Introduction to Mathematical Analysis I - Second Edition

124 4.5 TAYLOR’S THEOREM for some c in between ¯ x and x . Consider the function g ( t ) = f ( x ) − n ∑ k = 0 f ( k ) ( t ) k ! ( x − t ) k − λ ( n + 1 ) ! ( x − t ) n + 1 . Then g ( ¯ x ) = f ( x ) − n ∑ k = 0 f ( k ) ( ¯ x ) k ! ( x − ¯ x ) k − λ ( n + 1 ) ! ( x − ¯ x ) n + 1 = f ( x ) − P n ( x ) − λ ( n + 1 ) ! ( x − ¯ x ) n + 1 = 0 . and g ( x ) = f ( x ) − n ∑ k = 0 f ( k ) ( x ) k ! ( x − x ) k − λ ( n + 1 ) ! ( x − x ) n + 1 = f ( x ) − f ( x ) = 0 . By Rolle’s theorem , there exists c in between ¯ x and x such that g 0 ( c ) = 0. Taking the derivative of g (keeping in mind that x is fixed and the independent variable is t ) and using the product rule for derivatives, we have g 0 ( c ) = − f 0 ( c )+ n ∑ k = 1 − f ( k + 1 ) ( c ) k ! ( x − c ) k + f ( k ) ( c ) ( k − 1 ) ! ( x − c ) k − 1 ! + λ n ! ( x − c ) n = λ n ! ( x − c ) n − 1 n ! f ( n + 1 ) ( c )( x − c ) n = 0 . This implies λ = f ( n + 1 ) ( c ) . The proof is now complete. The polynomial P n ( x ) given in the theorem is called the n-th Taylor polynomial of f at ¯ x . Remark 4.5.2 The conclusion of Taylor’s theorem still holds true if x = ¯ x . In this case, c = x = ¯ x . Example 4.5.1 We will use Taylor’s theorem to estimate the error in approximating the function f ( x ) = sin x with it 3rd Taylor polynomial at ¯ x = 0 on the interval [ − π / 2 , π / 2 ] . Since f 0 ( x ) = cos x , f 00 ( x ) = − sin x and f 000 ( x ) = − cos x , a direct calculation shows that P 3 ( x ) = x − x 3 3! . More over, for any c ∈ R we have | f ( 4 ) ( c ) | = | sin c | ≤ 1. Therefore, for x ∈ [ − π / 2 , π / 2 ] we get (for some c between x and 0), | sin x − P 3 ( x ) | = | f ( 4 ) ( c ) | 4! | x | ≤ π / 2 4! ≤ 0 . 066 . Theorem 4.5.3 Let n be an even positive integer. Suppose f ( n ) exists and continuous on ( a , b ) . Let ¯ x ∈ ( a , b ) satisfy f 0 ( ¯ x ) = . . . = f ( n − 1 ) ( ¯ x ) = 0 and f ( n ) ( ¯ x ) 6 = 0 . The following hold: