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Sample Problem 2: Determine the Maclaurin series for the function f(x) = sinx. Recalling the Maclaurin series form, Σrin) (0) (x = 0)* f[n], (x-0)" n! n=0 we express the terms when n=0, n=1, n=2, n=3, n=4 and so on. (x-0)⁰ n=0: fᵒ(0) * -= f(0) = sin(0) = 0 (Note: x° = 1 and f(0) = f(0)) 0! x n=1: f¹(0)(x-0)¹ = = cos(0) = x 1! 1 x2 n=2: f²(0)(x-0)² = sin(0) sin(0) 2! 2 cos(0)/²=₁ =-2/²/20 n=3: f3 (0)(x−0)* 3! 6 n=4: f4(0)(x-0)4 = sin(0) / 4 = 4! 24 Therefore, the Maclaurin series of sinx is written in the form: x3 sinx(0). (x)n n! - + ... f¹² (0) ! 6 n! n=0 This is the fourth-degree Taylor series approximation since up to fourth derivative is solved. = = 0

Sample Problem 2: Determine the Maclaurin series for the function f(x) = sinx. Recalling the Maclaurin series form, Σrin) (0) (x = 0)* f[n], (x-0)" n! n=0 we express the terms when n=0, n=1, n=2, n=3, n=4 and so on. (x-0)⁰ n=0: fᵒ(0) * -= f(0) = sin(0) = 0 (Note: x° = 1 and f(0) = f(0)) 0! x n=1: f¹(0)(x-0)¹ = = cos(0) = x 1! 1 x2 n=2: f²(0)(x-0)² = sin(0) sin(0) 2! 2 cos(0)/²=₁ =-2/²/20 n=3: f3 (0)(x−0)* 3! 6 n=4: f4(0)(x-0)4 = sin(0) / 4 = 4! 24 Therefore, the Maclaurin series of sinx is written in the form: x3 sinx(0). (x)n n! - + ... f¹² (0) ! 6 n! n=0 This is the fourth-degree Taylor series approximation since up to fourth derivative is solved. = = 0

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Sample Problem 2: Determine the Maclaurin series for the function f(x) = sinx. Recalling the Maclaurin series form, 00 Σflnl (0)! fln) (0) (x - 0)n n! n=0 we express the terms when n=0, n=1, n=2, n=3, n=4 and so on. (x-0)⁰ n=0: fº(0) = f(0) = sin(0) = 0 (Note: xº = 1 and fº(0) = f(0)) 0! x n=1: f¹(0)(x- (x-0)¹ 1! cos(0) = x 1 (x-0)² n=2: f²(0) = -sin(0)/2 = 0 2! n=3: f3(0)(x→0)3 = −cos(0)/²= = 3! (x-0)4 4 n=4: f4 (0)(x- = sin(0) = 0 4! 24 Therefore, the Maclaurin series of sinx is written in the form: x3 xn (x)n sinx(0). = x n! —— - + ... ¹ (0) 2 6 n! n=0 This is the fourth-degree Taylor series approximation since up to fourth derivative is solved. =
Transcribed Image Text:Sample Problem 2: Determine the Maclaurin series for the function f(x) = sinx. Recalling the Maclaurin series form, 00 Σflnl (0)! fln) (0) (x - 0)n n! n=0 we express the terms when n=0, n=1, n=2, n=3, n=4 and so on. (x-0)⁰ n=0: fº(0) = f(0) = sin(0) = 0 (Note: xº = 1 and fº(0) = f(0)) 0! x n=1: f¹(0)(x- (x-0)¹ 1! cos(0) = x 1 (x-0)² n=2: f²(0) = -sin(0)/2 = 0 2! n=3: f3(0)(x→0)3 = −cos(0)/²= = 3! (x-0)4 4 n=4: f4 (0)(x- = sin(0) = 0 4! 24 Therefore, the Maclaurin series of sinx is written in the form: x3 xn (x)n sinx(0). = x n! —— - + ... ¹ (0) 2 6 n! n=0 This is the fourth-degree Taylor series approximation since up to fourth derivative is solved. =
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