MATH
Math_exam2913_Solution.pdf

# Math_exam2913_Solution.pdf - Mathematics I Master Solutions...

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Mathematics I Master Solutions Exam Fall Term 2013 Prof. Dr. Enrico De Giorgi 1 28 January 2014 1 Chair of Mathematics, University of St. Gallen, Bodanstrasse 6, 9000 St. Gallen, Switzerland, email: [email protected]

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2 Part I: Open questions Exercise 1 (a) ( 3 points ). We have: a n = n 2 + 1 n - 1 - n 2 + 3 n - 2 n + 1 = ( n 2 + 1) ( n + 1) - ( n 2 + 3 n - 2) ( n - 1) ( n - 1) ( n + 1) = n 3 + n 2 + n + 1 - ( n 3 - n 2 + 3 n 2 - 3 n - 2 n + 2 ) n 2 - 1 = n 3 + n 2 + n + 1 - n 3 - 2 n 2 + 5 n - 2 n 2 - 1 = - n 2 + 6 n - 1 n 2 - 1 = - n 2 + 6 n - 1 n 2 - 1 1 n 2 1 n 2 ! | {z } =1 = - 1 + 6 n - 1 n 2 1 - 1 n 2 . It follows that: lim n →∞ a n = lim n →∞ n 2 + 1 n - 1 - n 2 + 3 n - 2 n + 1 = lim n →∞ - 1 + 6 n - 1 n 2 1 - 1 n 2 = lim n →∞ - 1 + 6 n - 1 n 2 lim n →∞ 1 - 1 n 2 = - 1 1 = - 1 . (b) ( 4 points ). Bounded from below : b n = 3 + 7 8 n | {z } 0 3 .
Mathematics I: Master Solutions Exam Fall Term 2013 3 Bounded from above : b n = 3 + 7 8 n | {z } 1 4 . Since { b n } n N is bounded from below and also bounded from above, then it is bounded. Monotone : b n +1 = 3 + 7 8 n +1 = 3 + 7 8 n 7 8 = 3 + 7 8 n 1 - 1 8 = 3 + 7 8 n - 7 8 n 1 8 = b n - 7 8 n 1 8 | {z } 0 < b n . Since b n +1 < b n for all n N , then { b n } n N is monotonically decreasing. Finally, since { b n } n N is bounded and monotonically decreasing, then it is also convergent. (c) ( 6 points ). The series { s n } n N defined by s n = n X k =1 3 a + 1 a + 1 k , a R \ {- 1 } is a geometric series with q = 3 a +1 a +1 . The criterium for convergence is | q | < 1 . We have: | q | < 1 3 a + 1 a + 1 < 1 (3 a + 1) 2 < ( a + 1) 2 9 a 2 + 6 a + 1 < a 2 + 2 a + 1 8 a 2 + 4 a < 0 4 a (2 a + 1) < 0 a - 1 2 , 0 .

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4 If a ( - 1 2 , 0 ) , then X k =1 3 a + 1 a + 1 k = 3 a + 1 a + 1 1 1 - q = 3 a + 1 a + 1 1 1 - 3 a +1 a +1 = - 3 a + 1 2 a . (d) ( 9 points ). The following figure displays the borrowed money at time k = 0 and the payments at k = 1 , . . . , 30 . R 0 = 1 , 000 , 000 C 1 = C C 2 = C . . . C 20 = C C 21 = 2 C C 22 = 2 C . . . C 30 = 2 C k C k , R k To determine C we set the present value of payments equals to the borrowed amount R 0 = 1 , 000 , 000 Swiss francs. We have: 20 X k =1 C (1 + i ) k + 30 X k =21 2 C (1 + i ) k = 1 , 000 , 000 .
Mathematics I: Master Solutions Exam Fall Term 2013 5 It follows that: 20 X k =1 C (1 + i ) k + 30 X k =21 2 C (1 + i ) k = 1 , 000 , 000 C 20 X k =1 1 (1 + i ) k + 2 C 10 X k =1 1 (1 + i ) 20+ k = 1 , 000 , 000 C 20 X k =1 1 (1 + i ) k + 2 C (1 + i ) 20 10 X k =1 1 (1 + i ) k = 1 , 000 , 000 C 1 + i 1 - 1 1+ i 20 1 - 1 1+ i + 2 C (1 + i ) 21 1 - 1 1+ i 10 1 - 1 1+ i = 1 , 000 , 000 C 1 1 + i 1 - 1 1+ i 20 1 - 1 1+ i + 2 (1 + i ) 21 1 - 1 1+ i 10 1 - 1 1+ i = 1 , 000 , 000 C = 1 , 000 , 000 1 1+ i 1 - ( 1 1+ i ) 20 1 - 1 1+ i + 2 (1+ i ) 21 1 - ( 1 1+ i ) 10 1 - 1 1+ i i =5% 54 , 696 . 55 (Swiss francs) .

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6 Exercise 2 (a1) ( 3 points ).
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