Ma1502Te1Solns-HSpring2011

Ma1502Te1Solns-HSpring2011 - MATH1502 - Calculus II TEST 1...

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MATH1502 - Calculus II TEST 1 - H Group - February 3, 2011 NAME : ___________________________________ STUDENT NUMBER :_______________________ GROUP (e.g. H1) :_______________________ TEACHING ASSISTANT :_____________________ Write your solutions to the questions on this testpaper - you may use both sides of each sheet of paper. There are 52 marks on this paper. Full marks (100%) is 50 marks. You may NOT use a calculator or any notes. Question Points Ex 1 14 2 7 3 23 4 8 Total 52 ) 50 Question 1 Let f ( x ) = 1 + x 1 x x: (i) Compute the 3 rd degree Taylor polynomial P 3 ( x ) of f (about 0 ). (7 marks) (ii) Compute the Lagrange form of the remainder R 3 ( x ) and hence estimate the maximum error of j R 3 ( x ) j = j f ( x ) P 3 ( x ) j for x in [ 1 ; 0] : (5 marks) (iii) Use (i) to write down (without proof) the 9 th degree Taylor polynomial P 9 (about 0 ) to g ( x ) = x 3 1 + x 3 1 x 3 : (2 marks) Solutions 1
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f ( x ) = x + x 2 1 x ) f (0) = 1 1 0 = 0; f 0 ( x ) = (1 + 2 x ) (1 x ) ( x + x 2 ) ( 1) (1 x ) 2 = 1 + x 2 x 2 + x + x 2 (1 x ) 2 = 1 + 2 x x 2 (1 x ) 2 ) f 0 (0) = 1 : f 00 ( x ) = (2 2 x ) (1 x ) 2 (1 + 2 x x 2 ) 2 (1 x ) ( 1) (1 x ) 4 = (2) (1 x ) 2 + 2 (1 + 2 x x 2 ) (1 x ) 3 = 2 (1 2 x + x 2 + 1 + 2 x x 2 ) (1 x ) 3 = 4 (1 x ) 3 ) f 00 (0) = 4 : f 000 ( x ) = 12 (1 x ) 4 ) f 000 (0) = 12 : f (4) ( x ) = 48 (1 x ) 5 : METHOD 2: product rule (messy) f ( x ) = 1 + x 1 x x ) f (0) = 1 1 0 = 0; f 0 ( x ) = (1) (1 x ) ( 1) (1 + x ) (1 x ) 2 x + 1 + x 1 x = 2 x (1 x ) 2 + 1 + x 1 x ) f 0 (0) = 0 + 1 = 1 : 2
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f 00 ( x ) = (2) (1 x ) 2 (2 x ) 2(1 x ) ( 1)
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Ma1502Te1Solns-HSpring2011 - MATH1502 - Calculus II TEST 1...

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