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00:00 - 21:0021:00 - 00:00

00:00
Also, MVT FTW in analysis
the winterbash is on!
$A$ and $B$ being similar means they ahve the same eigenvalues. So they stretch/rotate any basis in the same way
Or to put it another way, they ahve the same effect on teh appropriate gridelines in $V$ and $W$
they may strech in different directions
It's so useful
@MATHASKER ^
00:03
wait what @LeakyNun
what wait what
@MatheinBoulomenos I mean in the appropriate pair of bases. Namely, choose a basis for $\{ v_i \}$ for $V$, then that induces the basis $P( \{v_i \})$ for $W$. These guys should be stretched/rotated the same way I think
so if it says rram of 8 which is subscribt is the n in the denominator to find delta x 8?
Hello. This limit will equal to zero, yes ?$\lim_{(h,k) \to (0,0) } \frac{h}{ \sqrt{h^2+k^2} } \frac{-k^4}{h^2+k^4}$
00:10
@PawełKusz yes, by geogebra
@PawełKusz Letting $h = \alpha k$ parameterize the line we're approaching the origin along, then the limit $ \sim -k^2\frac{1}{\alpha \sqrt{1+\alpha^2}}$
Try to bound it by using that numbers shrink as you raise them to powers near 0
That's how I would do it
Ok, Thank you :)
00:13
But I can't do these without actually writing it down and I'm on mobile
@Kevin that doesn't show that the limit exists. You have to be able to take any sequence $(h,k)$ that converges to $(0,0)$
Kevin no
^
@PawełKusz for each fixed $r$ the function is bounded
the bound goes to $0$ as $r$ goes to $0$
@MatheinBoulomenos that requires AoC :P
Also am/gm ineqs might hell
Help
00:16
@MatheinBoulomenos Can you give me an example that goes to 0 along every line, but the limit doesn't exist?
@MatheinBoulomenos honestly best response to when anyone mentions aoc
@KevinDriscoll take any real-differentiable function $\Bbb C \to \Bbb C$ which is not complex differentiable and consider the difference quotient
@Kevin $\frac{x^2y}{x^4 + y^2}$
If you let $y = nx$, the limit will be $\lim_{x\to 0} \frac{nx^3}{x^4 + n^2x^2} = 0$
I never saw anyone doing analysis without at least countable choice
But if you approach along $y = x^2$, the limit will be $\frac{1}{2}$
So rip in dead
00:21
I quite literally don't understand how this is possible
Yeah it's strange, I don't have a picture for why it works but... it does
the explanation is that if something can go wrong in real analysis, it probably does
I guess I spend all my time working with functions that are complex analytic
and so I never thought about it?
maybe the intuition is something like this: why should the limit exist if you approach the value in something like a parabola or a weird spiral if it only exists for straight lines?
Is @PawełKusz still here
00:27
this phenomenon is basically why real-differentiable doesn't imply complex analytic
Yes, I am :)
So here's one observation
Actuslly im not going to say anything because ill make mistakes without a pencil and paper in front of me
But in general, you want to find a bound for it that simplifies the expression using x^n<x^m for n>m, and the am-gm inequality
So, I started with $\lim_{(h,k) \to (0,0) } \frac{ \frac{h^3-h^2-k^4}{h^2+k^4}+1-h }{ \sqrt{h^2+k^2} }$
2sqrt|xy| <= |x| + |y|
I guess amgm is more useful if the qtys are negative
Nonnegative*
Okay got paper
00:54
Okay so how about this
Sqrt(x^2 + y^2) <= |x| + |y|. Do you see why?
I think thats enough
@KevinDriscoll $f(x,y)=1$ if the point $(x,y)$ is on the unit circle, $0$ otherwise
(@Daminark @MatheinBoulomenos)
(Take the limit to any point on the circle)
01:11
@Akiva trolololo
@AkivaWeinberger Thats not even continuous, I wouldnt have expected what I did ot work there
Polls are closed ya'll..... it has beeeeguuuuun
The hat business?
No, Alabama Senate race counting
@Daminark No trolling, that's legit the easiest example.
The continuous versions are all essentially smoothed out versions of that.
(Well, yours was a parabola instead of a circle.)
01:22
@Akiva wait what? It has a limit at 0 though
@Daminark Take the limit towards a point on the circle
Oh... so that's what you were going for, okay
Or shift the circle to the right one unit, either way
For a second I thought you were shitposting
Nah I was being legit.
@KevinDriscoll But yeah the continuous versions (like the rational function Dami showed you earlier) essentially work like that, just a bit smoothed out
(and Dami's was $1$ on a parabola instead of a circle)
(and by "continuous" I mean "continuous away from the bad point")
 
1 hour later…
02:56
I'm trying to figure out why one of my comments were deleted. Where would be the best place to ask this?
math mod chat
thanks kevin... i will try to find that :)
@randomgirl If you click site rooms in the upper right, on the page it takes you to you shoulkd see Math mods office
Thanks so much kevin
you were extremely helpful
I did find it
 
2 hours later…
04:48
Apparently people in Roy Moore's campaign don't understand how math works. Doug Jones has 49.9% of the vote. Roy Moore has 48.4%. There were 1.7% write-in votes. Moore's campaign was just arguing that the write-in votes has to be verified to make sure that they're legitimate votes for a person who could actually be a US Senator from Alabama.
And they're banking on that changing the margin of victory enough so that it shrinks from 1.5% to below 0.5% to trigger an automatic recount.
Except..... that's impossible by disqualifying write-ins. The margin is $\frac{\text{Jones vote} - \text{Moore vote}}{\text{Total votes}}$ if the denominator goes down by disqualifying write-ins, the margin goes UP, not down.....
I know you all know this I'm just kind of flabbergasted
Unless I'm being incredibly tired because its late and am overlooking something simple here
05:14
hopefully I'm just missing something obvious, but why does $p(p-1) = (a-1)(a^2+a+1)$ and $p$ is prime, mean $(a^2+a+1)/p \equiv 3 mod (a-1)$ ??
Additional context if necessary: https://math.stackexchange.com/questions/2561102/find-values-of-p-such-that-p2-p1-is-cubic https://artofproblemsolving.com/community/c6h35935p3629211
05:46
@PPenguin as mentioned in the comments, either (i) $p\mid(a-1)$, in which case $(a^2+a+1)\mid(p-1)$, or (ii) $p\mid(a^2+a+1)$, in which case $(a-1)\mid(p-1)$. The former is impossible as $p\mid(a-1)<(a^2+a+1)\mid(p-1)$ would entail $p<p-1$. Thus, $p\mid(a^2+a+1)$, and $p\equiv 1$ mod $(a-1)$. As such, $$ \frac{a^2+a+1}{p}\equiv\frac{1+1+1}{1} \mod{(a-1)} $$ since $a\equiv1$ mod $(a-1)$.
somehow I got all that but the last sentence and a half. But I think I can connect it now.
(p-1) = k(a-1) ... so p-1=0 mod (a-1) ... p = 1 mod (a-1)
similarly a = 1 mod (a-1) ... and then it all falls in place
Thanks!
 
1 hour later…
06:58
hats are here!!!
NANI???
Down, Balarka
Left, not down.
I live far left
#FOREVERRESISTSCAPITALISM
How can you be nti capitalist
and love hats?
07:14
R E k t
I believe hats should be available for everyone
we should all get a free hat
(this year's winterbash is too hard goddamnit)
Okay unrekt then
wait how does StackExchange make money>
@Kevin on the other hand, why are YOU propagating false claims about me being secretly a capitalist pigdog?
oh, now you are proposing SE is a capitalist site???
are you Wall Street Journal?
Because I am a capitalist pigdog, and thus am trying to undermine your pinko bona fides
at least thats what I might say if I were a lefty spy pretending to be a capitalist
07:22
you are a false news organization working for the right
this is a communist chat. go away, shoo
I refuse. You have no legitimate monopoly on the use of force here!
inb4 gulag
I will counquer your gulag with the power of freeze peach!
07:41
Hats are back ?
07:52
Yo @Eric
I'm not here @Daminark
Uh huh
08:40
Did I open the wrong chat?
2
No komrad this korrekt gul... khat!
09:00
For projective curves defined by a polynomial, how do you differentiate at infinity?
 
1 hour later…
10:21
Hi! I want to make sure about the meaning of "distribution of input". since we have possible choice of input and usually probability is distributed through these possible choice of input. So the distribution of input is the set of possible choice.
Do you have any argument!
I'm asking this question because I read this following statement: But the truth is that typically (e.g. in most practical applications), very little if anything is known about the input distribution.
10:48
Let $A,B \in M_n(\mathbb R)$ be such that $A+B=AB$. Then $AB=BA$.

$A+B=AB$ and $B+A=BA \implies AB=BA(\because $ Matrix addition is commutative). Am I correct? I am not confident with the argument. because It had appeared in the PhD Screening test.
Why does $B+A=BA$ hold?
Why is a semisimple Lie group $G$ its own commutator subgroup $[G,G] = \langle ghg^{-1}h^{-1} : g,h \in G \rangle$?
Semisimple meaning that $G$ is connected, nonabelian and has no nontrivial, connected abelian, normal subgroup.
11:05
@AlessandroCodenotti $A+B=AB$ given.
Ok, so $B+A=AB$, how does $B+A=BA$ follow?
I thought that $A+B=AB$. so, here first term is A and second term in in the summation is B. so multiply according to that.
@ManeeshNarayanan lol hi
@AlessandroCodenotti
$A + B = AB$ does not imply $B + A = BA$. You can't switch $A$ and $B$'s position. That's completely wrong.
11:17
@ManeeshNarayanan multiply what?
$A$ and $B$ are specific matrices.
@LeakyNun I am weak in pure mathematics. please don't tease me. I am not boasting. I am good in the calculus.
2
@ManeeshNarayanan no, that's a genuine question
you said "multiply according to that"
i ask you "multiply what" for clarification
btw
0
Q: Check whether the statement true or false

Maneesh NarayananTIFR-GS 2018-PhD, Int-PhD Screening Test. Check whether the statement true or false The set of nilpotent matrices in $M_3(\mathbb R)$ spans $M_3(\mathbb R)$ considered as an $\mathbb R$- vector space. I think the statement is false. Nilpotent matrices have zero diagonal entries. so it can...

"Nilpotent matrices have zero diagonal entries" — Kenny Lau 3 mins ago
@LeakyNun Really I don't know.
2
@ManeeshNarayanan but you said that yourself
11:23
@BalarkaSen thank you.
11:33
Any algebraic topologists out there who knows homological orientation?
All the math I knew I have forgotten
I need help ASAP
ASAP rocky
Ask in the very scary homological algebra chat over at MO
oh i'm not asking about homological algebra
this shtick's in Hatcher 3.26 a
but i have forgotten it and I never liked hatcher's exposition of it
I think there's a quicker proof
@AlessandroCodenotti I think the statement is false. Right? but I don't have counterexamples also. I am weak in proving. How to improve the problem-solving skills in Pure Mathematics? Some areas I am good. some areas I am bad.
Please suggest some elementary books that improve my proving skills. I have gone through a bad curriculum structure. Please help me.
A+B=AB is equivalent to (A-I)(B-I)=0, so you might want to look at zero-divisors to search for counter-examples
@Leaky you asked whether $\bar{\Bbb C(X)}$ is isomorphic to $\Bbb C$ the other day, right?
@Leaky you asked whether $\bar{\Bbb C(X)}$ is isomorphic to $\Bbb C$ the other day, right?
11:47
maybe
Did you work that out already?
@LeakyNun Thank u.
Say, does anyone know a method of finding the integral of $(x+1)^x$ ?
@AlessandroCodenotti I think yes
11:55
@LeakyNun $A+B-AB=0 \implies A(I-B)+B-I=-I \implies $(A-I)(B-I)=I$. I don't think it is the problem of zero divisors. please see.
oh heh
you've proved it already
Am I correct?
2
yes
(A-I)(B-I)=I
you can easily see the proof
(provided that you really have an MSc)
yes. I have M.Sc. I graduated this year. my convocation has over :) Being a graduate, that doesn't mean he/she has good knowledge and base. If it would be correct, Everyone will be a good researcher.
2
I had side business during my graduation. After completing the course, I was attracted to differential geometry and General Theory of relativity. I wish to master in the proof technique. I really don't know, how? I was lazy during my graduation. Please give advise, how to master in the proof techniques.
2
12:15
Anyway @Leaky the right approach is to think about transcendence basis, the "kill a fly with a nuclear bomb" approach is Steinitz's theorem that $\sf{ACF}_0$, the theory of algebraically closed field of characteristic $0$ is $\kappa$-cathegorical for all uncountable $\kappa$
sure, thanks
@AlessandroCodenotti is $\overline{\Bbb Q(\pi)}$ and $\overline{\Bbb Q(\pi,e)}$ isomorphic? :P
Is $\{\pi, e\}$ algebraically independent over $\Bbb Q$? :P
12:28
> a graduate doesn't necessarily have good knowledge and base
the most ridiculous thing I've heard today
@LeakyNun :)
3
to get a graduation. you need some cut-off. Right? 50%
2
not 100%
@LeakyNun Sadly, it's definitely true.
On the one hand, there are plenty of students that focus very hard on simply passing a course rather than truly understanding it; and who take as many "easy" courses as possible
12:43
sad
On the other hand, at plenty of institutions, especially with high tuition fees, there can be a mentality that since the since the student paid so much to study, he should graduate; and staff is discouraged to fail students
Hello will anyone tell me is tan x considered continuous if we remove points such as 90 deg from domain I think it will still be discontinuous due to limit as x goes towards 90 deg from both sides!
@samjoe It's discontinuous if it's discontinuous at some point in its domain.
@samjoe define continuous
If you remove those points from the domain, at which points of the domain will it be discontinuous?
12:45
@TastyRomeo I prefer to ask them definitions in order to check if they even know what they are asking
58 mins ago, by Maneesh Narayanan
Please suggest some elementary books that improve my proving skills. I have gone through a bad curriculum structure. Please help me.
Ugh, why or why it's always that indian education system
because many people don't
someone bomb the department of education in india
@Leaky isnt there already a definition.. oh maybe there might be many definitions, i was told that limits from both sides must exist finite, and that function must be defined at that point. Yeah limits of tan x as x goes to 90 deg becomes +- infinity
@Eric @Daminark yo
12:47
@samjoe there is, but you may not know it
if you don't know the definition of the terms in your question, then you don't know what you're asking
then why should we answer?
@Tasty But tan x is defined at all points if we remove points like 90 from domain. stlll isn't it discontinuous as limit becomes infinity
@leaky i told you no what I knew
FFS, had they not told us they are lazy, we will still have some sympathy on them being screwed by the education system
@samjoe at all points of its domain
@BalarkaSen sup
@Tasty I am giving an exam where tan x was said to be discontinuous still after removing points like 90 degree. I am searching for reason for that
12:53
@EricSilva what's a quick proof that $M$ is orientable implies $H_{top}(M) \cong \Bbb Z$ singularly?
I only remember the Hatcher proof which was some stupid unenlightening thing
I don't like Hatcher's exposition of the proof
@Tasty I reckon that the answer in test was wrong. So if 90 degree is removed from domain, we will not be considering lhl and rhl to 90 degree too. That makes tan x continuous on its domain mayb
I don't really like hatchers presentation of orientation in general
I love local/homological orientation
I know some people doesn't
but it's actually a lot intuitive
13:00
My problem isnt really with homological orientation
More I don't like the way this section is written as a whole
That's fair
Sorry i was lagging
If you're a Delta complex can't you just take the sum of the simplices as generator of top homology
ya
well
signed sum
Sure w.e. signs
13:04
I guess the proof idea is something like that. A generator of $H_n(M, M - p)$ is represented by a singular simplex $\Delta^n \to M$ with image of interior containing $p$
@TastyRomeo but I have studied in prestigious institutions in India. I thought that Maths means Calculus and Calculus means maths. I was interested in Physics. I thought that Entire mathematics required for doing physics. amateur thinking after schooling. :) I paid my 5 years for that. I luckly completed M.sc. :) I wish to do PhD in GTR only.
so you somehow triangulate $M$ like that
Yeah i don't think I wanna think about alg top for a couple weeks
After just doing an exam in it
And having a reading course in it coming up
My brain needs a break from alg top
13:12
@Maneesh Hello! My name is samjoe (in profile also). I am also interested in maths and current level of ed is high school (indian +2)
@Eric I think Hatcher's proof becomes somewhat less obscure when though it terms of sheaves
thought
the assignment $U \mapsto H_n(M; M - U)$ is actually a sheaf of abelian groups on your manifold $M$
the stalks are where local orientation lives
global orientation is a global section of the sheaf
... oh
$H_n(M)$ is exactly the group of sections of this sheaf I think
That should be the key thing in trying to prove existence of global section => Hn(M) = Z
correct
and the orientation double cover its etale space
(I should add that the espace etale is the orienta--- fucking hell)
ehehe
Sheaves are built to formulate local to global theorems and the orientation is just like the cleanest possible example
I never thought about this before
13:31
@samjoe are you in CBSE?.then it is hell. They don't teach basics of higher mathematics. I am not complaining the board. but, my calculus is very good :). It was my mistake. I should have read what is Higher Mathematics during my high school itself. I shouldn't judge from the curriculum framework. :)
@Maneesh Yeah cbse. Its customary for us to take up cbse and then appear for jee. Almost everyone seems to be doing that.
@Maneesh I know very little (read none) about higher maths, mostly i fascinate.
@MikeMiller I guess then the question boils down to patching up a consistent choice of element of each stalk to a section of the sheaf. I wonder if you can just do this "abstractly" or whatever
@Maneesh My calc is fine, but I hate analytic geometry (conics) and also don't like algebra much. So are you from cmi ?
Well, yeah, you can do that abstractly. I guess that's not the issue. I suppose the technicality lies in seeing $H_n(M)$ as a group of sections of the sheaf? There's a morphism $H_n(M) \to \Gamma$ given by sending each homology class $\alpha$ to the section $s : M \to \tilde{M}$ defined by $s(x) = \alpha_x$
But it's not clear if it's an isomorphism
I like this
13:40
JEE is the big curse for Indian mathematics :)(My opinion). I don't wish to give a negative impression to my institution. It might be my own mistake. I don't want to reveal my institution. sorry
if you have potential. institutions don't matter :)
I want to prove that $h$ is a parametrisation of hyperboloid $S$. To do this, I need to show that $h$ is surjective onto $S$. I see that $S$ is just a set of pairs of circles of varying radii. I see that for a given circle, I can always find some $(u,v)$ in the domain of $h$ such that it "draws" that exact circle. So then, is that enough to say that $h$ is surjective. First time I am trying this method. Is it sound?
@Maneesh Never mind about that! I was actually preparing for jee but thought about maths related institutions.
Hatcher's argument makes so much more sense now
@Maneesh Yeah thats very true.
Can you suggest some readings for high schoolers
13:50
I'm looking through the "matrix cookbook" and I'm somewhat confused by the notation in equation 37 there, since it's not explained earlier. The book I'm referring to can be found here (https://www.math.uwaterloo.ca/~hwolkowi/matrixcookbook.pdf)
What is meant by (∂X) Y there?
Yeah I was askin you only
read the set theory from the first chapter of Analysis and algebra book. I hope that will help you. ask your doubts to experts. group discussion is the best thing for Higher mathematics topics. please be sure that you are not deviating from the curriculum frameworks of CBSE.
:)
all the best
these are the things that i haven't done during my schooling. so, I am suffering now. :)
@maneesh you are doing bsc or msc maths
completed MSc
:)
see u later
@Rickyfox It's just $\partial X$ is a matrix and $Y$ is also a matrix so it is just matrix multiplication.
13:53
ok thanks for advice!
I almost don't dare to ask, but what is just $\partial X$? I mean, I know partial derivatives of functions, but now that I'm dealing with matrices I'm completely lost
@Rickyfox At a guess, it's element-wise differentiation
You take the derivative of every element
@Rickyfox Check en.wikipedia.org/wiki/Total_derivative and en.wikipedia.org/wiki/Matrix_calculus to get up to speed on matrix calculus.
the matrix calculus one I have open right now, the total derivative I haven't seen yet. Thanks, I'll have a look at it
@berrygreen thanks, will have a look at it
14:12
Whoa, is the word "plateaued" really spelled "plateaued"?
That's two many vowels.
What's wrong with it?
I'm working on an optimization problem for the first time and now my lacking knowledge of calculus comes to bite me. If I understand it correctly I need to get the derivative of my objective function with respect to the target variable and then get the gradient of that.
I'll post my (edited) version here later for anyone interested
14:41
0
Q: The smallest element without Unique factorization ? (Conjecture)

mickConsider a non-ufd ring $M = Z(\sqrt n)$ for a squarefree integer $n$. Denote the elements as $a + b \sqrt n$ ( $a,b$ are integers ) Define the quadrant as the elements with $a,b $ non-negative. Define the $m$ th element of the quadrant by $C(a,b) = m$ where $C$ is the cantor pairing function....

Any ideas ?
Hi pseudohuman. Can you beat alphazero ? :) @Pseudohuman
@mick I do not understand.
@Pseudohuman Recent chess engine developed by Google that was only given the rules of chess (no strategy), a reinforcement learning algorithm, lots of computing power, and four hours to train
@AkivaWeinberger I do not understand.
It plays chess
It's a computer
It figured out how to play chess on its own
14:59
Pseudohuman is a bot
15:13
0
Q: Problem with Two-valued Measure in Rudin's RCA

user193319 Let $X$ be an uncountable set, let $\frak{M}$ be the collection of all sets $E \subseteq X$ such that either $E$ is countable or $E^c$ is countable, and define $\mu (E) = 0$ in the first case, $\mu (E) = 1$ in the second case. Porve that $\frak{M}$ is a $\sigma$-algebra in $X$ and that $\mu$ i...

@AkivaWeinberger got trolled :P
15:36
Oh
Lord
15:59
What does "most general value of "$\theta$" mean?
@ManeeshNarayanan WHY do you call it a big curse?
16:20
lmao @Akiva tfw you talk about machine learning with a machine
5
epic conversation
I am reading a book, and there before proving that there are infinite primes, they ask this: "What is wrong with the following argument?-There are infinitely many positive integers. Each of them factors into primes by the fundamental theorem of arithmetic. Hence there must be infinitely many primes."
I can't see exactly where there is flaw in argument.
Consider the set of numbers of the form $2^n3^m$ and apply the same argument to this set
nice
though powers of 2 alone would seem to work just as well.
Oh, of course
17:07
Claim: $f_n : X \to [0,\infty]$ is measurable for every $n$, $f_1 \ge f_2 \ge ...$, $f_n(x) \to f(x)$ for every $x \in X$, and $f_1 \in L^1(\mu)$ implies $$\lim_{n \to \infty} \int_X f_n d \mu = \int_X f d \mu$$. Proof: Since the functions are nonnegative, $|f_n(x)| \le f_1(x)$ for all $x \in X$ and every $n \in \Bbb{N}$. Lebesgue's Dominated Convergence Theorem says that $\displaystyle \lim_{n \to \infty} \int_X f_n d \mu = \int_X f d \mu$. QED
Is it really that easy?
Hi. I get the idea of looking at the central binomial coefficient as in the top answer of mathoverflow.net/questions/258711/…, but what if, say, I have $\sum_{k=0}^K\binom{K}{k}^2 K^{-\alpha}$ for some $\alpha\in(0,2)$. Then because of $K^{-\alpha}$ the main contribution is from $k=0$ to some $x$. My attempts show that taking $x=C K^{\beta}$ with some $\beta<1/2$ might work, but I don't see how to show that.. any ideas?
@AlessandroCodenotti Wow! Never thought like that! Thanks a ton.
@Mathei @Balarka want to give me an hint on a Galois theory exercise?
So we have $F$ with characteristic $p>0$ and $E/F$ an algebraic extension with the property that whenever $e_1,\cdots,e_n\in E$ are linearly independent over $F$ also $e_1^p,\cdots,e_n^p$ are linearly independent over $F$
I'm asked to prove that $E$ is a separable extension
18:00
Given three solutions $y_1, y_2, y_3$ to an non-homogeneous 2nd order linear equation, how can you construct a fundamental set of solutions to the corresponding homogeneous equation?
I got ze hat
@Balarka nice
18:33
Hi all, I just asked a question on the main site about determinants:
0
Q: How can I calculate the determinant of this type of matrix?

overactorGiven the matrix $$ A = \begin{bmatrix} 1&1&1&1\\ a&b&c&d\\ a^{2}&b^{2}&c^{2}&d^{2}\\ a^{3}&b^{3}&c^{3}&d^{3} \end{bmatrix} $$ How can I prove that $\det(A)=(d-c)(d-b)(d-a)(c-b)(c-a)(b-a)$ ? And as a second question, what is the general formula for this type of matrix and how can I prove i? ...

one word google hint: Vandermonde @overactor
I'd very much like to solve it myself, so if anyone could give me a push in the right direction, without having to write down the full answer, I'd really appreciate that.
@Semiclassical I'll look that up
Thanks
simplest way, I think, is just gaussian elimination / elementary row ops
How's it going @Semi?
Ovi
Ovi
18:43
Can I ask somebody just a quick question please? I am writing a small project on the tower of Hanoi and I have to write an abstract. Should I explain the the abstract what the tower of Hanoi is?
Or just give the results relating to it?
@Ovi Depends who the intended readers are
Ovi
Ovi
@TobiasKildetoft The intended readers are some professors/students who may not be familiar with what the tower of Hanoi is
My initial inclination was to include an explanation, but I wasn't sure if that was normal
if you do say something in the abstract, make it short
no more than a sentence
If you do not expect them to be familiar with it, then I agree with Semi to include a brief sentence of what it is
Ovi
Ovi
@TobiasKildetoft @Semiclassical Hmm it's going to be extremely difficult to explain it in one sentence, I will give it a try.
18:57
don't try to explain it entirely. that's what the paper is for
Ovi
Ovi
Ah okay
"The Tower of Hanoi is an old puzzle in which one must move a tower made from different sized discs from one peg to another, all the time keeping smaller discs on top of the larger ones" or something like that
Ovi
Ovi
Thank you so much!
both :)
 
1 hour later…
20:22
hi
is there anybody online
i need help with PROOFS
Is there a topologist here?
eh ?
my question is Prove that, for all sets A, B and C, if A ⊆ C and B ⊆ C then A ∪ B ⊆ C.
@Isabella Ask; don't ask to ask. Depending on the question I may or may not be able to help.
Discrete MATH
where are all Mathematicians?
@BalarkaSen ok thank you. It turns out today my professor told us that if someone S is able to give a new proof for the Tychonoff theorem, then S will become famous, very famous. Is that really true??
20:29
There's lots of different proofs of Tychonoff theorem as far as I know.
Not really an expert on point-set topology, though. @Alessandro might be the authority here.
the wiki page for Tychonoff describes like 5 five different proofs
Any help will be appreciate for Discrete Math. :(
Don't spam. If someone wants to answer, they will.
There's probably one proof translated through the various equivalent definitions of topological spaces/compactness.
0
Q: Problem with Two-valued Measure in Rudin's RCA

user193319 Let $X$ be an uncountable set, let $\frak{M}$ be the collection of all sets $E \subseteq X$ such that either $E$ is countable or $E^c$ is countable, and define $\mu (E) = 0$ in the first case, $\mu (E) = 1$ in the second case. Porve that $\frak{M}$ is a $\sigma$-algebra in $X$ and that $\mu$ i...

20:40
I guess my professor refers to the wiki proofs, if someone gives the 6th proof then will become a famous mathematician
interesting:D
20:56
I doubt people care that much about Tychonoff
I've only ever seen the full blown one matter for Banach-Alaouglu
it's what makes your inverse limits compact
@BalarkaSen I've only seen one, I remember the general idea and I think I could work out the details if needed, but it's one of those results where the proof isn't very enlightening and you just remember the theorem
That ^
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