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9:00 PM
HAHAHAHAHAHAHAHAHAHAHA
-1
Q: pleaSE HELP NEED BY TOMMOROW

lalamy math question is how do you get 1428 out of 1,2,6, and 8 using only each digit once I need the answer by September 12,2014 or I will be disqualified from school

Anyway
 
:-))))
 
Topology is not my field sorry @JoseAntonio
@JoseAntonio You can always ask on MSE :)
 
jajaa ok ok @hippalectron, seems complicated my first idea was to use something similar to the discrete topology but of course immediately I realize that the idea doesnt work because might have to generate the same topology but makes the space complete....
 
Not many peole are reading the chat right now
Oh ok
 
I know, but I prefer to ask when I have an idea and only want to check
 
9:03 PM
I do the same :)
 
@JoseAntonio Try to find a homeomorphism with a closed subset of $\mathbb{R}^2$. The graph of a continuous function, maybe ...
 
@DanielFischer Ok let me think in an homeomorphism, thanks
 
Hi all
Hi @DanielFischer
 
Hi @JohnDoe.
 
@DanielFischer If you have a second at any stage will you have a look see at my post.
@DanielFischer I think the fact that it has a link to a set of notes is scaring people away.
@DanielFischer But it's just a quarter of a page that has to be read.
 
9:16 PM
@DanielF At first I thought the question was to put a complete metric on $\Bbb R^2 \setminus \{0,1\}$ which sounds
fun.
 
@JohnDoe But, one would have to download the pdf etc. It's nonzero work to do. More effective to scare me off is the name Galerkin appearing. I strongly disliked that stuff when I knew it, now I'm not in the least inclined to dig into it.
@MikeMiller What are $0$ and $1$ as elements of $\mathbb{R}^2$? Identify with $\mathbb{C}$?
 
@MikeMiller Yes is very fun.
 
@DanielFischer Right.
 
FISCHER!!!
 
@MikeMiller Any open subset of a polish space is polish. The same trick.
 
Bah! That's not a fun solution.
 
I'm happy :D
 
mm i don´t get it @DanielFischer is my first course
 
@DanielFischer The question is a small tiny thing that has nothing to do with Galerkin...I also strongly dislike it.
 
Hi, I have a few differential equations questions. First off I have a problem named as such: The field mouse population is modeled by the equation: $\frac{dp}{dt} = 0.5p-450$ My first question is what exactly is $\frac{dp}{dt}$ represent? I know it is basically the derivative of $p$ with respect to time, but does this mean it is the rate at which the population grows...? Can someone explain exactly what the equation is modeling and why? Especially why we use $0.5p$ instead of $0.5t$?
 
9:20 PM
@Hippalectryon Ignorance on both sides. It's just sick.
 
@Alizter ??
 
Why all the hate for lala's question...
 
@BalarkaSen Yeah go on
 
@MikeMiller If you want fun, push down the hyperbolic metric on the upper half-plane using the modular function $\lambda$.
 
I have been looking at combinatronics. It is interesting.
 
9:22 PM
@DanielFischer That was my plan, actually.
 
is probability considered a branch of statistics or combinatronics
 
Neither
also, the term is combinatorics, rather than combinatronics
 
no way
what
 
Sorry bud
 
Wow. Weird how the brain tricks you like that
I have been reading it like that for like a year
 
9:24 PM
I think it's a shame, because combinatronics sounds so cool.
 
@MikeMiller I have found a relevant paper which proves that if $R$ is commutative and that the semigroup $R^\times$ is finitely generated then $R^\times$ is finite.
it is a cool proof. left as an exercise for you.
 
OK.
 
=P
Have fun.
 
@Hippalectryon Not bad.
 
@MikeMiller I just went back to all these places where it is written....
 
9:24 PM
Unfortunately for you, @BalarkaSen, I'm not so sure I care enough to even try. :P
 
wat
 
Hi @DanielFischer would I be right in saying that a linear subspace cannot be bounded?
 
linear subspace of a vector space over what field?
 
@MikeMiller beh. then have the paper.
mumbles angrily
 
@BalarkaSen: Is it correct that my previous mindfarts where actually related to Arithmetic combinatorics?
 
9:26 PM
waaatt
 
@JoseAntonio Take a continuous function $f\colon \mathbb{R}\setminus \{0,1\}\to\mathbb{R}$, and consider $\Gamma(f) = \left\{ (t,f(t)) : t\in \mathbb{R}\setminus\{0,1\}\right\}$. The graph $\Gamma(f)$ inherits a metric from $\mathbb{R}^2$, and transporting that back, you get a metric on $\mathbb{R}\setminus\{0,1\}$ inducing the standard topology. Now you need to choose $f$ so that $\Gamma(f)$ is a closed (hence complete) subset of $\mathbb{R}^2$.
 
@Ropstah i dunno.
 
@BalarkaSen I've sent it to someone who cares more than me, so you've done a good deed for that fellow.
 
@MikeMiller hmm?
 
The paper.
 
9:28 PM
Ah.
 
@JohnDoe Well, there's $\{0\}$. And if you don't require the space to be Hausdorff, there can also be nontrivial bounded linear subspaces. But for Hausdorff TVSs (over $\mathbb{R}$ or $\mathbb{C}$), the trivial is the only bounded linear subspace.
 
Good.
 
@BalarkaSen Solution link or tell me
or hint
actually no hint
 
i am writing it. wait up.
 
I just found out that combinatronics is actually combinatorics.
 
9:29 PM
I want to learn counting by the use of structures (pyramids) using different bases
 
I think there is too much worry about those kids that come here for having the homework problems solved. If I were a professor I wouldn't see how MSE would possibly help any of them if they only want to cheat. Just think about it! You go to school with the lessons that are done here, but what's next? In a real test you fail!
 
@JohnDoe I see, so is that why you said that the only compact linear subspace is the trivial one the last time we spoke?
 
@DanielFischer wow very clear i understand
thanks
 
@Chris'ssis: You need mental visualization of the problem. Some get that through doing the math themselves, some get that through visualization, some get that through discussion and being taught
 
9:30 PM
@BalarkaSen hmm, this seems like a generalization of the fact that there are no rings with infinite cyclic group of units
 
I agree that copying an answer doesn't work
 
Keeel
 
what do you mean by R^\times though?
 
@MikeMiller Thank you for pointing that out. I feel like a total idiot.
 
@anon heh. i thought it was a generalization of the fact that fields with finitely generated multiplicative structure are all finite.
 
9:31 PM
Chill out, @Alizter.
 
@anon multiplicative structure of $R$
 
@BalarkaSen so (R,x) ?
R^\times means something else to me
 
yes, that.
@anon group of units?
 
yeah
 
pfeh algebraic number theorists
 
9:32 PM
or, if not that, then non-zero-divisors, and if not that, then nonzero elements
 
double pfeh
i am finding that for some reason i liked @blue better than @anon. funny.
 
you must unconsciously dislike orange
and like blue
 
i still occasionally get green and orange confused
 
Actually disregarding my last question, is this the proper way to integrate this? $\int\frac{dp}{0.5p-450} = 2\int\frac{dp}{p-900} = 2ln|p-900|$ however my professor says it should be: $2ln|0.5p-450|$ where am I going wrong?
 
@Link constant of integtration
 
9:35 PM
speaking of colors, whoever decided orange, poop brown and green would be he primary colors of mathoverflow should be slapped in the face
 
your answer and his answer are identical up to a constant
 
@MikeMiller Could you elaborate a bit? I took out a 1/(1/2) = 2, so shouldn't that factor out of the denominator?
 
@Link $\ln(2x) = \ln(x) + \ln 2$
 
um
 
shush
 
9:36 PM
(:
 
@Mike is right. $\log $ is additive. for example, $\log(1 + 2 + 3) = \log(1) + \log(2) + \log(3)$.
you can verify that numerically.
 
you're not funny
 
Hmm.. let me try and work it out.. $\int\frac{dp}{0.5p-450} = \int\frac{dp}{0.5(p-900)} = \int\frac{1}{0.5}\frac{dp}{(p-900)}= \int2\frac{dp}{(p-900)}=2\int\frac{dp}{(p-900)}=2ln|p-900|+C$
So where did I go wrong in this? It seems to be right, but evidently not.
 
that is right
that's also precisely your teacher's answer
 
But doesn't the fact that it's $p-900$ differ from $0.5p-450$ Doesn't this change the value?
 
9:41 PM
5 mins ago, by Mike Miller
@Link $\ln(2x) = \ln(x) + \ln 2$
 
Oh! I see what you mean now. I think? If I see then $2ln|0.5p-450|=2ln|p-900|+2ln2$?
 
@Alizter We want to prove that $\exists k$ such that there is an $x$ such that $2^x \in [n^{2^k}, (n+1)^{2^k}]$. Consider the interval $[\log_2(n), \log_2(n+1)]$.
Cutoff the two ends of the interval to produce an interval $[a/b, a'/b']$, (which is possible as $\Bbb Q$ is dense in $\Bbb R$) with rational limit points $a/b, a'/b'$. There exists an integer $y = bb'x$ inside $(ab'2^k, a'b2^k)$ for some appropriate choice of $a/b$,$a'/b'$ and $k$. Hence, $x/2^k \in (a/b, a'/b') \subset [\log_2(n),\log_2(n+1)]$. $\blacksquare$
 
yup, @Link, and because you have that $C$ there, that's accounted for
 
Oh! I get it, the $C$ is being folded into the $2ln(2)$, since it's just a constant. Thank you.
 
no prob
 
9:47 PM
Actually, @Alizter, you can just work your way though $[a/b \cdot 2^k, a'/b' \cdot 2^k]$ for some large enough $k$ to make $\lfloor a/b2^k \rfloor$ and $\lfloor a'/b' 2^k \rfloor$ differ as much as possible to fit an integer between those two.
 
heya @anon ... How goes the teaching?
 
Nonetheless, the whole problem is about proving that rationals with denominators a power of $2$ are dense in $\Bbb Q$, which is obvious/elementary/easy.
 
tries saying heya @Mike
 
hello
 
@BalarkaSen nice
 
9:51 PM
so when are the quals, @Mike?
 
8 days til the first
 
Yippee :)
 
well, that's a lie
it's 8 days til the 19th
:)
 
smacks @Mike
 
@Alizter now you can use this fact to prove that any nonempty set $S$ of positive nonzero integers such that if $x \in S$ then $\lfloor x \rfloor$ and $4x$ are in $S$ is all of $\Bbb N \setminus \{0\}$
have fun!
 
9:53 PM
@TedShifrin That's the response I aim for with my jokes
 
your students will rebel
 
I'll never tell them any of my good stuff.
 
you mean to suggest that what you share with me is "good stuff"?
 
did you do well in Jacob's review course? :)
 
9:56 PM
He wasn't at office hours today!
 
Hi @Ted.
 
guten Abend @DanielF
It's been ages!
That's not like him, @mike
 
Yes, we sort of missed (in the sense of "not having met") each other a lot in chat recently.
 
@TedShifrin I'll give him a hard time tomorrow, since he lectured the class a while back about how it was our job to come...
 
I'm sure you will, @Mike.
well, @DanielF, I haven't been around so much ...
 
9:58 PM
That could have contributed.
 
Not that I've been missed in the slightest.
 
@anon
 
Oh, @Ted, there was a certain scarcity of adults here.
4
 
That's part of my scarcity, @DanielF.
 
No comprende.
 
10:00 PM
He's saying it was a vicious cycle, @DanielFischer
 
My scarcity is due in large part to the observation you've made, @DanielF
 
I see.
 
Plus being busy with school and US Open tennis, which is now over.
 
I'm not watching tennis until Sampras makes his triumphant comeback.
Any day now, I'm telling you.
 
you'll be dead when that happens, @Mike ... he does play on the senior tour
 
10:01 PM
Still his old elegant self, I hope.
 
@MikeMiller I think there are rings $R$ such that $R^\times$ is finitely generated yet $R$ is infinite.
Hmm.
 
@BalarkaSen $\Bbb Z$
Does he really, @Ted? I'll have to watch that at some point
 
@MikeMiller stahp. $\Bbb Z^\times$ is not finitely generated.
 
The guy was my hero when I was younger
 
they show the matches on Tennis Channel from time to time, @Mike
 
10:02 PM
primes are infinite, @Mike
 
yes, great player ...
 
@BalarkaSen Please use standard terminology, then, @BalarkaSen
 
my dermatologist here in GA grew up playing tennis with Sampras in CA :)
 
Anon corrected earlier so I assumed you'd begun using $R^\times$ appropriately :P
@TedShifrin Ah, I don't have any channels.
 
OK, $(R, \times)$ then.
mumbles angrily
 
10:04 PM
@BalarkaSen Didn't you just find a paper that proved the assertion you're denying?
 
@MikeMiller for commutative rings.
 
I don't believe noncommutative rings exist.
 
Yes, @Mike, you've never met a matrix.
 
@MikeMiller O_O
What about our good old quaternions? I thought you were friends!
 
@TedShifrin I'm from Indian Wells, so I got to see Sampras and Agassi play a few times. It was great.
 
10:06 PM
I can never figure out where you're from, @Mike.
 
@TedShifrin I say "Palm Springs" because it's the closest town to where I'm from people know
 
@DanielFischer The phone company informed me that they're upgrading my internet service from 3 Mbps to 18 Mbps. They're getting rid of the old DSL service.
 
Your stay at Penn State confused me, when you were going to family in upper NY state, @Mike ...
 
@RandomVariable Lucky you.
 
@TedShifrin I've got a chunk of family from there - but I'm a CA native, and lived here most of my life
 
10:08 PM
well, I'm a CA native, too ... just from the good part of the state :P
 
Who won the men's singles?
 
US Open, @skull? Cilic.
 
Yes, thanks.
 
I wonder how much tickets cost to the IW open. Maybe I'll go again now that I'm down here.
 
if I move to San Diego, I'll definitely go, @Mike ...
 
10:10 PM
@DanielFischer 18 Mbps in now the cheapest service.
 
Well, you'll be free to watch the whole thing - I don't think I'll be allowed to take 10 days off in the middle of a term :P
 
no, you certainly shan't.
 
Hmm... the last few days are after final exam period ends for the winter quarter, @Ted. This bodes well.
 
LOL, except it'll cost you a year's pension.
 
I'm in LA, @Ted - a big city like this has plenty of banks to rob
 
10:14 PM
I didn't hear that.
 
HAHA
 
What does IW stand for?
 
Indian Wells
 
@MikeMiller They're pretty dry.
 
@TedShifrin pretty fun
@BalarkaSen ?
 
10:17 PM
@anon ?
 
@BalarkaSen You empty pinged.
 
Oh, right, I pinged you.
 
good, @anon ...
 
Oops I forgot.
Will empty ping you again if I recall what I wanted to say, @anon
 
@Balarka
 
10:18 PM
@anon what?
 
nothing
 
LOL @anon
 
mumbles angrily
 
If you motivate those students to work and learn, @anon, you'll be doing tremendously.
 
@Balarka ping with the question next time please :-)
 
10:19 PM
@IceBoy
 
Yes?
 
Nothing, nothing.
 
rolls all six eyes
 
@TedShifrin Do you really have me on ignore?
 
rolls eye
 
10:20 PM
your glass eye, @Mike?
 
That reminds me of Mad-Eye
 
@DanielFischer I'd give you half of the bandwidth if I could. I don't really need it all.
 
What is the smallest bandwidth?
Or is there even such a thing @RandomVariable?
 
I keep thinking that @RandomVariable needs a variance.
 
@IceBoy 3 Mbps used to be the cheapest service. Now it's 18 Mbps.
 
10:30 PM
@ted wouldn't that be a random variance?
 
it depends on the town covenants
 
hahaha
 
damn I get stuck again with the problem of how to find a complete topology for the linea without two poins. Someone knows a source to find similar problems. Sorry but we´ve seen in class very slightly the concept of topology and I'm crush trying to do the exercises in the book.
 
@DanielFischer I feel bad now.
 
@MikeMiller is the fun exercise. mmm I feel very stupid.
 
10:43 PM
@RandomVariable What for?
 
@DanielFischer Because my internet service is being upgraded and yours isn't.
 
@RandomVariable Don't take it wrong, but: Pfffffffffft. So what?
I mean, I have reason to believe that there are people in the world who don't even have running hot and cold water.
 
The quotient of two continuous functions is continuous, isn't it. :/
 
@Anthony Unless the denominator vanishes.
 
Well shucks.
I'm just dumb as a pickle.
 
10:57 PM
@DanielFischer first of all sorry for being annoying. Second with respect to your big hint, (I´m talking to the problem of a metric to make R \{01} complete and with th same topology as the relative topology) is not necesary that the map f: R \{01} \to R also tis inverse has to be continuous? sorry if is a stupid question...
 
@DanielFischer It's not a big deal. I just thought maybe it sounded like I was bragging, which I wasn't at all.
 
And, @RandomVariable, we recently had an offer for a better connection via the TV cable. Only one tiny problem: we don't have cable.
 
0
Q: Local truncation error of Euler method

user96402Wikipedia and this book say the local truncation error of Euler method is $O(h^2)$. But this book and A friendly Introduction to Numerical Analysis say it's $O(h)$. Which is correct? I have a similar problem with implicit trapezoidal method. EDIT My numerical investigation shows that the LTE of...

 
@JoseAntonio You use $f$ to get a map $g\colon t \mapsto (t,f(t))$. And $g$ is the homeomorphism. $g$ is continuous if and only if $f$ is, and the inverse of $g$ is the restriction of a coordinate projection, hence continuous (whatever $f$ is).
 
Anyone have any good resources for 3d differential geometry?
Links, and whatnot.
I can't find anything "nice" on calculating the curvature of a point on a 3D surface. I have been able to make a mental flow chart by looking through wikipedia, though nothing spectacular enough to appreciate what is happening.
 
11:08 PM
Ayo @DanielFischer could you help me out?
 
Could he? Probably. Will he?
 
sigh
I'll ask you Mike.
 
Hahaha
Could I? Maybe. Will I?
(Probably.)
 
I was proving that crap about the continuity of the stereographic projection. Epsilon delta proofs suck.
Could I have show it easier just working with open sets?
 
Probably not, since those are the same thing.
 
11:11 PM
Ugh.
 
@Anthony You can get the explicit formula and see that it is a rational function (either way).
 
But most of the functions were all combinations of continuous functions, I swear I didn't need to do anywork...
 
@Anthony $1/x$ is continuous on $\Bbb R\setminus \{0\}$.
So by doing a bunch of compositions with other continuous functions, your thing is continuous.
 
Yeah, I basically just ended up showing that.
Which was good/bad because I'm terrible at math.
:D
 
11:30 PM
@user96402 I've added a couple of comments to your question.
@Anthony but the stereographic projection rocks!
 
Yeah.
 
Thank you @robjohn.
 
@robjohn that's not a terrible pun is it.
 
@Anthony now show that the inverse map is continuous :)
 
@Anthony I don't see how, but I should look for one :-)
 
11:32 PM
@MikeMiller Woah woah woah mike.
Do I need to?
I mean they're all rational functions...
 
recall the definition of a homeomorphism
 
No, I know I need to, but do I need to.
 
@MikeMiller it is definitely not uniformly continuous :-) wait, the other direction; the stereographic map is not uniformly continuous.
 
Not if you can convince me it's continuous without doing any actual manipulation. But I am hard to convince!
@robjohn to clarify, we're showing that the punctured sphere is homeomorphic to the plane
 
The real thing I should be concerned about is if my professor is hard to convince.
But.
 
11:34 PM
@MikeMiller which is quite true.
 
@Anthony You should be concerned about the mathematical correctness of your arguments!
 
I am.
 
Convincing your professor and I are just corollaries.
 
But I also want to get all my homework done.
:P
But I agree.
Ugh
I mean, those functions are continuous.
In the numerator and denominator. They looks so nice!
 
They look a bit like polynomials, don't they?
 
11:37 PM
Yeah.
 
Can you cite a theorem that makes it immediate that polynomials are continuous?
 
I haven't worked with multivariable functions, really.
 
@Anthony
Is $x \cdot y$ continuous?
 
@MikeMiller Yeah.
 
The big thing has nothing to do with multivariable, it is more general. The only truly(?) multivariable thing needed is that the coordinate projections are continuous.
 
11:39 PM
Is $x+y$ continuous?
 
Yeah.
 
Are constant maps and the identity map continuous?
 
Why did you add in the identity map?
 
$x \mapsto x$ is the identity map... if that's not continuous, probably not many polynomials are :P
I promise you, continuity of the identity map is the easiest delta-epsilon proof you'll ever do.
 
Haha.
 
11:42 PM
Congratulations, now polynomials are continuous!
Oh, I guess $x \mapsto \lambda x$ should be continuous too, for real $\lambda$
But that's not so bad.
 
I was just wondering, it seemed unnecessary. x+y being continuous seems to imply that x would be continuous, or something of the sort.
 
Yikes. I meant the map $\Bbb R^2 \rightarrow \Bbb R: (x,y) \mapsto x+y$
 
But If I take y to be zero, isn't that what you said above?
 
hmm
you win this time
 
<3
 
11:45 PM
that also gives me the $x \mapsto \lambda x$ thing I needed (from multiplication)
 
Revolutionizing mathematics.
I think I should scratch all my epsilon delta work, then.
 
11:58 PM
Hi @Anthony
 
@TedShifrin Hola.
 

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