it's interesting. certainly any set of hypotheses in which the conclusion is true implies normality. i wouldn't say it is 'needed' in that a lot of hypotheses formally unrelated to normality might lead to the same conclusion. i share ted's confusion.
well, g induces a map from F[x] to F'[x] that you get by applying g to coefficients. so i guess, sort of.
but i have a large amount of distaste for writing down things that haven't been defined on the grounds that they can be figured out, or are "obvious."
if you're doing any kind of work with the map it would be necessary sooner or later to come up with a notation for it, which suggests that it's worth setting out explicitly.
koro yeah you will want a name for that map. maybe call it phi. it's a degree-preserving isomorphism, so if something factors in F[x], its image in F'[x] will factor too, and vice versa.
it's helpful to take one step back from the actual homomorphism at issue here and look at why it's doing what it's doing. what makes it capable of testing irreducibility seems to be degree preservation.
my daughter got up at 8:30 pm and demanded to listen to music. i put on some jazz. she said it was 'boring.' i put on some old r&b. she said that was 'boring.' i put on some rap, which she liked, and then some kraftwerk, which she liked. then she asked why the songs were so long and i said it had something to do with her perception of time.
then she tried to carry the cat into her bedroom and got clawed.
I think I found a way to solve twin primes, but it involves solving a complex-looking question about "disjoint, element-wise multiplicative, coverings of all composites in an interval".
After I have solved that, I can simply use an inductive argument
Question: when can we cover $[a,b] \cap C$ by disjoint intervals $[a_i, b_i]\cdot [c_i, d_i]$ where $C$ is the set of composite integers, and some other constraints are satisfied in relation to the counting formula I'm using for twin prime
but the underlying basic moves, e.g. paying attention to when things are positive vs. negative, separating a sum of related things into pieces that are estimated differently, those are general vibes.
rob: i'd be fired if i let something like that slip through at work.
but people can ponder the ongoing mystery of why i haven't been fired elsewhere.
@leslietownes Well, to me, analysis seems to be, you draw some diagrams, you get an idea, and you do some inequality algebra to get your proof done theoretically.
prithu this is definitely what you see early on, as i hinted at above. 'fun with the triangle inequality' is most of what people get. and it does have that flavor.
i don't think people are out there trying to press the triangle inequality into new results right now.
A lot of it is bad homework/pedagogy, too. Too many people are taught algebra as should symbol manipulation. Analysis is plagued by similarly bad teaching.
A turtle challenges a cheetah for a 10 km race. The cheetah gives turtle a head start of 1 km. Argh, I forgot this puzzle/question. In this the turtle wins and one is supposed to find the fault in the argument leading up to the win of the turtle. Does anyone remember that?
prithu: the triangle inequality is great, and it's great for 1-2 semesters of analysis. and sometimes (depending on application) that is literally all one needs. so it's not illegitimate to focus on that stuff. it's good stuff.
no, it was not half and half jumping. It was to test understanding of dx and was something like (not sure if this is correct): in time t cheetah travels some distance x and during that time the turtle travels x+dx.
Coming in to ask: I still see math.stackexchange.com in read-only mode (and it's the only SE I'm a member of that still seems like it's in read-only mode); is that just me, or do other people see that way?
if it's possible to edit the title of this chat to dynamically update whenever someone joins, so that the topic reads "PERSONAL BAN OF [new arrival] - VOTING CONCLUDED" i would support this. it would make waiting for the site to come back marginally funnier.
Someone cracked the integer factorization problem, they originated as a user from this site, and released codes in Python and C++ (on the MSE site). Then they wanted to prove it to us, so they cracked a bunch of our passwords.
ted: my personal campaign against you is beyond any stackexchange platform. expect your favorite wine shop to be out of your favorite sancerre. i am arranging all of this.
Despite the maintenance being listed as finished in the question body and also in this tweet, Math.StackExchange is still in read-only mode:
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I still don't see why they make us type posts vertically, the preview should naturally be on the right of the text entry (horizontal).... the posts are usually quite long after only a few paragraphs and you have to keep scrolling up & down
@Purple That is not a bad suggestion. Of course, once you post and start editing, there's no longer any preview whatsoever. I've never quite understood why.
There was something about that a few years ago on meta. People said you could have that feature, but you have to go to this and that link or something.. Some plugin maybe?
q.uiver.app for your CD needs, but you have to screencap the resulting diagram. MSE and Quiver should work together to allow embedding of just that
I would like to just set my phone up on a tripod and start drawing on paper, the camera then captures and converts all the diagrams etc into MSE postable data
It's no easy thing to understand how to use a calculator and there is still some understanding required to complete the problem
And as math grows in complexity, it's a question of who knows how to use the calculator, since some formulas - just writing them down - takes a horrendous amount of time.
Then if you make one sign change, you're doomed!
In your hand calculation I mean
Wear a certain eye glass that lets you see otherwise invisible ink on paper. They'll never see that one lol
They should only really ask unsolved problems on a test. Then grade you on the novelty and elegance of your approach.
@PurpleHaze I would personally start from simple problems, and slowly move to harder and tricker problems, and will show them the unsolved problems if they are interested.
Since this answer doesn’t seem to be drawing any attention from anyone who can do something about it, I’m posting it as a question. The Mathematics Stack Exchange has been in read-only mode for 7 hours now; it seems that the other Stack Exchange sites are back to normal, so perhaps Mathematics wa...
It's said here that Thomae's function is discontinuous at every rational point. Shouldn't it say it's discontinuous at every rational point, except at zero where it is continuous?
@robjohn I noticed that if you remove an opposite pair of octahedron vertices you can visit all the remaining edges in a single loop, although (of course) that loop does pass through the 4 remaining octahedron vertices twice.
Hamilton invented a game, the Icosian game, based on visiting the vertices of a platonic dodecahedron, which is (of course) equivalent to visiting the faces of an icosahedron.
no disrespect to hamilton, but it's a crummy game.
in his defense, there was nothing to do in the 19th century.
wolgwang, does 'identical' mean exactly the same? or something less than that. e.g. if you toss four coins, the outcome "exactly one heads" can occur in four different ways. are they 'identical'?
continuing my example, you could model that with the individual outcomes each in the sample space, with equal probability, or you could model it with a single "exactly one heads" outcome in the sample space, assigned the probability that it gets by the fact that it can arise in four equally probable ways.
talk me through the reasoning of why it might be 3/6? i'm not sure i see it. you definitely don't want to ignore the fact that there are three A's [and two I's]
here, yo ucould model it as "you draw the first A", "you draw the second A," "you draw the third A" as distinct events of equal probability, or "you draw an A" as one event with a probability that reflects there being three As.
there is an element of arbitrariness in the sample space. you get to choose.
probability books can sometimes get very simple formulas this way, because they can sometimes express things in a way that rolls up any arbitrary choices into the notation and gives the same result for each of them.
one sample space has 13 elements, where the jth element corresponds to drawing the jth letter. all of these outcomes are equally probable. so you'd answer the question by counting which outcomes give a vowel or not (= the total number of vowels in the word), and dividing by the total number of outcomes (= total number of letters in the word).
another sample space would be the number of distinct letters in the word, where each letter is given the probability it has as a result of its number of occurrences in the word.
same result in the end.
i am doing OK.
how are you?
sample spaces where computing probabilities reduces to counting a number of outcomes and dividing by the number of total conceivable outcomes are often conceptually easier to work with, but may involve things that seem unnatural in specific contexts (here, regarding 'drawing an A' as a composite of three distinguishable events).
@leslietownes Same thing will apply to other events like drawing a red ball from a box containing x red balls (distinct or identical doesn't matter?) y blue balls etc. , right?
yeah. if you want the simplest possible sample space (where everything is equally likely) you may want to distinguish the balls. if you want a small probability space (which only contains events relevant to you) you may want a space where different events have different probabilities. and then computing those probabilities is part of the task of defining the space.
I'm studying a paper, which I'll include a small piece here. And I'm struggling to calculate
$$C_n\|u_{m,n}\|^{\left(\frac{2*}{2}\right)^k\frac{2*-q}{(r_k)^k}}_{L^{2*}(\Omega)}$$
Where $\Omega$ is an open, bounded subset of $\mathbb{R}^N$. We also are considering values of $m$ and $n$ such that $m>
That looks like something a grad student would spend a week working out. A 50-point bounty for something that takes a week is too cheap. But it’s just way too intricate and involved.
@TedShifrin As a sanity check,if f in the definition of a homeomorphism $f:(X,A)\rightarrow (Y,B)$ is a homeomorphism such that $f(A)\subseteq B$ then it is also true that $f$ induces a isomorphism $H_n(X,A)\cong H_n(Y,B)$
the working through those inequalities - it's got to be wrong. better phrased as an email to the paper's authors (with 90% of that taken out) than a q on m.se.
> Out of 100 students two sections of 40 and 60 are formed. If you and your friend are among the 100 students. What is the probability that you both enter the same sections?
If the sections are A and B, I think S(sample space) is something of the form {(A,B,B,B,A... *100 times* ), ...$2^100$ times} But how do I make sure that only 40 students are in A?
This is not so obvious to me. You consider the class divided into a group of 40 and a group of 60. You count pairs from the group of 40 and pairs from the group of 60. What's the chance that you and your friend (the pair you care about) are in one of those two groups?
Despite the maintenance being listed as finished in the question body and also in this tweet, Math.StackExchange is still in read-only mode:
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