}\left( a \right)}{k!}{{\left( x-a \right)}^{k}}}\text{ } \\ & =f\left( a \right)+f'\left( a \right)\left( x-a \right)+\frac{f''\left( a \right)}{2!}{{\left( x-a \right)}^{2}}+\cdots \text{ }\left( \mathbf{1} \right) \\ & \text{First}\text{, we calculate some derivatives and their value at }x=0 \\ & f\left( x \right)=\cosh 3x\to f\left( 0 \right)=\cosh \left( 0 \right)=1 \\ & f'\left( x \right)=3\sinh 3x\to f'\left( 0 \right)=\sinh \left( 0 \right)=0 \\ & f''\left( x \right)=9\cosh 3x\to f''\left( 0 \right)=9\cosh \left( 0 \right)=9 \\ & f'''\left( x \right)=27\sinh 3x\to f'''\left( 0 \right)=27\sinh \left( 0 \right)=0 \\ & {{f}^{\left( 4 \right)}}\left( x \right)=81\cosh 3x\to {{f}^{\left( 4 \right)}}\left( 0 \right)=81\cosh \left( 0 \right)=81 \\ & {{f}^{\left( 5 \right)}}\left( x \right)=243\sinh 3x\to {{f}^{\left( 5 \right)}}\left( 0 \right)=243\sinh \left( 0 \right)=0 \\ & \\ & \text{Then}\text{,} \\ & f\left( 0 \right)=1,\text{ }f'\left( 0 \right)=0,\text{ }f''\left( 0 \right)=9,\text{ }f'''\left( 0 \right)=0 \\ & {{f}^{\left( 4 \right)}}\left( 0 \right)=81,\text{ }{{f}^{\left( 5 \right)}}\left( 0 \right)=0 \\ & \\ & \left. a \right)\text{ Substituting the previous result into the formula }\left( \mathbf{1} \right). \\ & \left( \text{Using the nonzero coefficients} \right) \\ & =\frac{1}{0!}{{x}^{0}}+\frac{9}{2!}{{x}^{2}}+\frac{81}{4!}{{x}^{4}} \\ & \\ & \left. b \right)\text{ Rewrite the power series} \\ & =\frac{{{3}^{0}}}{0!}{{x}^{0}}+\frac{{{3}^{2}}}{2!}{{x}^{2}}+\frac{{{3}^{4}}}{4!}{{x}^{4}} \\ & \text{Using summation notation}\text{, we obtain} \\ & \sum\limits_{k=0}^{\infty }{\frac{{{\left( 3x \right)}^{2k}}}{\left( 2k \right)!}} \\ \end{align}\]