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1:00 PM
@DHMO OOOooooooOOO! :D me dumb
 
@Koolman i would split it to 6+8 and 7+7 two cases
 
@DHMO If you have a canonical which looks like C = C - C to get to another one you have to make that C - C$^{-}$ - C instead of C - C = C, ri ... oh, wait, you're just breaking first the p-p bond and the free p-orbital of the middle atom overlaps with the free p-orbital of the last.
 
@BalarkaSen the allyl is quite stabilized by resonance
yes, the last sentence is correct
 
OK, I get it. I was horribly misunderstanding.
 
Hi!
just a curiosity
I'm not really a mathematician, so I don't have to deal with mathematics every day... if for like one month (or maybe even less) and I do not touch (i.e. read and study about) a certain topic, once I have again to deal with the topic usually I always have to revise the concepts...
for example, I was reading about the pseudoinverse of a matrix (which, btw, I had never heard about it) from the Wiki article, and I read there that the pseudoinverse can be computed using the SVD, which I had already heard about it (not a long time ago, like 3 weeks ago), and I also had understood it, but those concepts didn't come to my mind immediately, so I read the introduction of the Wiki article about the SVD... does this happen also to you?
 
1:14 PM
Yeah
I find that doing the exercises in textbooks helps me remember the topics @nbro
There are tons of things in math that I've read about but only vaguely remember, since I used to not bother with any of the exercises.
Which isn't so bad, in of itself, as long as I'm not being tested on the material or anything
but it does mean that I should go back to the books and reread them, doing the exercises this time, at some point.
Random question: Can an open set be written as the countable union of disjoint closed sets (in $\Bbb R^n$)? (Excluding trivial things like $\Bbb R^n\cup\emptyset\cup\emptyset \cup\emptyset\cup\dotsb$)
 
@AkivaWeinberger ok, of course, I expected that for some extent people forget about the concepts, but what's your background (if you want to share with us)? This helps me to understand if I should do more exercises or not...I usually don't do them because I don't really have time...
 
user84215
goodbye
 
@nbro High school. I just read a lot of books on math.
 
@AkivaWeinberger Lool, then you still have a lot of time to learn and revise!
 
Nah. Our death's coming fast
 
1:21 PM
@BalarkaSen At least he/she probably has more time than me, but it's interesting to hear your stories
 
@DHMO what mistake I have done
 
@BalarkaSen Any idea on the question I posed above?
 
@Koolman sorry, I don't want to think
 
Np
 
On whether open sets can be written as the countable union of disjoint closed sets in Euclidean space
 
1:24 PM
@AkivaWeinberger I'm dumb today but how do you decompose (0, 1) as a countable union of disjoint closed sets?
 
@BalarkaSen I don't mean "can every open set be written that way", I mean "can any open set be written that way"
 
I feel like I'm just missing something obvious
 
Interesting
Well impossibility of it boils down to (0, 1). If you have an open set near a point you can get a neighborhood. If it decomposes as a countable union of disjoint closed sets intersect with the neighborhood to get (0, 1) (homeom to that nbhd) as a countable union of disjoint closed sets.
So if you can prove it's not possible for (0, 1) you're done
 
Currently, I was thinking about entropy, and then for soame reason I get a continuum number of microstates for a relatively simple system
thus my entropy blow up
 
1:30 PM
@AkivaWeinberger isn't a countable union of disjoint closed sets always closed?
 
@DHMO How do you prove that?
I mean, it's not true. Look at the rationals in R.
 
@BalarkaSen I see
 
Union of $\{q\}$ for $q\in\Bbb Q$
 
@DHMO You can also look at the complement of the Cantor set, and replace each connected component with its closure
 
I think I have an algorithm
To create (0,1) from disjoint closed sets:
 
1:34 PM
This gives you the complement of the Cantor set together with the Cantor set's "pseudoboundary"
 
1. add [1/3, 2/3] to the list
 
What Akiva said.
 
2. this reduces the problem to creating (0,1/3) and (0,2/3)
3. iterate this ad infinitum
 
Doesn't work.
 
Yeah, if you're doing what I think you're doing, you're going to miss the points of the Cantor set's "pseudointerior" (for lack of a better word)
Like $\frac14$ and stuff
 
1:35 PM
@AkivaWeinberger oh...
 
Yup, you miss the points in the interior of the intervals of the Cantor set at each stage.
 
You're only hitting countably many elements of the Cantor set
 
So it's exactly Alessandro's example I guess
 
If you at least claim it's for complement of the Cantor set instead of all of (0, 1), I mean
You contain the pseudoboundary in the complement.
 
1:38 PM
By Baire category if aunion of closed sets is open (contains an open set) at least one of the closed sets has nonempty interior
 
How does one solve for x in Ax = B (matrices) when A is a 3x3 matrix and B is a 3x2 matrix?
 
I am increasingly skeptical about if it can be done. But I can't give a proof
 
I know that if B was a 3x1 matrix I could just append it to A and row reduce, but can I still do that if there are multiple columns?
 
I am quite certain it's impossible and there's a simple argument showing so, if only I could see it :P
 
Famous last words :P
@ROODAY Not sure, but probably.
 
1:50 PM
@AkivaWeinberger Ok
 
Today I have done one thing and currently doing another thing: I read the whole wikipedia article on surreals. Thanks to DHMO teaching me about dedekind cuts and that uncountable chain exericise, the surreals are manageable to me (except its multipication and division rules is a bit strange)
Currently, I am re-reading about ordinals, because I don't think I understood them good enough
My next step is to show that there is no bijection between the subset of the class of surreals (namely, the surreals with birthday $\leq \omega_1$) and the reals
Doing this should help give me a better feel on what sets of size $2^{\mathfrak{c}}$ with a linear order look like
 
2:15 PM
Could someone explain this to me? Explain why a set {v1, v2, v3, v4} in IR5 must be linearly independent when {v1, v2, v3} is linearly independent
and v4 is not in the Span{v1, v2, v3}
I understand that if v4 is not in that span, it is not a linear combination of v1, v2, v3
However I'm not sure what else I have to say to fully prove that v1,v2,v3,v4 is linearly independent
 
you have to show that v1 is not a linear combination of v2,v3 and v4 and so on
 
If it says that v1, v2, v3 are already linearly independent of each other, do I just have to show that v4 is not a linear combination of the previous 3?
 
hi chat
Suppose the four vectors were in fact linearly dependent. Then I could write $av_4+bv_3+cv_2+dv_1=0$ for some $a,b,c,d$ not all zero. @ROODAY
Since $v_4$ isn't in the span of the other three vectors, what would have to be true about $a$?
 
@Semiclassical it would have to be 0?
 
2:30 PM
Right. But if $a=0$, then we've got $bv_3+cv_2+dv_1=0$ with $b,c,d$ not all zero.
 
Since the others are linearly independent b,c,d would be 0, right? And to maintain that it all equals 0, a would also have to be 0?
Oh but you said suppose they are dependent
 
Right.
And the point is that if all four were linearly dependent, but v_4 isn't in the span of the others, then it'd have to be the case that the remaining 3 are linearly dependent.
 
Ah, and since its given that theyre independent, we have a contradiction so it proves that the set of all 4 is independent
 
Right.
 
Ok, thanks!
 
2:33 PM
It's obvious in 3D, of course: If two vectors are linearly independent then they span a plane. So if you've got another vector which isn't in that plane, then all three are linearly dependent.
(There's probably a simple way to phrase the argument that doesn't use contradiction. Try to see if you can present it more directly.)
 
Yeah. Would it be enough to say that if v1,v2,v3 are independent, and v4 is not in their span, v4 is not a linear combination of v1,v2,v3, and therefore independent of them, and therefore v1,v2,v3,v4 is linearly independent?
 
Hmm.
I don't feel like that quite works, but I can't put my finger on why.
 
Alright I'll stick to the contradiction example then
This exam is gonna be killer
 
1
Q: Find the coefficient of $x^6$ in a series

Koolman I can not understand how to solve it . Can anybody provide me a hint . Added: There solution is also given but I could not understand it .

 
@AkivaWeinberger I noticed you replied to my question regarding reimage. I'm told it's correct, and even seen it proved. But I don't quite understand why.
 
3:12 PM
hi
 
3:26 PM
@Sirmimer Are you sure you copied the statement correctly?
 
3:51 PM
@DHMO Replace x by -x and see what happens. The limit won't change.
 
@anonymous nice.
 
@DHMO Done ?
So fast!!
 
I have an basic idea
it's basically about conjugates
 
yeah
 
http://chat.stackexchange.com/transcript/message/35653634#35653634
DONE
 
4:15 PM
Hello
guys
Let $G$ be a directed graph that has a topological sort, is there a bound for how many sort can a graph have?
change "how many sorts can a graph have" with "how many sorts can $G$ have"?
 
what is a topological sort?
 
In the field of computer science, a topological sort or topological ordering of a directed graph is a linear ordering of its vertices such that for every directed edge uv from vertex u to vertex v, u comes before v in the ordering. For instance, the vertices of the graph may represent tasks to be performed, and the edges may represent constraints that one task must be performed before another; in this application, a topological ordering is just a valid sequence for the tasks. A topological ordering is possible if and only if the graph has no directed cycles, that is, if it is a directed acyclic...
 
4:31 PM
It says on wiki if you have a Hamiltonian path it is unique and it looks like in the other case you might me able to get all of the orders by certain swaps. How good of an upper bound do you want, and what is given, the directed graph and a linear order?
 
I basically want to know two things
1. If they're bounded (so they can be computed by a computer xD)
2. A way to compute all of them
You can assume you have a DAG and a linear order (so you have 1 topological sort)
 
can anyone help me with möbius map ?
f(3i) = 3 , f(1) = 2 and f(-3i) = -1
 
@KasmirKhaan god dag
 
Hi @DHMO been a while!
 
indeed
 
4:46 PM
How are you ?
 
busy as usual
 
Well that is life =p
 
indeed
@KasmirKhaan what is your question?
 
well i have to find a möbius map from C to C
f (3i) = 3
f(1) = 2
and f(-3i) =-1
 
@KasmirKhaan can you help me solve a system of linear equations?
 
4:49 PM
Yes
 
but he already has his system of linear equations to solve
 
am new on this topic so I dont even know how to start =p
 
3ia + b - 9ic - 3d = 0
a + b - 2c - 2d = 0
-3ia + b - 3ic + d = 0
@KasmirKhaan ^
@mercio por favor callate :)
 
gbwrk
 
how did you get that DHMO ?
 
4:50 PM
@KasmirKhaan I transformed from your question
 
Hmm we have 4 unknowns
 
@DHMO First remove the i by equating real and imaginary parts. You should get 4 variables and 4 equations. Then use Cramer's rule.
 
@anonymous tell that to Kasmir
and I never said that they are real numbers
 
@DHMO Oh
I see
Pretty interesting
 
2.cn-g9
@DHMO I came no where solving that ><
 
4:59 PM
@AlessandroCodenotti @BalarkaSen @DHMO I think I have it.
Take any closed ball not contained in one of the closed sets.
 
@KasmirKhaan hmm... you need to learn how to solve systems of linear equations then
 
Intersecting all of the closed sets with the ball, we get a way to write the closed ball as the union of disjoint sets, at least two of which are nonempty.
 
@DHMO I think I can do it. Treat d as a pseudo constant whle taking a,b,c as the variables on the left and solve it just like you do in Martin's method (using inverse matrix).
 
The remainder follows from here:
65
A: Is $[0,1]$ a countable disjoint union of closed sets?

NunoThe answer is no. In fact, as Steve D said, we have a theorem that holds for a wide class of spaces, which includes closed intervals, circles, balls and cubes. It was proved by Sierpiński in $1918$ $[1]$. You can find the proof in the book "General Topology" by Ryszard Engelking, but I'll post he...

(The answer generalizes it to all continua, including closed balls.)
 
Huh interesting
 
5:02 PM
You can skip Lemma 1 and the proof of Lemma 2, to be honest. The main ideas are in the statement of Lemma 2 and the final proof.
 
@DHMO I have found solutions but they are in terms of a
 
@KasmirKhaan good
they should be equivalent
 
What do you mean equivalent ?
 
your required transformation is f(x) = (ax+b)/(cx+d)
 
I get a very big expresstion for that
and both a and x are unknown
 
5:05 PM
show me
 
top = ax + (-81 +600i)/ 241
and bottom =
 
I thought b should be in terms of a
 
ax (53+18i )/241 + a(27+282i)/241
there are no b's
 
I mean the (-81+600i)/241
should be multiplied by a?
 
yes yes
 
5:07 PM
so cancel the a
 
i just did not put it by mistake
 
and multiply 241 throughout
 
okay
Top = 241x - 81 +600 i
bottom = (53+18i) x +27 +282 i
simplfied a bit
 
and can you verify?
 
yes with those 3 points given
 
5:10 PM
then we are done
 
I idi dnot verify yet but ill do it now :D
thanks DHMO
The first step i dont know what u did tho
are möbius maps some sort of quotion of 2 linear functions ?
 
yes
quotient..
 
Hmm is this speciell in some way ?
 
in what way?
 
I dont know =p why we study just it i mean =p
 
5:12 PM
In geometry and complex analysis, a Möbius transformation of the complex plane is a rational function of the form f ( z ) = a z + b c z + d {\displaystyle f(z)={\frac {az+b}{cz+d}}} of one complex variable z; here the coefficients a, b, c, d are complex numbers satisfying ad − bc ≠ 0. Geometrically, a Möbius transformation...
 
Thanks again @DMHO
I ll keep working on this topic :D
 
5:45 PM
@DHMO hello, please if i take $A=[1,2]\cap \mathbb{Q}$ then $\overline{A}=[1,2]$ and $\ovrset{\circ}{A}=\emptyset $ so $\partial A=[1,2]$
right?
and $\overset{\circ}{\partial A}=\{1,2\}$
right?
 
6:40 PM
Anyone there?
 
@ArmaGeddON 1) I'm guessing the middle term should be $b(r-p)$, just because of symmetry. 2) Maybe use the formula for the terms of an arithmetic series?
Note that\begin{align}&p(q-r)+q(r-p) +r(p-q)\\=&pq-pr+qr-pq+pr -qr\\=&0,\end{align}so that it's at least true for the special case of the sequence $(1,2,3,\dots)$. (Assuming we fix the middle term, that is.)
 
@AkivaWeinberger Thanks a lot, it was b there, and i got the answer :)
 
7:16 PM
0
Q: Prove that the rings $End(\mathbb{Z}^{n})$ and $M_{n}(\mathbb{Z})$ are isomorphic

JessyunBourneI need to prove that rings $End(\mathbb{Z}^{n})$ and $M_{n}(\mathbb{Z})$ are isomorphic. To start with, I let $A = \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots& & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{pmatrix} \in M_{n}(\mat...

The mult in showing the map is at least a homomorphism in one guy's answer is giving me trouble. Mult here is composition of functions but I'm not sure how to compose them.
 
7:29 PM
Hi guys. Let $G$ be an Euler graph with an even number of edges. Let $d_1,\dots,d_n$ be the degrees of each vertex. Show that $G$ has a subgraph with degrees $\frac{1}{2}d_1,\dots,\frac{1}{2}d_n$.

The solution says the following: Pick an Euler cykel in the graph. Then pick the first line, don't choose the second one, pick the third one, and so on (in this alternating fashion).

However, my issue is that I'm still only convinced that we've got total degree $\frac{1}{2}\sum_{i=1}^nd_i$, but what guarantees me that each $d_i$ is halved?
Oh I think I get it. I've written out a sequence of 10 points, and now I can see it.
 
Hey guys, need some help with a flow chart for decision mathematics
its cool, i realised i was being a massive idiot
 
7:51 PM
Determine the point which is symmetric to 3+i with respect to the circle abs(z-2)=2
Anyone help ?
 
8:25 PM
please
how to prove that if
$A\cup B$ is closed and $A\cap \overline{B}=\oveline{A}\cap B=\emptyset$
then
$A$ and $B$ are closed
?
 
hi
 
hi
 
hellow
 
hi @sha
how are you?
 
@Zach Pretty good! I'm visiting my parents, it's always good to be back at home :) how are you?
 
8:36 PM
tired, just studying a bit
 
On a Friday night? :P What are you studying for then? The camp thing?
 
no, just self-studying
right now i'm studying the spectral theorem and quadratic forms
 
Oh I think I've had that theorem too
 
which is just a little bit of the whole study of non-euclidean geometries
 
at least, in a linear algebra course
and I have a Linear Algebra grand finale test this monday :P so I have to know that theorem
 
8:37 PM
let's see if you know it. Recite it! :P
 
haha no wait, that's from a chapter that I skipped.
I have to be honest here now; it's a retake
but I will know it at the end of this weekend!
 
"The eigenvectors of an $n$-th dimensional symmetric linear map form an orthonormal basis of $\Bbb R^n$"!
 
omgoodness... I wanted to go to bed. But now you've excited me to finish some work XD
I'm going to finish some exercises for a course Introduction to Graph Theory
 
well i'm glad i motivated you :P
mmm graph theory
 
haha don't be!!
I need my sleep XD
 
8:40 PM
never studied it but it definitely looks interested
 
it's pretty simple!
it's one of the few courses at our uni that also allows non-maths students
 
Solved the 7 bridges of Konigsberg problem? :P
 
I haven't solved it yet
but I have come across it
it has to be an Euler cykel right
I still have to prove that it cannot be one
I guess
 
cycle?
 
@ShaVuklia Cycle in English
 
8:41 PM
Yea, I'm Dutch XD
 
'and hey @Tobias
 
my apologies
Hi! @Tobias
 
It's kind of like the jordan curve theorem
 
Determine the point which is symmetric to 3+i with respect to the circle abs(z-2)=2 ,Help anyone ?
 
8:42 PM
except apparently JCT is much harder to prove
haven't studied all that much general topology
 
oh I haven't heard of that:(
 
@KasmirKhaan what is the center of that circle?
@ShaVuklia it says every closed, non-self intersecting curve partitions the plane into 2 disjoint subsets
@KasmirKhaan (hint, when is that absolute value 0?)
 
z=2 is the center
 
good
also, what do you mean, with respect to the circle?
 
thats the problem i dont understand
we doing möbius transformation
 
8:46 PM
ahh
ok
you could invert around that circle
 
I dont think its that simple
 
Have you seen Sylvester's theorem @Zach? The one about the signature of symmetric matrices
 
uhh no, what does it state :P
@KasmirKhaan alright, i might know what to do
let me bring out geogebra
 
You're gonna run into it soon then, it's usually done after the diagonalizability of symmetric matrices. It's very useful in classifying quadratic curves too
 
@KasmirKhaan so we want to reflect A around this circle
lets ask ourselves, the is the slope of line AB?
@AlessandroCodenotti all symmetric linear maps are diagonalizable, right?
 
8:53 PM
What do they mean by point symmetric to a circle
 
they want to reflect it around the circle
let me show you
 
1-i you mean ?
 
@KasmirKhaan nevermind
youre going to have to invert A
that's what they mean
 
so 1-i by that logiv
 
wait
let me think
no, its not 1-i
 
8:57 PM
@ZachHauk Sure, but the spectral theorem tells you even more than that
 
@AlessandroCodenotti yeah i know :p
@KasmirKhaan let's think
draw a line through AB
we're going to construct the inverse
now, inversion tells us that we have to draw the line perpendicular to AB at A
 
(I'm assuming you're working in $\Bbb R^n$, you probably need hermitian stuff in the complex case)
 
call this line m
what is the equation of m?
 
We didn't talk much about complex vector spaces
 
well neither does Ted's book :P
 

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