@nerdy It's a phase portrait of $\varTheta_2^4(q)$ for complex $|q|<1$, where $\varTheta_2$ is one of the Jacobi Thetanull functions.

@nerdy You'll probably get feedback soon. If not, it may be because the question contains a number of sub-questions, and users hesitate to give only partial answers. Therefore: The more atomic the question is, the sooner it may get answered.

@meer2kat Nevermind, I just meant that the question you asked was clever.

Same mathematician from Jacobian matrix and determinant ?

what question...?

@matsgranvik

The are you a robot thing.

@ccorn yes , and its good that users hesitate to give partial answers :DD

@nerdy Carl Gustav Jacob Jacobi

oh i see :D

hmmmm okay i don't really remember hahaha

Folding letters, interesting.

@meer2kat Yes, reminds me of the "boiling water" joke which is along the same lines

@meer2kat Strictly speaking, that's more of a physical consideration. Math nerds would abstract the paper to zero width.

@ccorn if you read on, they did

and then they argued about if it's non-zero if it matter if it's an inch or a millimeter

lol

BTW, does anyone know how people made envelopes back in the olden days?

the conclusion was, in fact, no. it is either zero or non-zero

@ccorn is it different than now?

making envelopes is really easy

@meer2kat I'm under the impression that ancient letters were their own envelopes, but cleverly folded

(Don't take the ancient seriously)

i'm fairly certain if we're talking that ancient they were more like little scrolls

dangit LOL

and then

we moved on to tri-folded letters pressed with wax seals

what time period are we talking about here?

@meer2kat Let's say, enlightenment and later.

yay laid my clothes on the heater in a half-awake stupor when I woke up, now coming to consciousness nude and cold I see an optimally allocated pile of warm clothes :)

@ccorn Thusly?

bahumbug, i don't want to look up years for that. oh perfect, you linked it. yay. so you think they were folded in to envelopes? i really feel by that point it was just a mix of the scroll method and the tri-fold wax seal method. but perhaps that didn't originate until the 18th and 19th centuries?

@DanielFischer Wow, thanks. Reading...

lol @micklh

Heh I suck at history anyways, so I'll do my morning stuff brb!

i thought that said momming stuff at first

yayyy pinning locations

lol I probably suck at momming too, maybe I have a chance at dadding

@DanielFischer Great as always.

@ccorn Thanks. I try my best.

The von Mangoldt function is confusing. You can write Sum[MangoldtLambda[n]/n^(1/2+I*t), {n,1,k}], Discrete Fourier Transform it and you get a spectrum with spikes at Log[prime powers] that are proportional to MangoldtLambda[n]. You can also take the analytic Fourier Cosine Transform of the Sum[MangoldtLambda[n]/n^(1/2+I*t), {n,1,k}] and it will give you an answer that corresponds to each term in the sum. Still you know nothing about the Riemann zeta zeros when doing so.

@DanielFischer What space is $C(S^1) \times C(S^1)$ the continuous functions on?

@Mike $S^1\times S^0$. You always have $C(X)\times C(Y) = C(X\times\{-1\}\cup Y\times\{1\})$.

Hmmm.

Well, $=$ as in "canonically isomorphic".

Right.

@DanielFischer If one has a *-morphism $\varphi: \mathfrak A \rightarrow \mathfrak B$, can $\varphi(\mathfrak A)$, the image algebra, be canonically identified with some subspace of $\mathcal M_{\mathfrak B}$? If so, I think our problem's done. But I'm not sure it can.

(identified meaning "the continuous functions on some subspace up to homeomorphism")

hi

can anyone help with math.stackexchange.com/questions/742510/… ?

@Mike I don't think so. We should have a correspondence of $\ast$-morphisms $\mathfrak{A}\to\mathfrak{B}$ (with whatever are the required properties) to proper maps $\mathcal{M}_\mathfrak{B} \to \mathcal{M}_\mathfrak{A}$, so we have a subspace of $\mathcal{M}_\mathfrak{A}$ corresponding to a quotient of $\mathcal{M}_\mathfrak{B}$, and quotients typically don't correspond to subspaces.

Ah, you're right.

(I'm really just grappling with the fact that the image algebra is unital when $\mathfrak A$ is unital and hoping it goes somewhere.)

I think injections are going to correspond to subspaces, but there's no reason our $*$-homomorphisms should be injective.

But, wait. Say $\mathfrak{A}$ is unital. Then $\varphi(\mathfrak{A})$ is a direct summand of the unitalitizization $\widetilde{\mathfrak{B}}$. So it might lead somewhere still.

I'm not sure why you crossed out the unitalization (nor do I see how it's a direct summand in general, though some specific cases are easy to see)

If it definitely is such a direct summand, I'll try and figure that part out myself :)

Right, the direct summand was a thinko, just because a subalgebra contains a nontrivial idempotent doesn't mean it's the ideal generated by that idempotent. (For the crossing out, take a second look at what I crossed out.)

Oh, cute.

I'm stealing the word thinko, by the way.

@DanielFischer The counterexample I'm aiming for is to somehow codify the projection map $\Bbb R \rightarrow S^1$ as a $*$-homomorphism.

@Mike Counterexample for what? (You are aware that that's not a proper map, I suppose.)

That a $*$-homomorphism induces a proper map :) I want to find a $*$-homomorphism that induces that.

Obviously I'm not even going to induce a map as sets $C(S^1) \rightarrow C_0(\Bbb R)$ with that, I'm trying for the other way around (somehow).

Maybe that's a bit too wishful and I should be thinking of counterexamples in unintuitive spaces.

(removed)

I don't wanna be a chicken

I don't wanna be a duck

haha this is tricky. "As they say, beggars can't be choosers, in fact begger take what they can get. A begger on the street can make one cigarette out of every 6 cigarette butts he finds. After one whole day of searching and checking public ashtrays the begger finds a total of 72 cigarette butts. How many cigarettes can he make and smoke from the butts he found? "

@meer2kat without loss of generality let us assume that the beggar is smart .. hence he smokes 12 cigarettes (dosen't leave a butt when he is smoking on his own) :P LOL

nope

the answer is 14

think again

ah okay, you got the catch

well done

"At a party, everyone shook hands with everybody else. There were 66 handshakes. How many people were at the party? "

same answer as the last problem .. while no one was butting :P

@r9m Yay! Me going to a vacation after a long time this summer.

@Sawarnik where ?

@r9m leh and related places!

nice !!!

Can you arrange four 9's and use of atmost 2 math symbols , make the total be 100?

99 + 9/9

:P

that's one way to do it

which others? :)

IDK .. how ?

99/.99

:)

awesome :)

:14752071 the roots of the polynomial are $\displaystyle\pm\frac{\phi\pm i}{\sqrt{2\phi}}$

@robjohn I know you think of using complex analysis. :-)

Let me post it and let it here then.

@Chris'ssis how did you sum it?

@Chris'ssis yes, but I haven't done the partial fractions

i never thought i'd say this

but i miss related rates

What's the difference of a total function and a surjective function?

@robjohn one way is to use complex analysis, and another way is to make use of coth (x) in a clever way. (that way is pretty ugly)

@Chris'ssis but are you using one of the infinite sums for coth that are usually, but not necessarily, gotten using complex analysis?

@Sawarnik IDK .. how ?

@robjohn Yeap. This is the particular case of a general series studied by Ramanujan.

@Chris'ssis okay, once I had the partial fractions, then I was hoping to use $\sum\limits_{k\in\mathbb{Z}}\frac1{k+z}=\pi\cot(\pi z)$.

@Sawarnik whats that ?

@Sawarnik that guy crazy :P

:(

@r9m orly. i haven't checked :D

@robjohn I see.

@Sawarnik crazy is a positive sense ... put a lot into it !!

oh.

@Sawarnik I thought oj = orange juice

:P

@Chris'ssis It should work since the sum over all integers is twice the sum over the positive integers

@robjohn Yeah, it definitely works.

@robjohn The mind-blowing thing here is that I believe I'm going to finish it by Euler-MacLaurin formula. Maybe it sounds crazy, but I'll do it that way.

@Chris'ssis okay, I'd like to see that.

@robjohn OK

@Chris'ssis I answered a question using these methods this morning :-)

@robjohn Where?

@Chris'ssis all the work was in another answer, but I pulled out the relevant part to write this answer

@robjohn Nice work. I saw that "another answer" some time ago.

@robjohn are there less upvotes in the last period of time or it just seems to me so? (I'm referring to all users)

@Chris'ssis yeah, that has been around for a while and I've cited it in several other answers.

@Chris'ssis I think that votes seem to diminish since questions stay on the front page for far less time than they used to.

At any rate, it's unfair to make use of people knowledge and not to upvote.

@Chris'ssis I think there are just fewer people seeing each answer since the front page moves so fast. I don't know if people are using the answers hit-and-run style.

I think some don't realize how precious this work here is. Just an honest opinion.

@Chris'ssis The real problem are the people who ask homework questions and delete the question as soon as an answer appears, before it can be upvoted.

@robjohn Is that a significant proportion? (I rarely answer homework questions, as I am too slow against the competition anyway)

@robjohn I see.

@Chris'ssis: I had never thought about this until yesterday, but the number of ways to put $2n$ objects into $n$ pairs is $(2n-1)!!$

@ccorn No, we try to discourage that heavily and if it continues, suspensions follow.

@robjohn hehe, I think I met this one somewhere before.

@robjohn into $n$ unordered pairs, I think

@Chris'ssis It is interesting that it is always an odd number of ways.

@ccorn yes, that is what I think is meant unless it is specified as ordered pairs.

@ccorn in any case, that is indeed what I meant :-)

@robjohn yes, enlightening way to imagine $(2n-1)!!$

@robjohn Isn't this related to the Stars and bars combinatorial problem?

@ccorn other than in some simpler forms of series, I had never really seen a "real life" use of it so unadulterated.

@Chris'ssis Is it? I will have to think about that.

Watching Boyd he is good.

@JasperLoy Hello. How are you doing?

@ParthKohli I just woke up from a nap, drinking some coffee now.

@JasperLoy Nap? Isn't it the sleeping time in SG?

@ParthKohli I don't sleep regular hours.

@JasperLoy Why not?

@ParthKohli Well, it is related to my mental illness I guess.

@JasperLoy Oh, is it insomnia or is this just a symptom of the illness?

@ParthKohli Well, I think a lot, then I feel tired, but sometimes I can't sleep, so it is a consequence I guess.

@JasperLoy Ah, it's not a real mental illness; it's geekism!

@ParthKohli Well, I do have OCD. That causes me to think a lot about OCD-related stuff, I mean.

@JasperLoy Oh, this conversation is weird. From sorriness to relief to sorriness.

@ParthKohli All conversations in this chat are weird.

@JasperLoy The tiny me was of 5 years ago!

@Sawarnik I see. You must be huge now.

@JasperLoy He looks just like that.

@JasperLoy Of course not! But people say I am tiny still :(

I am quite short, about 165 cm.

But how lang is ur cack?

@JasperLoy Is it why girls don't like you?

@EnjoysMath OMG, lol.

@Sawarnik One reason is that I don't meet people nowadays.

O_______O <--- people who see Jasper

Here's Sawarnik's picture.

@ParthKohli Oh!

@ParthKohli Nonsense.

8=6=D - - O:

@JasperLoy Accept the truth, live freely.

@EnjoysMath I get the rest, but what's that 6 in the middle of the thing?

@EnjoysMath I am trying to figure out what you mean.

a hand

it's an ascii bj

Ah, well.

@EnjoysMath You must be drunk, lol.

no, i'm drinking tea

in the morning

lols

So am I, but I'm in complete senses.

you guys were talkin bout cacks and bals, just thought I'd share mein art

@ParthKohli How do you know? I think you are in no senses now.

@Sawarnik You're out of your senses for thinking that I'm out of my senses.

Infinite sequence starting in 3, 2, 1...

@Sawarnik It is a numeral representing a figure, you dirty mind.

@ParthKohli Its impossible to have that many gfs I think. You are lying.

maybe they're girls who he's friends with

otherwise he's got coconut-sized bals man

@robjohn Any idea on how to approach the inductive step in this question: math.stackexchange.com/questions/696707/… I asked a related question math.stackexchange.com/questions/734007/… but can't seem to incorporate the answers into something suitable for the first question.

Anyone, how can I prove the limit of $\sqrt(n-1)-\sqrt(n)$ goes to zero as n goes to infinity?

with patience

and thought

MIKE!

:D

Should I just use the definition of a limit?

Or is there more?

well, definition of a limit won't quite do it for you, since it's not obvious from looking at that (to me) that $\sqrt{n-1} - \sqrt{n} < \varepsilon$ for $n>N$ (with appropriate $N$)

@ccorn Hello, My first inequality in the following link is wrong?math.stackexchange.com/questions/737963/…

@robjohn: Should this Q be flagged? It is a duplicate of a non-answered question.

@Anant that is a weird question since the signs in the $n+1$ case don't match the signs in the $n$ case. I wonder if the problem really asks for a sum up to $2n+1$

@Anthony think about getting rid of the square roots somehow

you won't be able to get rid of them entirely, but perhaps you can make them a bit nicer

@ccorn if you find a question that you think it is a duplicate of, you can comment and, if you have enough rep, you can vote to close (as duplicate)

@robjohn Not sure, since that proposition turns out to be true for n=0, 1

@robjohn In any case, it's been a tough nut to crack, even offering a bounty didn't help :)

@Mike The only two operations I can really think of would to be just straight up square it, which gives you a cross term, or two conjugate it, which would leave you with the square roots again, albeit under something that grows linear with n.

Oh wait.

@user91500 Indeed, consider $k = n = 6$, then the divisor $6$ appears as $1\cdot 6,\, 2\cdot 3,\, 3\cdot 2,\, 6\cdot 1$ in the double sum.

@Anant but if you continue the plus and minus sequence, it doesn't end up the same way as the question on half the cases. It is hard to tell which is intended.

It would just be 1 on top, huh?

:D

<3

@user91500 But, $$\sum_{x\mid kn} x \geqslant 1 + \sum_{y\mid n} ky = 2kn+1 > 2kn.$$

@robjohn Right. Funny that the OP accepted an answer which claimed (unsubstantiated) that the proposition was false.

What do you guys use to drag diagrams in LaTeX?

Parth blocked me :'( :'(

I'm using the amscd package which is looking terrible.

@Mike Back to my confusion about connectedness, how would I show that the interval from [0,1] is connected?

Can I do it by showing it isn't the disjoint union of two open subsets?

@Sawarnik do you mean that Parth is ignoring you?

proper nonempty open subsets

but yes

(alternatively, show that it's path-connected ;) )

So could I say like, consider any two open subsets whose union is [0,1], we aim to show they are not disjoint?

path-connectedness is an almost trivial argument here so that's how I would go

but yes you can do that too

@robjohn Ignoring me here .. blocked on FB .. what should I do in revenge?

Agh there's all those easier ways of doing things but it's left me not even knowing how to show connectedness, openness, and closedness using basic arguments.

@Sawarnik why seek revenge? After all, you got the nice avatar...

Didn't Parth make that for you?

@Sawarnik eat tomato soup ..

@DanielFischer Thanks, Why $\color{red}1+\sum_{y|n}ky$?

@r9m the ultimate revenge!

true ..

@Mike It would, however, be kind of circular. That path-connectedness implies connectedness needs the connectedness of intervals.

Ah, you're right, I forgot that.

@user91500 Because for $k > 1$ - we have the assumption $k > 1$, don't we - $1$ is not of the form $ky$ for a divisor $y$ of $n$, but $1$ certainly divides $kn$.

@Anthony You can do this using a basic argument. You KNOW what the open sets in $[0,1]$ look like. What are they?

I mean just anything like (x , y)

No, there are more than that.

And you're forgetting the case where your set includes the endpoints of $[0,1]$.

Oh

Oh boy

Any collection of open intervals then, I guess. Right?

Almost.

Oh and like you said I need to include 0 and 1

@DanielFischer: OK, Do you like to answer this question or allow me to edit it?

@user91500 Go edit your answer.

@Anthony So assume $[0,1]$ is the disjoint union of two such sets.

Indeed.

Now find a contradiction! :D

@Mike Well if it is, then any element can be placed in one and only one of those sets, but then you have to be able to form a ball around that point and remain in the set, for all points.

Hnnggg.

@skullpatrol Wow, this effort is commendable. Keeping the room alive. chat.stackexchange.com/rooms/7610/openleisure

For some two points, for some smallest r, there is a point that is not in the set. Take such a point, and such an r. Then take the second point. there is no r that will leave your ball in the set, thus it isn't open?

Aghhhh

You're so close to the right idea.

I don't know why you say "for some two points".

Start by picking whichever open set contains $0$. Let's call it $U$. There is some least $r$ such that $r \not\in U$. (Prove this!)

Then since we've written $[0,1]$ as a disjoint union of two open sets, it must be in the other open set.

Find a contradiction from this fact,.

The first follows from least upper bound? But since this is the least r such that r is not in u, for every smaller r, it is in u. I mean, you can't form a ball around the other element. Which means the set isn't open.

@Mike The diagram package

That argument is a little muddy, and you'll want to clean it up before putting it on paper and turning it in :P And "the first follows from least upper bound" I would like some more words on.

Oh this wasn't for homework.

oh.

well clean it up for me then.

But yeah, thanks @Mike.

Er.

anyway, you have the idea.

I mean you can just take every distance from the point to another point outside the set

No, your argument is correct, it's just not written rigorously.

Actually I'm not too comfortable with LUBP arguments either. Also why is this depending on the LUBP? Shouldn't this be a topological argument only? I don't know why I need to invoke something about the Real numbers.

I mean I know we're working with the Reals, it just seems weird to me.

i'm not sure how you expect to get out of using facts about the real numbers, considering you're trying to prove something about the real numbers

I guess that was a dumb question.

@Mike: It's named pb-diagram and features an easy-to-use diagram environment. Description at CTAN. It's in my TeXLive distro, so the command line texdoc diagram shows the documentation there.

:P

try and prove that $[0,1] \cap \Bbb Q$ is connected and you'll run into trouble

I guess what was throwing me off was that when you view a subset of a metric space, the elements are now disjoint from the space. It's weird to me to think of [1,2] [3,4] as existing with the space gap between 2 and 3, but with no numbers to fill it, I guess.

why?

consider the 2-point metric space. is it weird to you?

Hi everyone

@user127001 Hi

Here is an easy question with an answer that can probably only be posted as a comment because the system requires some minimum of blurb, I suspect. Should we encourage F. Lemmermeyer to try it out?

Will Gaussian Mixture models still work if the number of components is greater than the number of datapoints?

@user127001 I see you have not changed your username, lol.

Hi @PedroTamaroff your legs still hurt?

@JasperLoy Nah. But I overslept.

@PedroTamaroff You can lie down and put your legs high up against the wall, that helps.

That is a trick I discovered myself.

I don't know why it works.

I get much fewer votes per answer here than on Eng

@JasperLoy True.

Drains the blood from the muscles?

@Mike @DanielFischer

@PedroTamaroff ?

@DanielFischer Does this look good to you?

I think making the diams go to zero is unnecessary.

Have not yet reached that tab, @PedroTamaroff, won't take long, though, if I don't get disturbed too much ;)

@DanielFischer Come again? =P

@PedroTamaroff I have a couple of tabs open. That among them. I have not yet reached that tab.

@DanielFischer Oh.

I have posted an answer.

And if you keep on ponging me, it will only take longer ;)

@DanielFischer OK, sorry. =/

Dang Daniel, you're being a meanie today.

If you ping me, you will get no answers for a math question.

I only answer when I please in chat.

@PedroTamaroff My rep is out of sync, and I can't trigger a recalc for it to readjust. Drives me nuts.

@DanielFischer There there, Daniel.

@DanielFischer It will adjust automatically after a few minutes

@PedroTamaroff I would take $$A(M) = \bigcap_{n\in\mathbb{N}} f_n^{-1}([-M,M]).$$ $\mathbb{R} = \bigcup\limits_{m\in\mathbb{N}} A(m)$. Baire says ...

@JasperLoy define "few"

@DanielFischer LOL.

@DanielFischer It varies, I think. Maybe 10.

@DanielFischer Yeah, sure.

@JasperLoy Half an hour now.

My proof works too, though, right? It's just a bit less elegant.

Anyone know a proxy that plays all flash videos flawlessly?

@DanielFischer Have you gone to math.stackexchange.com/reputation while logged in? Scroll to bottom and click "trigger reputation recalc".

@PedroTamaroff Yes, it works too. We don't need to shrink, since the intervals are compact.

@DanielFischer Yes, that's why I removed the condition.

@JasperLoy The recalc button is gone since a couple of weeks.

@DanielFischer Then there is nothing you can do. Just wait for a day, lol

@DanielFischer The only thing I might like about the "explicit" proof better is it highlights the jist of the matter a little more.

I prefer constructive proofs where possible for the existence of an object. One needs to know how to compute the object

Hi, I have one question if some can enlighten me : Density of $GL_n(K)$,it is valid in other fields as $\mathbb{R}$ or $\mathbb{C}$ ? Thanks..

@Nico Yes, as far as I can tell. A matrix can only have finitely many eigenvalues, and $A-\varepsilon I$ is invertible iff $\varepsilon>0$ is not an eigenvalue of $A$. By taking it small enough, we can make sure this doesn't happen, and hence we can find an invertible matrix $\varepsilon $-away from $A$.

@PedroTamaroff Thank you, you mean that $\varepsilon$ is not an eigenvalue of $A$ ?By taking it small enough ? I just don't understand what did you mean by away from A ?

@Nico $\lVert A-(A-\varepsilon I)\rVert =\lVert \varepsilon I\rVert=\varepsilon \lVert I\rVert=\varepsilon$ if your norm has $\lvert I\rVert =1$ (as I guess it has?)

@PedroTamaroff I see your point, thank you!!

i get to meet john thompson tomorrow :-O

@AlexanderGruber Is that the Feit Thompson?

@DanielFischer yes.

@AlexanderGruber Then he'll probably know what you're talking about.

@DanielFischer he'll probably have dreamed it 30 years ago, but found it unsuitable for publication

What, by the way, will you be talking about with him?

@DanielFischer i have no idea, he's giving a talk about differences of squarefree numbers, and then we're going for dinner

So I guess it's algebraic number theory?

(found the actual topic)

i guess so

Cool.

he's a diverse dude though it could end up being a lot of things

Hey folks, does anyone know for doing Inverse LaPlace transforms, is it strictly neccessary to full factor the bottom of an equation?

s/((s - 1) (s^2 - 1))

@VaughanHilts haven't seen you in a while

@MickLH I'm still active on Game Dev. SE, although not in chat. :)

Well it's mostly $\LaTeX$ in here

Yeah.. LaTex in chat isn't loading for me right now. :(

Everyone uses this bookmarklet: math.ucla.edu/~robjohn/math/mathjax.html

the one titled "start ChatJax"

Yeah, that page isn't loading for me right now.

I usually use that, too.

haha how (not) awesome of timing

I loaded it to test when I grabbed the link, it worked

refreshed when you said it's broken, it's gone!

How unfortunate lmao

pastebin.com/tctpbbb9

i grabbed it from googles cache

lol I grabbed it from the page I had open

Hang on.

@JasperLoy I am still thinking

@robjohn Can I use somehow \usepackage here (MSE) in question or answer ?

@Cortizol not that I know of. I'd look on the MathJax support page.

@Cortizol I don't see \usepackage Besides, that is a page level command, so I don't think it would be supported in MathJax

@robjohn All right. Then, do you (or anyone else) know how to make $\twoheadrightarrow$, but rotate $-90^{\circ}$, I need it for commute diagram

Hmm..one of my algebra books says that the ring has an additive identity but no multiplicative identity, but the other book says that a ring has the property of having a multiplicative identity but no additive identity..who is correct?

How does a proof of L'Hopital's Rule change if we change our domain from (a,b) to (a,$\infty$)?

hi Anthony

@Cortizol sorry... did you mean something like $\updownarrow$

@user127001 A ring may or may not have a multiplicative identity, depending on the author. But it is ALWAYS a group under addition (so it has an additive identity, in particular)

@Mike Hiiiiiiii

@KarlKronenfeld Hey

I'm about to do some AM exerciss

will you be around?

@PedroTamaroff Yeah

Thanks.

=)

@KarlKronenfeld

yo