« first day (1697 days earlier)   
00:00 - 12:0012:00 - 18:00

12:08 PM
Hi @robjohn
Could I ask you something about the complexity of an algorithm?
 
@evinda Go ahead
 
I have to use a Divide-and-Conquer procedure for the computation of the i-th greatest element at a row of integers.
I have tried the following:http://pastebin.com/rPnc74gd
The time complexity will be $T(n)=\Theta(n)+T(q-1)+T(n-q)$, right? But how can we compute the latter without knowing the value of q? @cirpis @robjohn
 
12:24 PM
Back.
@Semiclassical Your assumption was right.
 
indeed?
nice
hmm. since $n$ is an integer, you can drop the $2n$ from that power of minus one
 
@Semiclassical After you expressed that point I know there is such a way of getting the answer.
@Semiclassical Right. Done.
 
glad it pushed you in a useful direction, then
 
@Semiclassical Yeah, thanks.
Sometimes a simple detail makes the whole game.
 
12:29 PM
@Semiclassical It is related to the problem with the convergence. :-)
 
I hope this way is clear enough to avoid trivial solutions.
 
right
oh, was there an integral i showed you a while back?
 
Not sure what you mean.
 
yeah, that's pretty vague. let me write it in the form i have lately
 
12:35 PM
Just a second.
Here is the full problem (as I created it)
 
Does anyone have an idea about the following?
1
Q: The algorithm yields optimal ternary codes

evindaSteps to build Huffman Tree Input is array of unique characters along with their frequency of occurrences and output is Huffman Tree. Create a leaf node for each unique character and build a min heap of all leaf nodes (Min Heap is used as a priority queue. The value of frequency field is used t...

 
ah. so the challenge is to find a family of integrals in closed form
 
@Semiclassical Yeap. :-)
 
with the restriction to convergent being to avoid getting a nonsense answer :)
 
There are 2 points there. :-)
@Semiclassical It's interesting that $f_n(x)$ can have more solutions. Not sure how many yet.
The idea is to find a solution such that one can also compute the closed form in the generalization obtained. :-)
 
12:38 PM
eh, doesn't strike me as terribly surprising. once you find one convergent family, you can add a lot of different things to it without spoiling the convergence
 
Yeah.
 
what might be interesting is to ask for what family that integral is not just finite but moreover minimized
 
hi
 
since that penalizes adding extraneous-albeit-convergent terms
 
@Semiclassical Indeed.
 
12:40 PM
okay, my integral:
$$\frac{1}{\pi}\int_{\lambda}^\infty \frac{dz}{z}\sqrt{z+z^{-1}-\lambda-\lambda^{-1}}\left(\frac {d\ln p_N(z)}{d\ln z}\right)$$
where $p_N(z)=1+\dfrac{1}{z^{2n+1}}\dfrac{z-\lambda}{\lambda z-1}$
 
Is it possible to show a straight line from north to south with a function?
 
Interesting. How did you get that?
 
and with $\lambda>1$. (the case with the last fraction flipped is also of interest to me
 
3
Q: Derivatives of the commutator of flows (or, what are those higher derivatives doing in my tangent space?!)

Saal HardaliLet $p \in M$ be a a point in a manifold and let $\varphi^X_t$ and $\varphi^Y_t$ be the local flows of the vector fields $X$ and $Y$ respectively. Define the commutator of flows: $\alpha(t)= \varphi^Y_{-t} \varphi^X_{-t}\varphi^Y_t\varphi^X_t$. I'm trying to prove: $$\left .\frac{d}{dt} \right|_...

plz help me
my problem is explained in the comments to the current answer
 
from physics research. the $p_n$ function arises from the characteristic polynomial of some $N\times N$ matrix with free parameter $\lambda$
 
12:44 PM
So yeah, it got below freezing last night...
 
one is then interested in taking certain sums of a certain function evaluated at the roots of that characteristic polynomial. that can be written as a contour integral, and one can pull out the leading behavior (terms of order $N$ and $N^{0}$) pretty easily
 
@Semiclassical I think saw a lot of integrals, limits really amazing coming from physics, like Ahmed integral.
 
woops, denominator in $p_N$ should've had $2N+1$ not $2n+1$
now, what's intriguing is the following limit
suppose i take $N$ to be very large. then from numerical results i know i can approximate that function as $N^{-1} f(N(\lambda-1))$ for some smooth function $f$
the question is how that limiting behavior arises, and in particular if one can find a closed-form for it
 
Hello @Balarka
 
12:48 PM
hullo @BalarkaSen. new 'have you heard of this math concept' question for you
Group cohomology?
 
yes
well, group homology, more like.
 
ahah. i've been running into it somewhat lately, particularly re: nonabelian groups
 
You can study some of it from the chapters at the back of Dummit-Foote. I went through those a long time ago, just the basics.
Now what's amazing is that I recently found it mentioned on Hatcher that group homology of a finite group $G$ is actually just the usual homology of $K(G, 1)$.
 
i should probably know what $K(G,1)$ is off the top of my head. (the graph of the group?)
 
well, it's the space that has $\pi_1(X) = G$ and all higher homotopy groups being trivial
 
12:52 PM
ah.
is there an assumption on $G$ being abelian in this context?
 
nope
 
hmm, interesting
so that'd mean that the first group homology of a finite group is its abelianization, no?
since the first homology group is the abelianization of the first homotopy group
 
er... why should it mean that?
 
i'm probably misunderstanding
but, if i've got a space with first homotopy group being $G$, then isn't the first homology group the abelianization of $G$? (that's the hurewicz (spelling?) lemma, i think)
 
you may not be, but i don't get your logic.
@Semiclassical yes, true.
ugh. right. coffee :P
 
12:59 PM
so if the group homology of $G$ is actually just the homology of $K(G,1)$, i'd think that means that the first group homology of $G$ is the first homology of $K(G,1)$
heh
 
i find $K(G, 1)$ a pretty confusing notation ;)
 
hah.
a more involved question, i imagine, is the group homology of a Lie group
 
it wouldn't make sense in the homology-of-K(G, 1) language
 
there cannot be any finite lie group ;)
 
1:03 PM
i know people do talk about lie group co-homology, which is why it comes to mind for me
 
group (co)homology has a lot of other formulations, so does galois cohomology
 
true
really, though, what i'm trying to appreciate is what it means to have a cohomology taken with nonabelian coefficients, which seems different
 
Can I ask a question here ?
 
after reading this very interesting fact from hatcher, it seems to me that there should be some kind of connection between (etale?) (co)homology of Spec(Q) and galois cohomology of Gal(\bar Q/Q), even though i don't know any of the etale stuff, neither what Spec(Q) means.
 
well, if you don't know it, then i certainly don't :)
 
1:07 PM
@Semiclassical ah. well, homology with nonabelian coefficients wouldn't serve the fundamental property for which they are studied in the topological context -- the group won't be abelian.
 
it'd lose a lot of fundamental properties. are you sure it means what it should mean in the classical sense?
 
can't say i am
 
So I want to find the interaction between two spheres. My intuition is to calculate the interaction between one point of the sphere and the whole of the other sphere and then integrate over the first sphere.
but I don't know how to define the other sphere geometrically relative to an arbitrary point on the first sphere
 
well, you can make life easier by rotating things a bit. put one sphere at the origin, and the other on the z-axis
@BalarkaSen: the context I had in mind is the second paragraph i referred to before
which references "the bundle of first nonabelian cohomologies of some family of varieties"
 
1:17 PM
I hate when I have to talk for hours on the phone ... (back)
 
there is something here but i'm not sure how useful that stuff is
 
saw that, yeah
i may have gotten on the wrong footing there, since while searching for some physics instance of it i got onto the topic of "group cohomology" which has made some impact in condensed matter physics
e.g. the last bit of this section of Wikipedia's page on group cohomology
 
interesting. i knew algebraic topology is useful for studying world-sheets and stuff in string theory, but didn't know about that.
 
yeah, i'm not well-versed on it either
 
physics is weird [in a good sense, i.e., weirder than mathematics]
:P
 
1:24 PM
this paper is the one that introduced the idea of group cohomology in that context
with a whole appendix on group cohomology
i can't quite convince myself they're doing non-abelian group cohomology, though, rather than just abelian
 
the link is broken
at least i can't open it
 
bah
let me try again
not sure why it's not working, but the link is dao.mit.edu/~wen/pub/dDSPT.pdf
not going to pretend i can explain it, but
phrases in the table of contents like "Equivalent cocycles give rise to the same
SPT phase" is pretty neat
that and the fact that they evidently care about the distinction between linear representations and projective representations
 
apparently not something what i'd be able to understand
 
me either, at least not without a guide
 
@robjohn you're too silent these days. You might have missed some very nice questions I posted.
 
2:12 PM
Oh man it is getting late
I'll probably get to see Ted again hah
 
heya
Bah, struggling with some easy multivariable calculus. Bah, I hate calculations
 
Are you using Ted's book
 
Nah, I am TA and of course have to do their problems beforehand. I just keep getting the wrong answer ._.
 
Oh
 
A few days left, but it feels annoying :p
They are given an area $T$, restricted by $x^2/9 + y^2/4 = 1$, $z = x^2+y^2$ and the $xy-$plane. The problem is to calculate $$ \iint_{S_1} \hspace{-0.65cm} O \qquad F \cdot \hat{n}\,\mathrm{d}S $$ where $S_1$ is the cylindrical part of $T$ (eg. exclude the top and the bottom of the closed surface).
 
2:22 PM
Hello!
does anybody know if it can be proven that
a pole for a function is the same to the zero of 1/the function
and the orders remain the same
 
Ugh.... Was anyone taught this property in basic calculus or earlier? $$\frac{a}{(bx^2+c)^2}=\frac{a}{2c}\left(\frac{1}{bx^2+c}-\frac{bx^2-c}{(bx^2+c)‌​^2}\right),\; a,\,b,\,c,\,d\in\mathbb{R}$$
 
Yes, that can be proven @SebiSebi.
 
My gut tells me it's a trivial case of something I'm simply too tired to see...
 
That is wrong @teadawg1337...
 
Oh...
Why did I include a $d$...
$a,b,c\gt 0$ as well...
It has to be true, because it works when integrating the function and differentiating the result...
 
2:35 PM
@teadawg1337 Just basic partial fractions?
 
I never properly learned partial fractions, and I've never seen a second-degree polynomial in the numerator of a decomposed function...
No wonder I'm not acting like myself, I forgot to take my meds this morning... brb
How could I be so stupid....
 
2:58 PM
@teadawg1337 Never talk to you like that, it may affect your self-esteem.
 
Wait, partial fraction decomposition leads nowhere.... For example, solving $\frac{3}{(4x^2+5)^2}=\frac{Ax+B}{4x^2+5}+\frac{Cx+D}{(4x^2+5)^2}$ yields $A=B=C=0$ and $D=3$...
Yet $\frac{3}{(4x^2+5)^2}=\frac{3}{10}\left(\frac{1}{4x^2+5}-\frac{4x^2-5}{(4x^2+5)(‌​4x^2+5)}\right)$...
The square in the denominator wouldn't render, that's why I changed it to that.
@Chris'ssis My self-esteem is nearly nonexistent as it is, lol
 
@teadawg1337 But you always say you are above average
 
@ᴇʏᴇs I put on a "front" in social settings, so as to avoid ostracism. It's a lose-lose scenario; pretending to have a high opinion of myself lowers it even further, but the same happens if I perpetuate the low opinions of myself
 
morning, mr eyeglasses, @teadawg
good morning, mr @Pedro
 
@teadawg1337 I prefer to have a very good self-esteem.
 
3:13 PM
@Chris'ssis I don't choose to have low self-esteem...
 
@teadawg1337 But you can learn how to boost your self-esteem. I don't think that low self-esteem is of any help.
 
morning ted
 
Hello @TedShifrin
 
good night, @Mike
@teadawg: I don't wish to meddle, but my advice would be to just enjoy math and your friends as much as you can without thinking about what other people think ... obviously, don't go out of your way to be an asshole :) I've enjoyed interacting with you here.
BTW, @teadawg, when Owatch was doing partial fractions the other day, he was clearing denominators and saying equality of polynomials necessitates equality of corresponding coefficients. This gives a system of linear equations. This is the standard way we teach the material in Calc II. You wanted to be sneaky and substitute values of $x$ to make it easier; that works most of the time (and is sort of what we do with residues in complex analysis), but it doesn't work all of the time.
@Mike: The perils of the electronic era. I just noticed I had set the due date for the WeBWork assignment to be April 28, rather than March 28. I just changed it :P Interestingly, none of the students called my attention to that. I suspect they all knew :)
 
If $\tilde{x}$ represents the fourier transform of $x$, can I say that $\displaystyle\tilde{\frac{dx}{dt}} = \frac{d}{dt} \tilde{x}$?
 
3:20 PM
hehehe
 
Is $t$ a parameter, @user112495? Can you make your question more explicit?
 
@Ted It wasn't the way I was taught in high school, but I'm starting to realize that the standard method is superior in most cases
 
@teadawg: It's failsafe, though painful. But if you have irreducible quadratics or repeated roots, your way fails ...
without a bit more knowledge :)
 
@TedShifrin Sorry, yes. $x$ is a function of $t$.
 
I still don't understand. So $t$ is the independent variable? Write down explicitly what your definition of Fourier transform is.
Then differentiate with respect to $t$. And you won't get at all what you're saying, if I understand you correctly.
 
3:25 PM
@TedShifrin I'm trying to show that $\displaystyle\tilde{\frac{d^nx}{dt^n}} = (-i\omega)^n\tilde{x}$
 
@TedShifrin Heya dted, how is things going?
 
Right, @user112495. So that's NOT at all what you wrote.
Write down the definition and differentiate under the integral sign.
heya @n3B.
Well, folks, have a good day. I need to get going. Going to the big city for the afternoon and evening.
@teadawg: Hang in there.
 
@TedShifrin =( I was just going to ask you a quick multical q
 
Quickly, then :)
 
@Ted I'll try to
 
3:28 PM
Be well on your journey =)
@TedShifrin Hopefully you can help me clear up something :p It should be easy for you as it is multicalc. Trying to calculate the flux out of an area bounded by $z = x^2 + y^2$, $x^2/9 + y^2/4=1$ and thhe $xy-plane$. The question asks just for the flux out of the side, eg the cylindrical part
I tried to do it as follows
 
Bye, @Ted. I'm gonna climb Mt. Baldy.
 
Where's that, @Mike?
What a yucky question, @N3B. Do we know the Divergence Theorem?
 
@TedShifrin The method I was using was induction. We've shown it for the case n=1. Then $\displaystyle\tilde{\frac{d^{k+1}x}{dt^{k+1}}} = \tilde{\frac{d}{dt}(\frac{d^kx}{dt^k})} = \frac{d}{dt}((-i\omega)^k\tilde{x})$
 
@TedShifrin Yeah, but not here. It is just the side. Later one is to prove the flux out of the top by using the flux out of the side, and the divergence theorem
 
Opposite side of the world from Mt. Hairy, @Ted.
 
3:30 PM
No, @user112495, your first equality is wrong, I believe.
smacks @Mike
 
Alternately, about halfway between pasadena and san bernardino, go a little bit north.
 
Oh, ok, @N3B, so you're trying to do the direct calculation for the cylindrical sides. OK.
Is the force field nice? Presumably, you do the calculation by parametrizing by elliptical cylindrical coordinates.
 
$$
\int_0^{2\pi} \int_0^{(3\cos \theta)^2+(2\cos \theta)^2} [3r \cos \theta,2r \sin \theta,z] \cdot [\cos \theta,\sin \theta,0] 6r \,\mathrm{d}r,\,\mathrm{d}r
$$
 
@TedShifrin Oh... Is that because they're the Fourier transforms of the derivatives?
 
Ugh. @N3B. Do they know how to do general parametrized surfaces? If so, don't use plain cylindrical.
Yes, @user112495.
 
3:31 PM
@TedShifrin It is given as $[3x,2y,z]$. I used $x = 3 r \cos \theta$, and $y= 2 r \sin \theta$ to map the ellipsoid onto a plain cylindrical.
 
I don't have any shorts. Maybe I'll make some and bring jorts back into fashion,
 
Oh, so the force field is very special, @N3B. Can you say something nice about $F\cdot n$?
Still, I would use elliptical cylindrical coordinates to find the surface area, regardless.
Your normal vector is wrong, @N3B. You're thinking of a circle, not of an ellipse.
Oh wait, you're right. Hold on.
 
@TedShifrin =)
@TedShifrin I get zero, I think it is because of the symmetry. But not sure if that is what I should excpext.
 
OK, I apologize. You're right.
Well, symmetry does give $0$. Because the flux cancels when you reflect across $x=0$ or across $y=0$, right?
 
@TedShifrin Yeah
 
3:34 PM
I try to get my students to "exploit symmetry" whenever possible. You just didn't give me the $F$ early enough :P
 
@TedShifrin What a weird problem. Using the divergence theorem gives $117\pi$. So the field out the bottom and the top should be huge.
@TedShifrin I am just trying to working out the nitty gritty details =) Thanks.
 
Make sure you do the volume integral under the paraboloid :)
The numbers don't seem ridiculous, because the dimensions of the ellipse are reasonably large.
 
I'm assuming vector calculus is much more useful in differential geometry than in analysis? :P
 
Depends on what sort of analysis, @teadawg.
 
Real/complex
 
3:36 PM
People who do serious PDE need to use multivariable analysis/vector calculus stuff.
At the undergraduate level, you use some vector calculus in complex for sure, and multidimensional real analysis (the stuff in my course) is very important. But a first real analysis course is just single-variable.
 
I mean the divergence/curl and surface integral stuff, I know multivariable stuff is of vital importance to many fields of mathematics
 
@TedShifrin I did something like $\int_0^{2\pi} \int_0^1 \int_0^{(3 r \cos \theta)^2 + (2 r \cos \theta)^2} 6r \,\mathrm{d}r,\mathrm{d}\theta$ after using the jacobian and etc.
 
In PDE that stuff does get used, @teadawg, in a serious way.
 
@Ted: have any idea what this means?
 
@TedShifrin I am taking multivariable complex analysis now, it is really fin. Learning a bit about diff.geo too. Somewhat interesting =)
 
3:37 PM
You have a typo, @N3B ... $\sin$ instead of one of the $\cos$.
 
@TedShifrin Indeed
 
Several complex variables, @N3B? Very interesting and beautiful stuff, but a LOT of real analysis in there, too.
OK, I have to leave. You all have fun.
 
Cya @Ted
 
@TedShifrin Have a nice trip, thanks again for the help =)
 
Not even going to look at my link, @Ted? Harsh.
See ya.
 
3:38 PM
hibye @ted
 
Oh, hi @KarlKronenfeld
 
how goes it @MikeMiller?
 
@KarlKronenfeld hides all the analysis
 
s'alright.
 
@N3buchadnezzar ah much better
 
3:41 PM
^^
<^.^>
I love when you have worked somewhat hard on a problem, and come here and get confirmation that the workd you have done is good/correct =)
 
I'm ambivalent. Though, I don't take advantage of this resource that much (or.. at all :().
 
@KarlK I will check your work for you.
 
Ah, that's kind of you
@MikeMiller I think I've a proof that 1=0. Could you check it for me?
 
No.
 
Wop wop 12.99, 12.39, 15.50, 13.74, 14.93 ^^
 
3:48 PM
That's a problem I've been working on for quite a while though.
 
Sorry, I'm simply very busy.
 
It's been well-established that $1 \neq 0$ for some time now.
 
@DavidWheeler Is that well-established for very small values of $1$?
 
@DavidWheeler What!?
@MikeMiller (Field with one element reference, no I am not actually doing research in that)
 
Didn't need to clarify
 
3:55 PM
didn't need to do anything besides the basic necessities by the very defn of need.
 
Hi @ᴇʏᴇs @Chris'ssis @Committingtoachallenge
 
4:13 PM
Morning @MikeMiller
Where's everyone?
Anyone home?
 
4:37 PM
yeh
 
1
Q: What ring extension is this? (Dirichlet series values)

Enjoys MathLet $R \supset \Bbb{Z}$. Now consider $S = \{ \alpha \in R: \alpha = \sum_{n=1}^{\infty} \dfrac{a_n}{n^x}$ for some $x \in \Bbb{Z}, a_n \in \Bbb{Z}\}$. It is a ring because of Dirichlet convolution arguments which are easy. What can be done with this ring?

 
@JasperLoy Hey. How are you doing? :-)
 
Hey guys I am working on a problem and I could use some tips. math.stackexchange.com/questions/1209627/… Thnx !
 
4:54 PM
Between the usual metric on R and the absolute value function on R , is one of those more fundamental than the other ( is there an intrinsec hierarchy ) ?
Or are both notions equally as fundamental ?
By one notion A being more fundamental than the other notion B i mean that its more natural to consider notion B in terms of notion A , than the opposite./
For instance, we can see either of those in terms of the other
d(x,y) = |x-y|
and |x| = d(x,0)
But is any of those hierarchies more fundamental than the other in the sense that we should naturally choose that hierarchy in organizing the absolute value function on R and the usual metric on R ?
 
@Chris'ssis Not too good. Are you working on your book?
 
@JasperLoy I'm working on a proposed problem for AMM. Not good again?
 
I have some proof I don't understand in my algebra book maybe coz I just woke up
but I will ask away
Let N be a normal subgroup of a group G and let K be any subgroup of G that contains N. Then K is normal in G iff K/N is normal in G/N.
 
@Chris'ssis Well, it fluctuates due to various factors. I have made some plans to solve my problems in the next few months, but I am scared there might be obstacles hard to overcome.
 
I understand --> implication so I will ask in the --< implication
let a be any element of G and k be any element of K. we first prove a^-1ka in K. Since K/N is normal we have Na^-1ka = (Na^-1)(Nk)(Na)
I don't understand this part
how come Na^-1ka = (Na^-1)(Nk)(Na)?
anyone?
 
5:21 PM
@JasperLoy was life ever easy? Don't lie yourself. It's important to learn how to be powerful.
 
@Chris'ssis Life was pretty good before I got my mental problems.
 
@JasperLoy Life changes from a day to another. You never know what comes. So, you need to always be prepared and able to face anything.
 
@Chris'ssis Yes, it is so unpredictable. One's life can turn from heaven to hell or hell to heaven overnight.
 
@JasperLoy Exactly.
 
@Chris'ssis Maybe my hell will turn into heaven soon.
 
5:24 PM
You can make the hell, less of a hell with practice, independent of the external world
 
@JasperLoy Even the heaven might be hard to bear. I don't like that much these ends.
The cousine of Simona Halep killed himself last weeks. He had a life like in heaven.
 
@Chris'ssis What happened?
 
@JasperLoy I think he spent a lot of money in cazino, had many debts to pay.
 
Hello Math SErs
 
@everyone @chris's How should I prove that $f:\mathbb{N} \to \mathbb{N}$ where $f(n)=\lfloor \tan n \rfloor$ is not a polynomial? :|
 
5:34 PM
@Chris'ssis Actually, when I look at all the suffering in the world, it's really scary. So many people living with physical illness, mental illness, poverty, etc.
 
@JasperLoy Exactly. You cannot have great expectations. I'd better take advantage of the simple things.
@Sawarnik I think you can do that alone.
 
@Chris'ssis Sometimes, I ask, what is the meaning of life when there is so much suffering in this world? It's better the world does not exist.
 
@Chris'ssis Hints :D
 
@JasperLoy Well, as I told you many times, you can focus on the positive things, not on the negative ones. It depends on you how you wanna look at the world.
 
@Sawarnik Ask on the main site.
 
5:38 PM
I don't want to but :.
 
@Sawarnik can you write it as $$ a_n x^n +a_{n-1}x^{n-1}+\cdots +a_0$$? This is the hint.
The idea is to get a contradiction.
 
oo
 
I'm out for some jogging. Back in 30-60 min.
 
I am out for some sleeping. Back in a few hours.
 
@JasperLoy I thought of you stereogum.com/1700410/…
 
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