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11:00 PM
@ACuriousMind my drive home is bigger than your country
 
@0celo7 So it is. Why does that make it insane that we don't all own cars?
 
@BernardMeurer Hehe, also to me is hard to say on English everything :-) Yeah, it can be hard, but it only around the cities. If you travel between the cities, it is ok, even if you don't have a GPS.
 
@ACuriousMind I have actually entered and had lunch at Bielefeld
Lots of actors pretending to be residents
 
@ACuriousMind just...how do you get around?
 
@0celo7 Bahn, bus, tram, horse
 
11:03 PM
@0celo7 As I said, public transportation! Or by feet, or by bike.
 
@BernardMeurer You are in Brazil? Or in .de?
 
Horse is probably more expensive than a car
@ACuriousMind By feet?
 
@peterh Brazil, last week here
 
I haven't walked anywhere in a while
why bother?
 
@0celo7 Loosing the fupa
 
11:04 PM
I don't have a fupa
 
Right
 
@ACuriousMind Something I was working on: Let $B$ be $\tilde g$-bounded. Fix $x\in B$ and let $(y_n)$ be a sequence. Connect $x$ to each $y_n$ with a curve $\gamma_n$. Each curve can be made to have finite length, in fact, one can bound them above. Then the sequence $L_{\tilde g}(\gamma_n)$ has a convergent subsequence.
Now the goal is to get $f$ to blow up along this sequence, and somehow get a contradiction
i.e. choose $y_n$ so that $f(y_n)>n$.
 
What's going on in here?
@BernardMeurer, back to work!
 
Riemannian geometry.
 
@DanielSank Yessir! ::opens larson::
 
11:15 PM
::throws tomato, mozzarella, and basil at @ACuriousMind::
::leaves::
 
::munches happily::
 
@BernardMeurer is the fupa the roll right above the dick but below the gut
 
@BernardMeurer What is a "larson"?
 
@ACuriousMind Calculus book I told him to read
I was tempted to tell him to read Global Calculus but I figured sheaf/categorical diff geo isn't what he was looking for.
 
Also @count_to_10 has gone
 
11:20 PM
@ACuriousMind Aha
I figured something out.
The functional $h$ of $f$ has to be less than one, definitely.
Because (please correct me if I'm wrong) if one metric is always as large as another and is complete, then the smaller one is complete.
Because they share Cauchy sequences.
And a convergent Cauchy sequence for the big one is also one for the small one.
But our manifold was not assumed to be complete, so if the $\tilde g$ metric is larger and complete, then we have a contradiction
So it has to be something like $\mathrm e^{-2f}$.
but that makes the blowup strategy unviable.
 
@0celo7 You say "always as large", but why can't $h$ then be in some places larger than one?
I'm also not sure what you want to do with the "blowup" - wasn't the point to show that $f$ is bounded on $\tilde{g}$-bounded sets?
 
@ACuriousMind The way I've been trained to prove something is bounded is to assume the opposite and go for a contradiction.
 
going directly for a bound works sometimes.
i.e. "Cauchy sequences are bounded"
 
But then you don't "need to get " $f$ to blow up, you just assume it.
 
11:26 PM
right, but $\sup\mathrm{e}^{-2f}$ will always be 1, so there's nothing to blow up!
@ACuriousMind Hmm.
That would be a more sophisticated functional
 
@ACuriousMind Yep, it's the calc book
It seems quite good
 
@0celo7 Well...isn't that the proof, then?
I'm not following you really. If you're saying that $f$ cannot blow up because of the choice of $h$, then you're done
I'm not sure what the sup of $h$ has to do with $f$ blowing up, though
 
I'm not saying that.
The problem is linking the boundednes of the set to the blowup of f
and prove a contradiction in there
 
Can anyone tell me what the partial derivative symbol ∂ is called?
I'd imagine there's a name for it
 
It's a "del".
 
11:38 PM
@ACuriousMind huh
That's the gradient operator
 
@ACuriousMind One person called it "doh", another told me it was "dee"
The hell?
 
It's Dee.
 
Sigh...
 
@0celo7 In German, we do at least pronounce it "del", and I've never heard anyone call the nabla "del".
Although Wikipedia agrees with you on that usage
 
11:40 PM
delta
 
@peterh nope! $\partial$ vs $\delta$
 
@peterh ∂ looks nothing like Δ or δ
 
@SirCumference it looks a little like it...
 
well, really
 
@NeuroFuzzy I guess...sorta
 
11:41 PM
Makes me so mad
2
 
@SirCumference It's obviously derived from the delta, and in Greek handwriting both $\partial$ and $\delta$ could conceivably stand for a delta.
 
@ACuriousMind I thought it was derived from the latin letter d?
It honestly looks more like a curly d
 
Why would anyone star that
 
Star what?
 
Curly d is also an acceptable name
 
11:42 PM
Wait really?
 
I cannot believe you googled this and found nothing though.
 
Curly D sounds like a rapper name
 
You read partial derivatives with a regular d
 
All right, then what is the integral symbol called?
 
"The integral symbol"
 
11:43 PM
@SirCumference The correct pronunciation is "slash int"
 
It's derived from a $\Sigma$, btw.
 
Looks more like it was derived from this
Esh (majuscule: Ʃ Unicode U+01A9, minuscule: ʃ Unicode U+0283) is a character used in conjunction with the Latin script. Its lowercase form ʃ is similar to a long s ſ  or an integral sign ∫; in 1928 the Africa Alphabet borrowed the Greek letter Sigma for the uppercase form Ʃ, but more recently the African reference alphabet discontinued it, using the lowercase esh only. The lowercase form was introduced by Isaac Pitman in his 1847 Phonotypic Alphabet to represent the voiceless postalveolar fricative (English sh). It is today used in the International Phonetic Alphabet, as well as in the alphabets...
 
@SirCumference It's derived from the $\Sigma$ because of the integral's definition as the limit of a Riemann sum.
 
@ACuriousMind huh? Integral is defined in terms of primitives
 
11:45 PM
@0celo7 Stop trolling :P
 
@ACuriousMind Dude, sigma looks nothing like that
 
@ACuriousMind ...
 
@SirCumference So?
 
I thought Riemann sums were heuristics
 
@ACuriousMind It looks more like that Esh letter
More likely it's derived from the esh
 
11:46 PM
No it's from S.
 
@SirCumference lol, certainly not,
 
It's a big S
 
@ACuriousMind serious question, how does one know that the Riemann sum converges
 
Well it looks a bit more like an s depending on what region you are from
 
11:47 PM
@0celo7 By, uh, checking it?
 
That's English, German and Russian, respectively
 
There are general theorems on integration though
 
So I guess the Russian version of the integral sign looks like an S
 
@0celo7 You have to ask a more specific question if you want an actual answer
 
The German version is the best compressable. It has the smallest Kolmogorov-complexity.
 
11:50 PM
Well in Germany and Russia, definite integrals look more like this
$\int\limits_0^T f(t)\;\mathrm{d}t$
With the range on top and bottom
Unlike $\int_0^T f(t)\;dt$
So that's some interesting trivia
 
Uh, that will differ depending on the person typesetting the thing. That the "German" and "Russian" typographic traditions would have it that way doesn't mean it looks that way "in Germany" or "in Russia"
 
You guys are insane
 
Oye, made a mistake
That's right, ACM
 
However, sampling a few German versions of books I can access, it appears they indeed seem to mostly follow the German tradition
 
@ACuriousMind I don't really want an answer.
We did integration via primitives in my RA class because "Riemann sums are much harder to make precise"
 
11:55 PM
@0celo7 So what's your point in asking that question?
 
Wait, ACM
You good with astronomy?
 
@ACuriousMind Beats me!
 
@SirCumference Got no clue about it
 
But now that I'm back at a keyboard
 
11:56 PM
So, what does $B$ bounded wrt. $\tilde g=G$ mean?
$G$ so I don't have to mess with the tilde
it means that $d_G(x,y)\le C$ for all $x,y\in B$.
This means that given $x,y\in B$ I can find a smooth curve $\gamma$ so that $L(\gamma)\le C+\epsilon$, where $\epsilon>0$ is as small as I want.
The idea is to use the blowup to $f$ to show $L(\gamma)>C+\epsilon$.
Does that make sense?
 
o
 

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