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00:07
@DonLarynx And if I want to use a nxm array instead of lists, where n is the number of topological sortings and m=|V|, how could I do this?
@evinda It's the same thing, list is just the name of the 2-dimensional array. HINT: Go through the ith row and each jth column of the ith row.
for(int i = 0; i < numberOfT; ++i){
            for(j = 0; j < |V|; j++){
                 list[i][j] = "v, x";
            }
        }
@DonLarynx You mean like that?
@MaryStar As $A$ moves up one unit, $B$ moves down one unit.
@evinda "Variable names in C are made up of letters (upper and lower case) and digits."
@MaryStar thus, the distance between $C$ and $D$ decreases by two units.
@MaryStar that means the distance from $D$ to $E$ increases by four units.
@MaryStar thus, combining the drop of $B$ and the increase of the distance from $D$ to $E$, we get that $E$ drops five units.
Thus, the force at $E$ is $\frac15$ that at $A$
From there, it is pretty easy to get the rest of the forces by equalizing the torques
user147690
00:21
Have you guys heard of the button? It's a social experiment on Reddit where the button counts down from 60 seconds, and every time someone clicks it, it resets to 60 seconds left. 864,075 participants so far.
user147690
One month and one day have passed since it started, and it has never reached 0.
@alex so its kind of one of those "last person to post in this thread wins"
user147690
@JMoravitz Yeah I suppose haha. I think they just wanted to see how long it would last. It has heaps of statistics recorded.
Trending topics on the internet always seem to garner more participation than people would expect. Twitch-plays pokemon for example.
I wouldn't be surprised if it lasts a whole three months
user147690
00:30
@JMoravitz It seems to be dying haha. Only people with accounts older than the start of the button can click it, so there is a limited pool, and we are already at over 2/9ths of the whole population of people who can click, which many of which are inactive
I'm having a little trouble computing the flow of the vector field (x+1)d/dx -(y+1)d/dy
@AlexClark Are you sure that's real probabilistic analyzation
user147690
@ᴇʏᴇs It seems bad after I read it fully.
Yea it looked kind of sketchy to me
user147690
@ᴇʏᴇs What are you working on?
00:37
@AlexClark Topology homework
user147690
@ᴇʏᴇs Point-set or algebraic?
The vector field is on R^3. The ODE system x'=x+1, y'=-y-1, and z'=0 is easily solvable. You get x=aexp(t)-1,y=bexp(-t)-1, and z=c. So I would expect the flow to be (xexp(t)-1,yexp(-t)-1,z)...but that's not a flow. What am I doing wrong?
@AlexClark Point-set; I'll take algebraic next semester
00:50
Nevermind, I figured it out
01:09
Is it as followed?? @robjohn
We need one more relation to solve them, right??

Or have I understood it wrong?? @robjohn
@MaryStar Did you draw this in Microsoft Paint
Yes. @ᴇʏᴇs
01:28
@MaryStar $F_Z=20\text{ N }=\frac15\,F_L$
 
1 hour later…
user147690
02:51
Take a group $G$ acting on a set $A$, is the kernel of the action $\{g\in G| ga=1\}$ or $\{a\in A|ga=1\}$?
It should be the kernel of the cooresponding homomorphisms $G \to S_A$ (symmetric group on $A$), which would be the set of $G$ which would be $\{ g \in G \mid ga=a, \forall a \in A \}$ @AlexClark
$A$ does not, in general, contain $1$
user147690
Thank you @Paul
user147690
03:40
Say we have $(\Bbb Z,+) \circlearrowright \Bbb Z_5$(so left addition coming from the integers to $\Bbb Z_5$), the stabilizer of each element is the action of addition by any multiple of $5$ - and for $(\Bbb Z, \times) \circlearrowright\Bbb Z_5$ we have: stab(0) is everything, stab(1,2,3,4) is $\{1\}\cup \{6n|n\in\Bbb Z_{\ne 0}\}$
user147690
E.g. in both cases $e\pmod 5$
user147690
The identity group action, will always be a stabilizer, and thus we have identity. So the stabilizer of $a$ is a set of group actions that fix it
user147690
Oh nevermind, I just answered my own question that I was about to ask
Well multiplication doesn't form a group, you can probably come up with something which is analogous to action though. Stabalizer of 0 in that case is only 0, since multiplication acts on everyting (making them all 0), and for the other one no to that too. It looks like you are doing the same thing you did above, which I said was basically wrong.
user147690
Ahh crap, I seem to be bad at diverging from the content in the questions to explore
03:49
When you act on a set there is no distinguished element, so nothing should be special, like "1" or the identity, or 0, etc.
Looks like you applied the correct idea to addition though.
user147690
@PaulPlummer That's a relief :P
@AlexClark If you know about the dihedral group, you can look at some dihedral groups acting on regular polygons, and think about different points in the polygon (point on the middle of an edge, corner, center, outside the polygon, inside the polygon, off center of an edge, etc) and ask yourself what is the stabilizer of "this point", what is the orbit, etc. It is a nice exercise to get a grip on some of this stuff. Also it might be good to try, similar examples to what you tried
user147690
@PaulPlummer I'll give it a go, thanks
@TedShifrin
0
Q: Proof verification: ab = gcd(a,b)lcm(a,b) without use of prime factorization

Karim MansourI am trying to proof $ab = gcd(a,b)lcm(a,b)$. The definition of lcm(a,b) is as follows: t is the lowest common multiple of a and b if it satisfies the following: i)a | t and b | t ii)If a | c and b | c, then t | c. Similiarly for the gcd(a,b). Here is my proof: Case I: gcd(a,b) $\neq$ 1 ...

done
04:21
HI@KarimMansour
Hi @Rememberme
just gave you vote up
on your question
nice one
which one?@KarimMansour
representing groups geometrically
Oh okay... thanks
@TedShifrin please let me know what you think when you come back..
I am off to sleep cya guys.
04:53
math.stackexchange.com/questions/1261530/… @TedShifrin pls check this also also as well
Peace i am going to play some games now @TedShifrin pls look at the question
05:12
woo, got a research job for the summer
Congratulations! What are you researching?
looks like it'll be some graph theory
stuff related to something called the path partition conjecture
Interesting. Well, hopefully it's fun
congrats @SamuelYusim
@KarimMansour can you check my question
05:19
sure
1 moment
choose a basis for $F^2$ and then do what you did and your done
$F^2$ is some field right?
@KarimMansour Can you look at my question too ?
sure 1 moment
@KarimMansour will the standard basis suffice
05:27
okay fine then i am done
thanks a lot @KarimMansour
@karim you like music?
hey guys, I need your help. In the following link I submitted my problem. I hope you get interested in it. :-)
4
Q: My two assumptions look inconsistent. How can we explain it better?

NewbieBackground I am trying to solve the following problem and explaining it to my students. However, my students (as well as I) think of my two assumptions contradictory. > Given 2 distinct curves $C_1: y=f(x)=e^{6x}$ and $C_2: y=g(x)=ax^2$ where $a>0$. The objective is to find the range of $a...

@MathGod I did a problem related to this before 1 moment its about general prime divisors of fermat numbers.
yeah @Rememberme I like classical music and everytype
@Karim then you might have done perfect numbers also right?
05:30
yeah
@KarimMansour Can you tell me the problem or give a link to that problem ?
If you have then i have a conjecture on it and cubes that will literally blow your mind
lets speak about it tomorrow @Rememberme I am very exhausted
:D
fine.... :D
tomorrow though remind me
anyway good night guys cya tomorrow
 
3 hours later…
08:04
@MikeMiller OK, thanks.
@PaulPlummer I didn't figure that you were Disciple of Barney when we were talking about commutative diagrams :p
Changing usernames can be confusing...
Haha, @BalarkaSen Sorry :P
Getting any better?
(in reference to the stomach problems you were mentioning)
Yeah, feeling a lot better today. Having a buckload of antibiotics, so a bit weak. But other than that, never been better.
Thats good. Sometimes I like that "high" you get from changing from being awful to feeling good/normal because it feels amazing not being sick, so you feel extra good.
yeah, haha, that's true.
08:23
hi @PaulPlummer@BalarkaSen
Hello @Rememberme
sorry again
@BalarkaSen you know what is my new profile pic??
kissing spheres?
Boy surface?
08:25
i haven't bothered to look carefully
nope a twisted torus........
oh, it is RP^2, I guess.
@Rememberme dunno what a twisted torus is. klein bottle?
its basically a way to cut a torus such that the two tori which i get are interlinked
basically connected
i don't get what you mean, but nevermind.
there are hundreds of examples of twisted torus
08:29
@BalarkaSen I think he is talking about this youtube.com/watch?v=3_VydFQmtZ8
@PaulPlummer yes!!!!!!
Your pic doesn't look like one...
i am too lazy to open that youtube.
i guess you mean knotted torii, @Remember.
oh i dont want to go into a subject which i dont know thoroughly yet...@PaulPlummer i just thought of changing the pic and found this beautiful shape and got some half eaten knowledge about it :p
yes @BalarkaSen
Amazing shapes anyways
ok i think i got go know ..... peace!!!!!!!!
08:59
@robjohn I want to ask you something about your answer at physics.stackexchange.com/questions/179540/… . Could you explain to me further how you found $F_H=120N$ ?? I understand that it is equal to the sum of the forces $A$ and $E$ but how did you find that it is equal to $120$ ??
09:37
hi guys. i want to ask what is `BR` and how is it evaluated (or defined) in this question (http://stackoverflow.com/q/29996696/4481312). actually there is a link in one of answers to a wikipedia post (http://en.wikipedia.org/wiki/Abel%E2%80%93Ruffini_theorem).

i proposed a possible answer but have no idea how to finish it.
in the discrete fourier transform formula for FFT, what does 'i' and 'k' represent?
09:59
@Hippalectryon lol, I like that. :-)
10:16
@MaryStar $F_H=F_Z+F_L=20+100$
@edition what formula are you looking at?
A fast Fourier transform (FFT) is an algorithm to compute the discrete Fourier transform (DFT) and its inverse. Fourier analysis converts time (or space) to frequency (or wavenumber) and vice versa; an FFT rapidly computes such transformations by factorizing the DFT matrix into a product of sparse (mostly zero) factors. As a result, fast Fourier transforms are widely used for many applications in engineering, science, and mathematics. The basic ideas were popularized in 1965, but some algorithms had been known as early as 1805. In 1994 Gilbert Strang described the fast Fourier transform as "the...
thankyou for your question.
@edition $i$ is of course the square root of negative one.
thankyou
$k$ is just an index for the transformed sequence.
10:37
We have $F_H=2F_4$, $F_4=F_7+F_6$, $F_7=F_5+F_Z$, $F_2=F_5+F_6$

$F_4=F_5+F_Z+F_6=20+F_5+F_6=20+F_2$

$F_L=F_1+F_2\Rightarrow 100=F_4+F_2 \Rightarrow 100=20+F_2+F_2 \Rightarrow 2F_2=80 \Rightarrow F_2=40$

$F_4+F_2=100 \Rightarrow F_4=100-F_2 \Rightarrow F_4=60$

$F_H=2F_4 \Rightarrow F_H=120$

Is this correct?? @robjohn
user147690
@Guesswhoitis. J.M did you talk here much before leaving?
@AlexClark I do remember having not a few conversations with rob, t.b. and Asaf when he was still a habitue...
user147690
@Guesswhoitis. Yes I can't recall why he left - I also can't recall why you left
@AlexClark Have we talked before? My memory is faulty these days... about Asaf, he disagreed with some things done on chat, so he boycotted in protest.
@Guesswhoitis. Finally, the answer matches mine.
10:51
@robjohn $\Gamma$s all around... :)
@Guesswhoitis. Yeah... The question started out being a neat convergence problem, but it had a typo and the OP changed it and asked for the value.
I should compute the value of the originally intended series...
user147690
@Guesswhoitis. We have talked before yes, not on this account of mine[I wanted to become myself now that I am starting to do real math].
@AlexClark Ah, I see. :)
19 hours ago, by robjohn
A Question to show that $$\sum_{n=1}^\infty \frac{n!\cdot 3^n}{7\times10\times\cdots\times (3n+4)}$$ converged had a typo and was then deleted. I thought it was an interesting question.
Hmm, that @Remember guy seems to have a nice Wente torus on his Gravatar; pretty cool.
11:23
@Guesswhoitis. It is $3$
This looks like a cool question. I don't know how to do this.
Not at first sight, at least.
@BalarkaSen I could stare at it all night and have no idea.
@robjohn That's for the original formulation, yes?
ack, the answer there uses very heavy machinaries.
ah, the other answer seems to be what i want.
@Guesswhoitis. yes, with the typo corrected.
@Guesswhoitis. The terms decrease like $n^{-4/3}$
so it barely converges.
11:30
Yes, that decay rate is rather lazy.
Again, I used Beta functions to compute the sum. I wonder if there is a more elementary way to do it.
The sum being $3$ may be deceptively simple.
Or it may have been due to a fortuitous cancellation of nonelementary terms.
@Guesswhoitis. that's what I meant by deceptively simple
$$
\begin{align}
\sum_{n=1}^\infty\frac{n!\cdot 3^n}{7\times10\times\cdots\times (3n+4)}
&=\frac43\sum_{n=1}^\infty\frac{\Gamma(\frac43)\Gamma(n+1)}{\Gamma(n+\frac73)}\\
&=\frac43\sum_{n=1}^\infty\int_0^1(1-t)^nt^{1/3}\,\mathrm{d}t\\
&=\frac43\int_0^1(1-t)\,t^{-2/3}\,\mathrm{d}t\\[12pt]
&=3
\end{align}
$$
Thanks for the comment @Guesswhoitis.
@robjohn Hmm, in your first line, the general term is then $n!/((7/3)_n)$, right?
11:39
actually, i think it was false. i was being too hasty.
whatever, the author deleted the answer.
@Rememberme It is a Wente torus, right? Or did I err?
@Guesswhoitis. rising factorial, yes
@robjohn So it comes out as ${}_2 F_1\left({{1,1}\atop{7/3}}\mid 1\right)-1$. I think the simplification is only because it's a unit-argument hypergeometric function.
oh, i figured it out. one just needs to contract boundaries of the attatched cells inside other cells.
bleh.
11:55
See e.g. this.
(I think a simple proof of that identity is somewhere in CM; that could be used for this specific situation.)
Greetings
0
Q: Calculate in closed form $\int_{0}^{1}\int_{0}^{1}\frac{dxdy}{1-xy(1-x)(1-y)}$

Chris's sisCan we possibly compute the following integral in terms of known constants? $$\displaystyle \int_{0}^{1}\int_{0}^{1}\frac{dxdy}{1-xy(1-x)(1-y)}$$ Some progress was already done here http://integralsandseries.prophpbb.com/topic279.html but still we have a hypergeometric function. What's your thou...

If it's not a hypergeometric function with argument $\pm 1$, it usually can't be made more elementary.
12:18
Is the way I calculated the forces wrong?? @robjohn
lol, no one upvoted my question so far. Perhaps because it's a frustrating question ... (well, it's not that easy, that's true)
Nothing frustrating about it; I just said my take on your question in my previous comment.
@Guesswhoitis. You might like to write a full solution on main to learn from it.
@Chris'ssis The forum post you linked to already ended with the hypergeometric expression; my point was that it seems unlikely to me that you'll get anything more elementary.
@Guesswhoitis. OK
user147690
12:35
Showing $(\Bbb Q, +)$ is not finitely generated via the take $x_1,\cdots,x_n$ rational numbers and pick a natural number $k$ coprime to all of the denominators, then $\frac1k$ cannot be generated by $x_1,\cdots,x_n$ is fine. How do I approach this using the fundamental theorem of finitely generated abelian groups?
@Guesswhoitis.it is a knotted torii
@Rememberme Ah, I see. Did you render this yourself in software?
user147690
Nah he pulled it from here - listed as "Wente's Twisted Torus"
Oh no.....me and my brother were discussing about torus and I got to know about this torus in a video on YouTube
Its very famous you see you can find it in many places
Also try looking about it on numberphile there is a video by them on YouTube @Guesswhoitis.
@AlexClark Oh... I remembered rendering them myself in Mathematica, you see. The elliptic integrals involved took some amount of effort to simplify.
12:45
I didn't get it from there @AlexClark
@Guesswhoitis. you like algebraic topology?
It seems so
user147690
@Rememberme Well it's the same image so it matters not :-)
@Rememberme Not really, but I do like surfaces, and I like special functions. :)
Here's the Wente torus I did a few years ago:
Doesn't matter @AlexClark :) I see your love for the gamma function @Guesswhoitis.
@Rememberme It's... useful. :)
Wow....beautiful....they are just one of the eye catching structures in mathematics
12:50
...so I guess my picture is a fancier version of this. :)
Yes it is
...
Avengers were amazing.... ..pretty good movie
@AlexClark are you studying dihedral groups?
user147690
@Rememberme I am studying many things - right now, finitely generated abelian groups
user147690
@Rememberme Why do you ask?
Oh you were having a discussion with Paul Plummer and he told you something about dihedral groups so I thought you might be studying that because I am studying that from dummit and Foote
user147690
@Rememberme How is it going?
12:59
Fine...I can't day good butits going okay the questions in dummit and Foote are difficult....
user147690
@Rememberme They most certainly are, great book though
Yes amazing book
13:10
Hello
13:24
@Guesswhoitis. That is what Mathematica gives?
@MaryStar I don't know. It is too complicated. I would start the way I did and then figure how the forces get divided up by the torques.
@robjohn I don't have Mathematica on this computer; I just noticed that your sum's hypergeometric, and went from there.
@Guesswhoitis. Ah... That works :-)
Since I don't have my copy of CM with me either, I don't know where in that book is the proof of Gauss's theorem. But I know it's there. :)
hi @robjohn ... I see we have new names here ...
@Ted, like who?
13:38
@TedShifrin If you check the user id, you'll see that Guess who it is. is a rather older user... #70 in chat
I said new names, @robjohn, not new people ...
@TedShifrin Ah, then yes.
Not that I know who they are yet ...
But I have to skedaddle. They're showing my house in 15 minutes.
Well, it would seem you became chat-active for some time after my hiatus, so I understand. :)
@Remember: I think you need to work on mastering most of linear algebra proofs before you go anywhere near Dummit and Foote. Dummit and Foote is a serious graduate text.
Hmm, so who are you, @Guess?
Alternatively, there are arguments for going chat inactive :P
13:41
@TedShifrin I used to be the go-to special functions guy back at the main site. :)
Ah, has @Teadawg met you yet?
@Guesswhoitis. It is nice to see you again... I don't think I will ever see Asaf in chat again. I wonder about Theo.
I'm a Theo, @robjohn :D
@robjohn ah, I wonder about t.b. too...
I was dreaming about something math-related and when I woke up for some reason I said to myself "I have to ask TedShifrin about this" and then I immediately forgot what my dream was about or what I was going to ask lol
13:42
@TedShifrin methinks not.
LOL, I am sorry I'm responsible for your nightmares, mr eyeglasses. My students have blamed me considerably :P
I gather from previous snippets that you're now a professor emeritus, @Ted?
not quite ... still have a set of finals to give/grade ...
better to have merit in retirement than no merit in career? :D
@TedShifrin this Theo :-|
Ah, a last call to give your students the willies. Exploit it well. :)
13:46
ah, he was gone before I ever found the site, @robjohn.
ok, I should clear out ... I'll be back a bit later for a while.
Nice to have "met" you, @Guess.
@TedShifrin see ya.
@TedShifrin CUL8R
14:22
@TedShifrin Eh, I use to think that D-F is an undergraduate survey, but considering it's difficulty level, ok.
@Remember should definitely do some more mathematics first before doing D-F, however. It takes a great deal of mathematical maturity to digest it.
Morning @TedShifrin
morning @BalarkaSen aswell
hi, @Karim
i thought you were going to do some abstract algebra this summer?
I am
I am doing D-F
done some Sylow theory lately? those stuff are real good.
I did in my class but for this summer I want to start from the very begging and do problems as well as reading
and planning not to read the proofs for some of the theorem and try to proof it myself
theorems *
14:28
good!
yeah I want to increase my knowledge this summer aswell as my problem solving skills.
@BalarkaSen I was talking to ted about this yesterday very cool
2
Q: Proof verification: ab = gcd(a,b)lcm(a,b) without use of prime factorization

Karim MansourI am trying to proof $ab = gcd(a,b)lcm(a,b)$. The definition of lcm(a,b) is as follows: t is the lowest common multiple of a and b if it satisfies the following: i)a | t and b | t ii)If a | c and b | c, then t | c. Similiarly for the gcd(a,b). Here is my proof: Case I: gcd(a,b) $\neq$ 1 ...

but yea his method is much better than mine.
oh, yes, that one is good.
try to generalize this to a relation between $\gcd(a, b, c)$ and $\text{lcm}(a, b, c)$
not hard, but fun nonetheless.
in general I want to do it to gcd($a_1,...,$a_n$) and lcm($a_1,...,$a_n$).
today I will do that
14:32
ok, you can do that.
where are you studying number theory from?
this was a problem in my abstract algebra book
oh, ok.
but I studied algebraic number theory last semester from number fields marcus.
ooh. that's nice.
yeah very nice book
14:36
i never read it, though. i did a bit of (algebraic) number theory from Ireland & Rosen and notes taken during a class.
oh
ireland and rosen
very nice book
I want to cover if I have some time during this summer to cover elliptic curves
well, you may find it hard to really "cover" elliptic curves. you need a lot of algebraic geometry to know that stuff.
oh
I thought you could do it with relying on group theory only.
take what i say as a grain of salt though, as i never read seriously about ell. curves.
14:43
@KarimMansour introductory stuff about ell. curves you can understand with basic algebra, but i don't think you can really do anything with them unless you know algebraic geometry.
but then, as i said, i have never done that stuff, so don't let my presumably wrong opinions stop you :)
oh I see
so what are you doing this summer @BalarkaSen?
presumably would do some multivariable calculus and read up differential topology. would do algebraic topology alongside too.
Howdy @Balarka @Karim
14:56
You feeling better?
Hi @TedShifrin
loads, @Ted. going through a dozen antibiotics made me a bit weak, but i am quite fine right now. never been better, really.
I see you finally got my slant @Karim
yeah I did just woke up haha
just revising the linear algebra i did the previous week. bracing up to do some serious stuff (billinear forms, etc).
14:57
your method is also easier to generalize to any nth gcd or lcm that is lcm and gcd that is made of nth components not necessarily 2
Eat, Balarka 😀
Also to general PID, Karim
did you cover schur theorem @BalarkaSen ?
@KarimMansour hint : if you can do 2, you can do n.
no need to generalize the whole argument
@KarimMansour yep.
by Schur's theorem, you mean the decomposition theorem, right?
yeah
that any matrix A can be made into a upper triangular matrix
right, yeah.
it's easy.
of course, you mean a matrix with entries from C/from an algebraically closed field, right, @Karim?
15:01
balarka sen?
didn't know it was called Schur's theorem.
is somebody here who know the polynomials, I need some help please :D
the ways of the polynomials haha
karim can you help me please?
15:05
sure what is it
@Karim More interesting are Jordan forms.
yeah
but they are hard to deal with
why $\sqrt{2+i\sqrt{12}}+\sqrt{2-i\sqrt{12}}=2\sqrt{3}$?
well, not once you get used to them. they're just nilpotent elements in your matrix rings.
why don't you square both sides, @Lucas?
15:09
suppose we don't know the final result
and we need to find it is equal with $2\sqrt{3}$
then multiply by the conjugate
you can still square and then square-root :P
and see where it leads you
I think multiplying by the conjugate can reduce some steps @BalarkaSen
hahahaha, is work
thanks karim
2
Q: The complex roots of a biquadratic polynom

LucasIn my recent post I have a problem with the following function: $x^4-4x^2+16$, and what I need is to find the complex roots. Here is my answer: First step, I make the substitution $x^2=y$ which involving $y^2-4y^2+16$, with $x^2=2\pm i\sqrt{12}$. Therfore: $x^4-4x^2+16=(x^2-2-i\sqrt{12})(...

here I have another problem
Hello @AndresCaicedo.
We don't see you around here a lot.
15:11
lol somebody answered it
is incomplet
and is not what I want to sound
and you can answer too, if you know
why don't you try to find an answer
then do long division
15:26
very interesting question
28
Q: When are two proofs "the same"?

William StagnerOften, we find different proofs for certain theorems that, on the surface, seem to be very different but actually use the same fundamental ideas. For example, the topological proof of the infinitude of primes is quite different than the standard one, but the arguments are essentially the same, ju...

"Proof A is isomorphic. Proof B is not."
00:00 - 16:0016:00 - 00:00

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