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00:13
great
superb
excellent
does anyone know what's the definition of the quotient $X/A$ where $i:A\to X$ is a cofibration?
 
1 hour later…
01:35
the same as when it's not a cofibration
 
4 hours later…
05:36
0
Q: What is the difference between $[x:y:0]$ and $[0:0:1]?$

jasmineWhat is the difference between $[x:y:0]$ and $[0:0:1]?$ According to my understanding, points of the form $[x:y:0]$ are the line at infinity What is the meaning of $[0:0:1]?$

 
3 hours later…
08:43
@psie eggcelent
(a cofibration can always be identified with a subspace inclusion)
09:22
Okay, I've proved twin primes for the nth time
This time though, it's impeccable
Topological, ranges from modular arithmetic to direct limit in category theory for sake of proving continuity
It can also be used to prove infinitude of regular primes, 4-sep, 6, 8, ... separated cases
09:38
@psie you should rely less on this chat and more on yourself
 
4 hours later…
D G
D G
13:34
Is AAS and ASA the same because knowing 2 angles implies the third angle?
13:46
@DG Yes?
D G
D G
It should be yes. But I don't understand why many sites list both, for example: chitowntutoring.com/congruence-theorem-triangles
I'm not sure that I like the use of "because" there---the basic idea is that any two triangles which have the same angles are similar, meaning that one is a scaled copy of the other. If you then know that they have a common corresponding side length, the scaling constant is 1, so they must be congruent.
From that, you can deduce the AAS and ASA mnemonics.
There isn't a cause and effect here---there are two true statements: (1) AAS and ASA are basically the same theorem, and (2) if you know two angles in a triangle, you know the third. You can probably prove either statement from the other.
D G
D G
OK. Thanks.
D G
D G
14:12
I see. All theorems listed in the list given above can actually be "converted" to SSS by applying cosine or sine rules.
@DG If that is the theorem you find fundamental, sure.
14:25
Each one of the theorems about triangle congruence can be "converted" to each of the others (though I am not quite sure what is meant by "converted"---presumably, this means that you can prove one from the other).
D G
D G
I meant when ASA, for example, are given, we can find the other 2 sides and then apply SSS.
When HL (hypothenuse and leg) are given, we can find the third side and then apply SSS.
@DG If you have a hypotenuse and a leg, you have a right triangle. There are stronger theorems related to right triangles than to general triangles.
D G
D G
If "conversion" is allowed, all are correct.
I don't know what you mean by "conversion".
D G
D G
make the all sides and angles known.
by using cosine and sine rules
14:37
The question isn't about Law of Sines or Law of Cosines, though.
It is asking you which triangle congruence relation applies in order to show that the two triangles are congruent.
The triangle congruence relations predate the Laws of Sines and Cosines by at least a thousand years. You don't need to invoke those rules. It is overkill.
D G
D G
I see. I think the problem should be rephrased by adding "Without finding other angles or sides, ....".
I mean, if you can show that the triangles are congruent, then you know the remaining sides. The question is, how do you show that the triangles are congruent in the first place?
D G
D G
Make sense.
Does anyone know whether the following problem has been solved? mathoverflow.net/questions/22/…
The problem is about the existence of perfect squares composed only of digits $0$ and $1$ other than $10^n$
15:20
Hi
I'm having trouble with this integral $\int \frac{t^2}{\sqrt{16-t^2}} dt$ , I noticed that if there was no $t^2$ in the numerator then it's a $\arcsin$, but I can't figure out what to do in this case
@Pizza Consider a trigonometric substitution.
Draw a right triangle. The hypotenuse is 4 units, one of the legs is $t$ units.
In that right triangle, what does $\sqrt{16-t^2)$ represent?
Does this not suggest a substitution?
$t = 4 \sin(\theta)?$
@Pizza I have no idea. I haven't worked through the problem.
Ok, then I'll get to the end and then check, thanks a lot anyway!
@XanderHenderson LaTeX Is wrong
15:44
@XanderHenderson Ok,the substituion worked
you can also integrate by parts, $\int \sqrt{16-t^2}dt=t\sqrt{16-t^2}+\int \frac{t^2}{\sqrt{16-t^2}}dt$ @Pizza
or add and subtract $16$ in the numerator
15:59
I have a conjecture
Tell me if it is already a known result
My Conjecture: Let $\Gamma$ be a finite, connected, undirected graph embedded on an orientable surface $S$. Let $\{ T_{\gamma_1}, T_{\gamma_2}, \dots, T_{\gamma_k} \}$be a set of $\Delta$-actions along simple closed curves $\gamma_1, \gamma_2, \dots, \gamma_k \subset S$ such that each twist $T_{\gamma_i}$ preserves the equivalence classes of primitive cycles in $\Gamma$. Then the Ihara zeta function $Z(\Gamma; u)$ remains invariant under the composition of these twists:
$$
Z\left((T_{\gamma_1} \circ T_{\gamma_2} \circ \dots \circ T_{\gamma_k})(\Gamma); u\right) = Z(\Gamma; u).
@XanderHenderson Thanks now I'll compare it with what I did too
@SineoftheTime Oh ok, then I'll try that too to see if I get the same result
^^^Assume that $\Delta$-actions are generators of $\mathrm{MCG}(S)$.
They can be seen as similar to Dehn Twists
it's the same as Xander method the end of the day @Pizza
@Pizza It ends up being the same. But the advantage(?) of the integration by parts path is that the trig substitution is, perhaps, a little simpler.
16:16
In simple terms, I'm saying that certain types of self-homeomorphisms (which may be strengthened to diffeomorphisms) of $S$ don't change the equivalence classes of primitive cycles in $\Gamma$ (they just permute lengths of cycles) but the total class is preserved. Since this doesn't change the primitive cycle equivalence classes the zeta function remains intact and invariant under these transformations
 
1 hour later…
17:29
I have to post about this theorem again, because I had a look at the errata from Folland's book, and found out that in the above proof, he remarked that the first occurrence of rectangles should be changed to bounded rectangles, i.e. $\{T_j\}$ should be a collection of bounded rectangles. I can't figure out why we want bounded rectangles. Is this clear to someone?
Ignore the highlighted box.
18:09
I think if $m(T_j)=\infty$, then we can't find a $U_j\supset T_j$ such that $m(U_j)<m(T_j)+\epsilon 2^{-j}$. The inequality would be absurd.
18:19
@psie m?
@Jakobian yes? $m$ is Lebesgue measure. Though I don't understand how come we can find bounded rectangles that cover $E$. What if $E$ is unbounded?
@psie whats your definition of Lebesgue measure?
@psie unbounded rectangles can be made bounded at no cost to the definition
@Jakobian since we are in $\mathbb R^n$, we could just take e.g. cubes
@Jakobian let me type something
@Jakobian it is the completion of $m\times\cdots\times m$ on $\mathcal B_{\mathbb{R}^n}$, where $m\times\cdots\times m$ is the specific measure that is the restriction of the outer measure induced by the premeasure $\pi(E)=\sum_1^n m(A_j)m(B_j)$ on the algebra $\mathcal A$ of finite disjoint union of rectangles. Here $E\in\mathcal A$ and for simplicity I assumed $E\subset \mathbb R^2$.
But I think I understand now.
18:36
@psie a lot of what you said here makes no sense, but I get what you're trying to say. Its the completion of the product measure of copies of the one-dimensional Lebesgue measure
@psie in this case, sure, the inequality is probably the issue. The issue that's irrelevant because just replace $<$ with $\leq$
its basically a non-issue either way
alright
19:36
I'm considering developing a new type of geometry called: Dark Geometry.
IIRC Shohei Ohtani and the Dodgers are playing tonight - let's go Dodgers.
12-0 hopefully
20:02
@ModularMindset What is it?
Out of curiosity, do people know about Jordan content here?
I barely know myself about this, except that we are using cubes to approximate "volumes", and there is an inner and outer content, and then the Jordan content.
Limit as n approaches infinity of fractional part of zeta(1 + 1/n) = euler-mascheroni constant
I feel like the Jordan content is being bashed.
* tested for powers of 10
20:22
@psie Peano Jordan measure?
@SineoftheTime yeah
I think it's less popular than Lebesgue measure because it's less sophisticated
yeah, I guess
@SineoftheTime no its just not useful
20:42
hi everyone, could you please take a look at my proofs for these questions?
Is this lucid?
Do I need to elaborate more?
Its ungrammatical and unprecise
@CroCo You can label the theorems or tag the formulae in the definition to get a better outcome
21:00
@Almanzoris The short answer is: Dark Geometry is modeled on points that are specific isomorphism classes of algebro-geometric objects, hence it resembles a moduli space. In reality however it is a collection of moduli spaces where the points themselves are generalized to arbitrary dimensions. You can also do dark analysis (analogue to harmonic analysis) in these spaces among many other things.
@Almanzoris If you want the long answer read my 20 page paper which introduces the topic.
21:16
@Jakobian how can I improve it?
@SineoftheTime I don't understand what you're trying to say.
It is claimed the Jordan content is only meaningful for bounded subsets of $\mathbb R^n$, since otherwise the outer content equals $\infty$. Nothing is said about the inner content for unbounded sets though. This confuses me; if the set is unbounded, won't the inner content also equal $\infty$?
In other words, are there unbounded sets who have an inner content less than infinity and outer content equal to infinity?
desperately wishing one of my students would upload their homework early so I can procrastinate on doing a proof on one of my own homework sheets I don't want to do by correcting it
doesn't look like it's going to happen unfortunately
@CroCo instead of being vague and writing "above definitions", you can number the definition as you did with theorems
@SineoftheTime good point.
you don't need to reference the definitions here at all, they're right there
what else would you be referring to :)
21:26
yeah, I'd agree with Ben here
but also if you number theorems then you should also number the definitions, in line with what Sine is saying
Also make sure that the numbering actually make sense
why is theorem 1.2 followed by 1.4?
If the calculation are trivial, you can also refrain from explaining every step imo
to maybe make Jakobians point more precise, "By direct calculation as follows [calculation] [end of proof]" is rather poor form, if not outright ungrammatical
@SineoftheTime I think the level of detail is fine for something as elementary as this, but it depends on context
also move the end of proof symbol up into the equation environment if you end with it
\qedhere is your friend for that
they are probably using the \begin{proof} environment
How is that relevant? :)
21:32
@BenSteffan I'd just change that sentence to "But note, [calculation][end of proof]"
I'd write it this way: Let $H=\{x\in \Bbb R^n : c^tx=k\}$, where $c$ is a non zero vector in $\Bbb R^n$, be a hyperplane in $\Bbb R^n$. We want to show that for every $x_1,x_2\in H$, $x:= sx_1+(1-s)x_2$ is in $H$ for every $s\in [0,1]$
I would say that's rather ungrammatical as well. "But note that" would be better imo
and then put your calculation
@BenSteffan I agree, it depends on the context
@BenSteffan these are exercises taken from the book I'm reading. I'm worried more about the content of the proof itself.
The proofs are fine, content wise
not much to say there
21:39
@BenSteffan yes @psie was right. I'm using the proof package and I believe it places the square below the equation end.
8 mins ago, by Ben Steffan
\qedhere is your friend for that
^ don't make me tap the sign :)
@BenSteffan this is what I've got
I don't like it.
using \qedhere?
yes
Not sure what you're doing, but see overleaf.com/read/vvxsrcfvynjb#03e122
21:47
not sure why I'm getting this warning
I'm using same packages as yours.
@ModularMindset Sounds interesting, but my comprehension capacity and my knowledge are too bounded to follow the line. However, I would like to see the paper.
@CroCo neither am I, but remove the blank line after \end{align*}
@BenSteffan I did it didn't work.
not big deal at this time I guess.
@CroCo yeah, I wouldn't worry about it. It is not wrong as it is now, it's just a matter of style. Whatever you prefer.
@CroCo Have you tried putting \qedhere after the align instead? Like: "\end{align*} \qedhere".
21:56
@Almanzoris That will not put the qed where it's supposed to be.
@BenSteffan true.
Apparently, the package "amsmath" needs to be loaded before "amsthm" according to this answer and it worked.
https://tex.stackexchange.com/questions/436162/package-amsthm-warning-the-qedhere-command-in-documenta-style
@BenSteffan Noted.
do you guys like kakis?
@SineoftheTime The fruit?
22:04
yes
I haven't tried them at all.
@SineoftheTime I don't know what that is, and google is only giving me pants.
persimmons, in english
I guess kaki is more common nowadays; it's the japanese name
Oh, persimmons!
I love persimmons!
@XanderHenderson what browser you're using? :)
22:06
I ate the last one after dinner and now I want one :(
@CroCo Right now? Chrome.
Why?
@XanderHenderson chrome is giving me the fruit like a tomato. I haven't eaten this fruit before.
@SineoftheTime I should buy some
fond memories
Satsumas are also great.
A fruit like a tomato, jajaja.
22:08
'tis the season.
I'm a citrus man
@CroCo Okay...
Kumquats are criminally underrated
never heard about that
me neither
22:13
pity :(
if there's a turkish/middle eastern grocer near your place, they might carry it
Kumquats are great. The skin is bitter, but the fruit inside so very sweet. They are quite nice.
There were many kumquat trees which grew between the parking lot and the math department's building at my phd institution.
The skin is the good part :)
When they were ripe, I would often snack on them on the way to or from my office.
so much aroma
I'm jealous
@BenSteffan It is the combination of skin and interior that is good. You need the whole thing.
22:17
true, true
I will make irresponsible fruit purchases
@BenSteffan I live in a bit of a food desert. There is one grocery store within 30 miles of my home, and I'm lucky if there is more than one variety of apple when I go in. :/
:/
I would be very unhappy
It is one of the disadvantages of living where I do.
Fresh produce is a bit of problem.
On a good day, there are plantains. On a bad day, one kind of apple, a dozen very sad pears, and some limes.
(I'm exaggerating a bit, but produce here is very inconsistent.)
Also, anything vaguely "ethnic" is impossible to get here. I can't get kombu and dashi, I can't get matzoh, etc.
Spain without s
22:35
Though there is now an entire shelf(!) of pan-Asian foods. Nothing super amazing---Pocky, boba tea in a can, chow mien noodles, various sauces (e.g. Thai curry paste, ponzu, soy, etc), white miso paste, shrimp-flavored crackers, and... uh... that's it?
Oh, nori. There's nori on that shelf.
well that's a start
our supermarkets by and large have those now as well
@BenSteffan The grocery stores I shopped at in grad school had more stuff, but I was in a real city.
There are about 7000 people who live within 30 miles of my house.
but then there's usually plenty of asian grocery stores of various kinds around, if you're not living off in the countryside
@XanderHenderson that's not a lot,
The nearest Asian grocery store is a 99 Ranch in Phoenix, which is three and a half hours from here.
There are better grocery stores in Flagstaff, which is only 90 minutes away, but I don't got to Flag for groceries very often.
ah, Arizona?
eye-watering distances :')
22:38
@BenSteffan Yes.
@BenSteffan I mean, to a European, sure. :P
Or maybe even an East-coast elite.
But us real 'Merikans don't mind the distance. Isn't a thing.
to most people in the world I would hope ;)))
gas is free, right
or do you mean 90min by train :)))))
Fuel in the US is far, far cheaper than it should be.
And I drive a Prius, so I can go a long way for not a ton of money.
Gas was $2.99 / gal yesterday.
oh, okay
Not the 79¢ / gal I paid in high school, but still relatively cheap.
that's less than half the price here
$\pm$various conversions
and gas in Germany is also pretty cheap, compared to some of our neighbors
22:43
My sister did a Fullbright in Germany after she finished her BA.
Somewhere a bit to the west of Stuttgart.
Consider $\mathbb R^2$ and a lattice where we are approximating sets by (dyadic) cubes. We can speak of the inner content as the area of the cubes that fit into the set, or we can speak about the outer content as the area of the cubes that intersect the set. Now consider the complement of the unit disc. What would be its inner content?
@psie YOUR NOT MY REAL MOTHER! I DON'T HAVE TO CONSIDER ANYTHING!
*SLAMS DOOR*
now the door is closed, hence not open :D
@SineoftheTime WRONG!
MORE COWBELL FOR FU*KS SAKE!
22:52
It is clopen.
If I'll ever do a PhD, I'll put the sign "clopen" on my office door
@psie oo
@Jakobian ah ok, do you know if it is possible for a set to have inner content $<\infty$ and outer content $=\infty$?
@psie line in the plane?
$\mathbb{Q}^2$?
23:01
Anything with boundary of non-zero Lebesgue measure
Sorry, not just anything. The outer measure has to be infinite
Ok 👍
@BenSteffan yeah that works
My line example doesn't
@Jakobian what wouldn't work about your line?
it's inner content is $0$, no?
...I was also under the impression that the line would also work, but maybe I misunderstood what psie meant
23:44
when I think about $\mathbb Q^2$, I must think about the lower and upper Riemann integral of $\chi_{\mathbb Q^2}$. I guess this gives an inner and outer content. The lower one has value $0$, the upper $\infty$.

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