Midterm Answers Fall 2011 .pdf
Midterm Answers Fall 2011 .pdf

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School
McGill University
Department
Mathematics & Statistics (Sci)
Course
MATH 222
Professor
Christa Scholtz
Semester
Fall

Description
MATH 222 MIDTERM SOLUTIONS 1. SOLUTIONS Problem 1 Consider the sequencenfa g defined recursively by p a1= 11; a n+1= 2an+ 3 for n ▯ 1: p p [So 1 = 25 = 5, 2 = 13, etc.] Which of the following is true? (a) lin!1 an= 0: (b) lin!1 an= 11: (c) lim a = 3: n!1 n (d) lin!1 an= 2: (e) The sequence does not converge to any real number. Solution. The correct solution is (c). Notice that n!1 an+1 = n!1 an Let X = limn!1 an, then p X = lim 2an+ 3 p!1 = 2X + 3: We find that 2 X ▯ 2X ▯ 3 = 0: Therefore X = 3 or X = ▯1. The sequence is strictly positive, so we choose X = 3.▯ Problem 2 The value of 2x 2 3 3e ▯ 3 ▯ 6x ▯ 6x ▯ 4x x!0 x4 is (a) 1. (b) 2. (c) 3. (d) None of the above, because the limit is some other real number. (e) None of the above, because the limit doesn’t exist. Solution. The correct solution is (b). We can use L’Hopital’s rule, or we can substitute the 2x Maclaurin series for 3e . We know that X1 n x x e = : n=0n! 1 2 MATH 222 MIDTERM SOLUTIONS Therefore ▯ ▯ X1 (2x)n (2x)2 (2x) 3 (2x)4 (2x) 5 3e2x = 3 = 3 1 + (2x) + + + + + ::: n! 2! 3! 4! 5! n=0 2 3 4 96x 5 = 3 + 6x + 6x + 4x + 2x + + ::: 5! This becomes 2x 2 3 3e ▯ 3 ▯ 6x ▯ 6x ▯ 4x = 2 + 96x + :::; x 4 5! and now it is possible to calculate the limit. We conclude that 2x 2 3 lim 3e ▯ 3 ▯ 6x ▯ 6x ▯ 4x = 2 + 0; x!0 x4 and therefore the correct answer is (b). ▯ Problem 3 The interval of convergence of the power series 1 X 5 + 2 n n (x ▯ 2) is n=1 n (a) (▯:2;:2) (b) [▯:2;:2) (c) (▯1;1) (d) (1:8;2:2) (e) [1:8:2:2) Solution. The correct answer is (e). Use the ratio test. Let 5 + 2 n an= (x ▯ 2) : n Then ▯ ▯ ▯ ▯▯ ▯ ▯ ▯a ▯ ▯ 5 n+1+ 2 n+1 n (x ▯ 2)n+1 ▯ lim ▯ n+1▯ = lim ▯ ▯ n!1 ▯ a n ▯ n!1 ▯ 5 + 2 n n + 1 (x ▯ 2) n ▯ ▯ n+1 n+1 ▯ ▯ ▯ = lim ▯5 + 2 ▯▯ lim ▯ n ▯ ▯ jx ▯ 2j n!1 ▯ 5 + 2 n ▯ n!1 ▯n + 1▯ ▯ ▯ ▯ ▯ ▯1 + limn!1 2 n+1▯ = 5 ▯ ▯ ▯n ▯ ▯ 1 ▯ jx ▯ 2j ▯ 1 + limn!1 5 ▯ ▯ ▯ ▯1 + 0▯ = 5 ▯ ▯▯ 1 ▯ jx ▯ 2j 1 + 0 = 5jx ▯ 2j The ratio test implies that we have absolute convergence if 5jx▯2j < 1. Thus we must have 1 1 ▯ < x ▯ 2 < ; 5 5 4 MATH 222 MIDTERM SOLUTIONS Problem 5 The sum of the series
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