Ma1502Te1-06Solns - MATH1502 Calculus II TEST 1 February 2...

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MATH1502 - Calculus II TEST 1 - February 2, 2006 SOLUTIONS There are 54 marks on this paper. Full marks (100%) is 50 marks. Question Points/marks 1 14 2 8 3 23 4 9 Total 54 Question 1 Let f ( x ) = cos(2 x ) ° cos( x ) : (i) Compute the 4 th degree Taylor polynomial P 4 ( x ) of f (about 0 ) and also compute the Lagrange form of the remainder R 4 ( x ) . (8 marks) (ii) Estimate the maximum error as x ranges over [1 ; 2] : That is, estimate the maximum of j R 4 ( x ) j = j f ( x ) ° P 4 ( x ) j for x in [1 ; 2] : (4 marks) (iii) Use (i) to write down (without proof) the 8 th degree Taylor polynomial P 8 (about 0 ) to g ( x ) = cos(4 x 2 ) ° cos(2 x 2 ) : (2 marks) Solution to (i) f ( x ) = cos (2 x ) ° cos ( x ) ) f (0) = 1 ° 1 = 0; f 0 ( x ) = ° 2 sin (2 x ) + sin ( x ) ) f 0 (0) = 0 ° 0 = 0 : f 00 ( x ) = ° 4 cos (2 x ) + cos ( x ) ) f 00 (0) = ° 4 + 1 = ° 3; f 000 ( x ) = 8 sin (2 x ) ° sin ( x ) ) f 000 (0) = 0 ° 0 = 0; f (4) ( x ) = 16 cos (2 x ) ° cos ( x ) ) f (4) (0) = 16 ° 1 = 15 f (5) ( x ) = ° 32 sin (2 x ) + sin x: 1
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Then P 4 ( x ) = f (0) + f 0 (0) x + f 00 (0) x 2 2! + f 000 (0) x 3 3! + f (4) (0) x 4 4! = 0 + 0 ° 3 x 2 2! + 0 + 15 x 4 4! = ° 3 2 x 2 + 5 8 x 4 : (5 marks) The Lagrange form of the remainder is R 4 ( x ) = 1 5! f (5) ( c ) x 5 ; for some c between 0 and x . Then R 4 ( x ) = 1 5! ( ° 32 sin (2 c ) + sin c ) x 5 : (3 marks) Solution to (ii) We see that j R 4 ( x ) j = ° ° ° ° 1 5! ( ° 32 sin (2 c ) + sin c ) x 5 ° ° ° ° ± 1 5! (32 j sin 2 c j + j sin c j ) j x j 5 ± 1 5! (32 + 1)2 5 = 1 5! 33(32) : (4 marks) (( = 1 5 (11) 8 = 88 5 )) Solution to (iii) We see that we have substituted 2 x 2 for x to get g ( x ) from f ( x ) . That is, g ( x ) = f ± 2 x 2 ² :
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