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7:01 PM
Somewhat.
 
@DanielFischer Would be a indication of poor style to use both QED and the tomb stone together at the end of a proof?
 
@DanielFischer I have to write a function for the destruction of the planetary system "solid".
With the destruction of the planetary system, the asteroids where the gap between this one and the object, is smaller that "gap" will also be destructed. (so, they have to be deleted) The asteroids for which the gap is greater are converted to free-floating planets so they have to be inserted into the tree of free-floating planets. The total cost of the transport of all the asteroids that are converted into free-floating planets in the tree of free-floating planets should be $O(n)$, where $n$ is t
 
@skullpatrol Yes. One should use the Bourbaki-dollar ;)
 
:D @DanielFischer?
 
@MaryStar Does your code work? Well, first, does it compile without warnings?
 
7:07 PM
Whenever people write Q.E.D. at the end of their proof, I always take it as homage to Feynman and Quantum Electrodynamics, @skull. ^_^
2
 
@skullpatrol Write an S, draw two vertical lines through it close to the centre, and rotate a little clockwise.
 
@DanielFischer if you use both you are essentially saying "as was to be shown" "end of proof" which seems redundant to me, no?
 
@skullpatrol Right.
 
@DanielFischer thanks
 
Of course there are worse things, @skullpatrol.
 
7:14 PM
Not again.
 
sigh whoever is trying the hat, at least don't star the same people's messages more than once.
 
LEL yeah
 
@DanielFischer Agreed.
 
Whoever is trying for the hat should star only what I say.
 
Too many stars...
 
7:21 PM
I really dislike the use of QED in modern texts.
The square looks so much better. Why use anything else at all?
Some people use //// too.
 
13
Q: Why does drawing $\square$ mean the end of a proof?

ChanTo end a proof, I often write "as was to be shown" or "q.e.d". Both of these terms make sense to me as a reader. On the other hand, I feel a little strange to put down $\square$ although I saw it many times here and there. In fact, I learned $\square$ notation here. I wonder if anyone could give...

 
I like c.q.f.d.
 
@DanielFischer What does that mean?
 
@DanielFischer I don't get a segmentation fault... Is the way I deleted the nodes correct??
 
C'est qu'il fallait demontrer. Needs some accents, but I'm too lazy now.
 
Huy
7:28 PM
é
 
"Circular Quadrature, Futile Despair"
 
@alizter How is your cold?
 
@MaryStar It's too big for me to check the logic, that would take more time than I'm willing to put into it now, but it looks reasonable. If you don't get segfaults even with a few bogus calls in your programme, that indicates it may be correct. But better run it through valgrind or so to check for leaks.
 
@DanielFischer Ok... What is valgrind ??
 
@MaryStar A suite of debugging and memory-checking tools. It used to have some problems on OS X, I don't know if they are fixed now. If you're using some unusual OS like OS X or Windows, you may need to use other tools, but memory checkers and debuggers ought to be available there too.
 
r9m
7:41 PM
@Chris'ssis hmm .. Smooth !! :D
 
Huy
Time to rewatch A Beautiful Mind.
 
it's silly of you to downvote, whoever did that. it's a perfectly good answer.
 
One should get a Red Baron for finding a poorly received question, improving it, and giving a good, well-thought out question.
 
@Huy I just watched it a few days ago, together with Good Will Hunting.
 
7:45 PM
You did not improve that question and it will soon be closed.
(I did not downvote you.)
 
Huy
@JasperLoy: I watched Good Will Hunting again about two months ago.
 
@r9m Thank you! :-)
 
i don't see a point of downvote however @Mike
 
@Huy You didn't clarify. Did Bill leave voluntarily for a month?
 
@DanielFischer Ok!!
 
r9m
7:47 PM
@Chris'ssis :-)
I was wondering what's the minimal set actions I should take to get 3 hats together ! ^_^ (a.k.a. a Hat-trick !)
 
@r9m By the way, I've still been waiting for receiving good news for the Au-Yeung series proof ...
 
r9m
@Chris'ssis oh ! will it be out by the end of this year ? :D
 
@r9m As regards this article, all is dead, unfortunately, no news :-( Well, it's analyzed by a magazine but no feedback so far.
 
r9m
@Chris'ssis ugh ! :( then is there a sister of Au-Yeung ? ... (if one girl dumps you go after their sister =P) oh in your case its bro ! look if Au-Yeung has a bro ! ;)
 
@r9m lol, yeah, there is a sister. If you remember, I asked you in the past a series involving $2015$ ...
 
r9m
7:55 PM
@Chris'ssis ah ! Nice ;) Teach that a lesson then ! ;)
 
@r9m :-)
 
I got the handegg hat!
 
r9m
Nice ! :)
 
cool
 
Am I really the only one who sees this question as a possible attempt at cheating?
 
8:10 PM
@teadawg1337 Why would it be cheating? Any other question can be cheating too.
 
@Jasper It blatantly states that it's a question from a final exam. Couldn't there be other people taking the same class with the final, say, tomorrow?
 
Probably not, @teadawg1337. And if he was cheating he's doing a pretty poor job concealing it.
I think you're being needlessly paranoid.
 
you still didn't give me a map @Mike
 
Not my problem, @BalarkaSen
 
8:14 PM
@BalarkaSen There are many world maps online.
 
Stereographic projection, not the map I want
 
@MikeMiller I merely came to a false conclusion, it wasn't necessarily me being paranoid :/
 
Hmm @Mike. Let $p : X \to Y$ and $h : X \to X$. Then $p$ and $p \circ h$ are both covering maps $X \to Y$.
Then this should induce a map $h' : Y \to Y$ such that $h' \circ p = p \circ h$, not?
 
You're trying to convince yourself, not me.
 
I know. I am just thinking out loud...
I will not ping you anymore until I come up with some firm proof, @Mike
 
r9m
8:28 PM
 
@KhallilBenyattou talking about feynman here is an interesting article
 
Huy
@MikeMiller: I don't know for sure.
@JasperLoy: I forgot how beautiful Jennifer Connelly was.
 
@Huy Yes, she is beautiful.
 
@Huy Who is that
 
Huy
@Hippalectryon: The leading actress in A Beautiful Mind.
@JasperLoy: I'm very happy we eventually share an opinion.
 
8:37 PM
@Huy Well, we did not really disagree on anything, right?
 
Huy
@JasperLoy: I disagree.
 
@Huy What did we disagree on?
 
Huy
@JasperLoy: On not really disagreeing on anything.
 
That's pretty interesting, @skull!
 
@JasperLoy Cold
 
8:42 PM
@KhallilBenyattou Indeed
 
yo @Alizter
 
yo @BalarkaSen
@BalarkaSen What have you been up to as of late?
 
Mmmmm
I thought that if a function did not have isolated zeros / roots, then the function has to be constant. I learned that taking complex analysis.
Is not that correct?
 
Isolated ?
 
8:57 PM
@Hippalectryon Distinct. Say $x_1$ and $x_2$ are roots of the function $f$. Then $x_2$ is an isolated zero if there exists a ball $B_\varepsilon$ around $x_2$ with radius $\varepsilon>0$ such that $B_\varepsilon$ does not contain $x_1$.
 
r9m
Is it okay if I say sth like "Just a simple request, with the Winter-Bash going on in SE network and the crazy Naruto fan I am, please accept my answer if you feel it is helpful to you, but do not upvote ! (I want a Naruto Hat)" below my answer ? =P
 
Urm what about $f(x)=\delta_{x,1}+1$ ? @N3buchadnezzar
 
@r9m One does not simply become naruto, the road to hokage is long and painfull.
 
r9m
@N3buchadnezzar oh lord ! I just wanted a Hat (not a head-gear .. I'm not on the road to ninja :P)
 
@Hippalectryon What is your definition of $\delta$?
@r9m One does not simply drop of the road to becoming a ninja :p
 
8:59 PM
@N3buchadnezzar $\delta_{a,b}=1$ if $a=b$, $0$ otherwise
 
Fun fact there are currently 208 ninjas in this room.
@Hippalectryon That function has no zeros?
 
@N3buchadnezzar Where is it $0$ ?
 
@Hippalectryon Nowhere since you added the 1 at the back
 
r9m
@N3buchadnezzar LOL ! I won't become a ninja if I get a Naruto Hat for Christmas :P
 
@N3buchadnezzar So, doesn't it work ?
 
9:01 PM
@Hippalectryon No
 
?
9 mins ago, by N3buchadnezzar
I thought that if a function did not have isolated zeros / roots, then the function has to be constant. I learned that taking complex analysis.
That function doesn't have isolated $0$s but isn't constant
 
@Hippalectryon Of course the contrapositive is not neccecarily true....
I was doubting the statement itself
 
I'm not giving the contraposive
 
@r9m Santa will make you a ninja, whether you want it or not :p
 
@Alizter Studying.
@MikeMiller I think I have it.
 
9:03 PM
The contraposive would be "if a function isn't constant then it has isolated $0$s" @N3buchadnezzar
 
r9m
@N3buchadnezzar oh yeah ! even better if he hires me one winter ! ;)
 
Which is false too
 
r9m
OH Lord !!!! I was upvoted ! :((((((((((((((((( I feel like I was robbed ... X(
 
The map is $X \to Y$. We want to prove that a self-homeo of $X$ induces a self-homeo at the level of $Y$.
 
@Hippalectryon I see. the statement I gave was wrong
"Asume a function has zeros that are not isolated. Then the function must be constant"
 
9:05 PM
@N3buchadnezzar You need an analytic function.
 
@MikeMiller The map is $y \mapsto (p^{-1} \circ h \circ p) (y)$
 
A friend of mine gave me this one, and asked me to find it's zeros
$$
\phi(x) := \sqrt{x+6-4\sqrt{x+2}}+\sqrt{x+11-6\sqrt{x+2}} - 1
$$
 
Huh, my wireless card just randomly shut down and had to be manually reactivated as an administrator
 
@DanielFischer The function really puzzled me.
 
...
 
9:07 PM
@teadawg1337 Comcast ?
:P
 
@r9m You know his elves are just the naughtiest kids taken as slaves? ;p
 
@N3buchadnezzar That's false
I'd be an elf
 
@N3buchadnezzar $$\sqrt{(\sqrt{x+2}-2)^2} + \sqrt{(\sqrt{x+2}-3)^2}-1$$
 
@Hippalectryon Not my router, the wireless card inside my laptop randomly shut itself off
 
@DanielFischer I got that too, problem is. That function does not have isolated zeros.
 
9:08 PM
@teadawg1337 Maybe you pressed the wireless button accidentally.
 
@Jasper There is no external wireless "switch"
 
What's going on with the $xy^4$ and $xy^5$ inequalities? I keep seeing strangely worded questions of this kind. The very first one claimed it was for building a website....
 
@N3buchadnezzar Real or complex $x$?
 
@DanielFischer Real
 
@N3buchadnezzar Okay, then you have it globally defined, but it's not analytic.
 
9:10 PM
@Behaviour Isn't that just people speaking English badly ?
 
@DanielFischer Right
 
@DanielFischer Sanity check : $p : X \to Y$ be a covering map, $h : X \to X$ a self homeo of $X$ then $p^{-1} \circ h \circ p : Y \to Y$ is well defined, am I right?
Hmm.
 
@BalarkaSen No, sorry, deck transformation would mean you get the identity.
 
yeah. i think it's defined even if h is not a deck transformation
 
It must be a morphism of coverings, or, in other words, fibre-preserving.
$p(x_1) = p(x_2) \implies p(h(x_1)) = p(h(x_2))$
 
9:14 PM
well it maps a point to a point, right?
 
@Hippalectryon This is not about English. "i saw this equation posted yesterday and i wanted to ask if i could have help helping this kid solve it." (?)
 
Oh that.
 
@BalarkaSen The problem is that $p^{-1}\colon Y \to X$ isn't well-defined (in general).
 
sure, but that doesn't mean p^{-1} \circ h \circ p isn't does it?
 
@Hippalectryon Which is a broken English version of "asking for a friend"... and about as believable.
 
9:15 PM
ok take $y_0 \in Y$
 
So to get a push-down of $h$, the choice of the point in $p^{-1}(y)$ must be irrelevant.
 
fiber is the set $p^{-1}(y_0)$
 
@Behaviour All the more suspicious that the user is brand new
 
now transform to $h(p^{-1}(y_0))$
 
r9m
I lost a perfect opportunity for getting a Naruto Hat ! S** * :((
 
9:16 PM
@DanielFischer what we want to prove is that this ^ is a fiber of some point in $Y$
 
@BalarkaSen It would suffice that it is contained in one fibre.
 
WTF are these hats?
 
Huy
U JELLY BRO
4
 
@DanielF In case you don't know the situation, he's got a composite covering $X \to Y \to Z$. $h$ is a deck transformation over the composite.
 
irrelephant @Mike
 
r9m
9:21 PM
@AlecTeal Hats are like pants ! (but more of a fun than an obligation to wear) .. :) If you don't like them you can turn them off :-)
 
It's not at all irrelevant, @Balarka, unless you somehow think you can push down any self-homeomorphism of $X$ to one of $Y$.
 
that is what i am thinking
 
Sorry to hear that.
 
@Behaviour Could you by any chance make your chrome review plugin auto update the page ?
@Huy ^
 
r9m
9:28 PM
@Hippalectryon what does that phrase mean ? :O
 
Since I am not getting much upvotes on my alg topo answers I guess I should consider studying it thoroughly first before participating in answering questions on the mains. I presume my answers lack of quality.
 
@r9m jealous
 
r9m
@Hippalectryon ah ! ;) hehe
 
@Hippalectryon Funny. You cobbled up the trollface with a penguin.
 
r9m
Anyone with an idea for one liner for this one ? :-)
 
9:34 PM
@Hippa He has a bookmarklet that does just that, titled Review.
 
@MikeMiller I know, but as @Behaviour has a chrome plugin to enhance the review page, I thought he could as well include that in the plugin
 
It is not a natural addition as you might not want that enabled at all times.
 
Chrome allows settings for plugins
 
@Hippalectryon Refresh the task counts on the right, or the list of posts in the main field? For the list of posts, I'd rather not, because often I'm slow to go through the posts there (editing etc), and I don't always go in chronological order. Auto-refresh would cause some items to scroll past before I get to them.
 
@Behaviour The task counts
 
9:39 PM
Okay, I'll do that in the next version. Once per 2 minutes, maybe. To save bandwidth and avoid IP blocks.
0
A: Can we get an option to stop the global inbox from marking everything as read?

Jeff AtwoodWe generally don't do preferences, so this is unlikely. I'd much rather pick a sane default that works for 90% of folks than have a bunch of configuration options.

 
Thanks
 
I don't do preferences either, so once it's enabled, it's on. :)
 
Someone needs to clear up those stars
It bothers my eyes
 
The best way to clear it up is to star everything I say. Then the wall will be very beautiful.
 
There are now 5 people with 27 hats. I'd better get that Red Baron soon.
Luckily, I have a candidate in mind for when I finish grading.
 
9:49 PM
There are more than two weeks left in the bash, so I suppose opportunities will be there.
 
I think I see how to use the existence of Z @Mike
 
But I think the criteria of Red Baron were poorly chosen; the hat is more likely to be obtained through vote coordination than through actual use of the site.
 
Since Y \to Z is a covering map, every point y_0 in Y is in fiber of some point z_0 in Z
 
@Behaviour That perhaps applies to all hats, lol.
 
Yeah
 
9:51 PM
Since Y \to Z is a covering map, every point y_0 in Y is in fiber of some point z_0 in Z
 
@Behaviour I agree. The first Red Baron was obtained fairly in the intended fashion. I have a question I think is a good question that only attracted downvotes because of poor formatting; then the only trouble is hoping people read it.
 
Now lift y_0 to some point x_0 in X. Transform to h(x_0). Map by q \circ p, it returns to z_0. This means that p(h(x_0)) is in the fiber of z_0
 
I am going to sleep, good night.
 
I've got a question. If $g$ is continous and $f$ is continous, so that $gf$ is the identity. So $f$ is open since for every $V$ that is open, $g(f(V))=V$ and thus the source of $g$ must be open and that's $f(V)$. Right? Or am I doing something wrong?
 
hmm. That can't be right.
 
9:56 PM
@Studentmath $g^{-1}(V)$ can be a strict superset of $f(V)$.
 
it's right but it doesn't help. hehe.
oh god i can't do this one.
 
@Daniel Oh right.
But in that case, whatever is not in $f(V)$ will not be in $f(X)$ either, where $f:X\to Y$, right?
I mean, $g^{-1}(V)-f(V)$ is disjoint with $f(X)$
 

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