so now, about this comment i made: math.stackexchange.com/questions/120708/…, should that be taken down? somebody said the same as me right before me, but my explanation was much better (I thought).
because someone in here said the math.SE standard is to remove your comment/answer if someone beats you to it.
@HenningMakholm Hm. Let me think about it. You wouldn't believe me that its sole purpose was to amuse Matt, would you? It was just a piece of silliness...
@MattN No, it isn't. It's pretty serious actually. I'll come to the chat occasionally but I lost all motivation to answer the questions on the main site.
@BenjaminLim It's getting too repetitive to be inspiring and the quality of both answers and questions have dropped dramatically over the past half year.
@BenjaminLim that's something that is very hard to write down in a good way. It's best if you try to figure it out for yourself without giving too much thought on typos that I may have made. I gave you the idea and I hope that suffices.
@BenjaminLim I do. That's kind of what is still keeping me here.
To prove that piecewise function $x \sin(1/x), 0 (when\ x=0)$ is a continuous function, do I just take the limit from the right, the left, and the value at $0$ and show they're all equal?
@KannappanSampath Definitely. But I would do so anonymously. Otherwise it might (or might not) hail down votes on you, depending on whether he is the tantrum kind of guy like you-know-who or not.
@KannappanSampath If you really want to learn commutative algebra, AM is short and concise. I find Pete's notes are more for a second course in commutative algebra or something. Like once you have done AM, then you can look at Eisenbud's text or something. AM or Miles Reid's Undergraduate Commutative Algebra is a good place to start.
if it'll help, i was programmer for about 20 years before i decided to go back to school for math. i had no real math background and raced through a graduate program. it's not the most elite school around, so i feel like i'm still learning).
@Jeff I am sorry to have asked that. I thought you were in college in something. I was biased into thinking that all college students are young. But that is not true as even in my institution there are adults who learn. It is wonderful to see them there.
i hear you! that was part of my issue. i was always rushing through the courses trying to keep up and get a grade without always understanding everything. plus i didn't get the repetitions of having done it as undergrad and then again as a grad
@KannappanSampath When you get to snake lemma you will need some guidance on that. It is very confusing for the first time chasing a big diagram like that.
@tb well, my grad school was kind of uninspiring. most of the experienced students there thought it was a very weak school and made jokes about it. if it had been an elite school, i would not have been admitted and would not have survived.
@tb the graduate classes were the same as the undergraduate classes, but we were given more homework. we sat in the same class and had the same lectures
@BenjaminLim I meant to say that the uni I study at is a depressing place with a food catering company that feeds crap to the students because they wouldn't give a damn if they killed them (except for that there is a law preventing them from killing people).
@tb well, i guess that depends on how you look at it. i didn't have a lot of options, the school was close to home and offered me a graduate assistanceship (i'm unemployed). and i did learn a lot.
the glass could be half full or half empty, i guess
@MattN The teaching where I am now is not that good either. In fact I have not been to a single analysis lecture in 2 weeks. It is boring as hell and the lecturer confuses me even more.
@KannappanSampath Arturo posted a comment below it warning me to add a disclaimer since the question had changed but I couldn't be bothered, so I deleted it.
Because, you travelled into the world of ideals and they use a different system for measuring times. And, the up and down journey to that world is like 3 full hours in your place. :-)
@MattN well, it was having a flag as too localized. While the OP is one of the worst question askers on this site (excluding a few obvious exceptions), I don't quite get it.
Well, if I taught a course, I expect my students to understand the material because I take pain in writing my lectures, in writing problem sets, in making material more clear. So, I'd want some work to be put in by them. If Math was all done by someone helping you, we would not learn anything!
i think it's a complicated problem. the goal of taking a class is to learn the material. the trouble is, the effectiveness of this is adjudicated by assigning a grade. students get confused, and figure erroneously that "getting the grade" is the objective.
@MattN Depends. If the homework ends up affecting what's on the student's eventual diploma, then the school has an interest in making sure that they can stand behind the skills they certify to future employers, etc. But if the consequence is just that the student will fail his exam later, then I'd say it's his own problem.
On the other hand, many universites are punished financially if too many students fail their exams, so they still have an incentive to try to dicourage their students from using counterproductive studying "techniques" that will lead them to fail eventually.
@MattN if you want to know now, just send an email, I'll respond quickly... I'm not sure when I'll get up in the morning and if I'll have time to come here.
but even leaving aside the issue of "internet answer farming", it's still common practice for students to discuss homework amongst themselves...so homework performance is not a reliable indicator of information absorbtion
@BenjaminLim often GPA's determine what higher learning institutes (like graduate schools) or even potential employers are available...this gives students incentive to cheat
@DavidWheeler Not only reputation: around here the funding the universities receive from the government is directly proportional to the number of ECTS points their students pass. (Well, affinely related, at least). Creates a tremendous internal pressure to lower standards, of course, but also -- as a side effect -- an incentive to make sure the students learn.
my girlfriend is studying for a medical coder's certification...one of the ways her instructors deal with this is giving full credit for ALL homework turned in, but such work only counts as 10% of the grade awarded.
@DavidWheeler At my institution we have degrees where students need to maintain 80% to stay in them. This puts pressure on the students and affects the learning a lot.
OK. On that note, I'm going to try asking this questoin again: How do you find a limit on a (nowhere continuous) Dirichlet function? That is $F(x)=$ piecewise $x$, if $x$ is rational, or $0$ if $x$ is not rational. And the question is "Determine which values of a does the $\lim_{x->a}F(x)=F(a)$
So F(x) is not continuous anywhere. I was tempted to say it's $\lim_{x->a}F(x)=F(a)$ whenever $a$ is rational, but i'm not sure.
more seriously. i don't have the definition of conts in front of me now. but i have to use the def. of conts, pick $a \in Q$, pick a delta and show there is a non-rational number within $a \pm \delta$.
i suggest looking at $\epsilon = |a|/2$ for rational a. that leaves just showing the limit doesn't exist at irrational a. you might want to show there is a rational number b "close enough to a" that you can use.
unrelated to current conversation before I ask a simple question that gets rage-down-voted into oblivion - is there a relationship of any sort between a number and the length of its Collatz chain? I'm guessing no, since it seems that'd provide some sort of proof for the Conjecture, but hey, no stupid questions... right?
a look at the graph of numbers vs. length of their collatz chains indicates some "partial" relationships...perhaps there is a tighter relationship for certain KINDS of numbers
@ThomasShields That would be my immediate reaction too -- if there were any simple upper bound, that would effectively prove the conjecture. Of course the upper bound might be a bound of the length of the chain until some cycle was reached (which then wouldn't prove the conjecture), but it would be a very peculiar sort of result to prove that without proving that there's only one cycle.
@ThomasShields What I mean is that one could conceive of a result saying, for example, that the Collatz function iterated n!!! times on n will always yield a point that belongs to a cycle -- but didn't guarantee that this cycle has to be the 1-1-1-1-... one.
The reason I'm wondering is that I'm working on Problem 14 of Project Euler (find the number under 1 million that has the longest Collatz chain) and so far, brute-force has failed me, so I was wondering if there was some sort of connection I could draw that would rule out certain sets of numbers.
I think i'll just start graphing L(n) where L gives the Collatz length of n. Probably won't gain anything useful but it should be fun, if only because graphs are cool.
i don't know what the partial "radial lines" represent...but they seem to take care of "most" of the cases...and then a lot of the "exceptional cases" seem to have constant lengths (lying on a horizontal line).
@ThomasShields I think memoization and dynamic programming are roughly equivalent here (different ways of thinking of essentially the same idea). So even with memoization you still have two numbers that take so long time to get back into your memoization range that you don't know which takes longer?
@JohnSmith Oh, they are not the same in general -- I just think (based on not more than a few minutes of deliberation) that the effect of applying them to this problem looks like it will be roughly the same.
unfortunately, i'm not a programmer, but it seems to me, if you store calculated L(n) somewhere, and test output values (of the chain iteration) against previously calculated L(n), it should help with computation time...for example, we have L(5) = 5, so when we get to 10, we just add 1 to L(5).
@DavidWheeler yeah, that's what I'm doing. It still takes forever. It might just be my browser running JS slowly. I'm going to try the check against powers of 2 thing.
@JohnSmith thanks; I think the fact that i was using JS was part of the problem; but I don't just want to solve the problem, I want to learn, hence the inquiry into the problem :)
i saw a forum signature once: there are two ways to solve a mathematical problem: 1. reduce it to an easier problem. 2. turn it into a combinatorial problem, and have somebody smart solve it.
@David we're adding multiples of 3 from 1 to 1000, so it's 3+999, divided by two, times the number of (3+999)s you have. i'm guessing you have somewhere like 1000/3 of them. am i on the right track here?
the largest multiple of 3 < 1000 is 999 = 3(333), so for the multiples of 3, n is 333, so we have 3(333)(334)/2 = 166,833 contribution from the multiples of 3
and then since the intersection of (3) and (5) in Z is (15), we need to subtract the multiples of 15, since we've counted them twice (inclusion-exclusion principle)
