Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 10: Infinite Sequences and Series - Section 10.10 - The Binomial Series and Applications of Taylor Series - Exercises 10.10 - Page 633: 27

Answer

a) $ f(x) =\dfrac{x^2}{2}-\dfrac{x^{4}}{12}+\dfrac{x^{6}}{30}- ....$ $| Error| \lt 0.00052$ (b) $ f(x) \approx\dfrac{x^2}{2}-\dfrac{x^{4}}{12}+\dfrac{x^{6}}{30}+ ....+(-1)^5\dfrac{x^{32}}{31 \cdot 22}$ and $| Error| \lt 0.00089$

Work Step by Step

a) Integrate the integral with respect to $ x $ as follows: $ f(x)=\int_0^{x} \tan^{-1} t dt=\int_0^{x} [t-\dfrac{t^3}{3}+\dfrac{t^{5}}{5}-...] dt \\ =\dfrac{x^2}{2}-\dfrac{x^{4}}{12}+\dfrac{x^{6}}{30}- ....$ Now, $| Error| \lt \dfrac{(0.5)^6}{30} \approx 0.00052 \implies | Error| \lt 0.00052$ b) Now, $ f(x)=\int_0^{x} \tan^{-1} t dt=\int_0^{x} [t-\dfrac{t^3}{3}+\dfrac{t^{5}}{5}-...] \space dt \\=\dfrac{x^2}{2}-\dfrac{x^{4}}{12}+\dfrac{x^{6}}{30}- ....$ and $ f(x) \approx\dfrac{x^2}{2}-\dfrac{x^{4}}{12}+\dfrac{x^{6}}{30}+ ....+(-1)^5\dfrac{x^{32}}{31 \cdot 22}$ Now, $| Error| \lt \dfrac{1}{33 \cdot 34} \approx 0.00089$ or, $| Error| \lt 0.00089$
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.