Calculus (3rd Edition)

Published by W. H. Freeman

Chapter 11 - Infinite Series - 11.7 Taylor Series - Exercises - Page 589: 67

Answer

$$1-5x+\sin(5x).$$

Work Step by Step

Since we have$$1-\frac{5^{3} x^{3}}{3 !}+\frac{5^{5} x^{5}}{5 !}-\frac{5^{7} x^{7}}{7 !}+\cdots\\ = 1-\frac{(5x)^{3}}{3 !}+\frac{(5x)^{5}}{5 !}-\frac{(5x)^{7}}{7 !}+\cdots\\ =1-5x+5x-\frac{(5x)^{3}}{3 !}+\frac{(5x)^{5}}{5 !}-\frac{(5x)^{7}}{7 !}+\cdots$$ Then by using Table 2, we see that this is a Maclaurin series of the function $$1-5x+\sin(5x).$$

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