# X n taylor polynomial t n x n i 0 f i a i x a i f x t

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Chapter 6 / Exercise 8
Trigonometry
Larson
Expert Verified
x n ; Taylor polynomial: T n ( x ) = n i =0 f ( i ) ( a ) i ! ( x - a ) i . f ( x ) T n ( x ) . Taylor’s inequality: If | f ( n +1) ( x ) | M for | x - a | d , then on | x - a | d , | R n ( x ) | = | f ( x ) - T n ( x ) | M ( n + 1)! | x - a | n +1 . Taylor Theorem: If f ( x ) = T n ( x ) + R n ( x ), and lim n →∞ R n ( x ) = 0 for | x - a | < R , then f ( x ) is equal to its Taylor series for | x - a | < R . To this end, the following result is useful: lim n →∞ x n n ! = 0 . Series for composite functions; Multiplication and division of power series. Maclaurin series for some special functions Example. Maclaurin series for some special functions: e x = 1 + x 1! + x 2 2! + x 3 3! + ... , R = ; sin x = x - x 3 3! + x 5 5! - ... , R = ; 44
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Chapter 6 / Exercise 8
Trigonometry
Larson
Expert Verified
cos x = 1 - x 2 2! + x 4 4! - ... , R = ; ln(1 + x ) = x - x 2 2 + x 3 3 - ... , R = 1; arctan x = x - x 3 3 + x 5 5 - x 7 7 + ..., R = 1. Series for composite functions Example. Maclaurin series: (i) e x 2 , sin( x 2 ). (ii) e sin x = 1 + sin x + sin 2 x 2! + sin 3 x 3! + ... = 1 + ( x - x 3 3! + x 5 5! + ... ) + ( ( x - x 3 3! + x 5 5! + ... ) 2 2! ) + ( ( x - x 3 3! + x 5 5! + ... ) 3 3! ) + ... = 1 + x + x 2 2! + 0 x 3 + ... Binomial series If k is a real number and | x | < 1, then (1 + x ) k = 1 + kx + k ( k - 1) 2! x 2 + · · · = n =0 k n x n , here k n = k ( k - 1) · · · ( k - n + 1) n ! , k 0 = 1 . Application: Let f ( x ) = (1 + x ) k , then f ( n ) (0) = k n n ! = k ( k - 1) · · · ( k - n + 1) . Example. Maclauring series for f ( x ) = 3 1 + x . 45
Solution: 3 1 + x = 1 + 1 3 x - 2 3 2 2! x 2 + 2 · 5 3 3 3! x 3 - 2 · 5 · 8 3 4 4! x 4 + · · · . Example. Let f ( x ) = 5 1 + x 2 . Evaluate f (4) (0). Solution. Use the binomial series to find the Maclaurin series of f ( x ). f ( n ) (0) n ! = k n Hence f ( n ) (0) = k n n ! = k ( k - 1) · · · ( k - n + 1) . So f (4) (0) = - 0 . 8064. Taylor series at other centers Example. Find the Taylor series for f ( x ) = sin x at the center x = π 3 . Multiplication and division of Taylor series Example. Evaluate lim x 0 e x sin x - x x 2 . e x sin x = 1 + x 1! + x 2 2! + x 3 3! + ... x - x 3 3! + ... = x + x 2 + 1 3 x 3 + ... Therefore e x sin x - x x 2 = 1 + 1 3 x + ... Example. Let f ( x ) = x 0 t 3 e 3 t dt (a) Find the Maclaurin series of the function f . (b) Find the radius of the series in (a). 46
Solution: a) t 3 e 3 t = t 3 n =0 (3 t ) n n ! = n =0 3 n n ! t n +3 . Hence f ( x ) = x 0 n =0 3 n n ! t n +3 dt = n =0 x 0 3 n n ! t n +3 dt = n =0 3 n n ! t n +4 n + 4 | x 0 = n =0 3 n n !( n + 4) x n +4 b) | a n +1 /a n | = 3( n + 4) ( n + 1)( n + 5) | x | 0 as n → ∞ . Therefore R = . Example. Find the first 5 non-zero terms in the Maclaurin series for e x cos(3 x ). Solution. Note that e x = 1 + x 1! + x 2 2! + x 3 3! + x 4 4! + x 5 5! + ... = 1 + x + x 2 2 + x 3 6 + x 4 24 + ..., cos x = 1 - x 2 2! + x 4 4! - x 6 6! + ... Hence cos(3 x ) = 1 - (3 x ) 2 2! + (3 x ) 4 4! - (3 x ) 6 6! + ... = 1 - 9 x 2 2 + 27 x 4 8 - ... We have e x cos(3 x ) = 1 + x + x 2 2 + x 3 6 + x 4 24 + ... - 9 x 2 2 - 9 x 3 2 - 9 x 4 4 - ... + 27 x 4 8 + ... = 1 + x + ( 1 2 - 9 2 ) x 2 + ( 1 6 - 9 2 ) x 3 + ( 1 24 - 9 4 + 27 8 ) x 4 + ... 47
= 1 + x - 4 x 2 - 13 3 x 3 + 7 6 x 4 + ... Example. Find the first 3 non-zero terms in the Maclaurin series for tan x . Solution: tan x = x + 1 3 x 3 + 2 15 x 5 + ... . Applications of Taylor series. Example. 1 - 1 3 + 1 5 - 1 7 + ... = π 4 . Example. Calculate the sum of the following series, given that it is a known series evaluated at a certain value of x : 1 - 1 4 · 2! + 1 16 · 4! - 1 64 · 6! + 1 256 · 8! - · · · Sol: 1 - 1 4 · 2!