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Course: MATH 5525, Spring 2010
School: Minnesota
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5525. Math April 14, 2010. Midterm Exam 2. Problems and Solutions. Problem 1. Find the general solution of the equation xy - (2x + 1)y + (x + 1)y = 0. Note that one of two linearly independent solutions is y1 (x) = ex . Solution. By Abel's formula, the Wronskian W (x) = W (y1 , y2 )(x) satisfies W =- 2x + 1 p1 W = W = p0 x 2+ 1 x W, so that W (x) = W (x0 ) xe2x . On the other hand, 2 W (x) = y1 y2 y1 = e2x...

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5525. Math April 14, 2010. Midterm Exam 2. Problems and Solutions. Problem 1. Find the general solution of the equation xy - (2x + 1)y + (x + 1)y = 0. Note that one of two linearly independent solutions is y1 (x) = ex . Solution. By Abel's formula, the Wronskian W (x) = W (y1 , y2 )(x) satisfies W =- 2x + 1 p1 W = W = p0 x 2+ 1 x W, so that W (x) = W (x0 ) xe2x . On the other hand, 2 W (x) = y1 y2 y1 = e2x y2 y1 . Therefore, (y2 /y1 ) = const x, y2 /y1 = const x2 . We can take y2 = x2 y1 = x2 ex . Then the general solution has the form y = C1 y1 + C2 y2 = (C1 + C2 x2 )ex . Problem 2. Let y = y(x) be a solution of the problem y =1+y+ Show that y(x) - ln(1 - x) Proof. The function z = - ln(1 - x) satisfies z = 1 z2 = ez 1 + z + 1-x 2 for 0 x < 1, z(0) = 0. for 0 x < 1. y2 , 2 y(0) = 0. Note that the comparison theorem (Theorem 11.1) can be applied in the case of non-strict inequalities, if the function f (x, y) is Lipschitz. Therefore, we have z(x) y(x) for 0 x < 1. Problem 3. Let y = y(x, ) be a solution of the problem y = y 2 + 2 x-1 , Find the derivative z = z(x, ) = Solution. At the point = 0, we have y = y2, The function z(x) satisfies z = 2yz + 2x-1 = -2x-1 z + x-1 , Therefore, z(x) = Cx-2 + 1 = 1 - x-2 . 1 z(1) = 0. -y -1 = x + C = x, y = -x-1 . y y(1) = -1. at the point = 0. Problem 4. Solve the problem u = 2 4 3 6 u+ -8 0 , u(0) 1 = 4 ; where u = u(x) = u1 (x) u2 (x) . Solution. We know that the solution of the problem u (x) = Au(x)+b(x), u(0) = u0 , is represented in the form x eA(x-t) b(t) dt. u(x) = eAx u0 + 0 In our case, the characteristic polynomial p() = |I - A| = has roots 1 = 0 and 2 = 8. Denote p1 () = Then e We have eAx u0 = eAx eAx b = eAx x Ax 2 -2 -4 -3 - 6 = 2 - 8 p() = - 8, - 1 p2 () = -6 4 3 -2 1 8 = -10 5 -6 3 p() = . - 2 + e8x 8 e8x 8 2 4 3 6 18 27 -2 -3 , . = k=1 ek x 1 pk (A) = pk (k ) -8 1 4 -8 0 t=x = + , + e8x e8(x-t) -8 0 x -6x eA(x-t) b(t) dt = 3x 0 e8(x-t) dt = Finally, u(x) = e u + 0 Ax 0 x 1 e8x , =- + 8 8 t=0 1 2 e8x + + 8 3 8 -6x - 1 3x + 1 -2 -3 + e8x . eA(x-t) b(t) dt = 2 3 . Problem 5. Let A be a constant 2 2 matrix. Show that all solutions of the system u (x) = Au(x) satisfy the equality u(2) + c1 u(1) + c2 u(0) = 0, with constants c1 and c2 depending only on the matrix A, and not depending on the solution u(x). Hint. Apply the Cayley-Hamilton theorem to the matrix B = eA . Proof. Let c1 and c2 be the coefficients of the characteristic polynomial of the matrix B = eA : p() = |I - B| = 2 + c1 + c2 . By the Cayley-Hamilton theorem, p(B) = B 2 + c1 B + c2 I = e2A + c1 eA + c2 I = 0. Since any solution of u = Au has the form u(x) = eAx u0 , where u0 = u(0), we get u(2) + c1 u(1) + c2 u(0) = e2A u0 + c1 eA u0 + c2 u0 = p(B)u0 = 0. 2
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Minnesota - MATH - 5525
Math 5525: Introduction to Ordinary Differential EquationsSyllabus: Spring 2010Class Times and Location: 2:30 pm 3:20 pm MWF, AmundH 240. Instructor: Mikhail Safonov, VinH 231, tel: 625-8571, email: safonov@math.umn.edu http:/www.math.umn.edu/safonov Of
Minnesota - MATH - 3283
MATH 3283W. Sequences, Series, and Foundations: Writing Intensive. Spring 2009Homework 1. Problems and Solutions I. Writing Intensive Part 1 (5 points). Check whether or not each of the following statements can be true for some values (&quot;true&quot; or &quot;false&quot;)
Minnesota - MATH - 3283
MATH 3283W. Sequences, Series, and Foundations: Writing Intensive. Spring 2009Homework 3. Problems and Solutions I. Writing Intensive Part 1. Let f (x) be a continuous function on the segment [0, 1]. Show that f is uniformly continuous on [0, 1], which m
Minnesota - MATH - 3283
These notes by Mikhail Safonov serve as a supplementary material to the textbook by Weyne Richter &quot;Sequences, Series and Foundations. Math 2283 and 3283W&quot;Sequences, Series and FoundationsChapter 1. Truth, Falsity and Mathematical Induction1 Truth Table
Minnesota - MATH - 5615
Math 5615H. Name (Print)October 5, 2011.Midterm Exam.60 points are distributed between 5 problems. You have 50 minutes (2:30 pm 3:20 pm) to work on these problems. No books, no notes. Calculators are permitted, however, for full credit, you need to sho
Minnesota - MATH - 5615
Math 5615H. Name (Print)November 16, 2011.2nd Midterm Exam.60 points are distributed between 5 problems. You have 50 minutes (2:30 pm 3:20 pm) to work on these problems. No books, no notes, except for: Appendix A. Exponential and Logarithmic Functions
Minnesota - MATH - 5615
Appendix A. Exponential and Logarithmic FunctionsFor fixed b &gt; 1, the function bx was defined in Exercise 6 on p.22 in the textbook &quot;Principles of Mathematical Analysis&quot; by W. Rudin. It satisfies bx &gt; 0, and (E1). bx+y = bx by for real x, y. In particula
Minnesota - MATH - 5615
Math 5615H: Introduction to Analysis I.Fall 2011Homework #1 (due on Wednesday, September 14). 50 points are divided between 5 problems, 10 points each. #1. Prove that 6 and 2 + 3 are NOT rational. #2. Let A and B be bounded sets in R. Consider the algeb
Minnesota - MATH - 5615
Math 5615H: Introduction to Analysis I.Fall 2011Homework #2 (due on Wednesday, September 21). 50 points are divided between 5 problems, 10 points each. #1. Let F be a field. Show that there exist not more that two different solutions solutions of the eq
Minnesota - MATH - 5615
Math 5615H: Introduction to Analysis I.Fall 2011Homework #3 (due on Wednesday, September 28). 50 points are divided between 5 problems, 10 points each. #1. Let A := cfw_a1 , a2 , . . . be a set of real numbers defined as follows: a1 = 1, Find sup A. #2.
Minnesota - MATH - 5615
Math 5615H: Introduction to Analysis I.Fall 2011Homework #4 (due on Wednesday, October 5). 50 points are divided between 5 problems, 10 points each. #1. Let f be a mapping of A to B. Show that for each B1 B and B2 B, their inverse images satisfy the pro
Minnesota - MATH - 5615
Math 5615H: Introduction to Analysis I.Fall 2011Homework #5 (due on Wednesday, October 12). 50 points are divided between 5 problems, 10 points each. #1. Show that for an arbitrary sequence E1 , E2 , . . . of sets, the set ( n=1 k=n) Ek is contained i
Minnesota - MATH - 5615
Math 5615H: Introduction to Analysis I.Fall 2011Homework #6 (due on Wednesday, October 19). 50 points are divided between 5 problems, 10 points each. #1. Show that for an arbitrary set E in a metric space (X, d), the set E of its limit point is closed.
Minnesota - MATH - 5615
Math 5615H: Introduction to Analysis I.Fall 2011Homework #7 (due on Wednesday, October 26). 50 points are divided between 4 problems. You can use the following Theorem which was proved in class. Theorem. A subset K of a metric space (X, d) is compact in
Minnesota - MATH - 5615
Math 5615H: Introduction to Analysis I.Fall 2011Homework #8 (due on Wednesday, November 2). 50 points are divided between 4 problems. #1. (12 points). Let 0 &lt; x1 = a &lt; x2 = b be arbitrary real number, and let 1 xn := (xn-2 + xn-1 ) for n = 3, 4, 5, . .
Minnesota - MATH - 5615
Math 5615H: Introduction to Analysis I.Fall 2011Homework #9 (due on Wednesday, November 9). 50 points are divided between 4 problems. #1. (10 points). Using partial-fraction decomposition 1 A B C = + + , n(n + 1)(n + 2) n n+1 n+2 show that the series n
Minnesota - MATH - 5615
Math 5615H: Introduction to Analysis I.Fall 2011Homework #10 (due on Wednesday, November 16). 50 points are divided between 4 problems. #1. (12 points). The sequence cfw_an is defined by a1 = 0, and an+1 = 2an /2 for n = 1, 2, 3, . . . .Prove that cfw
Minnesota - MATH - 5615
Math 5615H: Introduction to Analysis I.Fall 2011Homework #11 (due on Wednesday, November 23). 50 points are divided between 4 problems. #1. (12 points). Find S :=n=1 n . 2nHint. One can try to rewrite 2S in terms of S, or use Theorem 3.41 with an = 1
Minnesota - MATH - 5615
Math 5615H: Introduction to Analysis I.Fall 2011Homework #12 (due on Wednesday, November 30). 50 points are divided between 5 problems, 10 points each. #1. Let aj and bj,k be real numbers defined for all j, k = 0, 1, 2, . . ., such that j=0|aj | A = c
Minnesota - MATH - 5615
Math 5615H: Introduction to Analysis I.Fall 2011Homework #13 (due on Wednesday, December 7). 50 points are divided between 4 problems. #1. (10 points, Exercise 2 on p.98). If f is a continuous mapping of a metric space X into a metric space Y , prove th
Minnesota - MATH - 5615
Math 5615H: Introduction to Analysis I.Fall 2011Homework #1. Problems and short Solutions. #1. Prove that 6 and 2 + 3 are NOT rational. Proof. If p := 6 Q, then p2 = 6, and we get a contradiction in the same way as in Example 1.1 in the textbook. If q :
Minnesota - MATH - 5615
Math 5615H: Introduction to Analysis I. Homework #2. Problems and Solutions.Fall 2011#1. Let F be a field. Show that there exist not more that two different solutions solutions of the equation x x = 1. Is it possible that there is only one solution to t
Minnesota - MATH - 5615
Math 5615H: Introduction to Analysis I. Homework #3. Problems and Solutions.Fall 2011#1. Let A := cfw_a1 , a2 , . . . be a set of real numbers defined as follows: a1 = 1, and ak+1 = 1 + ak for k = 1, 2, . . . . Find sup A. Solution. By induction, 1 ak &lt;
Minnesota - MATH - 5615
Math 5615H: Introduction to Analysis I. Homework #4. Problems and Solutions.Fall 2011#1. Let f be a mapping of A to B. Show that for each B1 B and B2 B, their inverse images satisfy the properties (i) f -1 (B1 B2 ) = f -1 (B1 ) f -1 (B2 ), Proof. (i) We
Minnesota - MATH - 5615
Math 5615H: Introduction to Analysis I. Homework #5. Problems and Solutions.Fall 2011#1. Show that for an arbitrary sequence E1 , E2 , . . . of sets, the set lim inf En :=n n=1 k=nEkis contained inlim sup En :=n n=1 k=nEk .Proof. For arbitrary
Minnesota - MATH - 5615
Math 5615H: Introduction to Analysis I. Homework #6. Problems and Solutions.Fall 2011#1. Show that for an arbitrary set E in a metric space (X, d), the set E of its limit point is closed. Proof. Let p be a limit point of E . Then r &gt; 0, the set Gr := Nr
Minnesota - MATH - 5615
Math 5615H: Introduction to Analysis I. Homework #7. Problems and Solutions.Fall 2011You can use the following Theorems which were discussed in class. Theorem 1. A subset K of a metric space (X, d) is compact in X if and only if every infinite subset E
Minnesota - MATH - 5615
Math 5615H: Introduction to Analysis I. Homework #8. Problems and Solutions.Fall 2011#1. Let 0 &lt; x1 = a &lt; x2 = b be arbitrary real number, and let 1 xn := (xn-2 + xn-1 ) for n = 3, 4, 5, . . . . 2 Show that the sequence cfw_xn converges, and find its l
Minnesota - MATH - 5615
Math 5615H: Introduction to Analysis I. Homework #9. Problems and Solutions. #1. Using partial-fraction decompositionFall 20111 A B C = + + , n(n + 1)(n + 2) n n+1 n+2 show that the series n=11 n(n + 1)(n + 2)converges and find its sum. Solution. We
Minnesota - MATH - 5615
Math 5615H: Introduction to Analysis I. Homework #10. Problems and Solutions. #1. The sequence cfw_an is defined by a1 = 0, and an+1 = 2an /2Fall 2011for n = 1, 2, 3, . . . .Prove that cfw_an is convergent and find its limit. You can use without proo
Minnesota - MATH - 5615
Math 5615H: Introduction to Analysis I. Homework #11. Problems and Solutions. #1. (12 points). Find n S := . 2n n=1Fall 2011Solution. Substituting n = m + 1, we get n m+1 ( 1 )m 2S = = =S+ = S + 2, 2n-1 m=0 2m 2 n=1 m=0so that S = 2. #2. (10 points).
Minnesota - MATH - 5615
Math 5615H: Introduction to Analysis I. Homework #12. Problems and Solutions.Fall 2011#1. Let aj and bj,k be real numbers defined for all j, k = 0, 1, 2, . . ., such that|aj | A = const &lt; ,j=0|bj,n | B = const &lt; for all j, n;and lim bj,n = 0 for eac
Minnesota - MATH - 5615
Math 5615H: Introduction to Analysis I. Homework #13. Problems and Solutions.Fall 2011#1. If f is a continuous mapping of a metric space X into a metric space Y , prove that f (E) f (E) for every set E X. (E denotes the closure of E). Show, by an exampl
Minnesota - MATH - 5615
Math 5615H. October 5, 2011. Midterm Exam 1. Problems and Solutions. Problem 1. (10 points). Let A and B be nonempty bounded subsets of R. Show that sup(A B) = supcfw_sup A, sup B. Proof. Denote M1 := sup(A B), M2 := supcfw_sup A, sup B. We have x sup A M
Minnesota - MATH - 5615
Math 5615H. November 16, 2011. Midterm Exam 2. Problems and Solutions. Problem 1. (10 points). Let A1 , A2 , A3 , . . . be subsets of a metric space (X, d). If B :=k=1Ak ,prove that B k=1Ak ,where B and Ak denote the closures of B and Ak . Show, b
Minnesota - MATH - 5615
Math 5615H: Introduction to Analysis I. Syllabus: Fall 2011(updated on September 8) Class Times and Location: 2:30 pm 3:20 pm MWF, VinH 211. Instructor: Mikhail Safonov, VinH 231, tel: 625-8571, email: safonov@math.umn.edu http:/www.math.umn.edu/safonov
Minnesota - MATH - 8601
Math 8601: REAL ANALYSIS. Fall 2010 Problems for Final Exam on Saturday, December 18, 4pm6pm, VinH 1. This Final Exam will be based on the material from the textbook, in the following Sections: 0.50.6, 1.11.5 ,2.12.5, 2.6 (Theorems 2.40, 2.41), 3.13.2, 3.
Minnesota - MATH - 8601
Math 8601: REAL ANALYSIS. Fall 2010 Some problems for Midterm Exam #1 on Wednesday, October 6. You will have 50 minutes (10:10 am11:00 am) to work on 5 problems, 2 of which will be selected from the following list. It is recommended to prepare solutions o
Minnesota - MATH - 8601
Math 8601: REAL ANALYSIS. Fall 2010 Problems for Midterm Exam #2 on Wednesday, November 17. This Midterm will be based on the material for the textbook up to (including) Section 2.3. You will have 50 minutes (10:10 am11:00 am) to work on 5 problems, 2 of
Minnesota - MATH - 8601
Math 8601: REAL ANALYSIS.Fall 2010Homework #1 (due on W, September 15). Updated on Sat, September 11. 50 points are divided between 5 problems, 10 points each. #1. Let F be a compact subset of Rn . Show that there are point x0 , y0 F , such that diamF :
Minnesota - MATH - 8601
Math 8601: REAL ANALYSIS.Fall 2010Homework #2 (due on W, September 29). 50 points are divided between 5 problems, 10 points each. #1. Let f (x) be a continuous function on [-1, 1], such that f (-1) &lt; 0 &lt; f (1). Show that f (c) = 0 for some c (-1, 1). Hi
Minnesota - MATH - 8601
Math 8601: REAL ANALYSIS.Fall 2010Homework #3 (due on W, October 13). 50 points are divided between 5 problems, 10 points each. #1. Let f be a real function on R1 . The image and the inverse image of a subset A R1 under f are correspondingly f (A) = cfw
Minnesota - MATH - 8601
Math 8601: REAL ANALYSIS.Fall 2010Homework #4 (due on Wednesday, October 27). 50 points are divided between 5 problems, 10 points each. #1. Let Rn be represented in the form Rn =Ik , where cfw_Ik are non-overlapping cubesk=1 with edge length 1. Let
Minnesota - MATH - 8601
Math 8601: REAL ANALYSIS.Fall 2010Homework #5 (due on Wednesday, November 10). 50 points are divided between 5 problems, 10 points each. #1 (Problem 30 on p.40. Let E be a Lebesgue measurable set in R1 with Lebesgue measure m(E) &gt; 0. Show that for any &lt;
Minnesota - MATH - 8601
Math 8601: REAL ANALYSIS.Fall 2010Homework #6 (due on Wednesday, November 24). 50 points are divided between 5 problems, 10 points each. #1 (Borel-Cantelli Lemma). Let (X, M, ) be a measure space, and cfw_An be a sequence of sets in M. Show thatifn=1
Minnesota - MATH - 8601
Math 8601: REAL ANALYSIS.Fall 2010Homework #7 (due on Friday, December 10). 50 points are divided between 5 problems, 10 points each. #1. Let f L1 (R1 ). Show that 1 Sn (x) = n in L1 (R1 ) as n , i.e. |Sn (x) - S(x)| dx 0 as n . Hint. Using Theorem 2.26
Minnesota - MATH - 8601
Math 8601: REAL ANALYSIS.Fall 2010Homework #1. Problems and Solutions. #1. Let F be a compact subset of Rn . Show that there are point x0 , y0 F , such that diam F := supcfw_ |x - y| : x, y F = |x0 - y0 |. Proof. By definition of sup, there are sequenc
Minnesota - MATH - 8601
Math 8601: REAL ANALYSIS.Fall 2010Homework #2. Problems and Solutions. #1. Let f (x) be a continuous function on [-1, 1], such that f (-1) &lt; 0 &lt; f (1). Show that f (c) = 0 for some c (-1, 1). Proof. Take c := supcfw_x [-1, 1] : f (x) 0 (0, 1). We claim
Minnesota - MATH - 8601
Math 8601: REAL ANALYSIS.Fall 2010Homework #3. Problems and Solutions. #1. Let f be a real function on R1 . The image and the inverse image of a subset A R1 under f are correspondingly f (A) = cfw_y : y = f (x) for some x A, f -1 (A) = cfw_x : f (x) A.
Minnesota - MATH - 8601
Math 8601: REAL ANALYSIS.Fall 2010Homework #4. Problems and Solutions. #1. Let Rn be represented in the form Rn =Ik , where cfw_Ik are non-overlapping cubesk=1 with edge length 1. Let Fk be a closed subset of Ik , k = 1, 2, . . . Show that the set F
Minnesota - MATH - 8601
Math 8601: REAL ANALYSIS.Fall 2010Homework #5. Problems and Solutions. #1. Let E be a Lebesgue measurable set in R1 with Lebesgue measure m(E) &gt; 0. Show that for any &lt; 1, there is an open interval I = (a, b) such that m(E I) &gt; m(I). Proof. Suppose that
Minnesota - MATH - 8601
Math 8601: REAL ANALYSIS.Fall 2010Homework #6. Problems and Solutions. #1 (Borel-Cantelli Lemma). Let (X, M, ) be a measure space, and cfw_An be a sequence of sets in M. Show thatifn=1(An ) &lt; ,then lim sup An = 0,n where lim sup An :=nAn .k=1
Minnesota - MATH - 8601
Math 8601: REAL ANALYSIS.Fall 2010Homework #7. Problems and Solutions. #1. Let f L1 (R1 ). Show that 1 Sn (x) = n in L1 (R1 ) as n , i.e. |Sn (x) - S(x)| dx 0 as n .n-1f x+j=0j S(x) = nx+1f (t) dtxProof. By Theorem 2.26, &gt; 0 there are a constant
Minnesota - MATH - 8601
Math 8601. December 18, 2010. Final Exam. Problems and Solutions. #1. Let A be an arbitrary set, and for each A, let an open ball B Rn be defined. Show that there is a finite or countable subset A0 A, such that B =A A0B .Proof. The open set :=AB =j
Minnesota - MATH - 8601
Math 8601. October 6, 2010. Midterm Exam 1. Problems and Solutions. Problem 1. Let (X, M, ) be a measure space with (X) &lt; . Show that for arbitrary A, B, C M, we have |(A B) - (A C)| (BC). where BC := (B \ C) (C \ B) = (B C) \ (B C) - the symmetric differ
Minnesota - MATH - 8601
Math 8601. November 17, 2010. Midterm Exam 2. Problems and Solutions. Problem 1. Findnlim1 1 1 + + + . n n+1 2n - 1Solution. After rewriting Sn := 1 1 1 1 + + + = n n+1 2n - 1 n1 n-11 1+ k=0k n,one can see that this is a Riemann sum for the integ
Minnesota - MATH - 8601
Math 8601/2: REAL ANALYSIS. Syllabus: FALL 2010Class Times and Location: 10:10 am 11:00 am MWF, VinH 1. Instructor: Mikhail Safonov, VinH 231, tel: 625-8571, email: safonov@math.umn.edu http:/www.math.umn.edu/safonov Office Hours: MWF, 11:15 am 12:05 pm,
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Ed Tell Sec. 002 Angela (Anqi) Liu 20415897 #8 Response: Delivery (Reading 15) The greatest challenge that lies in my way to an efficient delivery is perhaps my own fear towards speaking in public. More specifically, I am afraid that I might forget my wor
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Ed Tell Angela (Anqi) LiuSec. 002 20415897 #2 Response: Perception of Others (Reading 5) Socially constructed difference refers to the manmade distinction of people basic to various traits, such as ethnics, gender, and physical conditions (45). As the c
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Ed Tell Angela (Anqi) Liu #3 Response: Purpose and ResponsibilitySec. 002 20415897 Try as we may to reveal ourselves as much as possible when bonding with people, we are no longer free to disregard the context in which the communication is taking plac