University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Section 9.8 - Taylor and Maclaurin Series - Exercises - Page 536: 34


$1+\dfrac{x^2}{2}-\dfrac{2x^3}{3}; $ Converges absolutely for all values of $x$.

Work Step by Step

We have $f(x)=(1-x-x^2)e^x$ Now, distribute the polynomial $1-x+x^2+x-x^2+x^3+\dfrac{x^2}{2} -\dfrac{x^3}{2} +\dfrac{x^4}{2}+....$ or, $= (1-x-x^2)x+(1-x-x^2)(x^2/2)+(1-x-x^2)(x^3/6)+.....$ Now, we have the Maclaurin's series for $f(x)$ as follows: $f(x)=1+\dfrac{x^2}{2}-\dfrac{2x^3}{3}; $ Converges absolutely for all values of $x$.
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