@user2103480 Sorry, here's what I really meant to say instead of a lucid picture: If $X$ is a top. space with an equivalence relation $\sim$ and $Y\subset X$ such that $Y$ is saturated wrt $\sim$ and $\pi\colon X\rightarrow X/\sim$ is open, then a) the composition $X\setminus Y\rightarrow X\rightarrow X/\sim$ factors through a continuous map $X\setminus Y\rightarrow(X/\sim)\setminus(Y/\sim)$ and b) this map is an identification map, so that $(X\setminus Y)/\sim=(X/\sim)\setminus(Y/\sim)$.
Now $X=S^2$, $Y=$strip about the equator
@politeproofs the first equality cause $f,g$ are inverse, the inequality because $f$ is monotonic
Does anyone have a clever way of writing $(\prod_{1 \le i \le n} f_i)' = f'_1f_2 \dots f_n + f_1f'_2\dots f_n + \dots + f_1f_2\dots f'_n$ in summation notation?
Oh, ugh. Free = sliding, fixed means tail of the vector must be at the origin. For the others, you can put the base anywhere you want. Only physicists talk about this.
When you add $\vec a + \vec b$ you move the tail of $\vec b$ to the head of $\vec a$ and join the tail of $\vec a$ to the head of $\vec b$. We all do this, so we all allow ourselves to slide the vector over to do that.
Never ever in my 50 years of teaching have I used it.
When I write a vector in coordinates, of course its tail is at the origin. But I'm allowed to draw vectors $\overrightarrow{AB}$ whenever I want.
I taught out of a book that insisted on using parentheses for a point and brackets for the corresponding vector. Other than conforming to the book's notation when I taught that course, I never made the distinction. I identify a point and the vector from the origin to it.
I realize that I can add a vector to a point and not vice versa, but geez ...
Look, @123. I've spent plenty of time helping you. I do not want to discuss this topic, and I'm not going to change my mind. Did you read everything I wrote above? I doubt it.
well, i can look at a page or two, but i am not going to cycle through 17 pages looking for something. if you have a specific question i can try to answer. you should realise that notation used by physicists, mes, ees, etc, are not uniform and often not consistent with that used by mathematicians.
I am trying to understand vector rules for translation and rotation. Sometimes book used it as free vector in case of concurrent forces. Sometimes they used it as sliding vector move along the line of action.
@BalarkaSen This is my idea: consider the annulus with an intermediate circle, if I use this math.stackexchange.com/questions/1284674/… then due to intermediate circle we have two Mobius band each having the intermediate circle as the boundary, i.e. we are considering the identification of two Mobius band along boundaries
It is not immediately clear how to compute the fundamental group of the Klein bottle, so that would not answer OP's question. What they're doing is one way to compute $\pi_1(K)$
One can give an axiomatic definition of cup product from knowing cup product on the "universal object" $K(G, n)$. I'm not sure how computationally useful that will be.
Can someone give me an example of a proper homotopy equivalence of plane? A proper homotopy equivalence can be defined analog way as homotopy equivalence, but here one has to replace all maps, including all homotopies, with proper maps. Of course, a homeomorphism is a proper homotopy equivalence, but other than that.
Constant map on the whole disk. There's no issue with continuity, if $D$ is the finite radius closed disk this is the quotient map $\Bbb R^2 \to \Bbb R^2/D \cong \Bbb R^2$.
Something like this $\varphi:\Bbb R^2\to \Bbb R^2$ is defined as follows $$\varphi(x,y)=\begin{cases}0& \text{ if }x^2+y^2\leq 1;\\ \big(\sqrt{x^2+y^2}-1\big)(x,y)& \text{ if }x^2+y^2\geq 1.\end{cases}$$
you need to qualify it, i.e. it is the unique isomorphism that satisfies property X
and if you're asking whether any order-preserving automorphism (i.e. isomorphism with itself) of ordered fields must be the identity map (i.e. sending every element to itself), then this paper provides a negative answer
So, given a comm. ring $A$ and multiplicative set $S = A \ P$, $P$ a prime ideal, they denote $A_P := S^{-1} A$ the localization of $A$ at $P$, and then say that $S^{-1}P := PA_P$. Should I believe that $PS^{-1}A = S^{-1}P$?
I guess things on the right are in (and only in) the set $\{ p/s \vert p \in P $ and $ s \in S \}$ and things on the left are sums of products of things in $P$ and fractions in $A_p$.
Consider a single-step POMDP, where an agent receives a random belief state $b(s)$ over some fixed set of contexts $s$, selects a single action, and receives a single reward, before the episode ends. I understand that the belief state value function, $V(b)$, is piecewise linear and convex, with a...
you can write down inverse maps both ways by the universal properties like a normal human being, or you can note that we have $\operatorname{Hom}(S^{-1}A\otimes_AM,-)\cong\operatorname{Hom}(M,\operatorname{Hom}(S^{-1}A,-))\cong\operatorname{Hom}(M,-)\cong\operatorname{Hom}(S^{-1}M,-)$ on the category of $A$-modules on which $S$ operates bijectively and then apply the Yoneda lemma
the point is that if $S$ operates bijectively on $N$, homomorphisms $S^{-1}A\rightarrow N$ are uniquely determined by where they send $A$ and each such possibility is realized
ok ok fair. so for $S^{-1} M \cong S^{-1} A \bigotimes_A M$... the LHS has fractions, but the right has equivalence classes of pairs of (fraction with numerator in $A$, element in $M$)
I bet if $\{K_n\}$ is a Folner sequence, $\varphi_n : G \to S_{K_n}$ given by $g \mapsto (x \mapsto gx)$ whenever $x, gx \in K_n$ and extending in whatever way outside are the approximate homs
idk @Thorgott like could you send $(m,s) \to ((1,s),m)$ and then going backwards could you do $((a,s),n) \to (m,s)$ where $an =m$ which you could find since $M$ is an $A$-module?
idk I think I have soggy brain, might need to think about something else briefly
completely floored at how I wrote something that wasn't actually just crap. but yeah, there can only be one reasonable answer since this is a natural thing was my guiding thought
Note that Connes' Embedding Conjecture was resolved by quantum complexity theorists earlier this year. This is why operator algebras is the most amazing branch of maths---because it connects with so many other branches of math.
So a curve is nonsingular when its partial derivatives are not zero at all points in $Z_f(\overline{k})$. How would someone reasonable check this? $Z_f(\overline{k})$ could be pretty big. I am guessing you do not try to show $\{(a,b) \vert f(a,b) = 0, \partial f / \partial x (a,b) = 0 , \partial f / \partial y (a,b) = 0 \} $ is empty
I think it usually happens to be the case that the partials of $f$ don't vanish at most points and it only remains to check that the critical ones aren't zeroes
(which is more convenient than trying to first calculate all the zeroes and then check at each of them)
also lets say we got $f(x,y) = x^2 y - y^2 -2x + y$ and so we define a rational function $\alpha := \frac{x}{y-1} = \frac{y - x^2}{x-2}$. Book says you got a pole at $(2,1)$ "since $(y-1)$ vanishes and $x$ is inverible at $(2,1)$"
So yeah, I guess it makes sense that you get a pole since neither of the 2 versions of the rational function $\alpha$ are defined at $(2,1)$, but their rationale seems so odd. I guess my question is "how do we know $x$ is invertible at $(2,1)$ and how do we know there's not some other rational function equivalent to this one that is defined at $(2,1)$?"
@Thorgott coulda fooled me
I see that he refers to some points $(0,1)$ and $(2,4)$ as points where $\alpha$ is defined, and I noticed they are different from $(2,1)$ in that one or the other version of $\alpha$ is defined, meanwhile the other or the one version of $\alpha$ shows up as "0/0". When you put in $(2,1)$ you just get $2/0$, which is different.
maybe that's the crucial difference
I guess "invertible" is just "you get a constant that isn't zero on the numerator"...
Ok so I know this is not closely related but this is the only remotely related place I could find. Well I want to ask a question, it is about fuzzy logic, anyone here familiar with fuzzy theory, I wrote a basic fuzzy program in Java but I am getting unexpected results. It only has 2 input variables and 1 output variable.
If it's suitable to ask it, just tell me, I don't want to ask an off-topic question so I am asking to ask :D
Hey all, I asked a question really late in the morning without thinking about how it’ll definitely be like 3 pages down before the morning...I don’t want to repost the question, but can anybody help with this diff eq problem?
For the system of differential equations
$$\dot{x}=y-x, \dot{y} = \mu x - y$$
with Jacobian matrix
$$J = \begin{bmatrix}-1 & 1 \\ \mu & -1\end{bmatrix},$$
is a stable spiral for $\mu < 0$, a stable node for $0 < \mu < 1$, and a saddle for $1 < \mu$. However, how would the bifurcations between the...
What exactly is "the area element" in a surface integral? Until now I thaught the dx after an integral is just a notation that indicates that we are integrating "with respect to x". Now we had surface integrals in calc III and I seem conpletely lost. We have to do some calculations with those dσ named surface elements.
.. and I have no Idea what that means. I can write what exactly is the "definition" we had and what are the tasks when I'm at home, but maybe someone can give a general answer, I hoped
@BalarkaSen The norm is the Hilbert-Schmidt norm on $M_n(\Bbb{C})$, which is $||x|| = \tau(x^*x)^{1/2}$, and $\tau$ is the normalized trace on $M_n(\Bbb{C})$. Also, a group $G$ hyperlinear if and only if it can be embedded into the unitary group of the hyperfinite $II_1$ factor if and only if its group von Neumann algebra $L(G)$ can be embedded into the hyperfinite $II_1$ factor.
Hello!! Let $V,W$ be $\mathbb{R}$-vector spaces and let $\phi:V\rightarrow W$ be a linear map. Let $1\leq k\in \mathbb{N}$ and let $v_1, \ldots , v_k\in V$.
If $V=\text{Lin}(v_1, \ldots , v_k)$ then does it follow that $W=\text{Lin}(\phi (v_1), \ldots , \phi (v_k))$ ?
We have that $\phi (V)\subseteq W$ and since $\phi$ is linear we get that $\text{Lin}(\phi (v_1), \ldots , \phi (v_k))\subseteq W$, right?
@anakhro In the general case : The dimension of $V=\text{Lin}(v_1, \ldots , v_k)$ is $k$ (since all vectors are linear independent), then the dimension of $\text{Lin}(\phi (v_1), \ldots , \phi (v_k))$ is also$k$. But the dimesion of $W$ can be greater, as in the example you gave. Is that correct?
So basically we have 2 position vectors R1 = x1î+y1j and R2= x2î+y2j. If matrixA=( x1 y1) ( x2 y2) Then we need to prove that det(A) is invariant under rotation of an angle(theta) about x-axis.
My teacher says simply state that A'=RA and then apply determinant on both the sides, but that would have been possible only if the components of R1 and R2 were along the column of matrix A i.e if A were. (x1 x1) (y1 y2)
You are correct that the matrix should multiply the column vectors to get the new column vectors. And your teacher is correct, too, because the determinant of the transpose of a matrix is the same as the determinant of the matrix. But it's more natural to use columns as you want to.
The matrix equation the teacher stated only works if it's with the transposes.
I'm guessing the teacher was writing row vectors and you're writing column vectors.
@TedShifrin yes that's how it was given in the question. But the results A'=RA would hold only when the components of R vectors were along the column right?. I understand that in this case since we take determinant it hardly matters but saying A'=RA won't be correct right ?
If this is the convention that I am saying with row vectors typed, you're right. If she's typing the column vectors as rows (which is what I thought you wanted to do), she's right. I've asked you three times where the notation is actually defined.
@TedShifrin Oh ok so in the matrix A the elements are Aij where i is row j is colum and what I am saying is that the elements should have been A11= x1, A12= x2, A21=y1 and A22=y2
Yes, I agreed with that ages ago. What we have not established is what the parentheses your teacher/text are using mean. If they mean what I usually use, then you are correct. I am not going to discuss this any further.
Hello, everyone. This is a question I asked on the main site, but it never got any answers. We know that the real ordered field can be characterized up to isomorphism as a complete ordered field. However, that is a second order characterization. A natural question is, is the first order theory of the real ordered field axiomatized by the axioms for ordered fields and an axiom schema stating that every nonempty parameter-free definable set which is bounded above has a least upper bound?
