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12:15 AM
Was studying recently the usecase of stars&bars for counting partition sets of integers, unconsidering duplications (permutations of same sets)
for instance 6={1,2,3}={3,2,1}, i don't know any concise way to eliminate such dups
 
@Abra001 there's only recursive formulas for partition
the most efficient should be the one given by the pentagonal number theorem due to euler
 
I can generate a recursive notation, but i wondered if S&B could be helpful somewhow
pentagonal number ?
 
 
1 hour later…
1:35 AM
Puzzle my friend gave me: $x$ and $y$ are chosen uniformly from $(0,1)$. What are the odds that $x/y$, rounded to the nearest integer, is even?
(Note that it's rounded, not floored.)
 
that's a catch-mind
 
2:21 AM
@AkivaWeinberger Exchange x and y for psychological comfort. For $y/x$ to round to $2n$, one needs $2n-0.5 \le y/x < 2n+0.5$, i.e. $(2n-0.5)x \le y < (2n+0.5)x$. For $n=0$, one finds a triangle of length $0.25$. For other $n$, the triangle has the base on the top of the square, with starting point $1/(2n+0.5)$ and ending point $1/(2n-0.5)$, so the area is $\displaystyle \frac 1 2 \left( \frac 1 {2n-0.5} - \frac 1 {2n+0.5} \right) = \frac 1 {4n-1} - \frac 1 {4n+1}$
So, the sum becomes $\displaystyle 0.25 + \sum_{n=1}^\infty \left( \frac 1{4n-1} - \frac 1 {4n+1} \right)$
$\displaystyle 0.25 + \sum_{n=1}^\infty \int_0^1 (x^{4n-2} - x^{4n}) \mathrm dx$
$\displaystyle 0.25 + \int_0^1 \sum_{n=1}^\infty (x^{4n-2} - x^{4n}) \mathrm dx$ [fight me]
$\displaystyle 0.25 + \int_0^1 \frac {x^2 - x^4} {1 - x^4} \mathrm dx$
$\displaystyle 0.25 + \int_0^1 \frac {x^2} {1 + x^2} \mathrm dx$
$\displaystyle 0.25 + \int_0^1 \left( 1 - \frac 1 {1 + x^2} \right) \mathrm dx$
$\displaystyle 0.25 + 1 - \arctan 1$
$1.25 - \dfrac \pi 4$
0.4646018366
 
11 hours ago, by Secret
@LeakyNun Other than idempotents, what else will satisfy $f(x+y)=f(x)+f(y) \land f(0)\neq 0$?
 
2:40 AM
@LeakyNun can i ask where 1/4 came from, is it 1/2 squared ?
 
@Abra001 it's the triangle formed by the lines y=0x, y=0.5x, and the unit square
 
Zee
I don’t buy your proof @LeakyNun
 
@Zee are you drunk
 
Zee
I wish
 
I don't buy your unbuying of my proof
 
Zee
2:42 AM
Impressive...
 
well you don't point out where I'm wrong, so I'm just being equally unhelpful
 
Zee
this Last comment took away the cool points you gained
 
@LeakyNun something about geometric destribution ? well i couldn't get it visualized to a proper geometric area,nice done.
 
@Abra001 right, geometric interpretation of probability
@Zee I care about a fruitful mathematical discussion rather than cool points
 
Zee
Both are important
Ex: if you have lots of cool points then you don’t have to worry a lot about being part of the math community since you already get laid, that allows you to take more risks in math research, which has more rewards sometimes, see Grothendieck for example
 
2:46 AM
oh boy.
 
@LeakyNun Correct
 
@AkivaWeinberger nice
 
Zee
I mean it’s clearly 50% without doing all that computations
There is two options , either am wrong or right , so it’s 50/50
//kill me
//help
Where is the robot? :(
//awaken
//wakeup
//hello
//turnon
 
@MikeMiller Hey, I can't seem to get a straight answer to this: are all orientable 3-manifolds parallelizable/spin or is this only true for compact ones?
 
@bolbteppa I'm trying to think of a version of "don't feed the trolls" which is appropriate for dealing with crackpots
 
3:12 AM
I don't see how you cannot just ignore them?
 
@0celo7 I started writing down an argument and then felt like I read someone else write down this argument recently.
 
Oh, I think that's usually the right advice. I just wanted some different way of saying it, since crackpots usually are a different species than trolls
 
crackpots, however, are potentially more damaging than trolls in the long term when new users are involved in the conversation:
in The h Bar, 2 days ago, by Secret
@DanielSank the issue here is that new users get mislead, and new users don't necessary know what to do. That's one example of a loophole that the ignore function cannot handle
in The h Bar, 2 days ago, by DanielSank
It will be interesting to see if we need to amend policy to deal with "frequent stipulation of scientific misinformation".
in The h Bar, 2 days ago, by DanielSank
@Secret ah yes. I just had the same thought.
 
I suppose the difference is that trolls linger so long as they have a source of entertainment
whereas crackpots tend to nest down in a given spot
 
3:18 AM
SE don't have a shadowban option to restrict crackpots
 
@MikeMiller Thanks.
 
Sure
 
Yeah, trolls will be sustained as long they have entertainment, which is why ignore is such an effective option (unless the trolls happened to hate ignore, in which case they can potentially escalate and some other option is needed)
 
@LeakyNun This is the picture I made when I was solving it
 
Shadowbanning still remain one of the most effective countermeasure against a lot of disruptive behaviour in cyberspace as while it is easy to find out one is shadow banned, so far there exists no countermeasure nor loophole from them to escape unless they are master hackers of sorts
 
3:22 AM
@AkivaWeinberger same, just without the orange part
 
I imagine your mental picture looked similar @LeakyNun (EDIT: sniped)
 
and with x and y inverted
@Secret what being idempotent?
f(0) being idempotent? f being idempotent?
 
@LeakyNun An idempotent element is defined to be xx=x (an example being the projection operator). So in this case, since f(x+y)=f(x)+f(y). It follows it also holds for x=y=0 (since every mooned has an identity), thus we can substitute f(0)=f(0+0)=f(0)+f(0). Hence f(0) is an idempotent (which is common in monoids and semigroups)
 
Please give an example of a centralizer of a subgroup in a group that is not normal.
 
What's not normal, the centralizer or the subgroup?
 
3:26 AM
centralizer
 
@Secret then what are you asking
 
that's the only way I can think of that can prevent f(0) to become zero while having f obeying the morphism property
 
you proved from the axioms that f(0) is idempotent
 
@LeakyNun Is that the only case, or there are more exotic cases of f that I don't aware of that also satisfy the equation?
 
@Silent Remind me the notation? If $G$ is the group and $A\subseteq G$ is the subgroup, how do you write the centralizer of $A$?
$C_G(A)$?
 
3:27 AM
@Silent $G=S_3$, $H=\{(1 2)\}$
1 min ago, by Leaky Nun
you proved from the axioms that f(0) is idempotent
trust your logic
 
@AkivaWeinberger Well $A$ need not be subgroup of $G$, just a subset is enough.
 
ok I see, so I think that will include potentially pathological f s then
 
@Silent Right
Kenny's example works
 
@LeakyNun is H just a set or cyclic group?
kenny?
 
@Silent makes no difference
 
3:29 AM
ok!
 
just take $H = \{e, (12)\}$ if that makes you happier
everything commutes with the identity
@Silent ego sum kenny
 
oh yes.
 
@Silent Leaky Nun is an anagram of Kenny Lau (rearrange the letters)
 
Speaking about anagrams, I wonder if there exists algorithm to unscramble a given anagram?
I will imagine the algorithm need somehow have a dictionary of phrases
 
There wouldn't be a unique output
 
3:31 AM
Wow!
 
ah right...
 
My name is an anagram of Kiwi Bear Avenger. I don't know what to do with this information.
 
I don't know what anagram I can make with my username, btw
 
Could also be Bear Wiki Avenger
 
3:35 AM
Or Bear Avenger Wiki, the fan-made database for the classic and world-famous TV show Bear Avenger
 
lol
 
headcanon
 
Looks like my most scrambled anagram is a vitamin C supplement brand
Ester C
 
You need more letters
 
dunno if you know this game
but this is genius
 
3:39 AM
Or maybe just take the first several characters of its hash. When people ask you what it's a hash of, you can tell them it's a Secret
 
5ebe2294ecd0e0f08eab7690d2a6ee69
ooo -> bee
Meanwhile, the following (which is my full internet name) does not solve
 
@0celo7 yeah, i've watched quite a few videos
not sure about that commentator in particular
 
@Semiclassical I just found him randomly
 
3:58 AM
Hi
I will sit on IMO
But I have a confusion
In geometry if I use a formula,do I have to prove it ?Or,just write the formulas?
 
4:13 AM
Help with real analysis
Theorem: If $\{G_1,G_2,...\}$ is a countable collection of dense open sets than
the intersention $\cap_i^\infty G_i$ is non empty
But I can construct $G_i=(i,i+1)$, which are dense, open and countable whose intersection will be empty set.
What am I doing wrong here.
 
@PiyushDivyanakar How is $(i,i+1)$ dense o.O
 
$(i,i+1)$ is not dense in $\mathbb{R}$...
 
Yeah I figured. On the real line only dense and open set is $\
$\mathbb{R}$ itself.
 
no
take $\Bbb R\setminus \{pt\}$
 
Or, really, $\mathbb{R} \setminus F$, where $F$ is any finite set
You can actually do a little better than that, i.e. $\mathbb{R} \setminus \mathbb{Z}$
but describing the kinds of sets that you can remove becomes a little more complicated
a uniformly locally finite set might do the job...
of course, that isn't really the intention of the question, and I am probably not really helping at this point :\
 
4:26 AM
@XanderHenderson your description help me a lot. Multiple examples are good for understanding. Can you help with the following construction?
In order to prove this theorem I need to construct a nested sequence of subsets $I_n$of $G_n$, so that i can use the nested interval property. I am not sure how I should create this.
Hints?
 
Zee
4:38 AM
@PiyushDivyanakar Just use Baire catagory theorem
 
I have to prove that
 
Zee
Just cite it
If your working in R show the closure eats up the compliment , which is the union of the thin sets by demorgan laws
 
5:02 AM
On my way home I was taking a nap in the transport. I then have an idea or a dream (I cannot tell which) which goes like this:
 
Zee
Was it a wet dream ?
 
Since $(a,b)$ is not dense in $\Bbb{R}$ it follows that I can find some point $p \in \Bbb{R}$ that is not in $(a,b)$ or its limit point. The math chat then fade to blurry and then zipped to a white void which talks about how proving something about real numbers can fail if the natural numbers does not complete
I will figure out whether it make sense later
 
@PiyushDivyanakar $I_n$ has to be a closed interval? $G_n$ is an open subset of $\mathbb R$?
Every open subset $G_n$ of $R$ contains some open interval $(a_n,b_n)$. If $s_n=(a_n+b_n)/2$ is the center and $a_n=s_n-d$, $b_n=s+n+d$ then you have $(s_n-d,s_n+d) \supseteq [s_n-d/2,s_n+d/2] \supseteq (s_n-d/2,s_n+d/2)$.
 
One reason I like general topology is because it avoids the problem of dealing with inequalities directly, since my brain go BSOD if I see more than three inequalities in one line
and real analysis is riddled with inequalities
 
5:19 AM
@MartinSleziak thanks. I am understanding it as, we take progressively take middle half of $(a_n,B_n)$
But how do we gaurantee that $I_{n+1} \subseteq G_{n+1}$ when $I_n \subseteq G_n$.
 
That's quite difficult to answer since I have no idea what is the relation between $G_n$ and $G_{n+1}$.
We just know that they are dense open sets with non-empty intersection?
@PiyushDivyanakar If $G_{n+1}$ is dense, then $G_{n+1} \cap \operatorname{Int} G_n$ is a non-empty open set and it contains a non-empty interval.
Are you trying to prove BCT here?
 
@MartinSleziak Yes
 
So you just want to show that they have a non-empty intersection and to this end you want to have a sequence of nested intervals, right?
I see that I should write $G_{n+1} \cap \operatorname{Int} I_n$ rather than $G_{n+1} \cap \operatorname{Int} G_n$, since you want the close intervals to be nested.
Since you are using Cantor's intersection theorem I assume you have also some requirement on diameter of the intervals $I_n$. (You do not want to allow them to be infinite intervals, but once you choose some endpoints this is no longer a problem..)
In any case, I'll have to go. Baire category theorem is useful, so it is good to spend some time thinking about it.
If you are working on real analysis and topology, you might want to peek into general topology chatroom or calculus and analysis chatroom occasionally.
And there is also this calculus related room: Simply Beautiful Art's realm of calculus and analysis.
 
5:49 AM
@MartinSleziak
thanks
 
Out of curiosity: If I have a harmonic function phi satisfying Laplace equation and subsequently have another function f which is a function of phi such that f(phi). Is there anything interesting we can say about f?
 
Zee
6:18 AM
No
Let f=0
 
7:07 AM
Is $\mathbb{R}$ a finite field extension of something?
 
can't be the rationals, as they are countable. So any finite extension has to be something with cardinality continuum I guess...
 
Okay so I Googled this and the answer is no, for an interesting reason
 
13
Q: Is there a proper subfield $K\subset \mathbb R$ such that $[\mathbb R:K]$ is finite?

YZhouIs there a proper subfield $K\subset \mathbb R$ such that $[\mathbb R:K]$ is finite? Here $[\mathbb R:K]$ means the dimension of $\mathbb R$ as a $K$-vector space. What I have tried: If we can find a finite subgroup $G\subset Gal (\mathbb C/\mathbb Q)$ such that $G$ contains the complex conjug...

 
The Artin-Schreier theorem says that if $E$ is an algebraically closed field and $F$ is some subfield such that $1 < [E:F] < \infty$, then $E = F(i)$ where $i^2 = -1$
Then $[E:F] = 2$
Whoops
Yeah that's what this answer says, though I saw it somewhere else
Hey @Tobias!
 
@Daminark Hi
 
7:21 AM
How's it going?
 
The algebraic closure of the reals has to be the complex numbers, as otherwise I don't see how it can guarentee all polynomials have solutions?
(modulo isomorphism)
 
@Daminark Good. Got my postdoc extended for 7 months
 
Ah, that's fantastic!
@Secret yeah it is
 
@Daminark Yeah, our head of department gave me the offer to extend it on the condition that I teach algebra again, which I am happy to do.
 
Definitely not the worst class to teach. Seems like you did a good job last time, but yeah congrats!
 
7:31 AM
Hi there, I would like to know the general procedure to how WolframAlpha plot the following rational function. Any suggestions. see the following picture
 
@CroCo Do you just want to plot it as a function of $s$?
 
@TobiasKildetoft, similar to the picture.
 
@CroCo Then what is the issue?
 
I don't know how the points shown in the segment lines are computed for a general rational function.
I would like to design a plotter.
I'm looking for a numerical approach for this issue.
any suggestions?
 
 
2 hours later…
9:26 AM
In mathematics, in the field of topology, a topological space is said to be a shrinking space if every open cover admits a shrinking. A shrinking of an open cover is another open cover indexed by the same indexing set, with the property that the closure of each open set in the shrinking lies inside the corresponding original open set. The following facts are known about shrinking spaces: Every shrinking space is normal. Every shrinking space is countably paracompact. In a normal space, every locally finite, and in fact, every point finite open cover admits a shrinking. Thus, every normal metacompact...
In the mathematical field of set theory, an ideal is a collection of sets that are considered to be "small" or "negligible". Every subset of an element of the ideal must also be in the ideal (this codifies the idea that an ideal is a notion of smallness), and the union of any two elements of the ideal must also be in the ideal. More formally, given a set X, an ideal I on X is a nonempty subset of the powerset of X, such that: 1. ∅ ∈ I {\displaystyle \emptyset \in I} 2.if A ∈ I ...
Ideals really behave like black holes
once you are in there, there is no ecape
 
Does a linear subspace $U$ of the space $\mathbb{Q}^{n \times n}$ of square matrices contain an invertible matrix if its span in $E^{n \times n}$ for some field extension $E/\mathbb{Q}$ contains an invertible matrix?
I think it does, because the determinant $\det \colon U \to \mathbb{Q}$ is a rational function given as a polynomial with $\mathbb{Q}$-coefficients, and so if it vanishes on $U$, it vanishes on the span in $E^{n \times n}$. But I'm not really sure what I'm using here exactely.
 
The Baire category theorem (BCT) is an important tool in general topology and functional analysis. The theorem has two forms, each of which gives sufficient conditions for a topological space to be a Baire space. The theorem was proved by René-Louis Baire in his 1899 doctoral thesis. == Statement of the theorem == A Baire space is a topological space with the following property: for each countable collection of open dense sets { U n } n = 1 ...
hmm... the proof sounds interesting to be visualised later...
It will also be better to get a better grasp of dense sets and how BCT depends on the axiom of choice, given how I often think about choiceless stuff
 
 
1 hour later…
10:42 AM
0
Q: Potentials for Vector Fields on a Circle

ALannisterThis problem actually has three parts, and I have questions about each part. (What is in block quotes is the actual text of the problem.) Part (a): Consider the vector field on the circle given by $\dot{\theta} = \cos \theta$. (Note: $\dot{\theta} = \frac{d \theta}{dt}$) I need to show t...

I dunno if anybody's around who can help with this. @Semiclassical do you know anything about this stuff? Thx.
 
11:21 AM
0
Q: About $ A(f(x),g(x)) = h(x) = 1 + \int_0^x f(x - t) g(t) dt $

mickConsider the transforms $$ A(f(x),g(x)) = h(x) = 1 + \int_0^x f(x - t) g(t) dt $$ $$ B(h(x),g(x)) = f(x) $$ $$ C(h(x),f(x)) = g(x) $$ Where the functions on the LHS are given. Notice $B,C$ are the inverses of $A$. How to compute $B,C$ ?? Is there an integral representation for $B$ or $C$ ? ...

Any ideas ??
Hm
 
11:48 AM
@ALannister as a Physics guy I definitely should know about potentials. So let’s see what I know...
 
11:58 AM
What they’re getting at is that, if $\theta$ and $\theta+2\pi$ are to represent the same point, then we had better have $V(\theta)=V(\theta+2\pi).$ that’s what single-valued will mean, and that’s true for your first case
In the second case, by contrast, $V(\theta)\neq V(\theta+2\pi)$ despite them representing the same point on the unit circle
Hence that potential, viewed as a function of position on the circle, will be multivalued
Geometrically, the difference is this. As you move along the circle in the first example, the potential will change smoothly and return to its initial values at angle zero
On the other hand, if you think of your second case as just arc length on the circle, then going around the circle once increases the arc length by 2pi
 
If $\Gamma$ is a group acting properly discontinuously and freely on a manifold $M$, and $A$ is a subspace of $M$, does $\Gamma_A := \{ g \in \Gamma : gA=A \}$ act properly discontinuously on $A$?
 
It’ll be single-valued if you consider a particular trajectory as a function of time, but it won’t be single-valued as a function of position on the circle
 
12:26 PM
Not sure what they want for part (c) tbh
 
Any ideas on solving $ 2x+3[x]-4{-x}=4 $ ? Where [x] and {x} are Greatest integer and Fractional part functions respectively.
 
you’ll want to use \{, \} inside your $s to make curly brackets show up in mathjax
I’d start by noting that $x=[x]+\{x\}$
 
12:54 PM
any group $G$ is in bijection with $\operatorname{Hom}_{\mathrm {Grp}}(\Bbb Z, G)$
 
1:28 PM
change my mind
@Daminark do you have examples of three-fold adjunctions?
 
1:54 PM
Thanks @Semiclassical
When is the antiderivative of a $2\pi$ periodic function also $2\pi$ periodic?
 
$f(x+2\pi)=f(x)$
$f'(x+2\pi)=f'(x)$
$\int f(u) du= \int f(x+2\pi) dx = \int f(x) dx$
if $\int f(x+2\pi) dx = \int f(x) dx = g(x+a)$ then
$f(x+2\pi) = f(x) = g'(x+a)$
does not seemed to have any restrictions, probably can only be shown by something I have not learnt about
 
@ALannister precisely when the integral of the function from 0 to 2pi is 0
 
uh... what about ?
the cycloid is positive everywhere, thus $\int_0^{2\pi}$ will fail to give zero?
 
Is there a name for this thm? You can find the residue at an isolated singular or, pole of order m of f, iff f(z)=phi(z)/(z-z0)^m and phi is analytic and non-zero at z0
Then res(z=z0,f) = phi^(m-1)(z^0)/(m-1)!
It's very nice
 
2:18 PM
How can I use the fact that 'if $N\lhd G$ and $H$ any subgroup of $G$, then $N\cap H\lhd H$' to deduce that 'C_G(A)\lhd N_G(A)' where $A\subset G$? @LeakyNun
 
well you need to find the appropriate N G H
can you find any of them?
 
Zee
//puke zee
 
@Zee Maybe you should consider looking up the manual
 
Zee
//help
 
###################### Help ######################

==================== Commands
//about          | Let me tell you a little about myself...
//alive          | Used to check if the bot is working
//appul          | Apples.
//ban            | Bans a user from using the bot. Only usable by hardcoded bot admins
//ban-room       | Blacklists a room
//blame          | No description was supplied for this command
//declare        | Changes a commands status. Only commands available on the site can be edited
 
Zee
2:28 PM
//Knock knock
 
@Zee Maybe you should consider looking up the manual
 
Zee
//wave
 
@Zee Maybe you should consider looking up the manual
 
Zee
//hope
 
I once again procrastinated too much while I should be reading a 400 page user manual
 
Zee
//poo
 
💩 💩 💩
 
Zee
//flyaway
 
Zee
//unsummon
 
2:30 PM
@Zee 2 more votes required
 
Hi. I'm looking for an expression for a class of functions very similar to $f(x)=x^n$, $n\geq 1$; the only problem with those is that they are not symmetric wrt the $1-x$ line. $f(0)=0$, $f(1)=1$, continuity, being below the 45 degrees line, and varying curvature is, in short, what I need in addition to the symmetry. Feels like it should be something simple..
 
Any French speaking algebraists in the house?
 
french speaking = yes but not on, algebraists = yes but not on, both = no
 
Good enough: I've been trying to determine if Guy Renault is still among the living
problem is, I can only search decently in English :(
Could you do a quick scan to see if you see any emeritus/obituary information on the web?
The publication i was reading lists an address near the Universite de Poitiers
And he was active, at least, in the 60's and 70's
Doh, you said "yes but not on" well yeah, I know there are French speakers on the site in general :)
I just haven't managed to catch any here while I'm in
 
Dattier and vrouvrou are french speaking people, but they are not from algebra
 
2:42 PM
OK, I'll keep an eye out. I guess they don't really have to be algebraists.
I just figured it would be more likely they recognized the person.
thanks
 
@Secret I’m not sure what you’re getting at. The point is that, if you want to the antiderivative to be periodic, then the function itself had better be periodic but have zero integral from 0 to 2pi
And that’s true if your cycloid: it doesn’t integrate to zero over one period, so it’s antiderivative isn’t periodic
(The antiderivative_will_ be of the form $m\thera + f(\theta)$ with f periodic. So it’ll have a periodic part but will also have a linear (and thus non-periodic) part. Only way to avoid that is $m=0$ which is the stated condition.)
 
Hello can someone help me find gen solutions of this trig eqn
tan ^2 x + 2 root 3 tan x = 1
I tried to do it but ended up with tan pi/6 = cot 2x
please somebody help me with this :/ i have a test tomorrow
The options are A. x = npi/2
B. x = (2n+1) pi/2
C. x = (6n +1) pi/12
 
Note that cot(x)=tan(pi/2-x)
 
Where does it come from?
 
So you can rewrite your last equation to cot(pi/3)=cot(2x)
Sin(pi/2-x)=cos(x)
 
2:53 PM
never learnt that
 
I sorta doubt that. It’s just saying that the sine of an angle is the same as the cosine of the complementary angle
Which is obvious if you think in terms of a right triangle
 
I don't understand :/
like sin x = cos ( 90 - x)
ohhh i see
you just expressed it in terms of radians
 
Right
 
k lemme try
 
And then you just plug those into the definitions of tan and cot
 
2:59 PM
Hmm well I learnt it like this - tan x= tany -> x = n pi +x
 
I am not sure if it's the same fr cot
Is it?
 

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