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12:02 AM
@AlessandroCodenotti Happy to do so, demonic @Alessandro. Just a question of when :)
 
How can you express the following in $\mathbb{R}[X]$ this $\frac{1}{X-1}$? can I just chose $k_0 = \frac{1}{X-1}$ and all the other $k_i = 0$ ?
 
$k_n$ needs to be in $\Bbb R$
 
oh well yes I didn't tell you the whole story to be honest.
so I have a vector of (x+1,1\x-1)
out of R[x]^2
but I guess that doesn't change the fact that $k_n$ is in $\mathbb{R}$ so how do you express that component of the vector?
 
you can't
 
Are you suggesting that $\dfrac1{X-1}$ is a polynomial?
 
12:17 AM
i am not suggesting i am reading this out of my homework paper :D
 
Something's wrong. You can write a formal power series. But that's $\Bbb R[[X]]$.
 
I don't know. i have the two vectors, one of them is $(X+1, \frac{1}{X-1})$ and another one and the rest is "show that they are linear independent in the vector space of $\mathbb{R}(X)$ which is $\mathbb{R}(X)^2$
I am also confused by the question.
 
that's R(X)
not R[X]
 
Is there a difference_
 
sure
$\Bbb R(X) = \left\{ \dfrac{f}{g} \mid f, g \in \Bbb R[X], g \ne 0 \right\}$
 
12:24 AM
Oh ok, what is this new space called? so i can google it :)
 
the rational functions
 
Thank you ))
I see aha nice! .. you actually just need to read the definition to understand that
 
12:48 AM
Take a graph in the plane, and do the $\partial B_\epsilon$ operation (thicken it and take the boundary of the thickening)
How many connected components does that have?
Is it the number of faces plus the number of connected components (of the graph)?
(and $V-E+F=1+C$ where $C$ is the number of connected components)
 
1:05 AM
DogAteMy: You thickening on both sides or just on one side (assuming oriented)?
 
Both
It is $F+C$, I was just thinking out loud
 
So if your graph is a circle, the boundary has two components.
 
Oh, one of your remarks disappeared.
 
And if it's a discrete graph then it's as many circles as points
$F+C=2C+E-V$
 
1:07 AM
If I take a figure 8 graph, then I get 3 components, even though the graph is connected.
 
Yes - two faces, one component
$F=2$, $C=1$
 
Anyhow, I need to get going, but this is definitely Euler characteristic based.
Bye for now.
 
Yeah
So I amend my earlier conjecture
Let $G$ be a graph, and $H$ be a subgraph with the same vertices. Let $E_H$ be the number of edges and $C_H$ be the number of connected components
Then $|\sum_H(-2)^{2C_H+E_H}|$ should be a power of $2$
Conjecture^
("Spanning subgraph")
Hm… my old conjecture seems to be working and I'm not sure why?
No it isn't
 
2:02 AM
Are hausdorff spaces preserved under continuous functions?
 
what do you mean?
 
The image of a Hausdorff space is not necessarily Hausdorff, if that's what you mean
 
Yep that is what I meant, could you give an example please?
 
Let $S$ be the Sierpiński space, that is, $\{0,1\}$ where the open sets are $\emptyset$, $\{1\}$, and $\{0,1\}$
 
can $\ln(x)\ln(y)$ ever equal $1$??
 
2:08 AM
Define $f(x):[0,1]\to S$ by $f(x)=\begin{cases}0,&x=0\\1,&x>0\end{cases}$
This is continuous
(In fact, it is a quotient map - $S$ is homeomorphic to $[0,1]/(0,1]$)
 
let $x$ and $y$ be $e$? @Ultradark
 
(where by that I mean $[0,1]/{\sim}$ where $\sim$ is the equivalence relation defined by $x\sim y$ iff either $x=y=0$ or $x,y>0$)
 
@Shiranai yeah but besides that??
 
Thanks! @AkivaWeinberger
@Ultradark in general let $x=e^c$ and $y=e^{1/c}$
 
Is "shiranai" Japanese for "I don't know"?
 
2:12 AM
Yep
 
(or "I'm not familiar")
 
really love related phrases, eg socrates', ignoramus et ignorabimus, etc
 
(I'm not entirely sure of the difference between "shiranai" and "wakaranai")
 
wakaranai is more like "I don't get it" or "I don't understand", while shiranai is simply that you don't know (or aren't familiar with, depending on the context)
 
I've been told that wakaranai can mean "I don't know" as well
despite wakaru meaning "understand"
 
2:16 AM
hmmm maybe in some contexts, I don't think that it is a really common or natural way to use it though
shirimasen/wakarimasen are more formal ways of saying shiranai/wakaranai btw
also, whats the motivation between the function above? I get it works but how did you thought up that?
 
Just something I knew for a while that $[0,1]/(0,1]$ is the Sierpiński space
In fact all non-Hausdorff spaces are quotients of Hausdorff spaces, if I remember right
Ooh, here's a neat one
Define $f$ from $\Bbb Q$ to the natural numbers with the cofinite topology
by $f(x)$ equals the denominator of $x$
(Integers including $0$ are said to have denominator $1$)
(Write $x$ in lowest terms first)
And this is a quotient map as well (that is, $\Bbb Q/{\sim}$, where $x\sim y$ iff they have the same denominator in lowest form, is the naturals with the cofinite topology)
(Naturals not including zero. Positive integers)
 
Damn that is insane
got a reference book or did you thought that up?
 
Meow
Hate this... I wanted to prove by contradiction... The claim is that we can't have a filter on $\omega$ that is countably closed.
So what I'm thinking is that there exists an uncountable sequence and if there can't be a filter then it's an ideal but that also doesn't make sense either because ideals are small subsets of X. Bddkmabs
 
Whatever you can prove without contradiction, you can prove by contradiction. Just assume the negation of the consequence, prove that it is true, contradiction. In general though, if you can prove by contradiction, you don't necessarily can prove without contradiction.
 
Maybe $\omega$ is countably closed if it's an ideal... Since ideals are small subsets of X and countably closed may possibly mean that there aren't that many elements
 
2:30 AM
so you should be happy that it is without contradiction
 
¯_(ツ)_/¯
Maybe I can prove it directly since $\omega$ is countably closed (which by the way $\omega$ is the set of all natural numbers)...it can't be a filter because filters are usually large and ideals are small. Hmm
Dahhhhhhh
 
2:55 AM
@Shiranai I thought it up, but not on the spot
Like a few years ago I was trying to figure out how to visualize the cofinite topology, and non-Hausdorff spaces in general
and this is one of the things I realized eventually
and it stuck in my mind
 
3:09 AM
Nice, thanks for answering, cya
 
 
5 hours later…
8:35 AM
@usukidoll You can have one, as long as it's principal
For a nonprincipal filter $F$ you know that $\omega\setminus\{n\}$ is in $F$ since every nonprincipal filter contains the cofinite one, but if $F$ were countably closed their intersection, which is empty, would be in $F$, a contradiction
The same argument shows that no nonprincipal filter on $\kappa$ is closed under intersections of size $\kappa$
(and whether there exist a $\kappa$ with an $F$ closed under all intersections of size less than $\kappa$ is the statement that $\kappa$ is a measurable cardinal)
 
8:52 AM
Hi people, any fluid dynamics expert here?
I am a freshman wanting to publish papers on fluid dynamics.
My college starts from multivariable calculus , can anyone tell what are the other prerequisites for studying the subject?
 
 
1 hour later…
9:59 AM
hi all
Suppose that X, Y and Z are independent random variables, each uniformly distributed on (0, 1). How can you prove that (XY )^Z is also uniformly distributed on (0, 1).
 
10:24 AM
by performing a triple integral?
 
@LeakyNun Sounds interesting! But which triple integral?
 
what does it mean that a random variable is uniformly distributed on (0,1)?
 
I don't quite understand the question. Which part is unclear?
 
no I'm asking you whether you know the definition
if so, then you can just use the definition
 
oh I do.. the pdf is just 1 in that interval
but I definitely need help if you want to walk me through it
 
10:28 AM
now what is the pdf?
 
in what sense? I mean the integral of f(x) from -infinity to +infinity has to equal 1
and it's always non-negative
 
how would you find the pdf of (XY)^Z?
 
I don't know!
that's the whole question
 
what does (XY)^Z mean?
 
I don't think I have a clever answer to that
 
10:42 AM
then just give a regular answer
 
it is X*Y all raised to the power of Z where the X,Y , Z are the random variables defined above
 
well I know how to do that for real numbers
but what does it mean for random variables?
actually what's your background? i might adjust my answer based on your background
 
11:01 AM
If $F(D) = \Pi_{i = 1}^n (D - m_i) $, and Differential equation is $F(D) y = r(x) $ , then prove that, the particular solution of the differential equation would be $$ y = \sum_{i = i}^{n} A_i e^{m_i x} \int e^{-m_i x} r(x) dx$$
This problem seems wrong to me, is it?
Oh, sorry, it is not. But how one can prove this
 
11:21 AM
After I sit down and I compute the character table for $D_4$:
$$\begin{bmatrix}D_4&(1)&(r)&(r^2)&(sr)&(sr^2)\\\tau_1&1&1&1&1&1\\\tau_2&1&1&1&-1&-1\\\tau_3&1&-1&1&1&-1\\\tau_4&1&-1&1&-1&1\\\tau_5&2&0&-2&0&0\end{bmatrix}$$
I know that $\tau_5$ is an irrep of degree $2$, so $\rho(g)$ is a $2$ by $2$ matrix for each $g\in D_4$, and I know its trace, and I know since these matrices are finite order, they are diagonalisable, and their eigenvalues are roots of unity (k^th roots if it's order k), but since $D_4$ isn't abelian, I can't simultaneously diagonalise these matrices, so how then do I pick matrix representatives for $\tau_5$?
 
 
1 hour later…
12:34 PM
Hm. What's the simplest argument to show that if every element of a field $F$ satisfies $x^p = x$ for some prime $p$, then $F \cong \Bbb F_p$?
 
take $F = \Bbb F_2$ and $p=5$
 
Huh
Good point
 
take $F = \Bbb F_\ell$ and $p \equiv \ell \pmod{\ell-1}$
 
I was wondering if it's true; all I can say is basically that $F$ has to have order at most $p$
(Fundamental theorem of algebra)
 
I guess
 
12:39 PM
c is a column vector. What operation I can do on c to get a diagonal matrix with diagonal entries are that of c in same order.
 
diag(c)
 
@LeakyNun lol
 
not code. A matrix operation
 
it's not code it's a function
 
its not defined in math
 
12:41 PM
sure it is
ask any maths professor, they'll know what it means
also I just defined it so consider it defined
 
Oh I havent came across
thanks
 
I think leaky is trolling
in a friendly way though, because he's a nice guy
 
coughs loudly
 
thats why I politely ignoring him.
 
it's like those "find the next term in the sequence 1, 2, 4, 8, 16"
it's not maths
 
12:45 PM
On a glance, I think it's impossible to do what you want though, just using matrix multiplication, but I may be being dumb
 
and for the record I'm not trolling
 
If $k$ is char $p>0$ and $G$ is a $p$-group
Then $k[G]$ is indecomposable
This apparently implies instantly that $k[G]$ is a projective cover of the trivial $k[G]$-module
Don't they want irreducible, rather than indecomposable?
 
@tigre not necessarily multiplication. you can use any combinations of addition, multiplication etc...
 
Being an essential epi doesn't require us check only projective modules, where indecomposable probably rules them out since they can't be direct summands of an indecomposable guy
 
0
Q: How to convert a column vector into a diagonal matrix with same entries in same order?

Rajesh DachirajuAssume $c$ is a column vector. What mathematical operation or expression can produce a diagonal matrix with entries of that of $c$ in same order. Does such an expression exist? Basically I know that the sum of entries of $c$ is zero. I want to write it in an matrix expression/equation. I am hopi...

 
12:50 PM
nail the top entry to a wall and kick the bottom entry
 
@tigre look at $p=2$, $G=C_2$
be sure not to kick it towards the wall
 
@ÍgjøgnumMeg advice unclear, I now have a hole in my wall
 
Oh well that would have been good to know lol
 
cOmMon SeNsE
 
12:51 PM
@LeakyNun I am now looking at it
It has a submodule spanned by $\{1+\sigma\}$
and no complement
Right, it's reducible but indecomposable
Well, I think I'm not clear on the argument that makes it immediate that indecomposable implies $k[G]$ is a projective cover of the trivial $k[G]$-module (although for irreducible there seems to be an argument)
 
Question: let $M_k$ be the space of weight $k$ entire modular forms and $S_k$ the subspace of cusp forms. If $f \in M_k\setminus S_k$.. does that mean $f$ vanishes at NO cusp (by which I mean $s \in \Bbb Q \cup \lbrace \infty\rbrace$) or just that there are cusps at which $f$ does not vanish, or are these equivalent, since there's only one equivalence class of cusps for the action of $\operatorname{SL}_2(\Bbb Z)$ on the upper half plane?
I mean if $f(s) = 0$ for some $s \in \Bbb Q\cup \lbrace \infty \rbrace$ then there is a $\gamma \in \operatorname{SL}_2(\Bbb Z)$ such that $\gamma \infty = s$ so that $f(s) = j_\gamma(\infty)^kf(\infty) = 0$
and in fact $f(\gamma z) = j_\gamma(z)^kf(z)$ for all $\gamma \in \operatorname{SL}_2(\Bbb Z)$ so this is gonna hold for whichever $\gamma$ I pick, since I can just move them around with the action of SL_2(Z)
 
Is thre any name for this column vector with all entries equal to same number or equal to 1? (in linear algebra)
[1 1 1 1 1 1]
 
Not that I know of
 
Oh I just found is called J
 
$\underline{\mathbf{1}}$
 
1:02 PM
In mathematics, a matrix of ones or all-ones matrix is a matrix where every element is equal to one. Examples of standard notation are given below: J 2 = ( 1 1 1 1 )...
 
There are typically only names for things that require them (because they arise frequently enough, and/or are complicated enough to need a name)
I should make a wiki page called "Matrix of 87.4's"
And they are just matrices of various sizes, all of whose entries are 87.4
 
actually what I wrote is nonsense
I think
 
@tigre because projective indecomposables correspond to irreducibles
 
@LeakyNun You mean projective indecomposables correspond bijectively to irreps (which they projectively cover)?
 
yeah
 
1:13 PM
Hmm, how to prove this
I think I believe it
Projective envelopes are unique up to iso, so I just have to check that if the module I'm covering is irreducible, the envelope is indecomposable
 
2:09 PM
why is for a polynom $f(X) =0 $ then follows that his degree is $ -\infty $
 
Oh ok thanks just to clear up is a polynom $f(X) = k$ constant considered to be zero grade?
 
es heist "degree" auf Englisch
Konstantpolynom haben 0 Grad
 
Aha okay. danke )
 
3:12 PM
When people say $p(e)$ is a probability density, what is $e$ here? Is it a random variable, a variable, or something else? In an ML paper, I am reading, the authors say: "let $e$ be a random variable with probability density p(e)". Now, usually, random variables are written in upper case, so this is already confusing. Then they say "let $t(\theta, e)$ be a deterministic function".
Furthermore, sometimes, this notation and terminology confuse me even more, because people sometimes use p(E=e) or P(E=e) or P(E) or P(e) or p(E). What the hell are the differences? I've already read several questions on Stats, but it is still confusing
For example, in p(E=e), I suppose that E is a random variable and e is the actual observation or realization of the random variable.
ML people are so ignorant when it comes to notation, terminology and actual probability theory
In this same paper, the author says "suppose further that the marginal probability density of w, $p(w \mid \theta)$".
$p(w \mid \theta)$ should be a conditional prob. density, not a marginal
Of course, I know that, in this case, $\theta$ is considered a parameter, rather than a random variable, but how stupid is this?
 
 
2 hours later…
4:53 PM
Can someone tell me which calculus rule is being used here?
I understand that the derivative is first exchanged with the expectation. So, in the first line, we have the derivative of an expectation, in the second line, we have the expectation of derivatives.
I understand that this can be done, under certain conditions, which I assume are satisfied in this case
But the partial derivative of f(w, theta) with respect to theta, should just be the partial derivative of f with respect to w multiplied by the partial derivative of w with respect to theta. No?
 
5:12 PM
any body pliz help or any idea how to solve
 
5:30 PM
@Messififa use integration by parts and induction
 
5:43 PM
So I just conjectured something in an answer to my own question that turns out is false (forwards implication holds, backwards does not). Does etiquette say I should delete it or not?
 
A slick way to do that integral is to multiply by $t^n/n!$ and sum over nonnegative n
The resulting summed integral is elementary. One then differentiates n times w/r/t t and evaluates at t=0
 
6:36 PM
Socialism for the askers, hooray!
 
7:16 PM
@MatsGranvik I only just realised what you meant with that lol
 
@ÍgjøgnumMeg Great for me at least. If privileges continue to go hand in hand with reputation points, then I am going to get all kinds of undue influence that I can misuse.
Or maybe not so much on this site. The next reputation privilege is probably several points ahead for me.
 
7:50 PM
haha
Question: why might $e^{\pi i n^2 z + \frac{1}{2}\log(z) - \frac{\pi i}{4}} = e^{\pi i n^2\frac{-1}{z}}$
$n \in \Bbb Z$
 
8:17 PM
@ÍgjøgnumMeg where did you get this from
 
@Leaky I just multiplied $\theta(z)$ by $\sqrt{\frac{z}{i}}$ and naively ripped the root apart to show that $\sqrt{\frac{z}{i}}\theta(z) = \theta\left(\frac{-1}{z} \right)$ rofl
I think there's probably a better way
 
last time I checked $\theta$ isn't $e^{\pi in^2z}$
 
yeah it's a sum over those
 
so the summands don't need to be individually equal
 
sure
I was just hoping they were because then I wouldn't need to do anything else :D
 
8:20 PM
what
 
hmm so where does the factor of $\sqrt{\frac{z}{i}}$ come from
 
so you need to show that $\sqrt{\frac zi} \theta(z) = \theta(\frac{-1}z)$?
 
yeaaah
and $\sqrt{\frac{z}{i}} = e^{\frac{1}{2}\Log\frac{z}{i}}$
which is why I just dumped it into the summand
s
 
I'll classify your problem as an XY problem :)
 
I mean, that's how I interpreted the hint that was given
 
8:25 PM
what's the hint?
 
well it says "where $\sqrt{\frac{z}{i}}$ is defined for the principal branch of the logarithm"
guess I misinterpreted it
it's not really a hint lol
 
I don't think that's a hint
do you want a hint?
 
Yes please :P
 
reverse the equality
 
errrm
don't look
hahaha
right
$\theta(z) = \left(\frac{z}{i} \right)^{-1/2}\theta\left( \frac{-1}{z}\right)$ is what you're suggesting I look at?
 
8:32 PM
Why the requirement that $x$ and $y$ also belong to the same sets? (second last sentence in attached picture)
 
@ÍgjøgnumMeg are you sure it's true? if $z$ is real then the LHS is non-real
show me the sauce
 
(note $z \in \mathfrak{h}$ btw)
probably worth mentioning ;)
 
ah die Definitionen sind nicht gliech
 
wessen Definitionen? :D
 
mein Thetafunktion ist $\theta(t) = \sum_{n \in \Bbb Z} \exp(-\pi n^2 t)$
 
8:39 PM
Ah okey
Meine* btw
 
See, I saw theta and thought you meant Heaviside
 
danke
@Semiclassical :c
 
bitte :P
this is a cool exercise anyway
 
@ÍgjøgnumMeg yeah you need some dank Fourier stuff
 
You see theta(t) as the Heaviside step function way more in physics than the Jacobi theta function
 
8:41 PM
I remember vaguely from the course on L-Functions that you can just hit it with Poisson summation
 
heresy
 
and somehow get somthing out of that
 
@LeakyNun Could there be sets that are equal but belong to different sets?
 
correct @ÍgjøgnumMeg
@schn then what do you mean by equal
 
great
 
8:42 PM
@LeakyNun Could there be sets that contain the same elements, but that belong to different sets?
 
under that axiomatization, yes
(sets don't exist on their own)
 
Could you show this with an example?
 
no
well I guess if you don't have any other axioms
 
ergh now I need to remember if I can switch sum and integral
 
then you can just let a b c d be all the sets that exist, and then have a in b, a in c, c in d be the only relations
then b and c contain the same elements but belong to different sets
 
8:47 PM
Then b and c wouldn't be equal according to that axiom?
 
right
 
Is there some other axiom that says they are equal nevertheless?
 
depends on your axioms
 
In basic set theory one learns if two sets contain the same elements they are equal, no? Which axiom is used here?
 
the first one in the website i sent you
 
8:51 PM
Which one where you referring to with the example of the sets a, b, c and d?
 
the one you sent me
 
9:05 PM
@LeakyNun What is the difference between the one you sent me and the one I sent you? They both seem to specify that x and y also belong to the same sets.
 
9:19 PM
Is the quotient rule EVER useful?
 
@LeakyNun Is the one I sent you familiar to you?
 
well I've seen something like that before
@anakhro the answer is fries
 
NO NOT FRIES.
 
@LeakyNun Does it connect to the axiom of extensionality somehow?
 
read the wiki page
 
9:22 PM
Well if it isn't Leaky "Go google it" Nun.
 
Have any of you read "Road to Reality" by Penrose?
I wonder if you guys have any resources for better understanding one-forms and p-forms? I guess I have some understanding of it, but I would like to improve my geometrical intuition of it.
Also, any resources on Clifford and Grasmann algebras? Those sections in Road to Reality were a little confusing
 
@anakhro my calculus lecturer used to explicitly point out that he wasn't going to teach the quotient rule because it's made up
 
9:38 PM
@ÍgjøgnumMeg I have come to a similar conclusion after watching kids differentiate $f/g$ with the quotient rule where $g$ is a constant.
 
hahaha
 
@TanMath like differential forms?
 
yes
 
What in particular are you struggling with about them?
 
Their geometrical intuition. like what they mean geometrically/physically. I can "see" a vector and a vector field, but it is harder to do so with a one-form and p-forms
 
9:40 PM
Yeah, so it's quite hard to visualize them, but it's not important to be able to visualize them.
What is important is how you use them.
There are some tricks where you will visualize the forms in some particular way, but often it's not helpful.
 
@anakhro I'm telling Ted what you just said
 
@ÍgjøgnumMeg I think he'd agree with me if he heard me out about it.
 
haha maybe idk
 
@TanMath for 1-forms, you can view them by their kernel. This is the easy way.
The kernel of a non-trivial 1-form will give a hyperplane field in this way.
i.e. the dimension sent to zero at each point is the vector normal to the hyperplane defined at that point.
Exercise: try visualizing $\alpha = dz - y\,dx$ this way, in $\mathbb R^3$.
 
What do you mean by the kernel of a one-form?
 
9:44 PM
Well at each point $p$, $\alpha_p$ is a linear map $\alpha_p\colon T_pM\to\mathbb R$, right?
So it has a kernel as a linear map.
And this kernel, if $\alpha_p$ is not the zero map, is $\dim M - 1$ dimensional.
i.e. a hyperplane.
 
I have heard that for a one-form, since the scalar product of a one-form with a vector is a scalar (that's the definition after all), the one-form can be represented by a plate that the vector crosses to result in a scalar... is this similar to the interpretation you are describing here?
 
Are you describing the piercing one? That is, that the 1-form measures how many "level surfaces" the vector punctures/crosses/intersects?
Like this?
This is not what I am describing exactly.
This one is often used in physics books, but is problematic if you take it too far.
I had this example in some notes I wrote, but I took it out explicitly because I am now of the belief it is kind of harmful to focus on drawing a picture of differential forms, rather than just viewing how they are used.
 
@anakhro yes this is what I am thinking of. Where is this from BTW?
 
This is from the wikipedia page.
@TanMath Write out the kernel of $\alpha_p$ for $\alpha = dz - y\,dx$.
 
@anakhro sorry I am not following along. I don't know what you mean by kernel.
Oh wait by kernel you mean a null space?
 
9:57 PM
Null space?
Yes.
Null space.
Though that is less commonly used for maps, and more frequently used for matrices.
 
Well I am not used to maps, except as the concept that almost everything is a map lol
 
Well in any case, try computing the null space of $\alpha_p$. :)
 
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