Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 9 - Power Series - 9.3 Taylor Series - 9.3 Exercises - Page 695: 28

Answer

\[\begin{align} & \left. a \right){{10}^{2}}{{\left( \ln 10 \right)}^{0}}+{{10}^{2}}\left( \ln 10 \right)\left( x-2 \right)+\frac{{{10}^{2}}}{2!}{{\left( \ln 10 \right)}^{2}}{{\left( x-2 \right)}^{2}}+ \\ & +\frac{{{10}^{2}}{{\left( \ln 10 \right)}^{3}}}{3!}{{\left( x-2 \right)}^{3}} \\ & \left. b \right)\sum\limits_{k=0}^{\infty }{\frac{{{10}^{2}}{{\left( \ln 10 \right)}^{k}}{{\left( x-2 \right)}^{k}}}{k!}} \\ \end{align}\]

Work Step by Step

\[\begin{align} & f\left( x \right)={{10}^{x}},\text{ }a=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{Calculate some derivatives and their value at }x=2 \\ & f\left( x \right)={{10}^{x}}\to f\left( 2 \right)={{10}^{2}} \\ & f'\left( x \right)={{10}^{x}}\left( \ln 10 \right)\to f\left( 2 \right)={{10}^{2}}\left( \ln 10 \right) \\ & f''\left( x \right)={{10}^{x}}{{\left( \ln 10 \right)}^{2}}\to f\left( 2 \right)={{10}^{2}}{{\left( \ln 10 \right)}^{2}} \\ & f''\left( x \right)={{10}^{x}}{{\left( \ln 10 \right)}^{3}}\to f\left( 2 \right)={{10}^{2}}{{\left( \ln 10 \right)}^{3}} \\ & \\ & \left. a \right)\text{ Substituting the previous result into the formula }\left( \mathbf{1} \right). \\ & \left( \text{Using the nonzero coefficients} \right) \\ & =\frac{{{10}^{2}}}{0!}{{\left( x-2 \right)}^{0}}+\frac{{{10}^{2}}\left( \ln 10 \right)}{1!}\left( x-2 \right)+\frac{{{10}^{2}}{{\left( \ln 10 \right)}^{2}}}{2!}{{\left( x-2 \right)}^{2}} \\ & +\frac{{{10}^{2}}{{\left( \ln 10 \right)}^{3}}}{3!}{{\left( x-2 \right)}^{3}} \\ & \text{Simplifying} \\ & ={{10}^{2}}+{{10}^{2}}\left( \ln 10 \right)\left( x-2 \right)+\frac{{{10}^{2}}}{2!}{{\left( \ln 10 \right)}^{2}}{{\left( x-2 \right)}^{2}} \\ & +\frac{{{10}^{2}}{{\left( \ln 10 \right)}^{3}}}{3!}{{\left( x-2 \right)}^{3}} \\ & \\ & \left. b \right)\text{ Rewrite the power series} \\ & ={{10}^{2}}{{\left( \ln 10 \right)}^{0}}+{{10}^{2}}\left( \ln 10 \right)\left( x-2 \right)+\frac{{{10}^{2}}}{2!}{{\left( \ln 10 \right)}^{2}}{{\left( x-2 \right)}^{2}} \\ & +\frac{{{10}^{2}}{{\left( \ln 10 \right)}^{3}}}{3!}{{\left( x-2 \right)}^{3}} \\ & \text{Using summation notation}\text{, we obtain} \\ & =\sum\limits_{k=0}^{\infty }{\frac{{{10}^{2}}{{\left( \ln 10 \right)}^{k}}{{\left( x-2 \right)}^{k}}}{k!}} \\ \end{align}\]
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