Ma1502ExFinalEgroupSolns-Spring2009

Ma1502ExFinalEgroupSolns-Spring2009 - MATH1502 - H Group...

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MATH1502 - H Group Calculus II Final Exam Thursday April 30th, 2009, 8:00 - 10:50 125=100%. Name_______________________________________ Student Number ______________________________ Group G ___________________________________ TA _________________________ ______________ Question Grade Out of 1 14 2 15 3 22 4 40 5 47 Total 138 1
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Question 1 (a) New. Find the radius of convergence and interval of convergence of the power series 1 X k =2 k ln k (11 x ) k : (8 marks) Solution We use the root test applied to a k = k ln k (11 x ) k = k ln k (11 j x j ) k : We see that lim k !1 a 1 =k k = lim k !1 ± k ln k (11 j x j ) k ² 1 =k = lim k !1 k 1 =k (ln k ) 1 =k 11 j x j = 11 j x j : ( Recall lim k !1 k 1 =k = 1 . Moreover, lim k !1 (ln k ) 1 =k = 1 . They can assume the second limit, but give them extra credit if they give some proof. For example, lim k !1 ln k = 1 and lim k !1 ln k k = 0 ; so 1 ln k k for large enough k and then 1 1 =k (ln k ) 1 =k k 1 =k : By the pinching/ sandwich rule, lim k !1 (ln k ) 1 =k = 1 : ) 2
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By the root test, we have convergence for 11 j x j < 1 : And we have divergence for j x j > 1 11 . So the radius of convergence is 1 11 . (4 marks) Next, we test the endpoints x = 1 11 . x = ± 1 11 Here 1 X k =2 k ln k (11 x ) k = 1 X k =2 k ln k ( ± 1) k : Here we can use the Basic Divergence test: from class limits, lim k !1 ln k k = 0 ; so lim k !1 k ln k = 1 : (Alternatively, k ² ln k for k ² 1 , etc.). Thus the terms of the series do not approach 0 , and the series diverges by the basic divergence test. x = 1 11 Here 1 X k =2 k ln k (11 x ) k = 1 X k =2 k ln k : As in the case above, the terms do not approach 0 , and the series diverges by the basic divergence test. So the interval of convergence is ( ± 1 11 ; 1 11 ) : (4 marks) 3
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new Find the radius of convergence of the power series 1 X k =0 k !(3 k )! (4 k )! x k : (6 marks) Solution Because of the factorials, we use the ratio test, applied to j a k j = k !(3 k )! (4 k )! x k = k !(3 k )! (4 k )! j x j k ; with x 6 = 0 . We see that lim k !1 a k +1 a k = lim k !1 ± ( k +1)!(3 k +3)! (4 k +4)! j x j k +1 ² k !(3 k )! (4 k )! j x j k (2 marks) = lim k !1 (4 k )! (4 k + 4)! ( k + 1)! k ! (3 k + 3)! (3 k )! j x j = lim k !1 ( k + 1) (3 k + 3) (3 k + 2) (3 k + 1) (4 k + 4) (4 k + 3) (4 k + 2) (4 k + 1) j x j = lim k !1 k 4 (1 + 1 =k ) (3 + 3 =k ) (3 + 2 =k ) (3 + 1 =k ) k 4 (4 + 4 =k ) (4 + 3 =k ) (4 + 2 =k ) (4 + 1 =k ) j x j = 3 3 4 4 j x j : (1) By the ratio test, this converges for 3 3 4 4 j x j < 1 , j x j < 4 4 3 3 = 256 27 : (4 marks) Moreover, it diverges for j x j > 256 27 . So the radius of convergence is 256 27 . 4
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This note was uploaded on 02/09/2011 for the course MATH 1502 taught by Professor Mcclain during the Spring '07 term at Georgia Institute of Technology.

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Ma1502ExFinalEgroupSolns-Spring2009 - MATH1502 - H Group...

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