test

test

Good evening

is the preimage of a simplicial neighborhood under a simplicial map contractible?

what is a simplicial neighborhood?

is this related to graph theory?

no.. topology

I have no idea then

would you care to look at my post which can be found at:

@monoidaltransform by simplicial neighbourhood do you mean a neighbourhood homeomorphic to a simplex?

actually never mind, I don't know enough algebraic topology to answer this

@Jakobian I guess what I want to know is if $f:K\rightarrow L$ is a simplicial map and $x\in L$. Does there exists a neighborhood $N$ of $x$ in $L$ such that $f^{-1}(N)$ is contractible?

For simplicial maps, are preimages of contractible sets contractible?

Or can we subdivide $K$ and $L$ so that this is true

take $f$ to be the constant map at $x$?

good morning everyone, what is this norm $\| \cdot \|^2_M$ in this paper and why eq(7) is equivalent to eq(6)

@CroCo $M$ looks like a matrix, and $\|\cdot \|_M^2$ looks like a norm coming from an $M$ induced inner product... I might be wrong though, who knows

the name "weighted least squares" seem to confirm this

@Jakobian what does "an matrix induced inner product" mean?

If the dot product is $vw^T$ then the inner product I'm talking about is something like $vMw^T$

well, if we use column vectors I should have written $v^Tw$ and $v^TMw$ I suppose

@Jakobian $M$ is indeed a matrix (i.e. a positive definite one).

yes, this is all necessary in order for $\langle v, w\rangle_M = v^TMw$ to be an inner product

I still don't understand how they claim eq(6) can be rewritten as an optimization problem.

you're the one reading the paper

I'm the one that has to guess everything

your guess is as good as mine. they can not writing it clearly I guess.

I do understand their objective function

From eq(4) I can see $J\Delta \dot{q} = \Delta v$ but I still don't understand the rule of $M$ in minimizing the norm.

@Jakobian, Or may be from eq(6) $\Delta \dot{q} - M^{-1}J^T(JM^{-1}J^T)^{-1}\Delta v=0$, right? but I don't see it clearly in eq(7).

How to prove that any rank 1 operator on l^2 space is of the form S^m(I-SS*) S*^n ?

uh, as phrased, that's not true. but it might help to get a handle on what those operators are. if you index the ell^2 space starting at 0, and let e_k denote the kth basis vector, that operator is uniquely determined by the fact that it sends e_n to e_m, and all of the other e_k to 0.

so it's those particular rank 1 operators that have that form. there are tons of rank 1 operators that don't have that form.

are you trying to identify the cstar algebra generated by S? or at least show that it contains all compact operators (and so in particular all finite rank operators)?

i had an answer on that. see math.stackexchange.com/questions/4759702/…

where that operator is called "E_mn". those operators play a role loosely akin to the 'standard basis' for the set of all square matrices of some fixed finite size. the analog of E_mn in the finite dimensional case is the matrix having a 1 in row m, column n and 0s everywhere else (and any square matrix is a finite linear combination of those in an obvious way).

and they multiply the way you expect, and physicists sometimes like to use formal sums involving them even when they don't converge in any useful sense. some people would say that they form a 'system of matrix units' because of the relations they enjoy.

just some jargon for potentially helping you google or whatever to find what the actual question was

@leslietownes Nice:)

I struggle with understanding one direction in the equivalence between uniform convergence and convergence in sup-norm. In particular, why does $$f_n\to f\text{ uniformly on }E\subset\mathbb R\implies \sup_{x\in E}|f_n(x)-f(x)|\to 0\text{ as }n\to\infty?\tag1$$The other direction is immediate since $\sup_{x\in E}|\ldots|$ is an upper bound to $|f_n(x)-f(x)|$ for every $x$ (and using squeeze theorem), however, the implication in $(1)$ I don't see. Is this clear to someone?

isn't the definition of uniform convergence?

or maybe you saw the definition with epsilons and delta

right, I saw the definition with epsilons and delta

maybe one has to argue by contradiction

so I assume you saw this definition: $(f_n)$ converges uniformly to $f$ on $E$ if $\forall \varepsilon>0$ exists $n_{\varepsilon}>0$ such that if $n>n_{\varepsilon}$ then $|f_n(x)-f(x)|<\varepsilon \, \forall x\in E$

yup

if this is true for all $x \in E$, then is it true in particular true for $\sup_{x\in E}|f_n(x)-f(x)|$

and if you substitute this expression you get the definition of $\lim_{n\to +\infty}(\dots)$

ok, it took me some minutes of thinking, but now that you put it this way it is actually a lot clearer :)

thanks!

@SineoftheTime Since $\varepsilon$ is an upper bound to $|f_n(x)-f(x)|$, it also has to bound $\sup_{x\in E}|f_n(x)-f(x)|$, right?

yes, probably it should be $\le$

how to I check solutions of trig equations by graphing?

@FedericoRuck Generally speaking "check the solutions by graphing" means "graph the relevant function or functions, and check to see if they intersect / cross the $x$-axis where you think they should". But without knowing more about what it is that you are doing, it is hard to say anything further than that.

like how to check the solutions to sin^x*cosx-3=sinx^2*cosx

Graph the relevant function(s), and check the points of intersection.

ok thanks

Hi 👋

sup

I need to go get groceries for the week. What should I make?

(Also, I need to feed my starter, so that I can make bread tomorrow.)

@XanderHenderson banana bread

With sugar, it's good

$(x+x')+xx''=\cos(t) , x(0)=x'(0)=\frac{\sqrt{2}}{\sqrt{5}}, x(\frac{\pi}{4})=? $

Pls help

*$(x+x')^2$

@SineoftheTime

I think you need to start with expanding $(x+x')^2$

and then try with a solution of the form $x= A \cos(t) + B \sin(t)$

Mmm

@Pizza it's not linear

@BinkyMcSquigglebottom source of the exercise?

@SineoftheTime I don't understand, why couldn't it be done like this?

because it's not a linear equation

@SineoftheTime From an exam

with the term $\cos t$, I don't have in mind a possible strategy

Im thinking

@XanderHenderson White pizza

doesn't it become linear if you substitute $y = x^2$?

yep

@binky Jakobian found the trick

So y(t)=x^2(t)

Nice !!!

but why is it linear now?

Ah yes because there are no more products and squares ...

@Jakobian How did you find the right substitutions pls

@BinkyMcSquigglebottom just played around with the problem for a little bit

Thanks guys

@SoumikMukherjee do you understand why our conversation was moved to trash? (yesterday, in cafe and tavern)

A closed subspace of a compact topological space is compact. This is a theorem in Gamelin and Greene's book, but quite late into the book, in the second part of the book on topology. The first part of the book deals with metric spaces and nowhere in the book can I find this result for metric spaces; are the proofs of this result for metric spaces any different than that for topological spaces?

@psie no

ok, I guessed so

You take closed $A\subseteq X$, you cover it with open $U_i$, then $U_i$ and $X\setminus A$ form a cover from which you obtain a finite subcover, which then constitutes a finite cover of $A$

pretty simple

ok 👍

hey all

got a question,

if I use inverse trigonometric functions to solve a trigonometric equation when finding the result I kano that I must limit the first set of solution to the range of the inverse trig fucntion shoudl I limit it based on the domain of the non inverse trig fucntion itself too/

so basically my question is if we are finding the set of solutions for a trig fucntion and we use inverse trig functions to solve

do we ave to respect to limits to fidn the values for solution +cycle

1. the value must be in the first cycle

2.the value must be in the range of the inverse trig fucnctio

here is the immage of the problem

does it have to e int range of the original tire function: so like sinx=a where is a content x=sin^-1(a) a must bet in the range of both sin and sin^-1?

or does the value of a has to be in the range of sin and the vale of x in the range os sin inverse?

or mayby a has to obey both the range of sin and sin inverse?

@SineoftheTime No idea, but moving a normal conversation to trash is kinda insulting

Ultimately decided on a beef roast (with mashed potatoes and brussel sprouts) for a couple of nights, and chicken-and-waffles for another couple of nights.

What about the days?

I don't understand the question.

@XanderHenderson Chicken and what?

@冥王Hades Waffles.

@XanderHenderson Waffles with chicken?

Waffles with everything!

I feel like the words I have written are quite clear...

@XanderHenderson They are, I just can't fathom how that combination makes sense.

written typed

@SoumikMukherjee There is no reason other than amWhy didn't think your conversation fits the room.

@user20458579510081670432 No, I chose the word I chose.

It is a correct word.

They words were written via typing.

Not type set?

Ok. I'll accept that definition upon reflection of the phrase "type writer." :-)

merriam-webster.com/dictionary/write

See definition 4.

Yes.

"Type set" would be wrong, and it has nothing to do with the typewriter technology... :/

I don't type. I use smoke signals.

☁️☁️☁️

🌧️🌧️

🚬

@Jakobian hehe, funny that the description says a place to socialize

@XanderHenderson You wrote 'for a couple of nights', so I jokingly asked what about daytime

@SoumikMukherjee they lied in the description.

@SoumikMukherjee it is, under the eye of the authority

💪👀💪

I think the lie was actually "All Octupi and Dolphins welcome!!!" since its not a very welcoming place

best to leave children to their toys

Big brother is watching you

are the following equivalent:

did you reverse $A$ and $B$ in 1.

oh yes.

then yes

2. should ideally have an "for all $a\in A$"

I see. Thanks!