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12:20 AM
Hi, can someone explain "characterizations" in this from a textbook: The problem is to find a characterization for Hamiltonian graphs in the same way that we found one for Eulerian graphs with Euler trails.
 
Hi @meow
@Maks: No. No formula in general.
@Jeff: A numerical criterion that tells you yes or no?
 
man I haven't proved anything new in a few weeks and its really depressing
:(
 
Hi, @Forever ... Sorry to hear that. I often went months without proving anything new.
 
I only need little discoveries to keep me going
I guess it's possible that I've knocked off the easy things so thats why it's hard now
 
Does your adviser provide any guidance?
 
12:33 AM
kindof, but we sometimes just hit a wall together
 
I have been there :)
 
we just kindof stare at each other lol
come to think of it, a lot of math professors are good at staring
 
I don't stare ... well, not that way.
hi @MikeM
 
I hope he's thinking about math, and not "what is wrong with this guy, I wish he would leave my office"
 
@TedShifrin So they mean some numerical criterion which would tell us whether the graph has a Hamilton circuit or not. G has an Euler circuit if all the nodes have even degree.
 
12:38 AM
Right, that's the sort of thing they're asking.
 
@TedShifrin Understood. Thanks.
 
Sure.
Cute kitties, btw. :)
 
@TedShifrin Thanks, :D. The brown (younger) one passed on :(
 
aww, so sorry — I've lost a few myself.
 
yeah... i need some cute, soft, cuddly pets that have say, an 80 year lifespan :D
 
12:41 AM
um ... good luck on that one
 
as long as we're talking about it, is there a way for me to salvage that pic? my old comp crashed and I no longer have it
 
hmm, sure, go to your profile page and you can copy it there, I think
 
Well, i can see it there and get a screenshot. But it's small -- not the fullsize pic.
 
No, shouldn't be a screenshot. Hold on.
 
@Jeff try right-clicking and pressing" open image in new tab"
 
12:46 AM
Right. Or, if you're a Mac person, do option click. :)
 
That's better. Thanks.
 
Or if you're a linux person.... well, right click still :P
 
i think i must have cropped the pic when i loaded it on here :(
 
Yeah, most likely, @Jeff.
 
if you dont crop stack exchange pictures, they decide "oh lets just stretch it to the square"
 
12:48 AM
oh well.
yeah, that's probly why i did it
 
I can tell I am getting sick
 
I remember having to fiddle a bit with my cute spinning picture.
 
@ForeverMozart that's unfortunate
 
Take vitamin C and have lots of hot tea, Forever.
 
I have a dry cough and feeling a little woozy
 
12:49 AM
Hi.
 
so probably tomorrow I will have a fever
 
the worst is when you have a stomach bug and you can't eat because you'll throw it back up
 
luckily my stomach can handle anything
 
time for plain boiled rice, @meow
 
I always get colds this time of year though
 
12:51 AM
As I said, vitamin C and hot tea.
 
ok I think I have some
 
sunsets at 5:00 are the most depressing things :(
 
Well, I can think of things more depressing, but ...
 
@ted really likes to take things literally
 
Me too.
 
1:01 AM
Can some one tell me where i went wrong here: i.imgur.com/OioDKrl.png according to my book the radius is actually sqrt(5) but i don't know how they came to that conclusion. I got the origin correct how ever.
 
$4x^2-16x+\dots = 4(x^2-4x) + \dots = 4(x-2)^2 + \dots$
 
man I suck at chess late in the day
 
@Foreover: I suck always.
 
ah so its 4(x-2)^2 - 4 . . .
or 4((x-2)^2 - 4 )
 
I didn't read the rest @WDUK
 
1:04 AM
was referring to what you replied with
 
@ForeverMozart Lol
 
Well you need to correct: I didn't write out the constants.
$4x^2-16x = 4(x-2)^2 - 16$
 
okay i understand now! thank you :)
 
Sorry about that.
 
yeah I think my brain is tired. I did just beat some loser though. Took me 13 moves
 
1:06 AM
It's ok just to watch a fun movie, @Forever :)
 
Lmao!
 
@ForeverMozart would you like to play a game?
 
I'm afraid you would destroy me
 
hehe
 
dw :)
im not very good at chess
i just like to play for fun once in a while :)
 
1:10 AM
Im in a game now but maybe later
 
@meow: I'll play you in bridge when you learn that :P
 
you play online bridge?
 
I haven't, @Forever, because I play enough IRL ... but it's out there ...
I haven't checked to see if you can play against another person (bridge requires 4, of course)
 
@Ted do you play any instruments?
 
Many years ago I played piano.
 
1:17 AM
also, i updated my "mugshot"
 
Yes, I see your mugshot is updated. Much better :P
 
@TedShifrin what do you call donuts with glasses?
four-eye tori!
 
Yes, there are on-line bridge games where you can play with/against more people.
baddddd, @meow
 
;)
 
I play the piano
 
1:20 AM
When I retired, I sold my really good upright piano to a good friend who I knew would use it more than I. :(
 
Trying to play Maple Leaf Rag, but its quite difficult for me
Oh, sad
 
I did mostly classical stuff. Syncopation is difficult, and that's full of it :P
 
You have to practice the hands separately and then gradually put them together.
 
@Ted: Talk went well. Couldn't get to my favorite application at the end though.
 
1:22 AM
Oh cool, @MikeM. Is this the one I wanted notes from? :)
 
@Ted you ever browse vixra? some of the documents there are...
peculiar to say the least
 
Nope, @meow.
 
1:38 AM
@TedShifrin Which one did you want notes from?
 
LOL ... now I've forgotten.
Was it the four-color thing?
 
Oh, yeah, four color. No. This was uncountably many exotic $\Bbb R^4$s.
 
Ohhh, that's cool too. :)
I don't know that stuff.
 
@TedShifrin I did classical stuff for the most part too, but I've spent the last few years gaining a jazz and rock repertoire. Syncopation isn't my problem so much as intricacy.
 
"intricacy" = stuff that's hard? :D
 
1:48 AM
Haha, there's a definite thing I mean that I can't quite explain. Hard in a certain way, for sure.
 
I had fun with classical stuff with 3 against 4 ... and occasionally worse. Good thing I didn't want to be a serious pianist.
 
I have strangely large hands so fingerings were always awkward, especially Bach stuff.
 
In the Schumann piano quintet I've heard recordings where they get the rhythms off.
Oh, you and your large/tall stuff.
 
3 against 4 and 5 against 4 are fun to just tap around the house.
It's not like I can help it!
 
Well, the president would like to swap hands with you :D
 
1:49 AM
At least I get to play Rachmaninoff for my troubles.
 
Rachmaninoff ... far from my favorite. 2nd/3rd rate hack :P
 
I'm a Debussy guy myself.
 
Much better :) I'm a big French impressionist fan, love Fauré's chamber music.
And the German/Austrian crowd, of course.
 
There's little pre-contemporary music that I don't like.
 
1:51 AM
It's Taubes. Comes from a noncompact generalization of Donaldson's theorem.
 
@Fargle: Glad to see you're a talented guy. ... My dad was a contemporary composer, though, so watch out :P
Ohhh, right @MikeM ... Nope, don't know nothing about it.
heya @dsillman
 
I don't mean to disparage the contemporary composers at all--I'm just not familiar with them outside of, like, John Cage and the many movie score composers. Which is sad, I'd like to know more.
 
I was caught off-guard. Ko normally asks prodding, clarifying questions. This time he asked a technical detail I haven't thought about in a while.
 
There's much I really don't like, @Fargle (especially John Adams), but you should try some of my dad's stuff sometime. There are CD's ...
It's good for you, @MikeM.
 
I have a Jacobian Integration question
 
1:53 AM
Yeah?
 
I understand almost every part of the change of variables
However
When changing the integration limits, I seem to have a problem
 
Yes, that's the hardest part.
 
I made a question about it, but I don't know if its taboo to post that here
 
Scary, though, @Ted.
 
You can link to it, @dsillman.
 
1:55 AM
0
Q: How to set the limits for Jacobian Integration

dsillman2000I've been interested in properly calculating a Jacobian Integral (that is, an integral in which a change of variables occurs). I'm sure that, by one's reading this, you've all probably already heard of how Jacobian Integration was used to calculate $$\int_{-\infty}^{\infty}e^{-x^2}dx = \sqrt{\pi}...

 
hello all, im using Gram-Schmidt to find an orthonormal basis for an inner product of vectors $(a,b)$and $(c,d)$ defined as $ac+\frac12 (ad+bc)+bd$. Im only getting a ridiculous vector with even more ridiculous fractions... would one of you mind attempting this to see if they really are just ridiculous?
 
@dsillman: No, the polar coordinate trick only works if you integrate over the whole plane. Draw a picture. If you integrate over the unit square, then $r$ limits get very hairy as $\theta$ goes from $0$ to $2\pi$. Even if you pick one eighth of the square (by symmetry), you get $$\int_0^{\pi/4}\int_0^{\sec\theta} r e^{-r^2}\,dr\,d\theta,$$ which is impossible.
 
@Ted: This stuff is pretty interesting and probably has a lot of powerful untapped applications.
 
@Aksel'sRose: With that inner product, I expect things to be quite yucky.
@MikeM: I don't doubt you.
 
Ah I think I understand
 
1:58 AM
Pull back the H-bundle over G/H to G. What is this?
 
@dsillman: You have to draw the region of integration and figure out what it corresponds to in the other variables.
 
Yeah you're right that makes sense
 
@TedShifrin man, that stinks. I mean good because Im hopefully right, bad because I really dont want to write all of this out. Thank you for answering :)
 
It certainly wouldn't be very easy to integrate xD
 
LOL.
Impossible, @dsillman, for good reasons.
@MikeM: I assume you mean that you're pulling back $G\to G/H$?
Hmm ...
So presumably you get the trivial bundle.
 
2:03 AM
That's what I would guess...
 
All our functorial friends would tell us that this is the pushout diagram. Where is @AndrewT when we need him?
@MikeM: Seems right to me, even more generally. Consider $S^2\to\Bbb P^2$. The pullback of the tautological $\Bbb Z_2$-bundle is the trivial $\Bbb Z_2$-bundle.
But, embarrassingly, I don't have a general proof.
(Yet)
@MikeM: Yeah, it's right.
 
Why?
 
Because I can define the mappings so that the diagram commutes.
 
Ah, ok, I think I get it.
 
:)
Weird that I've never thought about this in 40+ years.
Or maybe I have and I've just forgotten.
 
2:16 AM
I have an equivariant map G -> EH, and the composite to BH is the classifying map to the pullback.
 
LOL.
 
But obviously the map to EH is (nonequivariantly) null.
 
Whoa. What?
 
-homotopic.
 
Oh.
I don't want to write the commutative diagram here ...
But I'm sending $(g,h)$ to $gh\in G$ horizontally and to $g$ vertically.
 
2:18 AM
No need. Our proofs are probably secretly the same.
 
LOL ... Only to you :D
 
I mean, probably literally, if you unravel the definitions. I don't believe something like this has multiple proofs.
 
The $S^2\to\Bbb P^2$ convinced me.
Anyhow, I'll now resign while I'm not behind :) Have a good weekend, and send notes.
 
2:40 AM
The putnam is tomorrow!
I'm too old to take it anymore :(
I'll have to wait until later this weekend to find out what everybody else suffered through
 
2:52 AM
Hello!!
Why is this not a $\mathbb{C}$-vector space? Which of the following properties is not satisfied?

- (V1) : $(V,+)$ is an abelian group, with the neutral element $0$.
- (V2) : $\forall a, b \in K, \forall x \in V : (a + b) \cdot x = a \cdot x + b \cdot x$
- (V3) : $\forall a \in K, \forall x, y \in V : a \cdot (x + y) = a \cdot x + a \cdot y$
- (V4) : $\forall a, b \in K, \forall x \in V : (ab) \cdot x = a \cdot (b \cdot x)$
- (V5) : $\forall x \in V : 1 \cdot x = x$ ( $1 = 1_K$ is the identity in $K$).
 
Heyo
Quick question, does modern day set-builder notation allow for unrestricted comprehension?
Just reading up on Russell's paradox
 
They way it's usually used in practice, yeah. But in theory you can't legally do that, ofc.
The*
 
@Jasch1 $\{\text{everything}\}$
 
Nvm I looked it up
Now we use restricted comprehension, which ias a bit weaker
 
Mary, you can't multiply by scalars. It is a real vector space, in particular.
 
3:03 AM
@Jasch1 You asked about the notation, not a particular set theory
 
3:24 AM
Ah ok... I want to check if the following are true:

- $V_1=\{a\in \mathbb{R}\mid a>0\}$ with the common multiplication as the vector addition and the scalar multiplication $\lambda \odot v=v^{\lambda}$ is a $\mathbb{R}$-vector space.
- $V_2=\{(x,y)\in \mathbb{Q}^2 \mid x^2=-y^2\}$ with the addition and scalar multiplication of $\mathbb{Q}^2$ is a $\mathbb{Q}$-vector space.


We have the following:

- We have that $x+y=x\cdot y$ and $x\cdot y=x\odot y$, or not?
For (V2) we have that for $a,b\in \mathbb{R}$ and $x\in V_1$
 
3:42 AM
Could somebody please explain Richard's paradox in laymans terms?
 
3:56 AM
hello anyone there? who knows probability?
@TedShifrin do you know covariance?
Show that $Y = a+bX$, then \\
\[
\rho(X,Y) = \left\{\begin{array}{lr}
+1, & \text{if } b>0\\
-1, & \text{if } b<0\\
\end{array}\right\}
\]

By the definition of the correlation coefficient between two random variables X and Y, we have\\
$\rho (X,Y) = \frac{Cov(X,Y)}{\sqrt{Var(X)} \sqrt{Var(Y)}}$\\
Since $Y=a+bx$, we obtain\\
$\rho (X,Y) = \frac{Cov(X,Y)}{\sqrt{Var(X)} \sqrt{Var(a+bx)}}$\\
Since $Var(a+bx)=b^2Var(x)$\\
$\rho (X,Y) = \frac{Cov(X,Y)}{\sqrt{Var(X)} \sqrt{b^2Var(x)}}$\\
$\rho (X,Y) = \frac{Cov(X,Y)}{\sqrt{(Var(X)^2b^2}}$\\
Unless I do
Cov(X, bX) = Cov(bX,X)
and then
Cov(bX,X) = b Cov(X,X)
and since Cov (X,X) = Var(X)
bVar(X)
 
4:12 AM
Ah @EricStucky you said that the above V is a real vector space. How can we find a basis ov V as a real vector space?
 
Choose some elements that are obviously linearly independent and show they span :P
(please don't ping me)
 
I got what was going on ... should've done
Cov(X,a) = E[Xa]-E[X]E[a]=0
0+bVar(X) = bVar(X) and then the Var(X)'s cancel and then proof by cases . and it's true when b = 1 and b = -1 the end yay
 
 
1 hour later…
5:39 AM
0
Q: Convergence in Orlicz space

Vrouvroui have that $f:\mathbb{R}\rightarrow\mathbb{R}$ continuous such that $|f(t)|\leq c [\phi(|t|)|t|+\phi_*(|t|)|t|]$ $(u_n)\subset W^{1,\Phi}(\mathbb{R}^N)$ such that $u_n\rightarrow u$ How to prove that $$\int_{\mathbb{R}^N} |f(u_n)-f(u)| |v| dx\rightarrow 0~\text{for all}~ v\in W^{1,\Phi}(\math...

someone have an idea ?
 
6:28 AM
24
A: History of the high-dimensional volume paradox

Steven LandsburgA related (and to me, when I first saw it, much more suprising) Fun Fact: Divide the n-dimensional cube in half in each of $n$ dimensions, to create $2^n$ smaller cubes of edge length 1/2. Inscribe a ball in each of these subcubes, and then construct the smallest ball tangent to each of those (...

Spoiler alert: the answer isn't $-1$.
 
 
2 hours later…
8:47 AM
$a\text{ log } b=c \text{ log } d\iff b^a=d^c$
?
 
@Null $a \log b = c \log d \iff \log (b^a) = \log (d^c) \iff b^a = d^c.$
 
@JesterTran thanks
 
9:09 AM
@Null you can use \log instead of \text{ log }, as demonstrated by Jester
 
@DHMO hehe, thanks for pointing that out
 
To calculate the radius of convergence of a series is the formula $\lim_{n\rightarrow \infty}\sqrt[n]{|a_n|}$ or $\lim_{n\rightarrow \infty}\sup \sqrt[n]{|a_n|}$ ?
 
@JesterTran the last "iff" is not lol
 
im a little groggy this morning haha
 
@DHMO Why? Logarithm is injective and it's assumed that $b, d > 0 \implies b^a, d^c >0$.
 
9:19 AM
@Null I know the feeling. Just cranked out a proof but it looks like a chicken brute-forced it.
 
@JesterTran alright, under your assumption it is correct.
 
@DHMO Why is it wrong? The assumptions are deduced from the question.
 
@JesterTran My fault, ok?
 
@DHMO All good.
Fun question - can the mean of any two consecutive prime numbers ever be prime?
 
@JesterTran only if you consider 1 as prime imo
 
9:34 AM
Let's say 1 is not a prime, @Null
 
then obv not
 
@Null Why not?
 
@JesterTran no coz it's between two consecutive primes and no primes are between two consecutive primes by definition
 
Nice @DHMO
 
@JesterTran primes above 2 are odd. therefore between two odd numbers there is an even one. an even one is not prime^^
 
9:36 AM
@Null 13 and 17 have odd mean.
 
(basicly dhmos argument before he deleted lol)
ah...
i thought about TWINS!
 
@Null so now you know why I deleted it
 
lol
 
hahaha
GROGG pls :D
lol @DHMO your proof is nice
 
@Null thanks
 
9:39 AM
can the mean of any two primes be prime again?
 
@Null do you want a math riddle?
@Null yes, (3+7)/2 = 5
 
ah, i see
@DHMO why not
 
@DHMO I'd like a math riddle.
 
@Null 3 and 7, 17 and 29, 31 and 43, to name a few.
 
@Null @JesterTran the only rule is don't actually calculate; just use your gut instinct
 
9:40 AM
ok
 
I'd guess it's open that there are infinitely many pairs of primes whose means are prime.
 
You can choose 3 options that you think might be correct.
Consider a n-dimensional hypercube. Divide this hypercube into 2^n smaller hypercubes as shown. Inscribe a n-dimensional hypersphere of diameter (1/2) in each hypercube as shown. Construct a hypersphere in the middle as shown such that it is tangent to all other hyperspheres. When n goes to infinity, which value does the diameter of the middle hypersphere go to?
A. -1
B. 0
C. 1/2
D. 1
E. 10
F. infinity
 
@DHMO saw that riddle already
so its not fair
 
@Null then let @JesterTran do it
 
k
altho i don't even get the riddle
 
9:41 AM
@Null the figure shows the case where n=2
in which case the hypercube is a square and the hypersphere is a circle
 
well, i have my 3 options, but i wouldnt know which one is right haha
 
Hmm, I think I have an answer. I'll wait though.
 
@Fargle hey, it isn't fair if you don't use your instinct
 
@DHMO I've done no calculating. It's all gut.
 
@Null post your 3 options here, i'll post the answer here after @JesterTran has also posted
 
9:44 AM
Ok I chose my 3 options
 
Alright, post your options here
 
B,C,D
 
My 3 options: B, C, F
 
0,-1,$\frac{1}{2}$
 
wow, congratulations @JesterTran it is indeed F.
 
9:45 AM
wow real? :s
 
Neat. Why?
 
The main trick is that while the side length is always 1, the diagonal of a n-hypercube is sqrt(n)
 
It was only gut so no understanding lol
 
and by considering the diagonal, one can derive the formula (sqrt(n)-1)/2
or in LaTeX, $\dfrac{\sqrt{n}-1}2$
 
How do you do that using purely intuition?
 
9:46 AM
@JesterTran the point is to show that intuition is unreliable
your intuition only goes to 3-dimension.
 
mmh, what is the volume of the hypercube?
 
Yes, consider having a look at the Cognitive Reflection Test. @DHMO
 
@Null 1
 
that doesn't even make sense haha
 
Volume is weird in $\Bbb R^\omega$.
 
9:47 AM
another mind-boggling result is that the volume of a n-hypersphere of diameter 1 is maximum when n=5, and then goes to 0.
 
yeah i guess
 
................. @DHMO lol
 
ignoring units.
 
$1 m^n$. this will be my answer to the policeofficer when i get pulled from the street
"ignore the units" :D
 
@Null 1x1x1x1x...x1 = 1
 
9:48 AM
"Ignoring units, I was only going 2! How can you pull me over?"
 
yes, but what is even $m^4$ with m for meter?
(or mile)
 
@Null Hypervolume. You know, that unit we use all the time to measure hypervolume.
Can't tell you how many times I've tried to figure out how many quartic feet of stuff I can take on vacation.
 
Why do we need to study objects of dimension $\geq 4$?
 
Consider n-hypercubes with side length 1 and n-hyperspheres with diameters 1/2.
When n=2, you can inscribe 4 hyperspheres into a hypercube
using the same pattern, one can inscribe 2^n hyperspheres into a hypercube
 
@JesterTran Spacetime is four-dimensional, and many physical systems exhibit symmetries that can only be realized in higher dimensional objects.
 
9:51 AM
however, you can insert one more hypersphere into the hypercube when n=4.
 
@Fargle that would somehow mean that we could travel in time forwards or backwards. As a dimension without direction makes really no sense to be called "spacial"
 
> Inscribe an $n$-ball in an $n$-dimensional hypercube of side equal to 1, and let $n \rightarrow \infty$. The hypercube will always have volume 1, while it is a fun folk fact (FFF) that the volume of the ball goes to 0.
 
@Fargle example?
 
@JesterTran Also, any data set with more than 3 variables can be turned into a sort of $n$-dimensional blob in $n$-space, and you can do topology-type stuff to that to find data connections.
 
@Null scientifically time is a mess.
your clock speeds up when you are near blackholes
etc.
 
9:53 AM
i really don't want to meet a blackhole
 
@Null Look up Minkowski spacetime. Physics takes exactly the convention I stated.
 
so cast your physics someplace else
 
@Fargle Ooh
 
how fast does our galaxy travel around it's neighbours?
 
@Null Do you know that it is more difficult to crash directly into the sun than to escape earth to infinity?
 
9:55 AM
@JesterTran Examples I'm a bit shaky on, but string theory requires ten dimensions to work properly.
 
@DHMO because the sun is so hot that my shuttle would melt before i reach the sun?
 
@Null no, but because you would miss it.
 
:s
 
@Fargle What is string theory? Is it the smallest objects to be analysed?
 
@JesterTran That's a question that I don't know enough to answer. I'd hate to lead you astray
 
9:57 AM
@Fargle I'll research on it! Thanks for making me curious about this.
 
@Fargle "but officer, you can't impose your 3d laws on me, i need at least 10 to work properly"
 
@JesterTran Of course! The topology thing was shown to me at a seminar at my previous school. That struck me as a really neat way of tackling the problem of finding connections in exceedingly rich data. Intuitively, holes (like that in a disk or torus) imply correlations.
 
@Fargle Mind is blown.
 
for a constant k, $n!>n^k$ as $n\to\infty$
 
10:05 AM
@Null Indeed. For $k$ constant, $n^k < k^n < n! < n^n$ as $n \rightarrow \infty$.
($k > 1$)
 
but $n!<n^n$ is not that big of a difference or? i mean you can atleast say what the difference is for explicit n right?
n(n-1)...(2)(1) vs (n)(n)...(n)
mmh
 
@Null I'd say it's a pretty big difference. $5^5 = 3125 \gg 120 = 5!$
 
10:53 AM
is the following true: b converges if and only if the log of b converges as well.
(b in R)
 
b is a sequence? Real numbers don't converge
 
ah, yes a sequence
 
Yeah that's true
hmm
okay you have to not do stupid things
 
yeah i mean for positive values at least
 
ah, no nvm even then
if it converges to zero, no dice.
 
10:54 AM
consider $\frac{1}{e^n}$
 
by log laws this is: $log(1)-log(e^n)$ @Alessandro i see
 

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