Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 11 - Infinite Series - 11.7 Taylor Series - Exercises - Page 588: 12


$$f(x) =x^2+ x^4+\frac{x^{6}}{2!}+\frac{x^{8}}{3!}+\frac{x^{10}}{4!}+....$$ Convergent for any value of $x$.

Work Step by Step

By making use of Table 2, we have the Maclaurin series for $f(x)=e^x$ as follows $$f(x)=e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...$$ Now by comparison with the function $f(x)=x^2e^{ x^2}$, we have the Maclaurin series as follows $$f(x)=x^2e^{ x^2}\\ =x^2(1+ x^2+\frac{x^{4}}{2!}+\frac{x^{6}}{3!}+\frac{x^{8}}{4!}+...)\\ =x^2+ x^4+\frac{x^{6}}{2!}+\frac{x^{8}}{3!}+\frac{x^{10}}{4!}+....$$ Moreover, again from Table 2, the Maclaurin series for $f(x)=e^x$ is convergent for any value of $x$. Hence, the Maclaurin series for $f(x)=x^2e^{ x^2}$ is also convergent for any value of $x$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.