University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 9 - Section 9.10 - The Binomial Series and Applications of Taylor Series - Exercises - Page 549: 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) We integrate the integral with respect to $ x $ taking a lower limit as $0$ and upper limit as $1$. $ f(x)=\int_0^{x} \tan^{-1} t dt=\int_0^{x} [t-\dfrac{t^3}{3}+\dfrac{t^{5}}{5}-...] dt $ or, $ f(x) =\dfrac{x^2}{2}-\dfrac{x^{4}}{12}+\dfrac{x^{6}}{30}- ....$ So, the first two terms of the series would give the accuracy. And the error can be calculated as: $| Error| \lt \dfrac{(0.5)^6}{30} \approx 0.00052$ (b) We integrate the integral with respect to $ x $ taking a lower limit as $0$ and upper limit as $1$. $ f(x)=\int_0^{x} \tan^{-1} t dt=\int_0^{x} [t-\dfrac{t^3}{3}+\dfrac{t^{5}}{5}-...] dt $ or, $ f(x) =\dfrac{x^2}{2}-\dfrac{x^{4}}{12}+\dfrac{x^{6}}{30}- ....$ So, the first 16th term of the series would give the accuracy. $ f(x) \approx\dfrac{x^2}{2}-\dfrac{x^{4}}{12}+\dfrac{x^{6}}{30}+ ....+(-1)^5\dfrac{x^{32}}{31 \cdot 22}$ And the error can be calculated as: $| Error| \lt \dfrac{1}{33 \cdot 34} \approx 0.00089$
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