matroid

Hi, mba05, Thank you for the concise but informative answer! I'm especially interested in NJ, USA (where you seem to be). How long is the IDP vaild there? Md.

Dear gigipiscu and nounou, Connectives AND and OR are monotone functions, i.e. if you change some of their attributes ("inputs") from 0 (NO) to 1 (YES), the result ("output") can change only in the same direction (remain the same, or change from 0 to 1; but never change from 1 to 0). Using this simple fact prove by induction (on the number of connectives) that the same holds for any function consisting ANDs and ORs only. Then check that functions like NOT or EXOR are not monotone functions (obviously). Md.

Thank you, swethans, Peter137 and Comb05! swethans: I was called on the phone. I applied to the Operations Research dept (Rutcor), they have a PhD program separate from both the Math and the CS departments. (Still I hope that I'll be allowed to attend courses of those depts, too! :)) Peter: Thank you for the info, I feel much more comfortable now! :) Cheers, Md.

Thank you, guys! Anybody about the financial part...? Md.

My most loved TMians, :) I've received a call (which I consider an unofficial admit) from Rutgers (New Brunswick, NJ) OR department that I'll be offered an admit with a fellowship of $20K (taxable) + no tuition. :grad: I'm very happy with the admit, but also a little bit concerned about how much that money really is. Will this amount cover all my expenses? (Only dorm is 7K per year.) Can anyone estimate the cost of living (which is more than just dorm+food) in the area? I'd also appreciate any piece of useful info, experience, story about the university. :D Thank you, Md.

Thank you for all the answers. I'm afraid, I'll have to e-mail them soon. :-/ I'm not happy... :( Md.

My dear TMians :), Has anybody got a decision from the CMU Math Dept. or Rutgers O.R. Dept? At the "whogotin?" site there are very few results from the former, and nothing from the latter. I can't believe they're still working on the applications, and I'm a bit afraid, that my inquiry about the result would result in a swift and simple rejection. I have to be careful since I don't have more shots left... Thank you for any help, Md.

I don't think so. I didn't try this time, but last november's test was also harder than most test takers (including me) expected. I guess it's always of the same difficulty -- always harder than the pretty easy practice test. :hmm: Cheers, Md.

Or marry an educated man. :p (Sorry, if it hurts...) Md.

Certainly not... but I begin to feel that I aimed too high :( (I don't have the neither the publications nor the contacts which are essential to get admission to a good program). I've got only 2 applications pending and I don't have much hope. What about you? I've read that you didn't have much luck with the Subject Test. Where have you been admitted?

If you prefer integral calculus than elementary geometry, I have no objections to your solution. :) I think, you practically did same calculations to mine using a more formal method. To answer your question: yes, I have already taken the GRE math test. I've got 92%, and I'm still waiting for my first admit. :( You'll surely have better luck, Md.

I second gstergia's opinion but disagree with Dingus. It's your contacts that matter: you or your recommendation letter writers should be known by somebody at the department you are applying to, otherwise you don't have much chance, if you're not a real genius. Even in TestMagic you will find several people with "stellar" admits with weaker scores and much less research experience than many of "The Rejected Mass". As for "research experience": you have to produce several "real" papers in English. Years of RAship, presentations, proceedings, anything in any language other than English etc. don't count much without a few serious papers. Md.

As far as I'm concerned, all the US univs are full of Indian graduate students. :) (Didn't mean to offend...) More seriously, what about the faculty? If there are many Chinese or Indian profs, it's quite logical that they trust the people of their country more. (They better know their univs or LOR writer profs, for example.) Otherwise, I don't know. You'll have to decide whether such an environment, which is not very diverse and also not full of your compatriots would be uncomfortable for you or not. Matroid

Hi Orpheus, I'm also waiting for CMU (had an interview, didn't went very well :-/) and Cornell (didn't heard anything from them. According to the WhoGotIn site the decisions aren't out yet.) Stay calm, and report immediately if you hear sg new. :) Cheers, Md.

Keep your faith, Vogon! (I'm telling this to you, without a single admit so far...!) I couldn't found your profile, could you post a link on it (or PM or whatever). I'm wondering whether your story fits in my "theory" (which is, in a nutshell, that at the best univs good LORs from profs with good contacts at the targeted departments is almost equivalent to an admit). Which univs' notification are you still waiting for? Md.

Please, please, I need an admit, too...!! :) Until Psymath's post I thought I'm the only one applying for AppMath/CS programs without a single admit. Well, knowing that I'm not alone doesn't make it any better... :( I've got 2 rejections (both are CS), and had 2 interviews (weeks ago, went very bad -- and still no feedback!) I don't even dare to pester them -- (I thought it'd result in an instant reject :rolleyes: ), but most of the programs I selected promised to send notifications in February... Now say sg clever, please ;), Md. That's really kind and tricky of her... :D

When it comes to PhD applications, IMO, contacts (or "networking" :hmm: ) are almost everything. Equivalently, your LORs must be written by someone who are well known at the department you are applying to. Publications and a stellar AGRE score also might help a bit. (A bit.) Cheers, Md.

At least you know the result and can stop biting your nails... :( This is going to be a devastating week for many of us, I'm afraid. Keep your chin up, Md.

Sorry, I'm still not ready with the figure... :whistle: So, let's see what we can do :hmm: : imagine a 1*1 (unit) square, each point of which represents a random (x, y) pair - that is, instead of picking 2 random numbers independently from [0,1] u can pick a random point from this square. What we are looking for, is the probability of abs(x-y) I'm sorry, I feel that I wrote practically the same as I did before... but I can't really put it in another way. :( Still in the hope it helps, cheers, Md.

I'd add only one comment. I think that in the case of the somewhat prestigious universities the most important is to choose at least one recommender who is well known at the university you're applying to. If your recommender is an old chap of the director of the program or sg, you're in, whatever your research experience or GRE scores are. If not, u have to be a true genius to be considered for admission at a "big univ". I don't know how it works for weaker univs, though. Cheers, Md.

No! (Draw more carefully!) It's pretty obvious, that the number of intersection points must be odd, because "in minus infinity" x^12 is greater but "in plus infinity" 2^x is greater. (Certainly this is not a proof, the curves could be tangent to each other, for instance, I'm just trying to give you a hint.) The solution is three (D) indeed, and it takes no time or exact calculation to see this. I think you missed the third intersection point (I mean the one with the largest x-coordinate.) Cheers, Md.

That's a nice explanation! I'd add only one comment: we don't even need Fermat Theorem (x^(p-1) is congruent to 1 (mod p) where p and x are relatively prime). Since f(p) := p^4 - 1 = (p^2 + 1)(p^2 - 1) = (p^2 + 1)(p + 1)(p - 1), and p>5, either (p-1) or (p+1) is divisible by 3. (Otherwise p would be, which is impossible.) In the same way: if neither (p-1) nor (p+1) is divisible by 5, the p must be congruent to +2 or -2 (mod 5), so (p^2+1) is congruent to 5 or 0 (mod 5). It's the same idea that Dragonfinity used to prove that f(p) is divisible by not only 4 but 8. Cheers, Md.

Oops! Sorry, I read it twice but I stilled missed it! Shame on me!! :o :o :o About the n*log n I don't know anything about the proof. (I learned number theory only at high school, actually, but I've learnt some complex analysis, that's where I've come across the Prime Number Theorem.) To get the n/log n formula of the Prime Number Theorem I don't know anything more simple than using the Zeta function and the Residue Theorem. But now that you ask, I'd be interested in it! :) Sorry again, Md.

I don't think I left out anything. What you write is correct, but if you write the triangle inequality the way I quoted, than the 2 "missing" properties follow from the other two axioms. For example, if we substitute b=c in ii) we get: d(a,b) + d(a,b) >= d(b,b) = 0, so 2d(a,b) >=0 which yields that d(a, b) >= 0 Also if we substitute a=c we get: d(a,b)+d(a,a) >= d(b,a), so d(a,b)>=d(b,a). Since this holds for all a and b, it also holds when we swap the two variables. So d(a,b)>=d(b,a)>=d(a,b), and so we have proven that the function is not affected by the order of the points. Md.

Wheew, I'm glad that the message "got through" despite the absence of the figure and my command of English... :) Md.