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A_PFNL_M2J_F09

# A_PFNL_M2J_F09 - Math 2J Lecture B Fall 2009(Instructor...

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Math 2J Lecture B Fall 2009 (Instructor: Professor R. C. Reilly) Answer Key for Practice Final Exam (1) Let A be the 2-by-3 matrix whose columns are the given vectors u , v , and w . That is, A = [ u v w ] = - 3 2 - 5 1 5 2 . It was shown in class that the condition for the given vectors to be linearly independent is equivalent to the homoge- neous system A x = 0 having only one solution, namely x = 0 . This, in turn, is equivalent to the general solution of the system having no free variables. However, since there are only two rows in the matrix A , the system A x = 0 can have at most 2 lead variables. But the system has 3 variables in all, so there must be at least one free variable, so the vectors are not linearly independent. To find numbers a , b , and c as required by the problem, let us actually solve the system A x = 0 by Gauss-Jordan Reduction: [ A | 0 ] = - 3 2 - 5 0 1 5 2 0 1 5 2 0 - 3 2 - 5 0 1 5 2 0 0 17 1 0 1 5 2 0 0 1 1 17 0 1 0 29 17 0 0 1 1 17 0 From this last matrix, which is in reduced row-echelon form, one sees that the general solution of the system has one free variable, namely x 3 ; and the formula for this general solution is x = x 3 - 29 / 17 - 1 / 17 1 NOTE: We are free to use any nonzero value of x 3 to get the desired a , b and c . To simplify checking our answer, let us get rid of the fractions by choosing x 3 = 17, which gives us the special solution x = - 29 - 1 17 . That is, we get a = - 29, b = - 1, c = 17. Clearly at least one of these numbers is not zero; indeed, this is true for all of them. Checking Our Answer : Note that - 29 u - v + 17 w = ( - 29) - 3 1 + ( - 1) 2 5 + 17 - 5 2 = ( - 29) · ( - 3) + ( - 1) · 2 + 17 · ( - 5) ( - 29) · 1 + ( - 1) · 5 + 17 · 2 = 0 0 (2) Notice that one has lim n → ∞ 12 n 3 - 7 n 2 + 9 n + 6 3 n + 2 = + and lim n → ∞ (4 n 2 - 5 n + 2) = + . Thus, trying to compute the limit of the given sequence ( a 1 , a 2 , . . . ) using only the ‘standard limit laws’ leads us to the indeterminante form ∞ - ∞ . However, if we do some preliminary algebra, we can simplify the problem. More precisely, write a n as a single fraction: a n = 12 n 3 - 7 n 2 + 9 n + 6 3 n + 2 - (4 n 2 - 5 n + 2) = (12 n 3 - 7 n 2 + 9 n + 6) - (3 n + 2)(4 n 2 - 5 n + 2) 3 n + 2 = (12 n 3 - 7 n 2 + 9 n + 6) - (12 n 3 + 8 n 2 - 15 n 2 - 10 n + 6 n + 4) 3 n + 2 = 13 n + 2 3 n + 2 = 13 + 2 /n 3 + 2 /n Thus, one easiy computes that lim n → ∞ a n = 13 / 3. In particular, the sequence does converge, and its value is 13 / 3. (3) The coefficient a 3 is given by a 3 = f 000 ( - 1) / 3!, so we need only determine f 000 ( - 1). To get f 000 ( x ) simply differentiate both sides of the differential equation f 00 ( x ) + 3 x 2 f 0 ( x ) + 4 f ( x ) = 0 to get f 000 ( x ) + 6 xf 0 ( x ) + 3 x 2 f 00 ( x ) + 4 f 0 ( x ) = 0 ( * ) 1

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But f ( - 1) = 2 and f 0 ( - 1) = - 3 (given facts); and we can substitute these values into the differential equation for f to get f 00 ( - 1) + 3( - 1) 2 · ( - 3) + 4 · 2 = 0. Thus, f 00 ( - 1) = 9 - 8 = 1. Substitute x = - 1 into the equation ( * ), and use the values of f ( - 1), f 0 ( - 1) and f 00 ( - 1) obtained above, to get f 000 ( - 1) - 6 · ( - 3) + 3 · ( - 1) 2 · 1 + 4 · ( - 3) = 0 and thus f 000 ( - 1) = - 18 - 3 + 12 = - 9. Thus a 3 = f 000 ( - 1) / 6 = - 9 / 6 = - 3 / 2. Thus the correct answer is ‘(A) - 3 / 2’.
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A_PFNL_M2J_F09 - Math 2J Lecture B Fall 2009(Instructor...

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