Chapter 10: Infinite Sequences and Series - Section 10.9 - Convergence of Taylor Series - Exercises 10.9 - Page 625: 16

$x^2-\dfrac{1}{2!}x^6+\dfrac{1}{4!} x^{10}-\dfrac{1}{6!}x^{14}+..$

Consider the Maclaurin Series for $\cos x$ is defined as: $ \cos x=1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+...$ Then, $ x^2 \cos (x^2)=x^2(1-\dfrac{(x^2)^2}{2!}+\dfrac{(x^2)^4}{4!}-\dfrac{(x^2)^6}{6!}+...)$ $\implies x^2-\dfrac{1}{2!}x^6+\dfrac{1}{4!} x^{10}-\dfrac{1}{6!}x^{14}+..$