1:03 AM
Anyone have any idea where I can download WebWork if all the links are down on the official website?
Or who I would contact to get those mirrors back up?

1:14 AM
@AlexGramatikov email the IT dept of university

8 hours later…
9:31 AM
I have many small questions regarding isomorphism and its meaning, should i post it here
For example do isomorphic vector spaces have same dimension
In LADW book by Treil it is written $M^{F}_{m\times n} \cong F^{mn}$ what does this mean.
Can I write basis of $M^{R}_{2\times 2}$ as $\begin{pmatrix}1&0\\0&0\end{pmatrix}$, $\begin{pmatrix}0&1\\0&0\end{pmatrix}$, $\begin{pmatrix}0&0\\1&0\end{pmatrix}$, $\begin{pmatrix}0&0\\0&1\end{pmatrix}$
The books says that if two vector spaces are isomorphic then I can define a linear transform to transform bases of one space to other, so how can we define matrix $A$ that transforms basis of $M^{R}_{2\times 2}\to R^{4}$ ??? Further this transform is invertible, I am already confused

10:30 AM
anyone here?

10:53 AM
sigma (r/nCr) is written as sigma (n-r/nCn-r). How come? I know about nCr being equal to nCn-r but just r in the sigma? Can we replace that too?

11:07 AM
@jeea yes: let $\varphi: V \to W$ be an isomorphism and $B$ be a basis of $V$. Then $\varphi(B)$ is a basis of $W$. Since $\varphi$ is bijective, $\# B = \#\varphi(B)$
@jeea the space of $m$-by-$n$ matrices is isomorphic to the space of coordinates in $F$ with length $mn$
@jeea yes
@jeea you are confusing between linear transformations and matrices. $m$-by-$n$ matrices represent linear transformations from $R^n \to R^m$
where $V$ and $W$ are vector spaces, a linear transformation $\varphi: V \to W$ is a function that preserves addition and scalar multiplication
in this case, we can define $\varphi \begin{pmatrix} a & b \\ c & d \end{pmatrix} := (a,b,c,d)$
this function respects addition and scalar multiplication and is bijective

1 hour later…
12:28 PM
hi, im trying to show that a finitely generated group with no proper subgroup of finite index can't be embedded in $GL(n,\Bbb C)$. can someone help?

2 hours later…
2:01 PM
0

Determine or find an upper bound for $$\limsup_{n \to \infty} \left|\frac{\frac{2}{n} (l(\beta) - l(\hat{\beta})) - h(\beta - \hat{\beta})}{h(\beta - \hat{\beta})}\right|$$ where $l$ is the log-likelihood function in a model of logistic regression with $n$ samples of the form $(y_i,X_i)$, where...

Hi, does anyone have an idea for the above question?

@infinity Is your intention to prove that all such groups cannot be embedded or to give one counterexample?

there is a trivial counter-example
but that may be oversight

2:32 PM
@Semiclassical you there?

for a little bit.

@Semiclassical Can we do something?

Context: Equilibrium of Fluid Mechanics, Sommerfeld Vol 2, Chapter 2, Section 6. $$\textrm{ The external force F cannot produce a moment either, this is certainly true when we assume a continuous distribution of lines of force; such a distribution can be always considered as parallel in any infinitesimal region (the field of gravity is a simple example for what is meant)}$$

huh
What's Sommerfeld's definition of "moment"?

2:38 PM
$$\textrm{ If , on the other hand, the lines of the field F have a singular point at our volume element (a point wher the field lines interest), then F must needs vanish there, so does the moment of **F}$$
@Semiclassical I can understand the normal meaning of what he is saying but I know there lies something very deep when he makes that argument of lines of force. Can you explain me why he chose that route? What’s the significance of that?

I mean, I don't know what he means by "moment" in this context

Torque, he is saying that torque must also be zero in Incompressible Fluids

ah, gotcha
the first part seems straightforward enough
i don't really get what he's saying in the second part, though

Do you mind explaining the first part in your style?
With ultimate rigor

not really tbh. if you restrict to the case of F being a conservative force field, then one definitely has curl F=0
but it's not clear to me how to make sense of that more generally.

2:46 PM
Yeah! ##Me too

actually, I think the point is this: If you look in a small enough region, then every continuous vector field is conservative.
hmm
i don't like what I"m saying
yeah idk

In a small region we can assume the force field to be very densely continuous that is acting on each and every point, ha?

Hi! Can anybody (with "authority") let me know if it is accepted to ask a question on Math SE/SO regarding possible places to do a maths research internship? I am not a student anymore, graduated two years ago, but I am returning to academia. Thanks!

3:02 PM

Thanks!

np :-)

1 hour later…
4:32 PM
Question: Is abelian group a cyclic group? I knew the opposite direction is true. But I believe the answer here is: not necessarily. Is it enough to answer this question or do I need to give a group where it is abelian but not cyclic
I'm looking for a group that have this property.
but I don't have any so far.

@user777 $\mathbb{Z} \times \mathbb{Z}$

Well, "not necessarily" is enough to answer the question in the sense that it is indeed the answer, but usually when you answer a question in math, you want an argument for why that answer is the correct one, in which case you would indeed need to provide a counter-example

@user777 So you have gotten to the definition of cyclic groups without really seeing any examples of groups?

A good way to cook up an example depends on what you already know about cyclic groups

@TobiasKildetoft I have many examples of groups and cyclic group or even non cyclic group, but I didn't know how to find an abelian group but not cyclic group.
@feynhat Thank you @feynhat I'm testing this group now

2 hours later…
6:49 PM
@user777 Try the symmetry group of a general rectangle (i.e,, not a square).

Hi @Ted

Hi Professor and Balarka

Hi @Ted @Balarka

@skullpatrol :-)

7:55 PM
@AlessandroCodenotti If I have an $L^2$ map $f\colon\mathbb R\to\mathbb C$, how do I use $|f(x)| = (|f(x)|\sqrt{1+x^2})\times\frac{1}{\sqrt{1+x^2}}$ and Cauchy-Schwarz to show that it is $L^1$?
This was an argument in a book and I don't see what the application is.

Isn't that claim wrong?

$1_{[1,\infty)} x^{-1}$ should be a counter-example right

Oh it could be relying on something else then?
Hmmm, I will have to have a closer look then.
So disregard my question for now!

@LeakyNun Thanks I understood!! :)

8:40 PM
@JackOhara hey
^_^

2 hours later…
10:49 PM
Hi, a @Balarka, demonic @Alessandro, @Leaky.

Hi @Ted!

I'm late.
@anakhro, There must have been decay conditions at infinity.

11:11 PM
The theorem I found in another book does indeed have decay conditions at infinity.