Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 9 - Infinite Series - 9.7 Maclaurin And Taylor Polynomials - Exercises Set 9.7 - Page 658: 33

Answer

False

Work Step by Step

By Definition 9.7.3, the sixth-degree Taylor polynomial about $x = {x_0}$ for $f$ is ${p_6}\left( x \right) = f\left( {{x_0}} \right) + f'\left( {{x_0}} \right)\left( {x - {x_0}} \right) + \dfrac{{f{\rm{''}}\left( {{x_0}} \right)}}{{2!}}{\left( {x - {x_0}} \right)^2} + \dfrac{{f{\rm{'''}}\left( {{x_0}} \right)}}{{3!}}{\left( {x - {x_0}} \right)^3} + \cdot\cdot\cdot + \dfrac{{{f^{\left( 6 \right)}}\left( {{x_0}} \right)}}{{6!}}{\left( {x - {x_0}} \right)^6}$ Taking the derivative gives $p_6^{\left( 1 \right)}\left( x \right) = f'\left( {{x_0}} \right) + f{\rm{''}}\left( {{x_0}} \right)\left( {x - {x_0}} \right) + 3\dfrac{{f{\rm{'''}}\left( {{x_0}} \right)}}{{3!}}{\left( {x - {x_0}} \right)^2} + \cdot\cdot\cdot + 6\dfrac{{{f^{\left( 6 \right)}}\left( {{x_0}} \right)}}{{6!}}{\left( {x - {x_0}} \right)^5}$ $p_6^{\left( 1 \right)}\left( {{x_0}} \right) = f'\left( {{x_0}} \right)$ Continuing with taking the derivatives and so on, we obtain $p_6^{\left( 4 \right)}\left( {{x_0}} \right) = {f^{\left( 4 \right)}}\left( {{x_0}} \right)$. Therefore, the statement is false.
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