@Derso An amateur is someone who is inexperienced. A rookie means someone who just started something, so also inexperienced. So a cookie made by an amateur is not good enough, so it's a rookie cookie.
@leslietownes I made a pilgrimage to Moe's today. Unfortunately no Folland, in fact slim pickin's really. I went to the ASUC 'bookstore' to find that they no longer carry actual books, all online now. I suppose that explains the paucity. It also means I can no longer get a feeling for what is being taught by looking at the class books. Depressing really, the effective demise of hardcopy.
coppper: and the kind of material that used to be easily browsed on "course websites" is now often students-only, behind some walled garden of course management software that even students at the uni who aren't taking the class might not be able to see
theres a cantor set like construction using more general sequences, of the type $a_0=1, 0<2a_n<a_{n-1}$ and so on. why would we do this, and not the usual case where interval length is taken to be $1/3^n$ at each juncture ?
Hi everyone. I am getting $\frac{5\pi}{3}(1+5\sqrt6-5\sqrt2)$ for the question below, but that is in none of the options. The correct option according to the book is (C). Can anyone help please?
how to group theory 101: given a theorem, barely check theorem for |G|=1, then assume theorem holds true for all groups of ord < G, find a proper normal subgroup, take quotient, apply the assumed theorem on the quotient, apply inverse of projection map on the quotient. Abuse lattice isomorphism theorem
@nickbros123 If you take out less than 1/3 at each step, You can make "thick" Cantor sets which have measure greater than 0 but still completely disconnected
Hi everyone. I would like to ask whether you think it is reasonable to receive a downvote in this answer without any reason being given in comments section. Is there some intrinsic feature of the answer (length, articulation, etc.) that makes an answer subject to downvotes? Thanks to those who would like to reply.
@BenSteffan No, I have expressed myself badly. I simply mean that the downvote could be -5, but in my opinion one should distinguish between two cases. 1) Bad questions/answers, no comments needed to justify the downvote; 2) Good questions/answers that received several upvotes, need to justify why the downvote. I do not find it fair for a user to see -5 in their reputation for a well-written answer.
@Bml This presupposes that there is some objective metric regarding what is a "good" question or anwer.
The whole point of voting is that there is not such an objective standard, and everyone with sufficient XP is allowed to express an option via their votes.
@BenSteffan Even if you write a super marginal answer, the expected value is positive. People don't downvote nearly as much as they upvote, and downvotes are worth less than upvotes, vis-a-vis XP.
there is some portion of down/upvotes that are misvotes due to the user hitting the wrong button and not noticing it. i don't know if the site design or any other UX considerations would bias this error rate up or down. i might expect people are less likely to notice mistaken upvotes because there is no effect on voter reputation associated with them. plus the bias on upvotes mattering more to the recipient's rep than downvotes...
as someone who does not work in this area, unitality at least sounds like the right answer, unicity sounds like a student who is unfamiliar with math and/or english aiming for "uniqueness" and failing
copper: not sure if you are asking something specific to a psie question, but per your recent bookstore visit, ~ 25 years ago, the book on the shelf for 202a would have been whatever edition was out at that time (new then). i don't think it has been updated since except for typos
derso: depends a tiny bit on what "matrices" means. that would hold if the basis with respect to which the matrix is taken is orthonormal
(which is common when an inner product is around, often so much that "matrix" is taken to mean that definitionally in those specific settings, but this is not automatic, and obviously sometimes the difference matters)
derso: math.stackexchange.com/questions/1490851/… contains some discussion of what the formulas for the matrix of the adjoint would generalize to if the bases involved are not orthonormal
@leslietownes sorry, a bit slow this morning, on a long boring work call. i just visited the asuc store for general interest, since my trip to Moe's was disappointing.
So if $A:I\to O(3)$ is a smooth curve of orthogonal matrices, we can write $\langle A'u,Av \rangle=-\langle Au,A'v \rangle$, for every $u,v\in \mathbb{R}^3$. :0
Just use that $A^TA=I\implies A'^TA+A^TA'=0$ and do the stuff with the inner product
In linear algebra, the transpose of a linear map between two vector spaces, defined over the same field, is an induced map between the dual spaces of the two vector spaces.
The transpose or algebraic adjoint of a linear map is often used to study the original linear map. This concept is generalised by adjoint functors.
== Definition ==
Let
X
#
{\displaystyle X^{\#}}
denote the algebraic dual space of a vector space
X
.
{\displaystyle X.}
Let...
See arxiv.org/pdf/math/0309036 On page 3, it says that "the action is proper if it is metrically proper for the above given pseudo-metric (i.e. for any sequence $g_n$ of elements in $G$ tending to infinity, then $d(g_n(p),p)$ tends to infinity in $\mathbb{R}_+$ for any $p \in Y$)." What does it mean for a sequence of group elements to tend or converge to infinity? My initial thought was that for any finite subset $E \subseteq G$, there exists $n_0 \in \mathbb{N}$ such that
for any $n \ge n_0$...But that would mean every infinite sequence of distinct group elements converges to infinity...I guess that's okay...I just wasn't sure if that would make the notion trivial.
homotopy theorists have been at war with the algebraic geometers over whose terminology is the coolest for the last few decades and I think we might be winning :)
@BenSteffan I skimmed this in under 10 seconds, truly inspiring
jokes apart, mathematicians putting a seemigly unrelated historical image before unloading 90 pages of the most complicated things ever written in human history will never cease to be comical to me
the term "telescope", in turn, comes from the fact that it is an incarnation of a mapping telescope, which is named because there is a canonical way to draw a sketch of what a mapping telescope is which looks a bit like a telescope :)
@user193319 If your space is locally compact, then we can take its one-point compactification and say that a sequence $g_n$ converges to infinity if it converges to the infinity point. In practice it means that for any compact $K\subseteq G$ there is $N$ such that $g_n\notin K$ for $n\geq N$
I think for most people the separation axioms are such that $T_s$ implies $T_r$ for $s > r$, and the named properties are reserved for the ones where we don't require $T_0$ or $T_1$ assumption
unless you are assuming beforehand things such as "every space is Hausdorff", this is what is most common that I see
I've never seen people use $T_4$ to mean closed sets can be separated. And if someone used that, I would cringe
A positive set $E$ is one for which $\nu(F)\geq0$ for all $F\subset E$, where $\nu$ is a signed measure. Why in the proof is there a sequence $\{P_j\}$ of positive sets such that $\nu(P_j)\to m$? At that point, we don't know yet that $m$ is finite, so we can't use the characterization of the supremum that says there exists a positive set $P_\epsilon$ such that $$m-\epsilon<\nu(P_\epsilon)\leq m$$to construct a sequence of positive sets $P_j$ such that $\nu(P_j)$ converges to $m$.
This exercise asks me whether the level set $L_0 = \{F = 0\}, F = x-y-e^{x+y}$ implicitly defines a function $y = g(x)$ on all of $\mathbb{R}$. I was wondering how to approach this? I know Dini's theorem (e.g. the Implicit FT) works locally on an open interval $I = ]x_0-a, x_0+a[, a>0$ when $F_y(x_0,y_0) \ne 0$, but in this case $F_y(x,y) \ne 0 \forall (x,y) \in \mathbb{R}^2$. Thus, I can extend $I \longrightarrow \mathbb{R}$?
@copper.hat I have the 2nd edition, 6th or later printings (here's the errata for that particular version by the way, useful at times). In one theorem in the earlier printings of the 2nd edition, I believe there was a bit of a flaw, which the 6th or later printings don't have.
@psie also, here's a typo. It should say $\nu$ does not assume the value $\infty$ in the first line of the proof.
@psie given the definition of m, if m = +oo, then for any n, there is some positive E_n with nu(E_n) >= n, and any sequence (E_n) thus chosen would have the property that nu(E_n) converges to +oo
it seems like you could even make the sequence E_n increasing (having chosen E_j as above, for any n the set F_n := E_1 union ... union E_n would also be positive by a short argument, and F_1 subset F_2 subset ... with nu(F_n) >= nu(E_n) >= n
im not sure if you would want that, but, you could easily arrange a kind of convergence of the sets themselves in that way and not just have them maybe arbitrarily skipping around your measure space
and sometimes in other arguments like this, you will see authors like folland omit this kind of simplification (hopefully expressly, by saying they are doing it, but not always)
Hello, I have this transformation. Without the homogeneous coordinate I understand that this is a change of basis vectors and that the rows are the basis vectors. If they are chosen to be orthonormal then the new coordinate system is also orthonormal etc. My question is what happens with the homogeneous coordinate. The combined transformation gives basis vectors that are not unit vectors. I want to know what are the basis vectors of the combined transformation in the previous coordinate system.
maybe I can restrict myself to $F(x_0,y), x_0 \in \mathbb{R} \text{ fixed}$. Now I can take, for instance, $x_0 =0$ and I have $F(x,y)|_{\{x=0\}} = \phi(y) = -y-e^y$
Maybe as a first more specific question, is it true that $$\hat{u}=\langle 1,0,0,0 \rangle $$ $$\hat{v}=\langle 0,1,0,0 \rangle $$ $$\hat{n}=\langle 0,0,1,0 \rangle $$ $$\hat{w}=\langle 0,0,0,1 \rangle $$ in the new coordinate system?
$O(3)$ acts as isometries on $\mathbb{S}^2$. What about $\mathbb{H}^2$? In the halfplane model they're essentially Möbius transformations, but I wished I could work with the hyperboloid (embedded in the Lorentzian space) model instead.
Do we have a "nice" representation for isometries in this model?
wait maybe I can work with a generic $x_0$ instead of setting it to 0. Now, if $\phi(y) = 0$ for some value $y_0 \in \mathbb{R}$. But this means $F(x_0,y_0)=0$ which is good because the point $(x_0,y_0)$ works fine for the local version of Dini's theorem. Now since $x_0$ is arbitrary I can span the whole real line
So all I need to verify is that $\exists y_0 \in \mathbb{R} : \phi(y_0) = 0$, namely that $0 \in Im(\phi)$...
@ShaVuklia uh, did they make you repeat undergrad or something? :D non jokingly those seem like "bad" boundaries for that problem, all but guaranteed to make the problem worse
i went through this about 100 times when i had to teach multivariable calculus twice during a postdoc
vector calc is tough because a lot of textbooks are drafted against a background reality that even "easy to work out by hand" problems might take more than "the usual amount" of computation