Paul August

- Forwarded Message - From: "Philip Liplip Vuong"  To: "Paul Ponmattam"  Sent: Thursday, August 19, 2010 11:55:08 AM Subject: Re: homework

Wrote the proofs on:

http://mathdojo.wikia.com/wiki/Paul%27s_Meetings

If tu tienes questionadas ask me por favor.

- Forwarded Message - From: "Philip Liplip Vuong"  To: "Paul Ponmattam"  Sent: Tuesday, August 17, 2010 3:48:49 AM Subject: Made you a page!

Edit summary pages por favor? No necessitas una account.

http://mathdojo.wikia.com/wiki/Paul%27s_Meetings

- Forwarded Message - From: "Philip Liplip Vuong"  To: "Paul Ponmattam"  Sent: Saturday, August 14, 2010 7:07:55 AM Subject: Re:

Good for your first test (most college tests average at around 50%). The linear algebra is ok and you owned the first half of the test. But we need to get continuity-Epsilon/delta proofs solid. Physics focuses on real world approximations and the reason why an approximation is valid is if it has arbitrary closeness (specifically epsilon delta closeness). For example, the infamous "small angle" approximation uses arbitrary closeness in its justification using a Taylor's series. Epsilon-delta closeness is the heart of analysis. For this week, I am going to postpone the new homework and we are going to go back to continuity of both multivariable and single variable. I will give you an updated homework assignment this evening.

Once again, really good job on the first half: when we do the BIg test, I know you will get the continuity questions.

First half: 47/50 Second half: 10/50 Total: 57/100

1. Good manipulation of terms/application of number theory. 10/10. 2a. Good understanding of intersections and unions. Correct. 5/5. 2b. Correct reasoning and correct choice BUT no counterexample given. 3/5. 3. Good use of complement laws. 10/10. 4. Good use of matrix properties. 10/10. 5. Correctly found an A, B and C and proved that A,B satisfied properties. C, however, is "incorrectly" proven to have no inverse. In the formula for a 2D inverse, you are correct that encoded in the formula is the statement "determinant is 0" which implies C has no inverse. But you did not cite this theorem. A more valid approach would have been to show either C maps to distinct points to the same point (failing injectivity) or C fails to map to a point in R^2 (failing surjectivity). I will take away 1 point (Thanks for finding an A such that A^2=A with A not 0 or identity. Whoops). 9/10. 6. Incorrect proof. Just because something in true for any finite case, it is not necessarily true for the infintite case (this is the basis of analysis, which is the center of physics. It is also where we separate the engineers from the physicts/mathematicians). For example, an intersection of finitely many open sets is still open (ex. (-infinity, 1/2), (-infinity, 1/3),...(-infinity, 1/100) are all open and when all intersected, yield (-infinity, 1/100). But the infinite intersection of (-infinity, 1/n) is the set (-infinity, 0], which is not open.

The correct proof: Let x be a point in the infinite union U.  Then x must lie, by definition of union, in one of the U_i for some i.  But U_i is open. So we have an open ball around x IN U_i. But! U_i is contained in infinite union U. So the open ball around x in U_i is ALSO a ball in U.  Since x was arbitrary (general case), we can always find a ball around x contained in U.  since this is the definition of open, U is open.

Please redo this proof in your own words since it is fundamental. You also forgot to answer 6b. 0/10.

7. Good, but you could have also argued straight from definition of closed into of using complement rules (which is sometimes easier). Specifically, a converging sequence of points by definition converges to a point in R^n. So duh. Also, that is correct the reasoning for the empty set, but I prefer an arguement using the words "vacuously true." Example: All girls in the math department at Stanford are pretty. Specifically, this means for any girl I find who is in the math department, I can say that girl is pretty. But THERE are no girls here! We can also say, at the same time,all girls in the math department at Stanford are ugly. 10/10

8. Last step untrue. Not sure what you are doing here. 9. Sorry. No. 10. Dude, continuity proof! It is the same proof as continuous in components. Ah well, at least we know what we should cover.

- Original Message - From: "Paul Ponmattam"  To: "Philip Liplip Vuong"  Sent: Friday, August 13, 2010 3:03:15 PM Subject: Re:

Here's the test -paul

On Fri, Aug 13, 2010 at 5:55 PM, Paul Ponmattam < windshrike@gmail.com > wrote:

hi, im scanning it in now, but it might be a few minutes late, i forgot to set aside time for scanning(computer is horribly slow)

On Thu, Aug 12, 2010 at 11:28 PM, Philip Liplip Vuong < pvuong@stanford.edu > wrote:

P.S, you can call me for question clarification and if you are really really stuck, I can maybe give a hint.

- Forwarded Message - From: "Philip Liplip Vuong"  To: "windshrike"  Cc: jmbrdck4@sbcglobal.net, "daniel niv"  Sent: Wednesday, August 4, 2010 9:59:42 AM Subject: Homework grading

Ok, I have a phone meeting with Paul in 1hr 30 min, so I will have to read through paul's work first. Then James, and finally Daniel Niv (this will take me the longest to understand). If you want to discuss the homework/my comments via phone or have questions let me know. Also please don't be totally offended (a kid wrote this session that she was offended by the way I graded and taught math and that I shouldn't teach kids).

Paul:

1.1 Question 2: Good! If you wrote only one point I would have been mucho upset. I am pretty sure you got them all.

1.1 Question 9: Also good: you noticed that it was not a complete circle but it had 1/4 of a circle for positive s and t.

1.1 Question 11: Ok, so you are right, but I want you to write out the expansion of ||x+y||^2 (i.e ||x+y||^2 = (x+y)*(x+y) = x*x + 2x*y+y*y = ||x||^2+2x*y+||y||^2 where * denotes dot product.

1.2 Question 8: Dude, I gave you this one? Beautiful problem. Especially after number theory. But dude. Dude. SO CLOSE. You set it up perfectly but the sum of the first n squares is n * (n+1)* (2N+1)/6. (Your final answer is good though, but the miracle of limits).

1.2 Question 1.3: a.(duh), b.good, c. (duh),  good, d. correct, but simplier argument is..when does the sum of square components equal 0? When they are all zero. e. Good, you used the reasoning I gave in d to contradict f.  g. good. h. BAD (please don't get offended like the kid), just because the vectors are linearly depended does not mean it does not SATISFY the properties of a subspace. YOU MUST check the properties directly hold!

1.2 Question 4: Good. This question is a standard one and I will guarentee it will be on a future exam.

3.1 Question 1: Yawn, computation. But yeah, you get the point.

I cannot find the book questions for 3, 8, 6, and 22: you write at the top Ch. 4 Lin alg...so I am not sure.

There are two GLARINGLY BAD statements though (sorry, please don't get offended), inverting a nx1 matrix does not make sense (you can talk about pseudoinverses, but that is not only not guarenteed to exist, but not useful in the context you are using them, which is to remove x from Ax=0).

Secondly, incredibly incredibly bad statement: MATRICES NEED NOT COMMUTE. (A+B)*(A-B) = A^2 - AB + BA -B^2. But AB need not equal BA.

Also, you are correct that A^2 = A can only be A = 0 or A = I. But your proof is bad in that an inverse need not exist. The correct proof would be to rely on distributivity and substraction: A^2 = A iff A^2 - A = 0 iff A(A-I) = 0 iff A = 0 or A = I.

Overall, good job. I am going to take a Starcraft 2 break now.