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8:12 PM
What I love about stew is that the leftovers get better with time.
 
8:24 PM
Hello there.
What a productive day : )
 
Quite. I bought a few things for house.
 
What?
 
A hook rail for the kitchen, a very late birthday present for my girlfriend, and I baked a cookie.
One cookie, at the size of Venus. Or at least a medium plate.
 
: D
 
Hang on, I'll put a photo online.
 
8:32 PM
anyone here know how to get arclength of a function of two variables within certain bounds and where the function = predefined z0?
 
@Matt What did you do today?
 
Futile Attempts.
I've done about 5 hours so far.
 
But... you said productive...
 
?
Good job I'm not holding my breath while you put up that photo.
 
Futile Attempts are so... futile...
Oh yeah, that...
 
8:38 PM
I wasn't looking sorry. Can you do that again?
: D
 
QED
"If you model some phenomenon with a polynomial, it's often of interest to determine when the polynomial evaluates to zero. One of the tools used in deciding when this happens is factoring."
what...
 
@QED what is this?
 
haha
naive answer to a naive question
 
QED
seems that way
 
8:42 PM
or a verbose tautology :P
 
I wonder how the cookie tastes.
 
anyone here read The Saga of Polly Nomial?
 
@AsafKaragila Hopefully not too dry.
 
never knew Math could be this raunchy. The audience here is the best to appreciate this thing

http://www.netfunny.com/rhf/jokes/88old/poly.34.html
 
9:01 PM
It feels good to downvote everything
Like everyone do to me
yes.... 30 a day. yyeees
goood
 
The challenge on this site to write correct answers in a way that even monkeys will understand to avoid downvotes, else write them so complicated they think they must be correct
 
Why are you on this site?
 
i like math
 
QED
*physics
 
9:13 PM
@QED He's also on math.SE
@JOHA It sounds as though you drink only Jack Daniels and you say that you love scotch.
 
I study physics, but I love math more
 
@AsafKaragila !
 
@Matt !
 
Love is a funny word. I think I've never used it when being serious.
 
It seems that I have to reboot now, to apply the new kernel version. I bid you a fair night... and whatnot.
 
9:20 PM
Bye Asaf.
 
QED
@Matt, I don't get this
 
What?
 
QED
あい love
 
QED
こい love
is one wrong? or what
 
9:21 PM
Oh and there are many more. I think this would be a good question for japanese.SE.
Both are right : ) Depends on what you want to say...
 
There are many ways to love!
 
QED
ah well that's good
as long as it's not a mistake
 
@AsafKaragila Congrats on the answer : )
 
QED
(removed)
2
 
: )
@QED And I don't know all of them nor their correct usage.
 
9:24 PM
@Matt Hah, thanks. I guess. It's an old answer... I forgot about it, too. Either way, my eyelids are dropping like frogs from the sky, and I'm not even in Egypt during the biblical period!
 
Well then good night Asaf. And whatnot.
 
@Matt you see, there are also complicated ways to prove continuity of norms...
what's going on here?
 
QED
nothing
 
$\color{lightgrey}{\text{(removed)}}$
 
@tb I'm a bit too tired at the moment -- why is he doing all this?
I think the first paragraph could be a one liner or am I missing something?
 
9:31 PM
@Matt he wants to use the preimages of open sets are open characterization of continuity. Why? Dunno.
 
@tb : )
How was your day?
 
Had a long session with yunone on extension of uniformly continuous functions in chat, enjoyed that it's still dry inside my house despite the thunderstorms and am freezing my feet off at the moment... Great day!
 
Socks? Or heating? : )
 
I'm wearing my mother's trademark woolen socks, I rest my feet on a hot water bottle and the heating is on maximum. No idea what's going on. :/
 
Oh noes, that sounds like you're coming down with something : (
 
9:40 PM
Maybe, I don't know. I'm just making tea now...
 
Do you have a way to find out the room temperature?
 
No, but I think it's around 21 or 22 at the moment. I don't have fever.
And it's just the feet, och, whatever, I'll survive :)
Heh, compare this to this
 
Same question, same answer?
 
jup :)
 
@tb If it hits full strength and you can't get out of bed and need something don't hesitate to send me an email with a shopping list.
 
9:48 PM
@Matt thanks, that's very kind, but it won't be necessary :)
 
That's what you want to think now. : ) Anyway, the suggestion was just in case.
I left some more writings in the other room for you btw, for when you run out of other things to do. : )
 
@Matt I appreciate it!
 
10:14 PM
@robjohn I'm sorry that I didn't check it. I went to the graduation of a friend that became a sommelier :-).
 
@JonasTeuwen No problem. I think that Didier and I have come to terms :-)
 
It looks very complicated.
 
@JonasTeuwen That's the problem with having all the details there. The main idea is really pretty simple.
 
10:35 PM
Good night folks.
 
@t.b. Should I be worried by your remark? =)
Hi @robjohn, do you want a sequences-and-series puzzle?
 
@Srivatsan Hello :-) what did tb say that you need to worry about?
@Srivatsan okay.
 
@robjohn This one isn't as nice, but what does it matter.
 
Tim
10:50 PM
@Srivatsan Thanks! I was wondering how you found out the link to a comment?
 
Sheesh, Tim you ask faster than I can hit enter
 
@Tim the way I do it is to look up the comment on the person's profile.
 
Define a valid path to be a sequence $p : \mathbb N \to \mathbb N^2$ such that (i) every point in $\mathbb N^2$ appears exactly once. (ii) The sequence is non-decreasing in the sense that for all $i < j$, either $p_i$ and $p_j$ are incomparable or $p_i < p_j$.
Of course, the order on $\mathbb N^2$ is given by $(a,b) \leqslant (c,d)$ if $a \leqslant b$ and $c \leqslant d$.
Definition clear?
 
@Srivatsan no you don't have to worry. But your construction of the bizarre topology doesn't make addition continuous...
 
@tb I never constructed a topology in my answer!?
 
10:54 PM
@Srivatsan the one you gave to Bill Cook
 
@Srivatsan and they are incomparable if $a\le b$ and $c>d$?
or vice-versa
 
@robjohn Something like that.
@robjohn Actually, one coordinate strictly smaller and the other strictly bigger.
E.g., (4, 3) can be compared with (4,5)
 
Where is JM when you need him? 8-).
 
yeah, sure.
@Srivatsan Unfortunately, I need to take my son to mail something off and to get some stuff at an electronics store.
 
Tim
@tb Okay, I will keep quiet. But I am wondering what you referred to "you ask faster than I can hit enter"?
 
10:58 PM
I will think about this on the way.
 
@robjohn Let me finish?
 
okay.
 
@robjohn :2907989 Now, given a doubly-indexed sequence $a = (a_{mn})$ and a "valid path" p, we can associate a sum $\sum_{i=1}^\infty a_{p(i)}$ provided this series converges.
 
Tim
@JonasTeuwen JM is sleeping. He will respond to you when he wake up.
 
@Tim I didn't mean to say that you should keep quiet. I was referring to our two exchanges in the comments today. The last comment came almost immediately after I posted it.
No offense intended at all
 
10:59 PM
@robjohn Question: does there exist $a$ and two valid paths $p$ and $q$ such that the two paths give different sums?
 
I didn't read the answers closely enough and I thought Srivatsan's topology he described to Bill was in his answer too.
As I said: addition isn't continuous there.
 
@tb Sure, that topology is bizarre. But (a) what do I do, however much I dislike it, that is the coarsest topology that makes the norm continuous. (b) I said that the topology is uninteresting in my comment. =)
 
Tim
@tb I see now. Your just deleted comment was supposed to address Sri's comment to Bill. No, I wasn't offended. I was making joke about keeping any secret that you might imply.
 
@robjohn Got the question? // I have an answer, but don't blame me if you don't like the answer. I didn't like it myself.
 
Okay I see. So no one is offended, I hope :)
 
11:02 PM
@Srivatsan I think so. As I said, I will have to think about this as I drive my son around.
 
Tim
I am not so easily offended.
 
@Srivatsan :-) okay
 
@robjohn Sure. Have fun. :=)
 
Very good, then. To see that addition is not continuous take a ball around zero and move it somewhere else. That set won't be open.
 
Tim
@tb Thanks! Does Sri's topology treat normally closed balls as open subsets?
 
11:14 PM
@Tim I'm not sure I understand what you're saying. Open sets are rings around zero or balls around zero.
So the sets $\{x : \|x\| \lt a\}$ or the sets $\{x : a \lt \|x\| \lt b\}$
 
Tim
Sorry, I should have said normally closed rings centered at zero are treated as open in Sri's topology.
He wrote "U(a,b) is the set of all vectors v such that a⩽∥v∥⩽b."
 
I think that was a typo. I'm pretty sure he meant what I just wrote.
(if he had really meant what he wrote, the set $\{x : a \lt \|x\| \lt b\}$ weren't open while it should be)
 
Tim
11:36 PM
@tb After reading some Wiki, I was wondering if the topology induced by a norm is the coarsest one that can make addition and scalar multiplication continuous?
 
@Tim No, not at all. There are the weak topologies which are much coarser. There's always the indiscrete topology, too.
I have to go now, good night!
 

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