Assignment4(1804)(Solution)(v2-31-3-11)

Assignment4(1804)(Solution)(v2-31-3-11) - MATH 1804...

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Unformatted text preview: MATH 1804 UNIVERSITY MATHEMATICS A Assignment 4 – Solution 1. (a) The third Taylor polynomial P 3 ( x ) of f ( x ) at x = 1 is P 3 ( x ) = f (1) + f (1)( x- 1) + f 00 (1) 2 ( x- 1) 2 + f 000 (1) 6 ( x- 1) 3 = 2- ( x- 1) + 0 +- 2 6 ( x- 1) 3 = 2- ( x- 1)- 1 3 ( x- 1) 3 , so f (1 . 1) ≈ P 3 (1 . 1) = 2- (1 . 1- 1)- 1 3 (1 . 1- 1) 3 = 5699 3000 ≈ 1 . 8997 . (b) The error committed is given by, for some c ∈ (1 , 1 . 1), | R 3 (1 . 1) | = f (4) ( c ) 24 (1 . 1- 1) 4 =- ( c 2 + 1) 240000 = c 2 + 1 240000 < 1 . 1 2 + 1 240000 = 2 . 21 240000 = 0 . 000009208 ... thus must not exceed 0 . 00001. 2. Let f ( x ) = sin x . For any x ∈ R , | f ( n ) ( x ) | ≤ 1 , since f ( n ) ( x ) is either ± sin x or ± cos x . The error when approximating f ( π 5 ) by the n-th Taylor polynomial of f at π 6 is given by R n π 5 = f ( n +1) ( c ) ( n + 1)! π 5- π 6 n +1 < 1 ( n + 1)! π 30 n +1 for some c ∈ ( π 6 , π 5 ). Then, direct computation yields R 1 π 5 < . 005483 ... ( > . 0002) R 2 π...
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Assignment4(1804)(Solution)(v2-31-3-11) - MATH 1804...

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