University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 9 - Section 9.9 - Convergence of Taylor Series - Exercises - Page 542: 26


$1-x-\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+..=\Sigma_{n=0}^\infty (-1)^n[\dfrac{x^{2n}}{(2n)!}-\dfrac{x^{2n+1}}{(2n+1)!}]$

Work Step by Step

We know that the Maclaurin Series for $\sin x$ and $\cos x$ is defined as: $ \sin x=\Sigma_{n=0}^\infty \dfrac{(-1)^n x^{2n+1}}{(2n+1)!}=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+...$ and $ \cos x=\Sigma_{n=0}^\infty \dfrac{(-1)^n x^{2n}}{(2n)!}=1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+...$ Then $\cos x-\sin x=(\Sigma_{n=0}^\infty \dfrac{(-1)^n x^{2n}}{(2n)!})-(\Sigma_{n=0}^\infty \dfrac{(-1)^n x^{2n+1}}{(2n+1)!})=(1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+...)-(x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+...)$ or, $1-x-\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+..=\Sigma_{n=0}^\infty (-1)^n[\dfrac{x^{2n}}{(2n)!}-\dfrac{x^{2n+1}}{(2n+1)!}]$
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.