@AsafKaragila: Here are some of my favorites, regardless of comic content: [one](http://math.stackexchange.com/a/90515/7850) [2](http://math.stackexchange.com/q/77563/7850) [3](http://math.stackexchange.com/q/77410/7850) My url skills are lacking
I hope you all perish in eternal flames of agony by the time I wake up, and when I will open my eyes in the morrow to come there will be nothing but bleak darkness and pure emptiness. Indeed it will be then that I will know that surely the empty set exists, zero is natural, and that ZFC is consistent.
@DavidWheeler Or if you don't want to change your name, you could just star my next comment.
Typing @ followed by the first few letters of someone's user name only works if there is no other user with a comment in the same room, whose user name begins with the same few letters, but alphabetically precedes the name of the user whom you are trying to ping.
@MarianoSuárezAlvarez My supervisor was saying finish some exercises on tensor product and direct/inverse limits in AM and then look at first 3 chaps of that book. That's why I asked :D
he was saying something like if we have $A \rightarrow B \rightarrow C$ and we tensor it by something, then in general the sequence is no longer exact on the left so there is a kernel which he denoted by Tor something.
@MarianoSuárezAlvarez One day I overheard some algebraists talking and then one guy was like ¨ take the injective resolution of this and this", man........
14x^2 -17x + 5 The negative factors of 5 are ( -1, -5) I want to know why I have to consider (-5, -1) also for negative factors of 5? But for the quadratic term you only use x, 14x and 2x, 7x? not 14x, x and 7x, 2x.
@MarianoSuárezAlvarez Do you remember when you first starting this stuff? I remember when I first started learning topology just writing down the sentence "Let $X$ be a topological space" sent shivers down my spine :D :D :D :D
@robjohn When you said "You need to try **all** combinations of factors of the leading and trailing terms." I want to know why I have to consider (-5, -1) also for negative factors of 5? But for the quadratic term you only use x, 14x and 2x, 7x? **not** 14x, x and 7x, 2x.
@Skullpatrol because what matters is which factors get paired with what. Once you've settled on 2x and 7x, that locks in what has to happen with the 1 and the 5.
If you had settled on 7x and 2x, you'd just get the same factors in the other order.
@Skullpatrol because you have to decide if (if you're using 1 and 14) "the x goes with the -5, or the x goes with the -1"
if you exchange the "quadratic" factors while holding the "constant factors" fixed, it's the same as holding "quadratic factors" fixed, and exchanging the "constant factors"
again, an algebraist would say: "unique up to a unit"
in this case (if we are just considering integers), the units are $\pm 1$
with quadratic polynomials, it's traditional to write the coefficients of x terms of linear factors as positive, but you could do the opposite, just to be "different"
Should I ask on one of the meta sites? I've just promised someone a bounty (for a fantastic answer), then discovered myself powerless to keep my promise.
my reputation is 51
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Well, I'll wait two days then see what happens. No big deal; this person can wait.
It's just that there's so much false belief about unit testing out there; this seemed like something I could do to make people aware that in this field, the true belief is a minority belief.
And therefore, when people answer questions on SO about unit testing, their answers are often wrong. And then people upvote wrong answers and accept wrong answers. And then more people do unit testing wrong.
OK, it's olympiad-like because there's a 4-digit number that's central to the problem, and that is a recent year. But maybe not so olympiad-like because any other number would do. The really special year-questions are ones where the year falls out by coincidence.
As far as Jeremy's comment is concerned, I don't think it was rude.
@DavidWallace I have had the feeling, but these days in my country, there are not very hard year-questions. I browse through Indian National Math Olympiad question paper and I always see Diophantine equations, geometry problems and one or two elementary combinatorial problems, and simple functional equations!
Sure, there are some easy year-questions out there, and the one that you posted a link to is one of them. Can I illustrate the distinction I tried to make about really speckal year-questions, with the paper from 1988 (the year I competed)?
Here, Q3 is kind of boring. 1988 doesn't really play a special part in the question; they just shoved it in. But Q4 is a different story - here, 1988 just happens to show up, which makes it a far more beautiful question.
I'm sure that the link that I posted isn't the original paper. I seem to remember that Q4 was written in sigma notation, not as a long sum. Let me keep looking.
Yes, I was right. Click the "download" link next to 1988 on imo-official.org/problems.aspx for the English version of the original paper.
Well, it's easy to give three separate formulas, one for each side and its production. But hard to give a formula that would give you all three sides, and then only the portion that's between the two intersections with the other sides.
Heron's formula tells you the area of a triangle. Is that what you want?
And for the triangle you mentioned, it's not the simplest way to get the area.
@DavidWallace $R^2=x^2+z^2$ is the circle formula, I want a similar formula for triangle. It is not an area formula so I think I need something else for triangle?!
Page 809 bottom here is the thing that I try to formulate for different shaped objects.
So, you're saying you want a single equation involving x and y that's true for any point that's on the perimeter of the triangle, and false for any other point. Do I understand you correctly?
It states that the solution of the boundary-value problem for circle and ball are he below equations $u(x,y)$ and $u(x,y,z)$ -- now what I want to do is to consider different objects such as the equilateral triangles (all sides of the same length).
If I can understand right, I need to formula a perimeter formula in terms of $x$ and $y$ for the triangle. I cannot see how to do that yet, thinking.
OK, you can consider them. But if I were to produce such a formula, a lot of the intelligence in it would be around dealing with the fact that there are three separate line segments, and ensuring that all three are dealt with, and that points on the line but outside of the line segment are not included.
Whereas, producing three separate equations for the three lines, and not worrying about the end points, is an absolutely elementary exercise.
@DavidWallace Yes, it is but I want one equation -- perhaps some absolute value trickery? Symmetry could be exploited here somehow, thinking...
If the center of the triangle is in the origin, then do we get that $x_{0}+y_{0}=C$ where $C\in\mathbb R$? If we consider some special norm (not just standard absolute value), can we come up with the formula? Thinking...
Yes, those are polar co-ordinates. Now subtract the appropriate multiple of $2\pi / 3$ from $\theta$ and you'll be cooking with gas.
I have to go now. Not sure whether I'll be back later. If you want to post this as a question on the main site, I'll answer it some time, or someone else can.
@MattN They do. Here I use descends to and factors over as synonymous. The situation is this: you have a group $G$ (here $\mathbb{R}$) with normal subgroup $N$ (here $2\pi\mathbb{Z}$). Given a homomorphism $f: G \to H$ I say $f$ factors over (descends to) $G/N$ (or more precisely factors over $\pi: G \to G/N$) if $f = g\pi$ for some homomorphism $g: G/N \to H$, in other words this happens if and only if $\ker{f} \supset N$.
@hhh I just posted what I think is the solution to your triangle problem in polar co-ordinates. Do you want to take over from here; or do you want me to try and put it back into rectangular co-ordinates?
Whichever you pick, please check first whether it's what you want.
@AsafKaragila How do the Garnir dream spaces do? For some reason this name makes me think of dragons swords and dwarfes. Maybe Garnir's dream space is the torture corner in the cave of a nasty dwarf?
@Ilya If you investigate the stars in the room the only pattern emerging is nonsense!
@KannappanSampath Oh, there used to be a way for room owners to move messages into other rooms. So a room owner could create a "trash bin room" and essentially delete messages from the room by moving them there.
He originally deleted his account from MSE, but it is likely that the global SE account remains, so when he registered again it was relinked and the badges were given.
I was wondering about that user before, actually. It makes a lot more sense now.
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