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final-practice-solutions

final-practice-solutions - MATH 31B/2 PRACTICE FINAL EXAM...

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MATH 31B/2 PRACTICE FINAL EXAM SOLUTIONS. Please note: the aim of this practice final is to give you several problems on the material not covered by the first two midterms and practice midterms. You should therefore treat the first two midterms and practice midterms as a part of the practice final. 1. Let f ( x ) = sin x . Find n so that Taylor’s polynomial of degree n around 0 approximates sin ( 1 ) to within 10 - 2 . Justify your answer. Solution. Using Taylor’s remainder formula, if we keep n terms in Taylor’s polynomial, the error is at most E n M ( x - a ) n + 1 ( n + 1 ) ! , where M is the maximal value of the n + 1-st derivative of sin ( x ) on the interval [ 0 , 1 ] . Hence M = 1 and E n 1 n + 1 ( n + 1 ) ! = 1 ( n + 1 ) ! . Hence we want n so that 1 ( n + 1 ) ! 1 100 , i.e., ( n + 1 ) ! 100. Since 4! = 4 · 3 · 2 · 1 = 24 < 100 and 5! = 5 · 4! = 120 > 100, we get that n = 4 works. 2. Let f ( x ) = 2 x + 4 x 3 - 1 . Express f ( x ) as a sum of terms using partial fractions. Use this to evaluate Z f ( x ) dx . Solution. Since x 3 - 1 = ( x - 1 )( x 2 + x + 1 ) , we will try to find A , B , C so that 2 x + 4 x 3 - 1 = A x - 1 + Bx + C x 2 + x + 1 . We then have 2 x + 4 = A ( x 2 + x + 1 )+( Bx + C )( x - 1 ) = x 2 ( A + B )+ x ( A - B + C )+( A - C ) . Equating coefficients, we get 0 = A + B , 2 = A - B + C and 4 = A - C . From the first equation we get that B = - A and from the last that C = A - 4. Substituting this into the second equation gives 2 = A - ( - A )+ A - 4 = 3 A - 4. Hence 3 A = 6 and A = 2. Thus B = - 2 and C = - 2. We conclude that 2 x + 4 x 3 - 1 = 2 x - 1 + - 2 x - 2 x 2 + x + 1 . To evaluate the integral, we need to evaluate the integrals R 2 x - 1 dx = 2ln ( x - 1 ) + C and R - 2 x - 2 x 2 + x + 1 dx . In the latter integral, we complete the square: x 2 + x + 1 = ( x + 1 2 ) 2 + 3 4 = 1
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MATH 31B/2 PRACTICE FINAL EXAM SOLUTIONS. 2
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