Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 9 - Infinite Series - 9.7 Exercises - Page 645: 43

Answer

$0.741$

Work Step by Step

$P_{4}(x) = \ln 2 + \frac{1}{2}(x-2) - \frac{1}{4 \times 2!}(x-2)^{2} + \frac{1}{4 \times 3!}(x-2)^{3} + \frac{3}{8 \times 4!} (x-2)^{4}$ $P_{4}(2.1)$, so replace $x$ with $2.1$ $P_{4}(2.1) = \ln 2 + \frac{1}{2}(2.1-2) - \frac{1}{4 \times 2!}(2.1-2)^{2} + \frac{1}{4 \times 3!}(2.1-2)^{3} + \frac{3}{8 \times 4!} (2.1-2)^{4}$ $P_{4}(2.1) = \ln 2 + \frac{1}{2}(0.1) - \frac{1}{8}(0.1)^{2} + \frac{1}{24}(0.1)^{3} + \frac{3}{64} (0.1)^{4}$ $P_{4}(2.1) = 0.741$
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