most of them are along the lines of "not enough context" even though the questions are well posed, and explain in at least half a sentence where the problem is coming from
astyx this is a grey area. simply identifying the source of a problem is probably not enough to avoid a vote to close. identfying context can be, depending. it is an ultimately subjective evaluation of, how much effort has been put into this.
if people find a question interesting they will upvote it perhaps despite general site guidelines. i wouldn't assume that the same level of review is applied to each post. some people flag topics of interest and only visit those things. and if a question is of some kind of general interest it might be seen by more people quickly.
sometimes people confuse downvoting with "i don't think this is an interesting question" and upvoting with "i do think this is an interesting question."
the trouble i have is that often a restated version of a textbook hint (not expressly described as such) can sound like effort, when it isn't. i can catch those pretty quickly in my field of expertise but am not so good outside of it.
@Astyx perhaps the users that scour the site for PSQs don't frequent those tags. There are many poorly posed questions (note that I did not say bad questions) that have escaped scrutiny.
@Astyx If you think the question is worthy (I am not giving my judgement), you could upvote it.
Yes, I confess that I get annoyed when people who are not remotely experts in differential geometry or manifold theory pass judgment on questions in that sort of field. I know enough algebra, for example, to judge up to graduate level algebra just fine, but I don't believe the police here are well-rounded in every field.
@robjohn I don't think it's a particularly good or interesting question. But I think the OP is honest, has given this problem some thought before posting it online, and is just stuck and need some hint to carry his on work on
This one has -1 upvote even though it shows effort from the OP https://math.stackexchange.com/questions/4346341/prove-inequality-for-vectors-in-mathbbrn/4346479#4346479
Then I gave him a hint about CS, and a few hours later gave a complete solution (had I the time I would have worked harder so he would have come up to the solution on his own)
Since we are told that we are approximating $f(x)$ with the linear function $\lambda x$. Why can't we just substitute that into $3\int_0^1 xf(x) dx$ and show that it equals $\lambda$?
@LearningCHelpMeV2 one needs to show that the error in the approximation is small enough that it does not interfere with the substitution you mention.
Oh, wait. That passage is showing how to compute what $\lambda$ is to minimize that square integral. We are not given that $f(x)$ is close to $\lambda x$. We are simply trying to find the best $\lambda$ for the approximation.
Bonjour, I'm trying to study the following DE: $\displaystyle \frac{{\rm dy}}{{\rm d}x}=\frac{x^2+3y^4(x)}{x-4y(x)}$. I think using the existence and uniqueness theorem we can conclude that solution there exists. But can we find the solution in closed form? Using change o variable, of techniques in homogeneous equation all seems does not work. How can I work in this problem?
Hello, does anybody know: We have equation 3x+5 = 2 in mod 10. I applied the extended10 euclidian algo on $3x+10k = 3$ and got to the equation: $10\cdot 1 - 3\cdot 3 = 1 \iff 10 \cdot 3 + 3\cdot (-3) = 3$.
However we written that $x = -3-10m$ and $k = 3+3m$, however I do not understand where we got x and k to be in that form (OK for -3-10m it is because of mod 10, but what about 3+3m ??)
I don't really get why use extended Euclidean algorithm in the first place
$3x+5 \equiv 2\pmod{10} \iff 3x \equiv -3\pmod{10}$ and we can use that $\gcd(3, 10) = 1$ to obtain, theoretically (you don't need to know what it is), some $y$ with $3y \equiv 1 \pmod{10}$ from which multiplying both sides by $y$ we get $x \equiv -1 \equiv 9\pmod{10}$
I came across this definition of a smooth mpa. A map f: U \subset R^m \to R^n is called smooth if it agrees with another smooth g an open set O, so g|(U\capO) = f|(U\cap O). So then how do we define smooth on g....? I thought smooth map meant it has a derivative everywhere.
Yes, but the point here is to define smoothness on domains which aren't necessarily open sets
We can't talk about the smoothness directly, so instead, we say that it extends to some smooth map in some small neighbourhood of every point of that set
So to define smoothness on g, we use open sets first and talk about g has a derivative everywhere and to find out if a function f on an arbitrary set U is smooth, we compare it to known constructions locally (with g)? And if this local comparison is true everywhere in U, then f is just smooth on U.
@VLC You an do $3x\equiv 4y \equiv -9y \pmod{13}$, divide by $3$, and then express the whole thing with respect to $x$ for example. Idk if this is what you're looking for.
Hey, I have a probability question: If you are rolling two die and you want to find the probability that none of the rolls will yield two faces with the same number, it's just $1-P(S)$, where $S$ is the probability that the faces will have the same number, correct?
yes. i'm a little unsure of the language here, 'none of the rolls' when there's just two of them, and 'the faces' when there's just two of them. if you mean just the two rolls, yes.
or alternatively two consecutive rolls of one die. oddly enough.
it gets subtler if you roll a die three, four, five, or six times because the event that 'at some point two values match' is different from simple equality between two fixed outcomes.
if you roll more than six times, you can be pretty sure that two of them will be the same. :)
unless the die falls off the table or vanishes in some kind of quantum incident
you sort of get to choose. i would say yes to that, with a caveat like, say one die is red and one is blue and you distinguish them. if the dice are indistinguishable you could choose to model it with a smaller number of outcomes because you don't have a way of seeing "this one is 3, and this one is 4" you just have "we rolled 3 and 4". with distinguishable dice the probability of each 'outcome' is the same. with indistinguishable dice, different outcomes may have different probabilities.
i find it's usually helpful to model these things with distinguishable dice. this results in a uniform distribution on the set of outcomes, and you can care about whether or not the distinguishability matters in whatever your event is in how you count them.
Let $(X,\Omega, \mu)$ is a $\sigma$-finite measure space $x$ is a bounded linear operator on $L^2(X,\mu)$. There exists an ascending sequence $\{E_n\}$ of measurable subsets of $X$ of finite measure whose union is $X$. If $1_{E_n}$ denotes the indicator function on, why does there exists a function $f : X \to \Bbb{C}$ such that $f 1_{E_n} = x 1_{E_n}$ for every $n \in \Bbb{N}$?
this doesn't seem right. consider X = R (which is sigma finite), you could take E_n = [-n, n], and x to be translation by 1 (say to the right). x 1_{E_n} is nonzero on the interval [n,n+1] but for any multiplication operator f, f 1_{E_n} is supported on E_n and cannot be.
We do know that $x$ is actually in the double commutant of $L^{\infty}(X,\mu)$, when viewed as left multiplication operators...Not sure if that makes a difference.
if you let g denote the function x 1_{E_n} then it is certainly an L^2 function. it is also supported on E_n because 1_{E_n} g = 1_{E_n} x 1_{E_n} = x 1_{E_n} 1_{E_n} = x 1_{E_n} = g. x commutes with multiplication by 1_{E_n} because it is assumed to commute with things that commute with multiplication operators, which include multiplication operators.
a priori i guess it's not clear why g should be in L^infty and not just L^2 but maybe he gets to that later.
long story short, the function here is just x applied to the constant function 1, which a priori is in L^2 and under further reasoning winds up in L^infty.
in my postdoc i helped a student who was stuck on his thesis. he had all of this pretty cool theory worked out, but it depended on something being in L^infty and not just in L^2, and he couldn't show that. it was not a von neumann algebras setting. he had been stuck on it for months. one day it just hit me and he copied my proof verbatim into his thesis without attribution.
which doesn't bother me, i think it was ignorance of scholarly norms and not hostility. and i am more famous than he is anyway.
under: i liked feller for self study although it is not the kind of book that you could really base a course around.
it's also somewhat more 'pure mathy' in its approach. not that its contents are useless and cannot be applied, but some books take an applied approach from the beginning and if that's why you're into probability you might not like feller.
my wife used resnick. it is a good book.
user: i am 'retired' from math now. the problem involved specific C* algebras.
I prefer the pure approach more than applications. The course is more applied, but I can still read it alongside the required textbooks. Either way, I just need to better solidify my understanding of what's actually happening.
@UnderMathUate Oh, I actually never read Billingsley, sorry. The book I was trying to remember is Probability with martingales by Williams. It has a small amount of exercises, but they're challenging and fun to do.
@VLC You could note that $5 \equiv -8 \pmod{13}$ and use that $x^3+8 = (x+2)(x^2-2x+4)$, then solve for $x^2-2x+4 \equiv 0 \pmod{13}$. Not sure if that's any faster
So basically we can say $x^3 = x'$ and then we use $x' -5 = 13k$, and $x'-13k = 5$, now we can use extended euclidian to find $x'$ and then we know that $x' = x^3$
@Hawk This kind of smoothness at the edges of the interval is seemingly different from the smoothness of functions in the theory of smooth manifolds, but you can show the following math.stackexchange.com/a/2162283/476484
You have $f:D\to \mathbb{R}^n$, and for all $x\in D$ there should exist some open $U_x$ and a smooth function $g_x:U_x\to \mathbb{R}^n$ such that $g_x$ is an extension of $f\restriction U_x\cap D$
what "smooth" means varies from author to author, sometimes it means $C^\infty$, sometimes it means $C^1$ and sometimes just differentiable
I was asking how "g" is smooth here. but apparently in book we only define maps from R^n to R^m on open sets where their components are calculus smooth.
@VLC Idk why would you do that or what would that achieve
What you could do, I guess, is note that $(-x)^3 = -x^3$ so you can just check $x = 1, 2, ..., 6$ to see if you get either $5$ or $-5 \equiv 8 \pmod{13}$
@robjohn Sorry this is late (fell asleep). But the approximation integral is the area between the graphs. Wouldn't the best approximation occur when $f(x) = \lambda x$. I.e. approaching zero area between the graphs
so you'd get $1, 8, 1, 12, 8, 8$ for $x = 1, 2, ..., 6$ respectively (note that to calculate $a^3$ you can always try calculating $a^2$, reducing mod $13$ and multiplying by $a$ again to make your job easier for bigger $a$)
so solutions are $x = -2, -5, -6$
Also, this hints at a convenient number theory trick, checking something mod some number. Here for example, we see that a cube needs to be $0, 1, 5, 8$ or $12$ mod $13$. So if in some equation you get a cube, and the other side is some other number mod $13$ than the ones listed, you know it's not solvable. And this trick usually works modulo other numbers, for example, a square needs to be $0$ or $1$ modulo $4$
i agree with jakobian. number theory might have slightly more structure to it and more obvious interactions with other fields as a beginner (i do not mean to criticize combinatorics, which is everywhere, but i think the connections come later on)
@leslietownes Wait, so for each $n \in \Bbb{N}$, we define $g = x 1_{E_n}$. But then $g$ depends on $n$. Are we supposed to take some sort of limit to get $f$?
the vibe of this is that if x is a multiplication operator (which it turns out to be) then its symbol has to be x 1. although it is not a priori clear that x 1 (belonging to L^2) also belongs to other function spaces in which one considers multiplication operators.
this comes up a lot in physics treatments of operator theory which ignore this. and they may be right to do so, there's nothing about it that's wrong in common settings, that i know of. it's just, if you want to pencil in the details, you do have to show by some route or another that multiplication by x 1 induces a bounded operator. that's not going to be true of general L^2 functions.
and if it does induce an unbounded operator that can be fine too, because we can work with unbounded operators. but that's not on page 11 of your notes. :) that might be page 90 of your notes.
90 chosen at random, not a pin cite. just, it comes later. often in assessing commutants of general von neumann algebras.
there's no route i know to understanding general von neumann algebras that does not involve unbounded operators, which is weird because they are by definition algebras of bounded operators.
tomita had a very deep understanding of this although some of his results were phrased oddly and confusingly. takesaki is the standard reference point although it would be in any book by now.
Hi, I know that if $z=f(x,y)$ then we can consider the parametrization for the surface as $x(u,v)=(u,v,f(u,v))$. Similarly if $x=f(y,z)$ then we can consider the parametrization for the surface $x(u,v)=(f(u,v),u,v)$. Then if the surface $S$ is given by $x=y+z^{3}$ then $x(u,v)=(u+v^{3},u,v)$ is a good parametrization for $S$? I know that the parametrization it not only, but my parametrization is correct?
Thank you leslie, my doubt was because when I try to calculate the Gaussian curvature, the principal curvatures become hard to calculate. So I had my doubts.
I am working on this problem, I have some nice topological spaces (metrizable & separable) $X_i$ which are Baire spaces, and I'm trying to prove that their product $X$ is also Baire
So I've tried taking $G_m\subseteq X$ which are open and dense, and $p_N:X\to X_1\times ...\times X_N$ be the projection, then consider $G_m^* = \bigcap_{N=1}^\infty p_N^{-1}p_N(G_m)$ but it doesn't get me anywhere tbh.
I already have this result for finite product of them
So I'm trying to somehow prove that $\bigcap_{m=1}^\infty G_m^*$ is dense, and I expect it to somehow magically imply that $\bigcap_{m=1}^\infty G_m$ is dense as well
idk
I could assume that $G_{m+1}\subseteq G_m$, that usually seems like a good property to have
No, I didn't say anything about a set being dense at x. I meant that: if X is dense in $\mathbb R$ then for every $y\in \mathbb R$, there is a sequence in X that converges to $y$.
maybe a language issue? what would the consequences of density, per the sequential criterion, be, at a chosen point a in R. i think this is what ted is getting at.
Ted, I'm afraid I don't understand how that defined dense at the point. Astyx: I said open interval. cl (X) containing an open interval means that there is an open interval (a,b), a<b such that $(a,b)\subset $ cl(X). If that happens then int (cl X) $\ne \emptyset$
You said "for every $a\in\Bbb R$, there exists a sequence $x_n\to a$ with $x_n\in X$. Thus, "$X$ is dense at $a$" if there exists a sequence $x_n\to a$ with $x_n\in X$.
If for every open interval $(a,b)$, there is a point $c\in (a,b)$ such that $c$ is not in X and no sequence in X converges to c, then we say that X is no where dense in R.
I have been thinking all day about a question I had in exam. I know intuitevely that it is true, but I am not able to show that a X ~ Binomial n=L, p=1/2, and L~Poisson(lambda), is equivalent to say that X~Poisson(lambda/2)
How can you know intuitively that this is true? Doesn't make sense to me. Anyway, how you do this is just calculate probabilities: $P(X = k) = \sum_{n=0}^\infty P(X = k | L = n)P(L = n)$
Anyway, I have a question
We know from descriptive set theory that there are $G_\delta$ subsets of $[0, 1]^2$ whose images under projection are not Borel
can we provide some simple example if we require the image to not be $G_\delta$
A prime example of a subset which is not $G_\delta$ I have in mind is $Z\subseteq [0, 1]^2$ defined as $Z =( \mathbb{Q}^2\cup( \mathbb{Q}^c)^2)\cap [0, 1]^2$
When we want to find the surface integral of a vector field $\vec A$ for a parameterized surface, lets say a sphere of Radius R, while $\vec r(\phi , \theta)$ it must be that $\vec A (\rho , \phi , \theta)$ . Is this assessment correct ?
You write everything in terms of the parametrization. Sometimes you can figure things out with geometry and symmetry and not need to do everything with parameters.
