Ma1502HW1SolnsFall2007

Ma1502HW1SolnsFall2007 - MATH1502 - Homework 1 - Fall 2007...

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Unformatted text preview: MATH1502 - Homework 1 - Fall 2007 due Tuesday September 11, 2007 Solutions Questions 1, 6, 7, 10, 13, 16, 19, 20 will be graded . Question Number Points 1 10 6 5 7 10 10 15 13 10 16 15 19 15 20 20 Total 100 Questions on Section 12.6 Question 1 (10 points) Find P 4 ( x ) for f ( x ) = (1 + &x ) 1 = 3 : Solutions f ( x ) = (1 + &x ) 1 = 3 ) f (0) = 1; f ( x ) = & 3 (1 + &x ) & 2 = 3 ) f (0) = & 3 ; f 00 ( x ) = & 2 & 3 & 3 (1 + &x ) & 5 = 3 ) f 00 (0) = & 2 9 & 2 ; f 000 ( x ) = 5 & 3 2 & 3 & 3 (1 + &x ) & 8 = 3 ) f 000 (0) = 10 27 & 3 ; f (4) ( x ) = & 8 & 3 5 & 3 2 & 3 & 3 (1 + &x ) & 11 = 3 ) f (4) (0) = & 80 81 & 4 : Then P 4 ( x ) = f (0) + f (0) x + f 00 (0) x 2 2! + f 000 (0) x 3 3! + f (4) (0) x 4 4! = 1 + & 3 x & 2 9 & 2 x 2 2 + 10 27 & 3 x 3 3! & 80 81 & 4 x 4 4! = 1 + & 3 x & 1 9 & 2 x 2 + 5 81 & 3 x 3 & 10 243 & 4 x 4 : 1 Question 2 Find P 5 ( x ) for f ( x ) = e & 2 x cos x: Solution f ( x ) = e & 2 x cos x ) f (0) = 1; f ( x ) = & 2 e & 2 x cos x & e & 2 x sin x ) f (0) = & 2; f 00 ( x ) = 4 e & 2 x cos x + 2 e & 2 x sin x + 2 e & 2 x sin x & e & 2 x cos x = 3 e & 2 x cos x + 4 e & 2 x sin x ) f 00 (0) = 3; f 000 ( x ) = & 6 e & 2 x cos x & 3 e & 2 x sin x & 8 e & 2 x sin x + 4 e & 2 x cos x = & 2 e & 2 x cos x & 11 e & 2 x sin x ) f 000 (0) = & 2; f (4) ( x ) = 4 e & 2 x cos x + 2 e & 2 x sin x + 22 e & 2 x sin x & 11 e & 2 x cos x = & 7 e & 2 x cos x + 24 e & 2 x sin x ) f (4) (0) = & 7; f (5) ( x ) = 14 e & 2 x cos x + 7 e & 2 x sin x & 48 e & 2 x sin x + 24 e & 2 x cos x = 38 e & 2 x cos x & 41 e & 2 x sin x ) f (5) (0) = 38 : Then P 4 ( x ) = f (0) + f (0) x + f 00 (0) x 2 2! + f 000 (0) x 3 3! + f (4) (0) x 4 4! + f (5) (0) x 5 5! = 1 & 2 x + 3 x 2 2! & 2 x 3 3! & 7 x 4 4! + 38 x 5 5! : 2 Question 3 Find P 8 ( x ) for f ( x ) = x 2 e & x 3 : Solutions The "brute force" method would be to calculate f ( j ) (0) for j = 0 ; 1 ; 2 ; ::: 8 . This can be done but is quite messy. We can simplify the calculations with two observations: (i) The Taylor polynomial P 8 at for f is the Taylor polynomial P 6 for e & x 3 but multiplied by x 2 ; (ii) The Taylor polynomial P 6 at for e & x 3 is the Taylor polynomial P 2 ( x ) for e & x with x replaced by x 3 . Let us make life easier by using these observations. Firstly let us compute the Taylor polynomial of degree 2 for e & x at : g ( x ) = e & x ) g (0) = 1; g ( x ) = & e & x ) g (0) = & 1; g 00 ( x ) = e & x ) g 00 (0) = 1 : So the Taylor polynomial P 2 for g ( x ) = e & x at is g (0) + g (0) x + g 00 (0) x 2 2! = 1 & x + x 2 2! : (Of course, we could also have taken P 2 for e x and then substituted & x for x ). Then the Taylor polynomial P 6 for g & x 3 = e & x 3 at is just 1 & x 3 + & x 3 2 2!...
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Ma1502HW1SolnsFall2007 - MATH1502 - Homework 1 - Fall 2007...

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