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00:01
So Ted apparently Danny Calegari Is teaching a low dimensional topology class in winter
that should be good :)
And André Is teaching minimal surfaces
You saying you have a surfeit of courses to take? Wait 'til grad school starts!
Which I will almost definitely take since that sounds lovely
I'm trying to take it easy but the course catalogue is making it hard on me
I know, I know ... You want to be a full-time cook.
00:09
speaking of i have just completed a nice brazilian dish
did you save me some?
of course
ah benson farb is also teaching a three quarter sequence called geometric literacy :(
too many things
geometric literacy? whats dat?
you need to think about levels of things, too ...
00:25
it's a course about general geometry techniques
yeah i would drop any course i wasnt ready for
no, I meant it the other way ... don't they teach anything for undergraduates not at the level you're at?
sorry what do you mean exactly? I'm having trouble parsing that
double negative
I meant that some course might be too low-level for you
oh yeah, ive been told by a few profs that all the non-algebra undergrad classes are a waste of time at this point
OK, now you know what I meant.
00:34
the first year grad classes ive taken were also generally like... too easy
ive heard the first year algebra sequence is a killer though
at UGA some of the grad students told grad students never to take a class from me because I was too demanding ... well, do you want to learn something or not?
(In fairness, I think they admitted I was a good lecturer/teacher, but ...)
i think the only class here i would say gave me a "hard" time was galois theory cause it was waaaay outside what I think comes "naturally" to me
but that's still great
but it still was one of the easier classes ive taken in college
oh i loved that class
see, that's just no good
00:38
@Ted is this right? As you said, I'm a bit careless with my conjectures :p
You mean $k\ne 0$? Well, this one has a good chance of being right. I'm waiting for a proof.
my prof for algebra last year gave very good problems
The case $k = 2^m$ is pretty direct
Why is that?
I don't know if I can do some sort of direct-reverse induction
@TedShifrin $z^2 = (a^2 + nb^2) + 2ab\sqrt{n}$
00:41
Sure ...
the important part $2ab\sqrt n$ shows that if $a, b \ne 0$ then $z^2$ can't be an integer
Of course.
the same way as $z^4, z^8, z^{16}, \ldots, z^{2^m}$
OK, I guess I believe that.
is it any good to choose $2^m < k < 2^{m+1}$?
00:44
I have no idea.
I've seen this in a AM-GM inequality proof
ok, so the proof was, basically, Newton's binomial
I got it
How do you see there can't be cancellations in coefficients?
we've shown that case $k = 2^m$ is true. now suppose that $k = p2^m$, where $m$ is a non-negative integer and p is odd.
it's clear that every integer is of this form
00:53
sure
although you have to worry about the $m=0$ case.
@TedShifrin we're getting there. hahaah
I still don't see it.
So you raise such an expression to the $p$.
You could get all sorts of cancellations.
$z^k = (z^p)^{2^m}$. so $z^k$ is integer iff $z^p$ is also an integer
oh, OK, different order of exponentiation. But now what?
now we use Newton's binomial and care only when we have an irrational number
00:56
what's "Newton's binomial"?
Academia might be a bit of an uphill battle on a family front as well, it seems
Are you talking about you, Demonark? "as well"?
@TedShifrin probably it's not the conventional naming there in US, I must say
Explain it, Lucas !
it's $(x+y)^n = \Sigma \binom{stuff}{} x^j y^{n-j}$
00:58
I mean, as well meaning, on the family front along with just being uphill in general
We call that the binomial theorem, @Lucas. But I have no idea why you can't have cancelling terms.
Demonark, you're doing fine, aside from our little spats. Of course it's uphill. And it will get steeper. But if you're passionate and truly want it ... go for it.
Oh I mean, those I take to be healthy, thinking about/discussing pedagogy and such is good
Right!
Again using that $z = a + \sqrt(n)b$: $z^k = \Sigma \binom{stuff}{} a^j (\sqrt n b)^{k-j}$
Your parents might worry about your future, Demonark. That's only normal for parents.
@Lucas: But there could be cancellations if $a$ and $b$ have different signs.
01:02
there comes the cool part
we won't care about the rational part
It's more like, I dunno, there's some resistance on the family side since I think most of an academic career as being a minimal upgrade of a high school teacher, and that the hell required to get this far would have just been thrown down the drain. So it'll take some time to get everyone to realize that it's actually different
we'll look only for the irrational part
Plus there truly are jobs in industry for Math Ph.D.s if it comes to that, Demonark.
sure, Lucas.
so we choose $j$ even
fuck. that won't do it. HAHAHAH
sure
covers ears/eyes
01:04
man, I'm truly disappointed.
Yeah I mean, I definitely want to at least attempt grad school, I think the concern is more, after that. Which can wait but you know, they're from Morocco, whose education system is modeled after Europe
I forgot that $b$ can be a negative integer.
Hence, you should know more or less how your life is gonna go by ~13 :P
I did mumble about signs.
Well, Demonark, give them time.
Professor, do you have any idea of how to approach this problem?
01:05
But do think about contingencies, because academia is becoming a harder road with the current views in the US (and I'm not meaning just the idiot president).
@Lucas: Honestly, I haven't tried to do it. I might know some algebra you don't that I might try to use (and I'd let $a,b\in\Bbb Q$ to apply that). But I don't know a proof yet. You might need to learn up through chapter 7 stuff in my book, though :P
or at least chapter 5.
Huh, I didn't actually think Trump had much of an effect along those lines
well, his administration is anti NSF and anti-education and anti-intellectual, so it won't do any good for us.
@TedShifrin so it's at least a picky problem?
I'll need some theory.
it's a good question, Lucas. Not super advanced, but I don't see it yet.
Oh, yeah NSF hits are not gonna be good
01:07
Demonark, you mean they haven't chosen a wife for you yet? :D
What's exactly the problem? To show that no power of $a+b\sqrt{n}$ is an integer?
Right.
No nontrivial power.
@TedShifrin apparently not shrug
It should be stupid easy (generalizing to $a,b\in\Bbb Q$) with field theory, but I don't see it immediately.
LOL, Demonark. Sorry, didn't mean to add stress :D
01:09
Oh it's fine, like they're not dictating at all
Then, stay cool and keep doing well in college, etc.
Even along career paths
Just don't be stooopid naïve. If you have a chance to get some practical computer skills, etc., take the opportunity.
I think it's also smart to have a sense of what your 'plan B' is.
It's more just like, there's definitely a bit of side-eye going on and the "this and then 6 years of grad school to teach?"
01:11
Semiclassic speaks from very personal perspective.
Well, if it's teaching + research, that's different. If it's even teaching at a more sophisticated level, they should have more respect for teaching.
@Ted do you think that this is a question worth posting?
I don't think it's a question of respect so much as, most of the people they know associated with teaching struggle financially
On the other hand, Demonark, I don't look at you and see someone who wants to dedicate himself to a life of teaching college kids. :P
not at the college/university level they don't, Demonark, relatively speaking. I mean we're not corrupt Donald Trumps, but ... I did OK for myself.
01:13
Why of course I do! I just want to teach first years Hartshorne!
:P
Then you'd better be a strong researcher :P
But yeah I mean, they are starting to realize that things are different than they thought
Lmao
I think what Semiclassic said is worth thinking about seriously. Get some computer skills. Investigate biomath/bioengineering (which is super-hot still). Whatever.
Plan C is important too @Semiclassical
The other thing worth keeping in mind is that, after a certain point, the title of 'grad student' is a bit of misnomer.
01:15
skull, by induction, ...
Because you're no longer a student.
I'm still a student?
In the sense of taking classes etc.
You're still an apprentice to your adviser, so I disagree on that one.
Yeah, I've looked at stuff like graph theory and theoretical compsci, it seems like that's got somewhat better prospects
01:18
Hmm.
That's fair.
But when you get to plan Z you're gonna have to learn Greek :P
Hello
Did people math
I guess my point is that, when you're in undergrad, the 'apprenticeship' model is not what you think of when you say 'student.'
Hi pal
It's that sense of being a student that you leave behind in grad school.
01:21
Yeah we decided that the answer to "Is the Hodge conjecture true or false?" is "Yes"
Good answer.
Semi-student :P
@LucasHenrique So what's the problem, raising $a+b\sqrt n$ to a power won't kill the irrational part?
@AkivaWeinberger yeah.
01:25
@skullpatrol lolyes
Oh I think I have an idea
So we want $(a+b\sqrt n)^k\ne(a-b\sqrt n)^k$
but that's always true since $|a+b\sqrt n|\ne|a-b\sqrt n|$
or, rather, $|a+b\sqrt n|=|a-b\sqrt n|$ implies either $a=0$ or $b=0$, and we've assumed they're not
Does that work?
@AkivaWeinberger that'd be interesting, but why?
I mean, how are the statements equivalent
@LucasHenrique Because $(a+b\sqrt n)^k$ is always the conjugate of $(a-b\sqrt n)^k$
and a number in $\Bbb Z[\sqrt n]$ is an integer iff it's equal to its conjugate.
So if one is purely integer than so is the other.
So we just need to prove that $(a+b\sqrt n)^k$ does not equal its conjugate.
01:29
holy crap. that's ridiculous...
damn. you're right.
That's slick, yeah.
good one, @AkivaWeinberger.
DogAteMy always is cleverer than I am.
OK, I'm gone to cook dinner.
01:34
Aaaand I'm gone to do homework.
Later pal
On an unrelated note, it's strange how hours can feel longer than days
See you @Ted!
And @Lucas!
like, a boring hour will go by very slowly, but you can look back at a boring week and feel like it rushed by
'cause I guess that, for long periods of time at least, the body assumes more experiences implies more time
I don't think I've often had that feeling, surprisingly enough. Everyone talks about it
01:36
like it's measuring time in terms of experiences, as opposed to something shorter like an hour which is measured in like seconds
Sure, things appear to going in slow motion sometimes.
Consider though that when you feel an hour dragging, that's in the moment
But when you feel like a week just zipped by, it's in afterthought
True dat.
There's a baby crying right outside my room
Go have a look bro
01:42
The parents are dealing with it I think
"It" referring to the situation and not to the baby
@Daminark I while ago I learned that there was a star such that the relative size of the Earth to this star was the same as the relative size of a second to two days
now compare the relative size of the earth to the distance to the next inhabitable star system.
which kinda broke my mind because while obviously I have an intuitive understanding of a second and of a two-day duration, I don't know if I feel them on the same scale
@Semiclassical Distance to Alpha Centauri is almost exactly a century IIRC
if Earth's diameter is a second
Are we talking about wrt diameter, mass, volume?
Oh fuck diameter
well, earth's diameter is apparently 0.0425 lightseconds
whereas the distance to Alpha Centauri is about 4.39 lightyears.
2 days to a second is somewhere near 200,000
01:48
Yeah the conversion rate is roughly a twentieth of the speed of light
So we'll say that scales to about 7500 light seconds
Alternatively, the first number I gave is about 1/20th of a lightsecond.
2 light hours
Which is intense
Sounds about right
You could fit Jupiter's orbit inside it, IIRC
Replacing the sun with this star would swallow Jupiter
I rescale that to one second, I get 1 second : 103 years.
And that's the nearest one.
01:50
:O
Nah the nearest one is 8 light-minutes away :P
...lol
yay, you win.
Multi-lel kekking
(which translates to somewhere over 2.5 hours I guess?)
If you restrict yourself to systems with observed exoplanets: en.wikipedia.org/wiki/…
01:53
And to think, the most crucial part of the universe is so tiny :P
The nearest system after the Alpha Centauri system is Wolf 1061. That's 13.8 lightyears.
So that'd be about three centuries on the "earth's radius = one second" scale
Yeah a quick jog yeah?
@Semiclassical We were doing diameter
right, that's what I meant.
I find the star thing more surprising
It's stuff like this that makes me skeptical of us ever travelling to other star systems or having meaningful contact with alien species.
01:56
Distances between objects is huge, that's less impressive than knowing there's a huge sphere out there
The universe is big enough that alien life can both exist and be entirely irrelevant.
Relativity is a harsh mistress.
But if we ever do figure it out, and we figure out how to harvest matter from stars, that's a nearly inexhaustible resource
And we haven't even talked about blackholes yet.
orders of magnitude larger than what's possible just from earth
Meh. "If."
01:59
Huge "if"
Hm. Imagine it were possible to send a radio wave out into space in such a way that, when it came into contact with whatever planet we were aiming at, it would assemble the nearby atoms into a clone of you
which I'm sure is impossible for a variety of reasons
But would that count?
deep question
02:28
Might be opening myself up to more downvotes, but in case anybody can at least suggest something, I'd be more than grateful.
You know what somebody just did? They wrote "The North remembers" after downvoting me.
02:43
Hi -- can someone kindly shed any light for this Q:
https://math.stackexchange.com/questions/2376054/satisfying-the-following-determinant-inequality
thank you!
would appreciate any assistance on this
Northerners have been so touchy ever since the Red Wedding...
03:02
Can anyone shed some light on interpretation of a conservation law for me? If I have a conservation law C = stuffhere and I move to a different jet space (in different space of variables) what is the significance of C = 0 = stuff here in the new jet space?
 
1 hour later…
04:21
2 hours ago, by Akiva Weinberger
Hm. Imagine it were possible to send a radio wave out into space in such a way that, when it came into contact with whatever planet we were aiming at, it would assemble the nearby atoms into a clone of you
Even if the radio wave is 100% efficient, and not deflected nor degraded as it pass through interstellar dust, without the knowledge of the atomic composition and distribution at the destination, it is impossible to program the wave in a way such that it can induce assembly of atoms into a clone
for those who don't like raw numbers and want to focus on the patterns
I wonder if there's a structure that describes numerical errors and their propagation...
ray
ray
04:51
Hi. I'm aware that (x+y)^2 = x^2 + 2xy + y^2 and also that you can use Pascal's Triangle and the Binomial Expansion to find (x+y)^n with its factors. I'm trying to confirm whether the same is true when using subtractions, i.e. (x-y)^n ?
Thanks in advance!
To clarify, I'm not asking if (x-y)^2 = x^2 + 2xy + y^2; I know it's x^2 - 2xy + y^2; I'm asking if the same logic can be used and if there's a pattern for the + and - signs in the factors retrieved from Pascal's Triangle.
05:13
@ray x-y=x+(-y), so (for instance) (x-y)^2 = x^2+2(x)(-y)+(-y)^2=x^2-2xy+y^2. So the negative terms are the ones where y shows up as an odd power.
ray
ray
@Semiclassical That makes sense. I was seeing it as x+(-y), but had not tried expanding larger powers and wasn't 100% sure what the + and - pattern would be. Thanks.
hi pal
...
What's going on?
05:22
@AkivaWeinberger Except that the norm is $x^2 + by^2$ when $x + \sqrt{b}y$ is the element of a polynomial ring.
@skullpatrol Collatz Conjecture.
Just have to solve the cases of 12n + 3 and 12n + 7 in the context of an inductive step for n
im tempted to brute force it
(as operations on pairs of xn + y)
it's like a hydra
try to solve 2n + 1 and you must break it into 4n + 1 and 4n + 3
4n + 1 is trivial
4n + 3 then becomes 12n + 3 and 12n + 7
@ray let z = -y. It's trivial to see.
@skullpatrol Any thoughts?
0
Q: $ U(x+y) = U(x) V(y) W(y) + U(y) V(x) W(y) + U(y) V(y) W(x) + U(y) V(x) W(x) + U(x) V(y) W(x) + U(x) V(x) W(y) $

mickAre there nonconstant solutions $U,V,W$ such that for all real $x,y$ we have $$ U(x+y) = U(x) V(y) W(y) + U(y) V(x) W(y) + U(y) V(y) W(x) + U(y) V(x) W(x) + U(x) V(y) W(x) + U(x) V(x) W(y) $$ How to decide existance or uniqueness ? Can we express the solutions with closed forms or differential...

???
Anyone ?
@mick where is your attempt?
@mick define "closed form"
I assume it can either be solved by exp cos sin type solution or not at all.
05:33
that doesn't define what closed form means
No
That was Just a remark
From the looks of his avatar he spends too much time in the gym :P
indeed
not enough time doing math
Exactly.
^^^
cough piecewise-constant cough
05:34
Closed form : standard functions
sorry
I choked
Lol. Yeah but Maybe analytic solutions too ??
what have you tried?
Are you ok?
@mick what are the standard functions? Be specific. Otherwise, someone will answer your question and you'll just say "no, because it isn't a closed form solution"
@skullpatrol I choked on the lack of math skills here.
@mick For all I know, that is an unsolved conjecture. Way outside of my league.
05:38
No i Will not say that
and the collatz isn't?
@skullpatrol Yes, but literally a 10 year old can at least attempt to solve it without the equivalent of throwing mathematical darts.
10 year olds are naive
(If a 10 year old ends up solving it I'd be shocked though)
@skullpatrol my point is that it is trivial to conceptualize.
Elementary , trig , exp , log , theta , hypergeo , ei li si Bessel gamma beta zeta Lambertw clausen polylog Mathieu , the compositions and inverses , those are standard functions
05:42
@mick Fair enough. Just recommending an edit so people don't get confused later. Ummm... I have no clue man. I can only throw guesses and test them. I have no experience with that. I presume piecewise constant works but that's just a guess considering constant works.
I'd say that's beyond standard. There's a reason they're called special functions, after all.
2
@Semiclassical true, but he can define closed form to be whatever he wants without contradiction.
@Typhon Maybe you can show that it must be associative
It's actually used as a teaching tool on 10 year olds to build one's respect for trivially easy looking questions @Typhon
05:44
@mick I said, I don't know. I have no clue how to solve it. All I can do is guess solutions.
@skullpatrol is that last comment about my question
maybe someone clever will write a program to approximate solutions. idk.
No @mick
Wait a sec
The collatz
05:45
@skullpatrol fair enough. My point is that I can make a decent attempt towards it. This question on the other hand is beyond my knowledge level.
The special functions which I have a respect for are the Painleve ones. Them be crazy.
@skullpatrol I believe an inductive proof can be done for elements of the forms 12n + 3 and 12n + 7. It's just a matter of splitting the cases in just the right way. It might be that it only works if you consider things like 1024q + r.
Are you trying to prove collatz ???!
Have fun.
05:48
I do not think it is possible
actually it is impossible to not be able to prove or contradict it
it's a decidable statement
No
you're wrong
why do you think it is "impossible"?
Interestingly, there is a generalization of the Collatz conjecture which is provably undecidable.
Look at the literature.
05:50
Even If you Find a potential counterexample , you need to decide a halting type problem
wat
Also semiclassicals comment
@mick no you just prove that either an element diverges to infinity or falls into a cycle
There's a quick mention of that fact in this answer: math.stackexchange.com/a/1156016/137524
That is halting @Typhon
05:51
@mick and we can decide whether or not things halt in some cases...
@Semiclassical I know of that.
That we are unable too @Typhon
@mick you're claiming it is impossible to resolve the conjecture.
Yep
05:52
that's ridiculous
It is harder than prime gaps for x sake
I don't think it's been proven either way.
so?
I have a question: is this book the original one or has it been changed to be more similar to the film? amazon.com/Film-tie-Stephen-Kings-ebook/dp/B003BKZW7C/…
@Semiclassical how can it not be resolvable? If it is false, we eventually find a counter-example. If it is true, then there exists a proof such that it can be shown true.
05:54
Why spend sooooo much time on it?
Also , unlike most math , it does not relate to other things ; giving lack of attack
the issue of undecidable sequences comes from generalizing it.
@skullpatrol it's fun?
You're equating "This statement can be proven true" with "This statement is true"
@Semiclassical yes.
I am.
If we had a proof that Collatz was decidable, then that'd be fine.
05:55
By using cantors paring function , we show that the number of indices does not matter for halting
But I don't think we even have that much.
we can decide collatz for any expressible number.
@skullpatrol tell me please
So
we just iterate upon it until it increases to a point that we know it cannot ever reach 1
05:56
what a waste of a discussion
total trash
(well it is trickier than that, but you know what I mean)
@BalarkaSen what a waste of a comment, bashing what other people like to talk about.
We can not even prove it for a single number If it requires more iterations than our computer can handle !
The example of "probably true, but no one has ever managed to prove it" which comes to mind for me is whether there's any odd perfect numbers.
Sorry @Twink I have neither read the book nor seen the film.
Riemann hypothese :)
4
Q: Conjecture about arcsin and $\sqrt{\quad}$

mickLet $r(a,b)$ be a rational number depending on $a,b$ and nonnegative. For every $b$ there is an $a$ such that $r(a,b)$ is not $0$. Let $C(a,b)$ be a squarefree positive integer depending on $b$ and different for every $b$. Consider for positive integers $I,J$ : $$ S_J = \sum_{j=1}^J \sum_{i=1}...

Try that one
05:59
@mick I don't trust that one as much.
too powerful
Jeeze. Odd perfect numbers have been checked for up to 10^1500.
in my experience such "sledgehammer" statements don't tend to work out.
@Typhon What tou do not trust ? Define trust ? Define powerful ?

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