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5:19 AM
Wait meh you'd need to prove that real/im parts are harmonic so that'd give you second partials anyway
 
Some users suggested recently that it might be good idea to revive c.r.u.d.e chat room.
C.r.u.d.e chat room. is intended to help with various "janitorial" task such as closing, reopening, (un)deleting, editing and improving post. I am posting here mainly to make more users aware of that room - which might help to make it more useful.
 
5:41 AM
Morning everyone!
 
5:53 AM
Demonark: I'm not always after the cheapest solution. I'd deduce this directly from the Cauchy integral formula (which, of course, also gives you analyticity).
 
Well, the reason I'm trying to do it this way is to see if I can make the Stokes' theorem proof of Cauchy's theorem completely general
 
What do you mean?
 
Well, the proof of Stokes' theorem assumes the function is $C^1$, right?
 
function?
 
Hey! @TedShifrin
 
5:56 AM
hi @Perturbative.
 
Well yeah, like if $f$ is a holomorphic function, presumably you have to know that $f$ is $C^1$ in order to do the whole business with saying $f(z)dz$ is a closed form, and thus we can homotope our curve without changing anything
 
@Ted, I have a somewhat soft question to ask, is there any commonly accepted notation used in Differential Geometry?
 
Basically, Stokes's Theorem for a $1$-form on a rectangle is just the usual FTC, Demonark, so you have to think through what you need. I usually prove everything for smooth forms, but you need to be able to apply Fubini to the integral $\iint d\omega$.
 
I've looked through a few books and almost all of them have had their own notation
 
First, @Perturbative, people use the term differential geometry in lots of ways. I mean differential geometry (things with connections and curvature), not the general background on manifolds ("differentiable manifolds"). What do you mean?
 
5:58 AM
Oh well then I mean Differentiable Manifolds, definitely not connections and curvature
 
Well, in FTC, you have to have that $F$ is the antiderivative of a continuous function $f$, and then say that $F(b) = F(a) + \int_a^b f(t) dt$
 
e.g Introduction to Manifolds by Tu writes a vector in a $n$-dimensional vector space as $v = (v^1, v^2, ..., v^n)$
 
Well, you don't quite need that, Demonark.
Most of us use upper indices for coordinates, yes, @Perturbative. When you start doing things with tensors, it makes things work out better.
At any rate, Demonark, if $f$ is integrable and $f=F'$, then it works.
And if you use Lebesgue stuff you get a stronger result.
 
That's how I've seen it at least. Though I guess you can get an a.e. result?
 
Spivak proves the version I just said with (Riemann) integrable.
 
6:01 AM
Ahh okay, thanks Ted :), I guess I was just looking for a reason to motivate the use of what seemed to be nonstandard notation to me
 
So my point is to think through this stuff, divorcing it from the complex analysis, I guess.
 
Oh, then I'm gonna check that out, actually
 
@Perturbative: It ties in with what's classically called the Einstein summation convention. And it makes sure you're thinking of the right sorts of tensors. But there's not wildly different notations.
Demonark, to be honest, it's not stuff I usually think about much, but it's worthwhile to think it through.
 
Ah, you're right, well this is nifty
 
Demonark: You might check out Lang's real analysis. I know he states some more general versions of Stokes's Theorem (allowing singularities in the forms, even).
 
6:04 AM
I thought I remembered it strictly relying on continuity
 
Nope.
This will also get into the stuff Eric's working with ... the difference between forms and currents, things I've mentioned before.
 
Well, as for using Fubini, I guess we can invoke the power of Lebesgue and say that if we're doing stuff on some box/compact set, we should be able to pull it off, no?
 
Well, you need hypotheses for Fubini to hold (if you're not assuming $d\omega$ is continuous).
anyhow, I'll check in tomorrow sometime ...
 
Well, I think you need that $X$ and $Y$ are $\sigma$-finite and that $f\in L^1(X\times Y)$
 
do you ever sleep
 
6:17 AM
Almost never before 2-3AM, sometimes afterwards
But yes
 
Thanks for that info @TedShifrin
I've got another quick question
This is taken from Lee's Introduction to Smooth Manifolds
When Lee says 'we need only verify that each transition map $\psi \circ \phi^{-1}$ is smooth' what does he mean by "each transition map"?
As he defines above, for any two charts $(U, \phi), (V, \psi)$ there is only one such transition map
 
@Perturbative the idea is that you're doing across all pairs of charts
 
6:33 AM
Ah okay, the wording he used got me slightly confused
Thanks @Dami
 
No problem!
 
6:49 AM
ohi chat
 
Yo
 
What's up ?
 
Not much, how about you?
 
Just got confirmation I'm going to Polytechnique
Apart from that not much
 
Woohoo, congrats!!
 
6:51 AM
Thanks :p
 
@Astyx Which one?
 
French one
 
More precisely?
 
Nice, congrats
 
6:55 AM
The one that was created by Napoléon
Thanks
 
Hey @TobiasKildetoft!
 
@Daminark Hi
 
@arctictern I understood it now, I think. The point is that indeed $Sp(n)\subset U(2n)$ but $Sp(1)\cap U(2n)=U(1)$, basically $e^{i\varphi}$ where $i$ is the complex structure.
 
multiplying elements of H^n on the left by exp(i phi) is different from multiplying by it on the right, unless exp(i phi) is real
Sp(n) is a subset of U(2n) where both are acting on H^n from the left (to commute with scalars coming from the right), no?
 
Yeah
even of $SU(2n)$
@arctictern The intersection of $Sp(1)$ with $U(2n)$ must be $U(1)$ though (implied by claims in a paper I'm reading)
I think multiplication with $e^{i\varphi}$ from the right is the same as from the left
but only if $i$ is the (almost) complex structure defining the identification $\Bbb R^4=\Bbb C^2$
So then $e^{j\varphi}$ is not going to be unitary, but $e^{i\varphi}$ is
isn't that right?
 
7:07 AM
not sure what almost complex structure you're talking about, unless you mean multiplication of C^2 by i as usual
 
How's it going?
 
@arctictern Yeah, tha'ts what I'm talking about
 
@Daminark aight
multiplication of H on the right by i doesn't match multiplication on the left by i
 
I'm thinking about the space of all possible almost complex structures on $\Bbb R^4$ compatible with a certain other ($Sp(n)\cdot Sp(1)$-)structure.
@arctictern I realize that
But if you look at it like $\Bbb C^2$ then it does
I start from $\Bbb R^4$ and want to understand the stabilizer of a complex structure on it, but not by the action (by conjugation) of $SO(4n)$ but rather of its subgroup $Sp(n)\cdot Sp(1)$
 
if your complex structure is right multiplication of H^n by i, then that's an element of Sp(n)Sp(1), and its centralizer is Sp(n)U(1)
 
7:13 AM
Exactly
The standard identification of $\Bbb R^4$ with $\Bbb C^2$ is the one induced by (right- or left-)multiplication by $i$, regarded as an element of $\Bbb H$
and I just wanted to understand why precisely it is preserved by $Sp(n)U(1)$
in particular that $U(1)$ factor comes from the units in $\Bbb H$ of the form $e^{i\varphi}$, right? This is what what I was trying to say earlier.
 
yes
 
Sorry for being so unclear
 
Since Sp(n) and Sp(1) commute in Sp(n)Sp(1), to find the centralizer of i (the element of Sp(1)) it suffices to compute its centralizer purely in Sp(1), which will be U(1)
 
right
 
but Sp(n)Sp(1) intersect U(2n) (where U(2n) acts on H^n=C^2n from the left) will just be Sp(n) rather than Sp(n)U(1) since right multiplication of C^2n by i will not match the original left multiplication by i
for instance when n=1 we can write H=C+jC, with C acting on the right. then the original left multiplication by i will act as the nonscalar matrix diag(i.-i).
 
7:22 AM
Okay. Thanks for the clarification.
 
7:36 AM
Can someone confirm my thoughts that the $\mathbb{C}$-rank of $SO(2,\mathbb{C})$ is one (its isomorphic to $\mathbb{C}^\times$ via $\begin{pmatrix}a & b \\ -b & a\end{pmatrix} \mapsto a+ib$), while its $\mathbb{R}$-rank is zero (its group of real points is isomorphic to $S^1$)?
 
7:58 AM
@Astyx Congrats on getting into École Polytechnique!
7
Does that also mean you're not going to ENS? Or are you still waiting for your recount and you have Polytechnique as backup?
 
8:11 AM
Thank you ! I'm still waiting for my recount
 
Hey @Alessandro!
 
Hi @AlessandroCodenotti
 
Meh
 
8:32 AM
$S=\{x+\frac{1}{y},x\in \Bbb{R},y\in \Bbb{N}\}$
 
That's just $\mathbb{R}$ :^)
 
In terms of set builder notation, this looks very similar to sets of the form $a+b(stuff)$ where $stuff$ are elements such that they are incompatible with $a$ (e.g. \Bbb{Z}[\sqrt{5}] written in set builder notation). It is easy to check that $S$ gives $\Bbb{R}$ as the harmonic sequences that is implicated in the set have its infimum ranging through all reals, thus it is complete. However, I am wonder whether the natural topology of $S$ is the same as $\Bbb{R}$ since in $S$, we basically use just one type of sequence to generate all of $\Bbb{R}$
More generally, the natural topologies of this family of sets may be interesting:
 
@Secret You are adding precisely nothing to the reals by adding those inverses of natural numbers. So I have no idea what you are trying to do
 
$S_a=\{x+\frac{1}{y},x\in [0,a),y\in \Bbb{N}\},a\in \Bbb{R}$
 
@Secret That is just the set $[0,a+1)$
no naturals other than $y=1$ contribute anything
(well, assuming $a > 1$)
 
8:46 AM
Ah I am sorry, I miswrote the sets, let me rewrite them (I don't intend to terminate the harmonic series at 1
 
@AkivaWeinberger Thanks for that. I didn't realize it's possible.
 
Ok, so it is clear that $S=\{x+\frac{1}{y},x\in \Bbb{R}, y \in \Bbb{N} \cup \{\frac{1}{z},z\in \Bbb{N}\}\}$ is just $\Bbb{R}$ with nothing new and hence its natural topology is the same as $\Bbb{R}$

Now consider: $S_a=\{x+\frac{1}{y},x\in [0,a), y \in \Bbb{N} \cup \{\frac{1}{z},z\in \Bbb{N}\}\}$, then something interesting should happen here since for fixed $a > 0$, if we go far enough, gaps will start to appear. Hence the natural topology cannot be the usual topology of $\Bbb{R}$
(NB I don't know how to better write the sequence ...,10,9,8,7,6,5,4,3,2,1,1/2,1/3,1/4,1/5,.. in set builder notation)
 
Ok, so clearly this set is only interesting for $a < 1$ as otherwise you just get all the reals.
 
Right, so for $0<a<1$ its topology cannot be the usual topology as there will be gaps in the set. But how those gaps behave requires figuring out its natural topology
 
@Secret I don't know what you mean by the usual topology.
 
8:57 AM
The usual topology of the reals is the open interval topology, i.e. the base is consist of open intervals (a,b)
 
@Secret And the topology of this set is the set of the intersections of such sets with the subset
So I suppose the questions is which additional sets this gives compared to only taking open intervals with endpoints in the set itself
 
right, I guess I can figure this out by computing the subspace topology of $S_a$ wrt $\Bbb{R}$
 
9:15 AM
Hmm, so after computation, the explicit form of $S_a$ will look something like this: $(0,a) \cup \bigcup_{i \in \Bbb{N}} [i,i+a)$, therefore the base for the subspace topology is $\{(0,a), [i,a),i \in \Bbb{N}\}$, I think
so we have half open intervals being open sets in $S_a$, unlike in the open interval topology of the reals
Hmm, now to explore $T_a = \{x+10^y, x \in [0,a),y \in \Bbb{Z}\}$. This one should be even more interesting as we cannot get the positive reals unless $a=\infty$
Naive inspection suggests everytime $a$ gets bigger than a positive power of 10, we must obtain the interval (0,a) as part of our base (and thus any gaps that are < a will be filled in completely, and all gaps has to exist at > a)
 
user84215
9:36 AM
How can I create an event in a room?
 
user84215
for the room which I am not its owner
 
3
Q: Disconnectedness of the rationals with the subspace topology

Alex PetzkeI have tried to prove that $\mathbb{Q}\subset \mathbb{R}$ equipped with the subspace topology is a disconnected space. I'd like to make sure my attempted proof is correct since topological properties of $\mathbb{Q}$ seem like nice things to be familiar with. First, a base for the subspace topol...

The rationals are very weird neither sets under the usual topology of the reals. But the transcendentals are even weirder. It has gaps that are not disconnected, thus it is insanely hard to visualise such set
 
@aminliverpool I do not think you can. At least that's how I understand the FAQ: What are chat events and how do they work?
Quote: "Any room owner of a chat room can create an event on the Schedule tab of the room."
 
user84215
If I have a topic that I want others to speak about it, what should I do?
3
 
9:51 AM
@Secret "it has gaps that are not disconnected" wut?
 
Sorry I misremembered. I need to check again
 
You can just reuse that argument from the question you linked.
Between any two transcendental numbers, there will be a rational number (hence algebraic)
 
How should a book with a lot of authors (eight) be cited in the bibliography at the end of a document?
 
Depends on journal/institution
 
user84215
use et al.
 
10:01 AM
If it's for something official like a thesis, ask if you university/department has guidelines.
 
nah, it's not that official
 
Most likely you'll end up using "et al." like amin says, but different institutions have different rules for the number of authors up to which you should list all, etc
Choose your favourite integer between, like, 3 and 10, and let biblatex sort it out for you I guess.
 
Yeah that makes sense, thus the transcendentals are also totally disconnected
 
Ok, thanks. I think "et al." will be fine here
 
(friendly reminder to use biblatex with biber as backend for optimat LaTeX bibliography)
 
10:07 AM
I don't know what is, time to find out I guess
(I don't know LaTeX very well, just enough to write simple documents)
What that is*
 
I'm still using BibTeX, maybe I should switch eventually.
 
user84215
Because I think it is not good that only a room owner can create an event for the room, I have created the room, Discussing Specific Topics . In this room, people can discuss the topics they have specified before. Each topic lasts in this room for at most one day; It depends on its popularity among others. Its duration also can be specified before beginning.
 
10:28 AM
Does anyone in here know about algebraic groups?
 
10:42 AM
hello.
there is a problem says there are n objects being put in a line, we want to know how we can group them in groups of different sizes.
 
user84215
Hello. I think it is better to clarify your question.
 
I'm adding to it
the solution is : ("r"for number of groups , "n" for number of objects) there are n-1 gaps in between so $ n-1 $ choose $ r-1 $ is the combination of how they can be put in groups of size r.
it's said that it's the number of **POSITIVE** group sizes.
to obtain the **NON NEGATIVE** group sizes, we do as bellow :
$ x_i +1 = y_i $ so $ y_i = n+r $.
we are actually adding "r" to both sides of the equation for the positive group sizes to gain the nonnegative.
why adding r to both sides?
 
11:28 AM
I suppose it's because it is adding to the selection items, form "one" to "at least one" which contains "zero" too
 
If $x+y+z=1$ then $(x-1)^2+(y-1)^2+(z-1)^2=x^2+y^2+z^2+1$
 
11:48 AM
@LucasHenrique Hm. So we want to show that $(x-a)(x-b)(x-c)$ has a value greater than $8$ at $x=a+b+c$ given $e_1e_3=3$
That is, $x^3-e_1x^2+e_2x-e_3$ has value greater than $8$ at $x=e_1$ given $e_1e_3=3$
Hm that doesn't seem to help
We need to use the fact that that has three positives real roots
 
12:09 PM
in The h Bar, 56 mins ago, by ACuriousMind
@Secret Then we already have variables that are of greater "size" than the reals - fields. The set of all real functions of real numbers is of greater cardinality than the reals.
in The h Bar, 7 mins ago, by Dawood ibn Kareem
The number of paths from where I am now to where I'll be in 5 minutes is $\aleph_2$, right?
Plausible $\aleph_2$ physical models
 
$(a+b)(b+c)(a+c) \geq 8$, huh?
First prove that $(a+b)(b+c)(a+c) \geq 8abc$ and that $\frac{a+b+c}{3} \geq \sqrt[3]{abc}$
Then use the second part to prove that $abc \geq 1$, and combine with the first part
 
12:33 PM
[Random]
Conjecture: The set of things which cannot be found on the internet is empty or has finite support
Meanwhile, let me check MSE whether one can create a set with no gaps with countably infinite things alone... (obviously, if that is possible, it cannot be subset of reals)
the closest thing we have towards this direction are the rationals, since it is not well ordered (in the usual ordering), but it still have gaps
I suspect the notion of gaps can be made more rigorous as infimas and supremas that are not located within the set (except infinity of course, though we can avoid this problem by picking a set that is bounded by a minimum and a maximum)
 
12:50 PM
an oat is hindering my shift key on the keyboard. any advice+
Oh now it is working!
 
In mathematics, the Cantor set is a set of points lying on a single line segment that has a number of remarkable and deep properties. It was discovered in 1874 by Henry John Stephen Smith and introduced by German mathematician Georg Cantor in 1883. Through consideration of this set, Cantor and others helped lay the foundations of modern point-set topology. Although Cantor himself defined the set in a general, abstract way, the most common modern construction is the Cantor ternary set, built by removing the middle thirds of a line segment. Cantor himself mentioned the ternary construction only in...
arguement invalidated
 
user84215
Because I think it is not good that only a room owner can create an event for the room, I have created the room, Discussing Specific Topics . In this room, people can discuss the topics they have specified before. Each topic lasts in this room for at most one day; It depends on its popularity among others. Its duration also can be specified before beginning.
 
Any reason why you intentionally typoed argument? o.O
 
that's a typo, when I type fast, I tend to forget the e
(O nvm, I really misspelled argument, casue I always thought the one with no e is that thing related to complex numbers)
 
Lol
 
12:59 PM
In mathematics, arg is a function operating on complex numbers (visualized in a complex plane). It gives the angle between the positive real axis to the line joining the point to the origin, shown as φ in figure 1, known as an argument of the point. == Definition == An argument of the complex number z = x + iy, denoted arg(z), is defined in two equivalent ways: Geometrically, in the complex plane, as the angle φ from the positive real axis to the vector representing z. The numeric value is given by the angle in radians and is positive if measured counterclockwise. Algebraically, as any r...
 
I found it really weird how you spelled it correctly and then edited it to misspell it :P
 
Ok I just realised today: They are both argument, and arguement does not exist
bleh.... my poor english...
no matter, I won't spell it wrong anymore now that I knew
 
Details of the argument that is invalidated
in The h Bar, 9 mins ago, by Secret
which means the question is really not physically interesting even if in the highly unlikely case that map=territory, because we cannot really distinguish between the cardinals $\aleph_0$ and $\aleph_n$ based on whether something has gaps, and we must use bijective maps to compare between them
that's back to the drawing board for me about cardinals. I cannot believe I even get countable vs continuum wrong
 
Hi guys! Is it true that any abelian Lie Algebra is nilpotent? I would say yes... Let's consider a Cartan subalgebra of a semi-simple Lie Algebra. Then, it is made of commuting generators that can be simultaneously diagonalized in the adjoint representation; let's diagonalize them. Then, the Cartan subalgebra is also nilpotent (how can I see this?) and the Engel's theorem states that (Ad(X))^n = 0 where Ad is the adjoint representation and X is in the Nilpotent Lie Algebra.
How can this be possible? We have diagonalized Ad(X) and I don't understand how (Ad(X))^n can be 0
What's wrong in my argument?
 
Ummm...
I don't know why you're making this so incredibly difficult?
A Lie algebra $\mathfrak{g}$ is abelian if $[\mathfrak{g},\mathfrak{g}] = 0$.
A Lie algebra $\mathfrak{g}$ is nilpotent if the lower central series $\mathfrak{g} > [\mathfrak{g},\mathfrak{g}] > [[\mathfrak{g},\mathfrak{g}],\mathfrak{g}] > \cdots$ eventually becomes zero.
For an abelian Lie algebra, the lower central series becomes zero, well, instantly. So of course they're nilpotent.
 
1:10 PM
Well, I didn't have doubts on this.. I just asked for be sure
What I don't understand is the rest of the question
I had to split the question in two parts, sorry
 
I'm having trouble gaining intution behind, the following Proposition
$$\text{Proposition (0.1)}$$


For every $\text{Partition}$ $P$ of $[a,b]$ in $(1)$,

$(1)$
$$m(b-a) \leq L(P,f) \leq U(P,f) \leq M(b-a)$$

$\text{Remark}:$

Where $m := \inf\big\{f(x)| a \leq x\leq b \big\}$ and $m := \sup\big\{f(x)| a \leq x\leq b \big\}$
I manged to make things a little bit more clear by doing some substutions, and breaking it up into cases
$$\inf\big\{f(x)| a \leq x\leq b \big\}(b-a) \leq \sum_{i=1}^{n}\inf \big\{f(x)|x_{i-1} \leq x \leq x_{i} \big\}(x_{i}-x_{i}-1) \leq \sum_{i=1}^{n}\sup \big\{f(x)|x_{i-1} \leq x \leq x_{i} \big\}(x_{i}-x_{i}-1) \leq \sup\big\{f(x)| a \leq x\leq b \big\}$$
Case $(1)$
$$\inf\big\{f(x)| a \leq x\leq b \big\}(b-a) \leq \sum_{i=1}^{n}\inf \big\{f(x)|x_{i-1} \leq x \leq x_{i} \big\}(x_{i}-x_{i}-1)$$
Case $(2)$
$$\sum_{i=1}^{n}\inf \big\{f(x)|x_{i-1} \leq x \leq x_{i} \big\}(x_{i}-x_{i}-1) \leq \sum_{i=1}^{n}\sup \big\{f(x)|x_{i-1} \leq x \leq x_{i} \big\}(x_{i}-x_{i}-1)$$
Case $(3)$
$$\sum_{i=1}^{n}\sup \big\{f(x)|x_{i-1} \leq x \leq x_{i} \big\}(x_{i}-x_{i}-1) \leq \sup\big\{f(x)| a \leq x\leq b \big\}$$
Bassically Case $(1)$, of Proposition $(0.1)$, we draw a line at the $\inf f(x)$ and compute the area underneth
Rigoursly speaking: $\sum_{i=1}^{n}\inf \big\{f(x)|x_{i-1} \leq x \leq x_{i} \big\}(x_{i}-x_{i}-1)$ would be a bound for $\inf \big\{f(x)|a \leq x \leq b \big\}$
and Case $(1)$, bassically reads: the smallest value of f(x) over a subinterval must be less than or equal to the largest value of f(x) over that same interval
For Case $(2)$, In the second, you are adding the areas of many different rectangles, whose top sides are always above the one of the previous rectangle.
I'm having trouble visualzing what's happening to the apporxmiation of our function
 
1:33 PM
@Zophikel since your question is so long, why don't you take into account the option to ask a question on SE?
 
@apt45 i'll have to reedit my question I made it too breif :>(
 
1:48 PM
Problem: Let $X$ be some locally compact space, $Y$ some generic topological space, and $f : X \to Y$ a continuous open map. Show that $f(X)$ is locally compact. Here is my proof: Let $y \in f(X)$. Then there exists an $x \in X$ such that $f(x) = y$. Since $X$ is locally compact, there exists $C \in X$ compact such that $x \in U \subseteq C$, where $U$ is some open nbhd of $x$. Then $y = f(x) \in f(U) \subseteq f(C)$. Since $f$ is open, $f(U)$ will be open;
...and since $f$ is continuous, $C$ will be compact. This shows that $f(X)$ is locally compact.
 
@ap45 edited it:
3
Q: Intution behind $m(b-a) \leq L(P,f) \leq U(P,f) \leq M(b-a)$

ZophikelIn the text "An Introduction to Measure and Integration by Rana" i'm having trouble gaining intuition behind the following Proposition in $(1)$ $(1)$ $$\text{Proposition}$$ $\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \,\, ...

 
@Zophikel very good. BTW I am a physicist ;)
 
@apt45 mathematical or theoretical plz put some pictures and answer my question i'm almost there but I need some help :>(
 
2:12 PM
@SteamyRoot Oh silly me
Yeah that totally does it
Okay that's nifty
 
@Daminark Do you know a little topology? If so, would you mind checking the proof I gave above?
 
Sounds good to me
 
Are all well ordered sets necessary totally disconnected?
in particular, is $\omega_1$ totally disconnected. I tried to make an interval $(0,\alpha) \cup (\alpha,\infty)$ but that will mean I will always miss $\alpha$?
 
Hello people i have a thought about complex numbers is this correct that:

For plotting any function of a complex variable *z* we actually require 4 dimensions, one for Re(z), other for Im(z) third for Re(f(z)) and fourth for Im(f(z)) ??
 
2:31 PM
there are 8 women and 6 men. for creating a group of 3 men and 3 women, in a case that two men refuse to be together in a group, how many ways are possible to create such groups?
I answered by thinking about once considering one man , the other time considering the other man.
so : ( 5 choose 3 * 8 choose 3 ) + ( 5 choose 3 * 8 choose 3)
but here what i'm doing is separating the situation in which there is no condition or limitation, just as if I write : 6 choose 3 * 8 choose 3
answers are the same
 
@Daminark this is false. :/
Hey there, guys
 
you can write them as : u(x,y) + iv(x,y) , u for re part and v for im part @samjoe
you can not say there are 4 dimensions, we have x,y,z as dimensions , means 3 dimensions at most.
complex numbers won't add any dimension.
complex numbers can have 1,2,3,... dimensions just as real numbers, it's up to the question or problem.
 
@lucas @parvin i mean in order to fully represent a function f(z) of a number z
 
look just imagine you have a real number p, you put it in a function f, gives you f(p)
same goes with complex numbers.
I assume you think complex numbers are functions, isn't it?
@samjoe
or am I making mistake?
 
No parvin
 
2:38 PM
ok
 
I mean in real world we need two axes
for representing a function
of one variable
one for f(x) and other for x
Now complex world, we must represent Im(z) and Re(z) on separate axes, separate dimensions
am i wrong? i am confused a bit
 
@user193319 I will just remind this about locally compact spaces: "There are other common definitions: They are all equivalent if X is a Hausdorff space (or preregular). But they are not equivalent in general." BTW feel free to post also in general topology chat room - perhaps you could help to bring new life there; it was rather inactive for some time.
 
@LucasHenrique By AM-GM, $\frac{a+b+c}{3} \geq \sqrt[3]{abc}$
 
@parvin I believe you've counted the cases in which neither of the two are selected twice.
 
look there are only one dimension for a complex number @samjoe you just show them in two axis
 
2:43 PM
Not directly related, but someone told me here in chat a while ago that you can get the real part and the imaginary part for a function like the Riemann zeta function without using the Im() and Re() operators like this:
N[Zeta[1/2 + I*2]]
N[Re[Zeta[1/2+I*2]]=(Zeta[1/2 + I*2] + Zeta[1/2 - I*2])/(2)]
N[Im[Zeta[1/2+I*2]]=(Zeta[1/2 + I*2] - Zeta[1/2 - I*2])/(2*I)]
 
@samjoe look at this, 2nd paragraph in wiki :en.wikipedia.org/wiki/Complex_number
 
Ok, I now know that all ordinals < $\omega_1$ are totally disconnected, but what about ordinals after $\omega_1$ (under the order topology), are they also totally disconnected?
 
@SteamyRoot given the elementar symmetric polynomials (just to write less :p), you get that $e_1 e_3 = 3 \implies e_1 (\frac{e_1}{3})^3 \geq e_1 e_3 = 3 \implies e_1 \geq 3$.
therefore $\frac{1}{e_1} \leq \frac{1}{3} \implies \frac{3}{e_1} \leq 1$
 
@Mats Thanks for that, not that i understand those things
 
and that's just the opposite of what we wanted
 
2:46 PM
@Secret What's your first language?
 
I have an separate olympiad group and even the best students couldn't solve it
 
@AkivaWeinberger Cantonese
 
just one guy solved it using homogeneous inequalities and mean inequality
 
@Secret Your English is very good - I had thought you were a native speaker this whole time
Cantonese, so Hong Kong?
 
yup
 
2:48 PM
@samjoe
f(z) has two parts, a RE and a IM,
a complex number is also separated into two parts of RE and IM
you have such number : 2x+2ix this is one dimension
you have 2xy+2yi this is two dimensions
 
@samjoe if your function is like z^2 you will put the number z inside the function and the function finds its own RE and Im.
 
haha ! happy birthday :) @AkivaWeinberger
@TimTheEnchanter yes I did why shouldn't I and why is it wrong?
 
@AkivaWeinberger hbd akiva
 
2:57 PM
@parvin thanks i understand!
 
@parvin when you want the number of ways it's generally assumed you don't want to count some of them twice... or am I misunderstanding the question?
 
imagine m5 and m6 don't want to be together
we have :
m1 m2 m3 m4 m5
m1 m2 m3 m4 m6
and for each we have combinations of women 8choose3
then we add them up
that's what i did
@samjoe thanks god!
@TimTheEnchanter you're right thats it ! thank you.
 
@parvin ok good luck
 
Just ordered your book, @Ted!
 
3:30 PM
hi chat
 
hi Semi
 
Sneak peek of new MSE question:
 
I think I linked this yesterday, but it's still good: dieselsweeties.com/ics/355
 
@Secret Totally disconnected for what topology?
 
I think it is the most natural ones for each example "e.g. open interval for $\Bbb{R}$, discrete for $\Bbb{Z},\Bbb{N}$, subspace of $\Bbb{R}$ for $\Bbb{Q}$, (I don't know what I used for the cantor set $\Bbb{C}$, but since it consists of only limit points and endpoints, it is totally disconnected), and order topology for ordinals"
 
3:39 PM
Cantor set with trivial topology is not totally disconnected.
 
I suspect I am most likely using the subspace topology wrt $\Bbb{R}$ when I am analysing the cantor set, otherwise, yes under the trivial topology, it is not totally disconnected
 
i feel like the term cantor set almost assumes it has the usual topology inherited from $\mathbb{R}$.
at least it seems to me that the common usage is to assume that "a cantor set" is something homeomorphic to the standard one sitting in $\mathbb{R}$
 
I've never seen it used with a different topology
 
yeah exactly
 
But claiming a set is totally disconnected, and at the same time claiming you don't know what the topology on the set is, seems very fishy to me.
 
3:46 PM
oh sure
i moreso meant a Cantor set with a trivial topology is kind of only disingenuously a Cantor set
 
I just have the right mix of topology background to be able to analyse and construct this question,and when I analyse for total disconnectedness, I am busy thinking about how to form partitions from open sets using what I learnt from the rationals
so it only become clear later on to me that I implicitly used the subspace topology of the reals
More generally, given any maths topic that I try to investigate, I have a haphazard set of knowledge that I know enough for me to do investigations and even answer questions as if I am an someone in the field. It is only when the questons or comments thrown in that requires answers that lie outside of my knowledge base do my non-expertise become visible
Using a mathematical analogy, if each point in [0,1] is a piece of prerequiste knowledge for the investigation in question, then an expert should be a continous function in [0,1]
but mine is more like a nowhere continous function
It is only when in moments that I have sufficient time to read for months that my knowledge base become more stable
in conclusion, my thinking is messy
 
The above haphazard background is one reason why when I ask my MSEs, I never request the answerers to dumb down the content to my level. This is because I know with time, if I don't understand the terminologies in the answer, I will understood them soon enough after enogh reading
this, interestingly, had the side effect of making the answer quite useful to the experts out there and future readers, since it is written in expert language most of the time
 

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