## Calculus (3rd Edition)

Published by W. H. Freeman

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

#### Answer

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

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