I just think it is absurd how some teachers will ask questions on tests that don't even test the material but obscure rules you were suppose to learn sometime
sure but just for this test I had to know some obscure log rule, strange trig identities, some weird unused second derivative rule verbatim, and that isn't including the normal test stuff I had to know
oh and then I had to know some really hard geometry stuff that I still don't know
if we have two points on a parabola $y = x^2$, say, $(a,a^2)$ and $(a+h,(a+h)^2)$ then the slope of the line passing through both of them is: $((a+h)^2 - a^2)/h = (2ah + h^2)/h = 2a + h$
well the thing is, it's not all that helpful to know what the y-intercept is, we're interested in what is going on "near a", not the y-axis (x = a, NOT x = 0).
sometimes competition brings out the worst in people. unlike in other fields, it's rarely ok to be "wrong" in mathematics, and this carries over to mathematicians' social views.
Let $\{a_{n}\}$ a sequence such that $a_{n +1}=2^{a_{n}}$, $a_{1}=1$
show that $\{a_{n}\}$ diverges to $+\infty$
hint:
It would have to prove by induction that:
$a_{n}\geq 2^{n-1}$, $n = 2,3, ...$
Using the inequality
$2^{n-1}=(1 +1)^{n-1}=1+(n-1)+\cdots\geq n$ (if $n\geq 2$)
Could they ple...
@mixedmath can you tell me in this instance what this means in concrete terms so i can understand by example? i don't know what sections we are considering
@mixedmath would you look at X.10 here and tell me why condition (1) is required? This is a problem from Lang. I thought I did the problem correctly but I can't understand condition (1), and why surjectivity does not suffice and why an isomorphism is required cems.uvm.edu/~voight/Sp2003-250B/2003-250B-HW07.pdf
I only meant in that he rarely shows that he puts effort into solving the questions himself, when asked for clarification never provides it, always using the imperative, that kind of thing.
Anyway, a lot of math terms users could look them up on WP or whatever but instead deign to ask on MSE. It just means they're un-researched questions arguably deserving downvote, not spam.
@Gigili If you're talking about narges's question, I disagree. There's clearly a significant language barrier, and the questions are very elementary, but they appear to be genuine questions, and narges is interacting with the commenters. They are certainly not spam, and only an incompetent moderator would treat them as such.
The thing about new users not putting in a lot of effort is they have very little exposure to our community and its workings, so how we go about our business is going to be of little encouragement or discouragement.
@Gigili No, you are aspiring to become an incompetent moderator :-)
joke
@BrianMScott I think Gigili is right though. Good moderators would close this question, to put it in "time out" until the OP (or other users) can revise it and make it better.
A good example of this is Yannis Rizos, on Programmers.SE - he frequently closes poorly worded questions, and asks the OP to revisit them.
@BrianMScott The primary purpose of all the SE sites is to put together a battery of high quality questions and answers. Questions that don't fit this description have no business remaining open.
You point of view is a user who likes to answer such question and see it as good question (after a minor edit!), a moderator is deeply concerned about the quality of the questions.
If you encourage a question like that, because it "could" be a good one, even though it is not now.. There'll be a bunch of questions "plzzzzzz help me sooooon".
Closures are going to be interpreted as permanent by OPs (unless otherwise informed), and few people tend to drop in and explain that a closure is contingent on a question being improved to a satisfactory quality. (Indeed, reopenings are rare.) I also do not think moderators should be quality police. That's the community's job alone.
What is really required is for someone with 2000+ to get in there and make the question better. It's ironic that one such user has already done an edit, but didn't go nearly far enough.
@DavidWallace If that were done it would be much better, but I think it would be safer to allow a window of opportunity for a question to be improved, with help from others, before a moderator or others resort to closure.
@anon but "closure" doesn't remove the window of opportunity.
I used to think the same way as anon and Brian here; I had a long argument with Yannis about it, and he eventually convinced me of the error of my ways.
I'll see if I can find a link to some of our conversation.
@skullpatrol Thanks for suggesting it. Adding it is one of the few things I can do as room owner, so I thought I would try it out. If it annoys people, I will consider removing it.
I am a firm believer in minimal moderation. In particular, I don't want the moderators making by fiat decisions that the community is perfectly capable of making on its own. I want them to deal first and foremost with administrative problems, like merging accounts, and secondly with personal interactions that get completely out of hand.
@DavidWallace If the MSE community actually came to operate strongly on that premise, I'd be strongly tempted to find another way to help folks with math questions.
@BrianMScott I find that a bit sad. After all, the body of questions and answers that we build up helps many people, basically forever. Isn't that more worthwhile than helping just one person right now?
@BrianMScott I couldn't agree more, that the goals are compatible. Which is why the site works so well. But sometimes, we have to favour one or the other.
But the immediate user has his/her answer, now - in fact more than one. So now we should stop and think about what's best for the future users. And what's best would be for this question to EITHER be improved by editing, OR removed. Closure is a kind of limbo between these two, from which the question may either be resurrected or destroyed.
@DavidWallace Leave it to the individuals answering the questions to favor one or the other, just as it is now. Compare Bill Dubuque's answers with mine: he's at one extreme, and I'm towards the other end of the spectrum.
@robjohn So have I, for that matter: I've occasionally added an answer to a question that's already been well answered, simply because I thought that a different approach or way of looking at a problem might be useful to someone else.
I'm not sure whether my edit will override Brian's once it's approved, or whether it will just go in the trash bin. It doesn't matter too much; my edit was very similar to his.
First off, let me say I love the community moderation system. I think it's tops.
But I feel like too many questions that could elicit interesting answers (specifically, interesting as a programmer) are being closed prematurely.
I get that polling and extended discussion are Bad Things. I also g...
Professional programmers (like me) search for answers to questions several times a week, and find them on these two sites. They're extremely valuable, because of the quality of their questions and their answers. The value for "future users" of any question far outweighs the "here and now" value to the individual who posts the question.
But maybe the same is not true of Mathematics.SE - this site is less "ongoingly valuable", and therefore the needs of the immediate user should maybe be held in higher esteem.
Hmm, I edited one of Gigili's answers yesterday too, and the edit hasn't shown up. In fact, the fact that I can't see it any more probably means it was rejected.
\begin{align*}
x - \alpha y &= 1\\
\alpha x - y &= 1
\end{align*}
By simplifying, we have:
$$\frac{x-1}{\alpha}=\alpha x-1$$
Which is:
$$(\alpha^2-1)x-(\alpha+1)=0$$
$$x=\frac{\alpha+1}{\alpha^2-1}$$
Assuming $\alpha \neq1$, $x=\frac{1}{\alpha-1}$ is your unique solution. For $\alph...
There's a potential issue in one of my answers I realize. An invertible matrix isn't necessarily diagonalizable over an algebraically closed field, is it?
@robjohn: Apparently you ask for feedback on the rss-feed thingie: I find it very annoying since it provides me with unsolicited information. Two immediate reactions and proposals for improvement 1) I would like to have the option to turn it off. 2) It shouldn't show me questions belonging to tags I do not want to follow (those on my ignore list).
@tb: I've corrected this answer of mine just now. I wanted full generality on $k$ and an algebraic argument that didn't require topological considerations, but I'm not sure how or if that's possible. Do you have any input?
I have no idea what CIE Lab colour space is, except that maybe it's some new mathematical parametrization of colors for computer graphics. I was pointing out I just saw an icon I liked on WP and decided to nab it for myself, with no other considerations involved.
@skullpatrol Yes, the main page has a counter of added to the window title, so whenever that number changes I know there was new activity. That's more than enough information.
Yes, it's a parameterisation of colours, but not specifically for computer graphics. Anyway, I just noticed the similarity, and wondered whether you had chosen it for that.
Like, what is continuity in a finite field? I think in Q's answer I linked to we have reals and the continuity of the functions involved allowed us to extend to the degenerate cases.
"It follows that ... for all diagonalizable A, hence all A by continuity"
@anon I see. I think it is generally true that the diagonalizable operators are Zariski dense in the linear operators over a field. Since you're having an identity of polynomials which hold on a Zariski dense subspace and since polynomials are continuous in the Z-topology, the identities must hold everywhere. That's one way of establishing the Cayley-Hamilton theorem, for example.
@tb Thanks. I've heard a lot of topological Zariski talk without ever actually learning about it, but it sounds like the way to go to understand that situation.
@anon The point is that the set of matrices with repeated eigenvalues is the zero-set of the discriminant of the charateristic polynomial, hence it is closed in the Zariski topology. Its (open) complement consists of diagonalizable operators, and Z-open sets are Z-dense.
@anon So: in order to prove that a polynomial identity holds for all matrices, it suffices to prove it for the diagonalizable ones since they contain the dense subset of the diagonalizable ones. Technically, you need to work over the algebraic closure but that doesn't change a thing since you can always restrict later on.
Aha! In my previous version I mentioned the possibility of going up to $k^{\rm alg}\otimes_k V$ and then restricting back down. I had a feeling that was useful.
@tb That was my impression when I saw it on Mma and ELU chats. Those points should be brought up on meta (individual control, as with sound, and respecting the ignore list). If those are addressed, then it would be a lot better.
@robjohn For the record, I think that I could have got used to it, but I'd prefer either to be able to turn it off or to have it appear in the sidebar.
@anon He can get them by your technique even without showing that the equation is already homogeneous: $t-(-t)=2t$ is a solution to the homog. eqn. so $t$ is, and then $t^2-t$ is, so $t^2$ is.
Let $\{a_{n}\}$ a sequence such that $a_{n +1}=2^{a_{n}}$, $a_{1}=1$
show that $\{a_{n}\}$ diverges to $+\infty$
hint:
It would have to prove by induction that:
$a_{n}\geq 2^{n-1}$, $n = 2,3, ...$
Using the inequality
$2^{n-1}=(1 +1)^{n-1}=1+(n-1)+\cdots\geq n$ (if $n\geq 2$)
Could they ple...
\begin{align*}
x - \alpha y &= 1\\
\alpha x - y &= 1
\end{align*}
By simplifying, we have:
$$\frac{x-1}{\alpha}=\alpha x-1$$
Which is:
$$(\alpha^2-1)x-(\alpha+1)=0$$
$$x=\frac{\alpha+1}{\alpha^2-1}$$
Assuming $\alpha \neq1$, $x=\frac{1}{\alpha-1}$ is your unique solution. For $\alph...
