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12:16 AM
sup chat
Inf chat
12:40 AM
(inf + sup)/2 chat ... Howdy, Eric.
1 hour later…
1:40 AM
@TedShifrin Is it useful to define a metric as scaling the length of a unit vector
and the metric would be the length of the vector
such that the length of the blue vector defines the distance metric
so if you called the blue vector $s$ and the horizontal axis $x$ and the vertical axis $\phi$ you'd have the metric as: $ s^2=x^2+\phi^2$
and if you made the distances small, then you'd have $ds^2=dx^2+d\phi^2$
3 hours later…
4:29 AM
MSE T-Shirt!
4:56 AM
@geocalc: A Riemannian metric (as opposed to a metric on a metric space) is a (smoothly varying) dot product on tangent spaces as you move from point to point. It allows you to measure angles, not just lengths of tangent vectors. But then, by integrating, it allows you to find lengths of paths.
5:18 AM
Basically it's like a 2D surface that lives in 3D space, except it's actually an $n$D surface that doesn't live anywhere
5:57 AM
@AkivaWeinberger I think the more important question for that picture is: "Why has it been so dry?".
It's not a picture, it's a video
Or, rather, the video came first, and then a thumbnail was chosen
It has been hot
@geocalc33 it's a giraffe!
6:55 AM
Does $\exp(a \wedge b)$, where $|a| = |b| = 1$, yield the rotation matrix that sends $a$ to $b$?
7:13 AM
What's $a\wedge b$? The matrix with $a$ and $b$ as columns?
(Are $a$ and $b$ two-dimensional vectors?)
7:26 AM
I guess $a \land b$ lives in $\Lambda^2 V$ which is also the space of skew-symmetric matrices
and this is consistent with its exponent being a rotation matrix
Fell asleep wanting to do math. Woke up wanting to do math.
Life = Walking from 0 to $\omega_1$
Haha! Yeah.
7:52 AM
Prove that:$$(x-\sqrt2-\sqrt3)(x-\sqrt2+\sqrt3)(x+\sqrt2-\sqrt3)(x+\sqrt2+\sqrt3)$$has integer coefficients when you expand it.
@OskarTegby If you're wanting to do math (and haven't seen it before)
(Mostly) unrelated fact: If a triangle has side lengths that are square roots of rationals, then the area is also a square root of a rational.
hi @Akiva, can you see your own "profile pic" in this chat or is it just some weird "missing picture" junk?
I can see it
It's this (an "impossible" soccerball)
@Akiva yis, for some reason I can only see it on some of your messages
and the same for some other people
I'm definitely interested in seeing the proof. I don't really know how to make an actual proof, but the reason when one explicitly expands it is that the square roots are multiplied with each other and take each other out.
What browser are you using?
Google chrome
7:57 AM
Indeed, maybe I just need to restart the browser, but I have a lot of tabs open and I'm scared
Have you refreshed the page?
Yeah, it's only people who have profile pics that are different to the one generated by MSE
I wish there is an easy way such that given:
Okay. It's probably just something with them not being loaded by an error.
7:58 AM
the criteria of surds a to f such that the cross terms in the expansions are integral
Where did you find the question, @Akiva?
I don't remember where I first saw it
@AkivaWeinberger galois theory, lalala
8:09 AM
I didn't
A sequence of Reddit posts I spent way too much time writing
Ah! Of course it's Galois theory.
(in response to someone posting on the 3Blue1Brown subreddit saying "You should do a video or series on Galois theory")
I see
@Akiva the "small puzzle" is cool in that it is accessible to someone who doesn't know Galois theory
8:10 AM
There's a pretty simple solution, but there's also a cool (and simpler) solution if you know about automorphisms, incidentally
I especially like part 3
The third post, or the third question in the first post?
Third question in the first post
Yeah, it occurs to me that it might require some machinery that not everyone knows
but I dunno, maybe it's not required
In any case, you can pretty much skip that one
just some euclidean division and a contradiction argument or smth?
8:13 AM
Something like that yeah
I think you might need Euclid's algorithm (or at least the fact that you can take GCDs of polynomials)
when you're doing this in general, euclidean division is what you need
and the fact that $K[X]$ has such an algorithm for $K$ a field
For $K$ a Euclidean domain, even
Oh, wait
Hm, for $\Bbb Z[x]$, do you need *monic polynomials?
(Or at least that the smaller one be *monic)
monic division algorithm works for any ring
8:16 AM
Still true in the question, though, anyway
so you also have monic remainder theorem and factor theorem
Fair, just.. when you see this for the first time in a fields course you probably just use the fact that $K[X]$ is a PID
The "automorphism" version of the first question is, incidentally, basically "It doesn't change when we replace $\sqrt2$ by $-\sqrt2$ and it doesn't change when you replace $\sqrt3$ by $-\sqrt3$"
@ÍgjøgnumMeg circular argument
right because there's a $\Bbb Z/(2) \times \Bbb Z/(2)$ lying around
8:18 AM
Did someone say Galois?
The other version is, $(x-\sqrt2-\sqrt3)(x-\sqrt2+\sqrt3)$ is the difference of squares (it's $(x-\sqrt2)^2-3$) so the $\sqrt3$s are gone
and similarly for the other half of the polynomial
and expanding those out you get difference of squares again which kills the $\sqrt2$s
@Daminark See here:
@Leaky Why? I'm just saying this because you have a division algorithm in $K[X]$ so you can use this for Akiva's question
26 mins ago, by Akiva Weinberger
Prove that:$$(x-\sqrt2-\sqrt3)(x-\sqrt2+\sqrt3)(x+\sqrt2-\sqrt3)(x+\sqrt2+\sqrt3)$$has integer coefficients when you expand it.
Reddit thing I wrote when I had lots of time on my hands
8:19 AM
Ah nice
just expand it
difference of squares
So, fun fact, I'm three hours late to class and an hour late for my test
but it doesn't matter because my grades don't count for anything
Rofl fun
3 < 4 though so it could be worse
(I'm taking Hebrew classes in Jerusalem)
That's very cool
8:25 AM
"$3 < \infty$"
I'm in level 4 (out of 6)
@Daminark A fine remark
This reminds me of something I saw in a paper once
But $8\cdot i=\infty$
8:26 AM
a tight bound indeed
Reminds me of the proof of Goodstein's theorem, where you use various types of infinity as upper bounds
it's quite usual to see things like $[K:F] < \infty$ I would say
@AkivaWeinberger ordinals
(specifically, each step is bounded above by an infinite ordinal)
Let me type OK? Yes.
8:28 AM
my apologies
OK, you know what, I've decided
screw school today
Just going home
Or maybe not
The non-commital behaviour of a kind-of-rebel
Akiva is a rebel now
Oh my God Ígjøgnum
Nah, the thing is, I left my family Tel Aviv this morning with the hopes of only being sort-of late
If I don't go to class at all, then I could have just stayed in Tel Aviv until the end of the day
(My family is vacationing in Israel and leaving soon)
is in vacation
8:32 AM
You know what Tel Aviv has that Jerusalem doesn't? A beach
And also my family
OK, final decision: go to my dorm, rest for a bit, use the bathroom, drink, and then go to class
how long is this f*ckin class
7 hours
Until 1:30 (it's 11:36 now)
8:37 AM
Seven hours of class?
Oh! Daminark wrote that, not Akiva.
Hey everyone
It's 8:30-1:30, so five hours.
@AkivaWeinberger maybe you should learn yiddish
When is the lunch break?
8:41 AM
Oh, God, the languages I have now are enough for now
Why are you learning Hebrew?
I'm gonna stay in Israel for the year
Exchange year?
It's fairly common for American Jews to spend their gap year between high school and college in Israel
8:42 AM
I'm taking Hebrew classes for a little bit until Eretz HaTzvi (the actual program) starts
Hebrew verbs, incidentally, don't actually make sense
I have a friend who did theoretical physics, but who dropped out one year before finishing to start learning sanskrit and music theory. He also wants to work in a flower store. Needless to say, I'm lost as of why.
@Daminark My logic professor wanted to show that $\bar{Q}$ is countable and used $\aleph_0$ as an estimate for the number of roots of a polynomial
(That's not true, but the conjugation rules clearly weren't made by a sane person)
8:44 AM
Is there any sane person doing mathematics, though?
@AlessandroCodenotti I see it
@AlessandroCodenotti sounds reasonable
Like, on the one hand, lol, but on the other hand, I see how the proof went
@Oskar I think it's really good to broaden your interests tbf, I think it's rare to find someone who doesn't also do something artistic/musical/lingustic here
@AkivaWeinberger Yeah it's much faster to estimate everything is a rough manner and use some cardinal arithmetic
In this case it's good enough so why worry about preciseness
8:46 AM
Yeah! Sure, but dropping out from theoretical physics when you have done already passed four out of five years.
@Oskar Oh I didn't read properly, that is definitely weird hahah
@Alessandro I am now probably delaying my masters for 6 months until the summer semester -_-
keep changing my plans lol
Haha! Yeah. He's very smart, but I think that it came with a bit of insanity too.
@Oskar är du svensk btw?
We have the Cauchy problem

$u_t+xu_x=xu, x \in \mathbb{R}, 0<t<\infty$

with some given smooth ($C^1$) function $g$ as initial value.

I want to check if the problem is will defined for each time.

We know that a problem is well defined if the solution exists, is unique and depends continuously on the data of the problem.

I have found that the solution is $u(x,t)=g(xe^{-t}) e^{x(1-e^{-t})}$.

So we have that the problem is well-defined if $g$ does not take two different values for some specific t, right?
@ÍgjøgnumMeg Ja. Du också?
9:00 AM
@Oskar neeeej men du har ett svenskt namn och jag försöker prata svenska varje gång möjligheten uppstår hahaha
@ÍgjøgnumMeg Är du norsk?
@Oskar nej jag är engelsk
@ÍgjøgnumMeg Hur kan du svenska?
@AkivaWeinberger Sure it has been hot, but the main problem this year still was the drought, at least in west/central and northern Europe.
@Oskar hade en svensk flickvän och lärde mig svenska för att prata med hennes föräldrar
eller med sina*
vet inte
9:02 AM
@Rudi_Birnbaum It's a neat video anyway. Did you watch it?
@ÍgjøgnumMeg Why?
Not yet, OK, I'll check it out!
"Hennes" är rätt i den meningen. :)
@Alessandro I am too poor to fund it, so I will work for 6 months and apply for some scholarships (whose deadlines I missed before -.-)
@Oskar juhuuu
Hey @LeakyNun
Do you have an idea?
9:04 AM
@Oskar och jag har svenska kompisar i min stad så kan jag snacka med dem lol
Haha! Jag har en amerikansk flickvän. Hon driver hela tiden om "Jaha!", "Oj!", "Jaså?", "Jajamen!". Varenda dag. xD
Technically $\infty \cdot i = 8$ will give you the trivial object, as proved below:
@Evinda nein
I sometimes wonder whether Evinda and Marystar are the same person. Their posts have the same fomat
9:07 AM
@ÍgjøgnumMeg Ah, I see, I'm sorry to hear that
$\infty\cdot i=8$ can't be true. Right?
@Oskar it's meant to be a joke (think what complex numbers do when you multiply!)
It's a joke in that multiply by i will rotate something by 90 degrees
for both of you
But actually taking this as an axiom in the extended reals, it can actually collapse everything into the trivial object ,similar to trying to divide by zero in semirings
9:09 AM
Haha! Yeah! Okay. I was getting a bit confused there.
Ah ok @LeakyNun
@Alessandro me too! But I think I have a strong application for a lot of scholarships so hopefully this will just make things easier
and i can ask a lecturer at my current uni (who researches in alg. geo) to give me a project to do for 6 months or smth
@AkivaWeinberger Akima yes I agree its quite good in its explanations and formulations I also like the vizualizations!
ok I cannot find that proof again, grr
@OskarTegby yeah a bit sick ...
9:14 AM
$x \cdot \infty = \infty$ for all $x \neq i$
$x \cdot (\infty \cdot i) = \infty \cdot i \implies x \cdot 8 = 8$
$(x \cdot \infty) \cdot i = \infty \cdot i \implies \infty \cdot i = 8$
congradulations, you just f88888 associativity of the extended reals
gr8 m8 i r8 8/8
Oh! Now I get it. xD
Damn. I'm slow sometimes.
well to be fair, you might be able to define a pretty interesting complex projective space structure with that joke, one where all multiples of 8 are points at infinity
plus some nonassociative geometry
but basically, because all multiples of 8 becomes absorbers in the reals, you also loses the multiplicative inverses for these elements, thus you definitely will not end up with a field
My suspicion is this will be a nonassociative ring
(to be explored later)
As usual, we can once again apply the explosive generalisation operator to the above and give us the following general question:
Let $W$ be an axiom which has the following form:
"the explosive generalisation operator" nice
where $T$ is in general a nonlinear map, $a$ is some element in an algebraic structure and $b$ is an element in the algebraic structure which is the result of the symbol for $a$ geometrically transformed into $b$
What is the common property for any algebraic structure $A$ that contains such axiom
We can clearly see that the rotate by 90 joke is a subset of this general formulation where $T$ is multiplication by $i$, $a=\infty$, $b= 8$
What happens in the joke is basically you have a map that maps an element generated by induction, and is invertible to an absorber. Thus in order for such rule be compatible, some symmetry breaking is needed to accommodate this discrepancy. In our case, associativity breaks
The above logic also explains why division by zero is in general problematic:
Let $Z$ be the map that maps $0 \to 1$ such that there exists an inverse $Z^{-1}$ that undoes it
Thus you have $Z(0)=1$ and $0=Z^{-1}(1)$
You can see the insane number of axioms have to be broken for such rule to be compatible
9:36 AM
Someone should write a book called "Breaking math" or something like that.
Btw since I mention this alot, it helps to formalise what exactly is "The Explosive Generalisation Operator" (ver 1)
The Explosive Generalisation Operator
So say I have the tensor of type ${1}\choose{1}$, with $T \in T_p(\mathbb{R}^n) \otimes T_p^*(\mathbb{R}^n)$ where $$T = T^{a}_{b} \frac{\partial}{\partial x^a} \otimes dx^b$$ summed over all $a, b$. Wouldn't $T$ take as an input (when identified with it's corresponding multilinear map) in its first argument a one-form and in it's second argument a tangent vector?
Let $P$ be a proposition with some interpretation $I$, procedure or whatever well founded sentence. Then define $💥$, the explosive generalisaton operator to be

where $\mathscr{P}$ is the family of objects such that for each object $R \in \mathscr{P}$, $I$ is contained in them

In set theoretic language, let $P$ be a set. Then $\mathscr{P}$ is a collection such that

$$\bigcap \mathscr{P} = I, P \in \mathscr{P}$$
Cause I had this one problem on my pset that was like "Find $$T\left(\frac{\partial}{\partial x^k}, dx^l\right)$$"
Recall that any tensor maps elements from a tensor space into a scalar, thus the first argument has to be paired with a tangent vector and the second argument has to be paired with a one-form
so that when you sum them up, they give inner products each and become a scalar
but again, my knowledge of tensors is mostly from physics, thus I don't know how deep I know about this issue
9:49 AM
Thanks for that @Secret
I'll ask on main and see what they other people have to say too
I wish more symbols and operators were colorful.
So basically... the explosive generalisation operators takes the minimal thing that defines something, and then query all possible examples of it. As an example:
$💥(\sum) = \mathscr{P}$ and $\int, \oplus, \cdots \in \mathscr{P}$
One can see that this is a lot more explosive (pun intended) than the powerset operator, because it generates not just a powerset, but often proper classes or even categories
In fact, a conjecture is that often 3 successive applications of $💥$ and you are within abstract nonsense territory
(have not figure out how to prove that yet)
In theory also, the nature of $I$ is important, as depending on which property we want to specify, the output can be different
For example $💥(0)$ can give you a lot of things such as zero objects, multiplicative identities, identity maps etc.
But if we want to restrict this a bit, and querying all possible mathematical objects that behaves like an identity element, we might want to specify that in $I$
Now, there are also a couple of interesting cases for $💥$ such as:
$💥(\varnothing)$, I have not quite work out what it is yet
Also once can easily see that if we rejects essentialism, en.wikipedia.org/wiki/Essentialism
then for such an object $K$, then $💥(K)=K$ because there is nothing else like $K$ itself
In more extreme cases when the law of identity is broken, there are some object $L$ such that $💥(L)$ is undefined because it cannot be certain whether $L$ is the same as itself
Finally, as a joke, the following will destroy mathematics as we knew it:
$$💥\left(\frac{1}{0}\right) = ?$$
It is also unclear what happens when:
$$💥(💥) = 💥^2 = ?$$
probably we can treat it like any map, in that we can group them together. but perhaps $💥$ works more like a deductive system thus associativity of maps can be broken and the order will matter. Have not studied that in detail yet
10:19 AM
Oh and one more thing, outputs of $💥$ are not only uncomputable except for finite number of members (similar to the busy beaver function), it is indescribable or even inconceivable. To get this idea, imagine how hard it is to ask people back in stone age to conceive the idea of building airplanes
It is basically a super crazy operator where most of its members are unknown unknowns, thus represents the limit of human imagination
10:34 AM
What is the busy beaver function?
Is this what happens when drinking and doing math at the same time?
11:11 AM
The busy beaver game consists of designing a halting, binary-alphabet Turing machine which writes the most 1s on the tape, using only a limited set of states. The rules for the 2-state game are as follows: the machine must have two states in addition to the halting state, and the tape starts with 0s only.As the player, you should conceive each state aiming for the maximum output of 1s on the tape while making sure the machine will halt eventually. The nth busy beaver, BB-n or simply "busy beaver" is the Turing machine that wins the n-state Busy Beaver Game. That is, it attains the maximum number...
It is an uncomputable function that has applications in computation theory
11:35 AM
Hello. If i have a function s.t its integral over $R$ is finite can i say that the function is bounded over $R$? i think yes, but cant see why.
@Shobhit $$\displaystyle m \left({b - a}\right) \le \int_a^b f \left({x}\right) dx \le M \left({b - a}\right)$$
Q: Why does the integral of 1/x from negative infinity to infinity diverge?

user135330I'm confused as to why $$\int_{-\infty}^\infty\frac{1}{x}dx$$ diverges. If $\frac{1}{x}$ is an odd function shouldn't the area to the left of origin be the opposite of the area to the right of origin, resulting in a net area of 0?

ok it bypasses
11:52 AM
@MatsGranvik What i meant was if$$\int_{-\infty}^{\infty} f(x)$$ is finite then can i say that $f(x)$ is also bounded in $(-\infty, \infty)$? I cannot use what you have wrote for $a$ and $b$ as $\infty$ and $-\infty$
@secret drawing pictures of arbitrary graphs seems to agree, but can't put it into words.
I cannot think of an integral where it has an integrable singularity that returns a finite value yet
Ok, wacky counterexamples can exists
@Shobhit I copied the inequality from here: proofwiki.org/wiki/Upper_and_Lower_Bounds_of_Integral
think of a continous probability mass function
m is the minimum of f(x)
M is the Maximumum of f(x)
so basically, the counterexample makes the infinith step into a stick thus avoiding the blowup
12:01 PM
@Secret like $f(x) = e^{-x^2}$
no the gaussian is bounded
@Secret nice counter-example, i proved something supposing this, will have to work again now.
$e^{-x^2}$ is the gaussian function, which is bounded between (0,1]
Q: Can an unbounded function have a finite integral?

john melonI am wondering whether there exists a function such that: $$\lim_{x \rightarrow a}f(x)=\infty$$ at some point $a$ on the real axis but yet, $$\int_{-\infty}^{+\infty}\left|f(x)\right|\ dx<\infty$$ Does the fact that a function is unbounded imply that it has no finite integral?

also more wacky examples here
12:16 PM
lol you can even get a complex number out of singularities
@Secret o.O
@Secret Can you prove that?
no idea, I am still terrible at integration, but I suspect there is an integration by parts going on as there's a gamma function
me too
Still a great find, thanks @Secret
Also if you try to find, $$\int_{-\infty}^{\infty}\frac{e^{-x^m}}{x^{\frac{1}{n}}}$$, it only converges for even m and diverges for odd.@Secret
Integration by Hogwarts:
(*integration by parts*)
Clear[x, n];
Part1 = Exp[-x^2];
Part2 = 1/x^(1/n);

u = Part1
du = D[Part1, x]
v = Integrate[Part2, x]
dv = Part2
aa = FullSimplify[u*v - Integrate[v*du, x]]
bb = Integrate[Part1*Part2, x]
x = 5;
n = 4;
$$💥\left(\int_{-\infty}^{\infty}\frac{e^{-x^m}}{x^{\frac{1}{n}}}\right) = \text{don't have enough time to analyse yet}$$
12:31 PM
@Secret try for different values for m, i just came to this conclusion (without proof) :P
12:55 PM
Anyone interested in probability and real analysis?
Q: Multivariate extension: convergence in distribution implies uniform convergence

kevinIs the following theorem true for multivariate case? Convergence in distribution plus continuity of limiting distribution implies uniform convergence. Remark: 1) In univariate case, one can find A such that F(x) vanishes outside [-A, A], and make use of uniform continuity of F on [-A, A](cont...

1:20 PM
Where are the community wikis found?
@OskarTegby Means?
1:55 PM
Hi! Is anyone alive here?
2:32 PM
@OskarTegby yedidia-aaronson had an interesting paper on busy beaver recently some may enjoy arxiv.org/abs/1605.04343
2:43 PM
I feel moped.
2:56 PM
I just found this.

@Kevin Do you know, girl?
$11^{13} +13^{11}$ is divisible by?
I am thinking of writing 11 as 1+10 and 13 as 3+10 but that will be so longggg
3:17 PM
Seeervus @Mathein
3:32 PM
@ÍgjøgnumMeg servus
@Leaky hoi :)
logs and mods are not helping
@ÍgjøgnumMeg wie geht's dir
@Secret @Fawad I would probably note that $(11, 13) = 1$ and that these are prime and then some CRT/FLT magic
@Leaky Guuuut, aber ich hab gerade mein Fahrrad kaputt gemacht :(
3:34 PM
duh just ask wolframalpha
Or @Leaky's suggestion also works
@ÍgjøgnumMeg brauchs du et taeglich?
@ÍgjøgnumMeg Was ist denn kaputt?
@Leaky ja leider, ich tu jeden Tag zur Arbeit radeln hahaha, aber da die Schule meiner Mutter gerade Ferien hat (sie ist keine Schülerin, sondern Lehrerin, bevor du irgendwelche Witze machst ;)) kann sie mich zur Arbeit fahren
@Rudi derailleur hanger :(
Schaltauge heißt das auf Deutsch anscheinend
@ÍgjøgnumMeg Was ist damit? abgebrochen?
3:41 PM
@Rudi ja völlig halbiert halt
@ÍgjøgnumMeg Hats dirs ins Laufrad reingezogen? Das ist blöd. Ist es ein teures Rad?
@Rudi ja genau so ist es passiert -.- Aber ich war zum Glück nur so 2 minuten vom Haus entfernt (obwohls trotzdem bei diesen blöden Fahrradschuhe weh tut so weit zu laufen hahaha)
Ja geht schon so 1000€
aber es kostet nur so 10€ oder so für eine neue Schaltauge
@ÍgjøgnumMeg Ja genau, eigentlich kein Beinbruch, aber schau Dir die Speichen genau an die könnten auch einen Pecker bekommen haben. Fährst Du zum Sport rad?
Ja tu ich aber nicht so seriös, ich tu oft am Wochenende so "long distance riding"
I feel like we've reached some kind of critical mass in German speakers so that the chat is turning into a German language chat
gibt's nur zwei
3:52 PM
@ÍgjøgnumMeg cool. I do it for more than 30 years now!
Oh nice! :D
Its a great activity but unforntunately a bit weather dependend ...
Yeah exactly, also I'm technically still a little over the weight limit for my bike so I have to dodge a lot of holes in the roads until I lose some more weight :)
@Rudi_Birnbaum Auslautverhärtung :P
Q: A question on definition of Sobolev norm

Rajesh DachirajuDefine $\|.\|_{T^k(\Omega)}$ as $$ \|f\|_{T^k(\Omega)} = \|f\|_{L^2(\Omega)} + \|(\sum\limits_{i=1}^d(\frac{\partial^{k}f}{\partial x_i^{k}})^2)^{\frac{1}{2}}\|_{L^2(\Omega)} $$ Is this norm equivalent to Sobolev norm, $\|.\|_{W^{k,2}(\Omega)}$, under two cases $\Omega$ is a bounded open ...

3:59 PM
@ÍgjøgnumMeg oh ok thanks (dependent :-).
@Rudi also the derailleur hanger is really specific and for some reason NOWHERE sells it hahaha
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