Explanation Given: The function is, f ( x ) = ln ( 1 + x 2 ) . Approach: If the series f ( x ) = ∑ n = 0 ∞ a n ( x − x 0 ) n has a positive radius of convergence ρ , then f is differentiable in the interval | x − x 0 | < ρ and term-wise integration gives the power series for the integral of f as, ∫ f ( x ) d x = ∑ n = 0 ∞ a n n + 1 ( x − x 0 ) n + 1 + C for | x − x 0 | < ρ Calculation: The Taylor series for ( 1 + x 2 ) − 1 is, 1 1 + x 2 = 1 − x 2 + x 4 − x 6 + ... + ( − 1 ) n x 2 n + ... Multiply the above expression by 2 x to get, 2 x 1 + x 2 = 2 ( x − x 3 + x 5 − x 7 + ... + ( − 1 ) n x 2 n + 1 + ... ) 2 x 1 + x 2 = 2 ∑ n = 0 ∞ ( − 1 ) n x 2 n + 1 … ( 1 ) Now, ∫ 0 x 2 t 1 + t 2 d t = ln ( 1 + x 2 ) From equation ( 1 ) ,

Fundamentals of Differential Equations and Boundary Value Problems

7th Edition

ISBN: 9780321977106

Author: Nagle, R. Kent

Publisher: Pearson Education, Limited

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Chapter 8.2, Problem 38E

To determine

To find:

The Taylor series for the given function about $x0=0$ .

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Chapter 8.2, Problem 37E

Chapter 8.3, Problem 1E

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