Math 5210-1
|
|
Links:
|
Homework is posted here!
Due January 16
2.1 3, 5, 6, 8, 9;
Class exercises 1, 2, 3, 4 from jan13.pdf class notes.
hwsols1.pdf solutions
Grading: 10 points total, distributed as follows:
Class exercises 1, 2, 3 = 1 point each, 4 = 2 points.
2.1: 3, 5, 6, 8, 9 = 1 point each.
Due January 23
2.2 1, 5, 8a, 10, 11;
2.3 1, 2, 3, 7, 9, 10;
hwsols2.pdf solutions
Grading: 12 points total, distributed as follows:
2.2: 1 = 1 point, 5 = 2 points, 8a, 10, 11 = 1 point each.
2.3: 1 = 1 point, 2 = 2 points, 3, 7, 10 = 1 point each.
Due January 30
3.1 1, 2, 4, 5, 7, 9, 11;
9.1 4, 7, 10, 14;
9.2 2, 3, 14, 15; Do #3 for each p-norm less than infinity, the proof is no harder than just doing it for p=1. (7 and 9 are postponed.)
Class exercises 1abc in jan23.pdf notes. Note, there was a typo in 1a), the function phi(t) actually has a minimum (not maximum) value of zero, at t=1.
hwsols3.pdf solutions
Grading: 29 points total, distributed as follows:
3.1: 1abc, 2, 4 = 1 points each, 5 = 2 points, 7, 9 = 1 point each, 11 = 2 points;
9.1: 4, 7, 10 = 1 point each, 14 = 2 points;
9.2: 2 = 2 points, 3 (as modified) 2 points, 14 = 1, 15 = 2 points;
Class exercises 1abc = 2 point each.
Due February 6
9.2 8-13, 17, 18;
9.3 1-3, 6, 10;
Class exercise 1 in jan30.pdf notes. (Fun and challenging!)
Note: the section 9.3 hw and the class exercise are continued until next Friday.
hwsols4.pdf
Grading: 10 points total, distributed as follows:
9.2: 8, 9, 10, 11 = 2 points each; 13, 18 = 1 point each. (12, 17 not graded.)
Due February 13
Class exercise 1 in jan30.pdf notes.
9.3 1, 2, 3, 6, 7, 10, 15, 16, 20, 26; In #1, show that the "inverse image of closed is closed" statement is equivalent to the "inverse image of open is open" statement in the usual definition of continuous (since we showed in class that the definition using closed sets is equivalent to sequential and epsilon-delta continuity). In #10, expand the problem to give a complete proof that the space C(M,N) of continuous functions from M to N is complete, under the hypotheses that M is compact and N is complete....we did a special case of this theorem in the notes jan27.pdf. In #15 also show that if you use closed intervals, i.e. f:[0,1]->[0,1] continuous, then f must have a fixed point.
hwsols5.pdf
Grading: 24 points total, distributed as follows:
Class exercise 1abcd = 1 point each.
9.3: 1, 2, 3, 6 = 1 point each; 7 = 2 points, 10 = 4 points, 15 = 2 points, 16 = 3 points, 20 = 2 points, 26 = 3 points each.
Due February 20
Class exercise 1ab on page 3 of feb11.pdf notes.....1a) Show Hausdorff distance is a distance function on the collection of compact subsets of a complete metric space, and 1b) show the compact sets together with Hausdorff distance is a complete metric space.
Class exercise 2 on page 6 of feb17.pdf notes.
hwsols6.pdf
Grading: 6 points, with up to 4 points of extra credit:
Class exercises 1a = 3 points, 2 = 3 points. Exercise 1b up to 4 points of extra credit, at the grader's discretion.
Due February 27
Class exercises 1,2,3 on page 3 of feb20.pdf notes.
Class exercises 4,5 on page 4 of feb23.pdf notes.
11.1 #2c, 3, 5.
hwsols7.pdf
Grading: 22 points, graded as follows:
11.1 # 2c, 3, 5 = 2 points each.
Class exercises: 1abc = 1 point each, 2a = 2 points, 2b = 1 point, 3ab = 1 point each, 4ab = 2 points each, 5ab = 2 points each.
Due March 6
7.3.4 2, 5, 7, 12;
7.4.5 4, 7;
7.5.5 4, 8, 10, 15.
hwsols8.pdf
Grading: 12 points, graded as follows:
7.3.4 7, 12 = 2 points each
7.4.5 4 = 3 points
7.5.5 4 = 1 point, 10, 15 = 2 points each
Due March 27
7.6 1, 2, 3, 5, 9;
12.1 1, 4, 5, 6, 13;
12.2 2, 12.
hwsols9.pdf
Grading: 22 points, graded as follows:
7.6 1 = 2 points, 2 = 1 point, 3 = 2 points, 5 = 1 point, 9 = 1 point;
12.1 1 = 3 points, 4 = 4 points, 5 skip (I gave some people bad advice), 6 =2 points, 13 = 2 points;
12.2 2 = 2 points, 12 = 2 points.
Due April 3
Royden p. 56 6, 7, 8;
Royden p. 62 10, 11, 13, 14b;
hwsols10.pdf
Grading: 11 points, graded as follows:
p. 56 Royden: 6 = 4 points, 7, 8 = 1 point each;
p. 62 Royden: 10 = 2 points, 11 = 1 point, 14b = 2 points.
Due April 10
Royden p. 56 13;
Strichartz 14.2 2, 3, 5, 8.
Class exercise: Prove the Hausdorff dimension of Sierpinski's triangle is at most ln(3)/ln(2). For extra credit, prove that this is exactly the Hausdorff dimension.
hwsols11.pdf
Grading: 25 points, graded as follows:
p. 56 Royden: 13a,b,c = 3 points each.
p. 654 Strichartz: 2 = 4 points, 3 = 3 points, 5 = 3 points, 8 = 4 points (2 for each inequality).
Class exercise = 2 points. (Up to 4 extra points if anyone attempted the extra credit.)
Due April 17
Royden p. 69-70 22, 24, 25. Problem 28 is either postponed or canceled....I'll decide later.
hwsols12.pdf
Grading: 12 points, graded as follows:
p. 69-70 Royden: 22abcd = 2 points each, 24, 25 = 2 points.
Due April 29 Note - this is WEDNESDAY.
Royden p. 86 3, 4, 7, 8;
Royden p. 89-90 10, 15, 16, 17, 18, 19. : 17, 18, 19 won't be graded.
hwsols13.pdf