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00:02
The decreasing ones should have cardinality $\aleph_0$, though, right?
@DanielFischer
Yes, and you just proved it @Pedro
Earlier.
@PedroTamaroff Right.
I'm pretty tired.
I should see a doctor about being tired all the damn time.
Guys
I heard that Euler conjectured that there are no natural numbers $x, y, z$ such that $x^3 + y^3 = z^3$, really?
take x=y=z=0 and that's false
00:09
Not a natural number, but I was thinking the same thing with $1$ ;)
^5 Fernado, you define the natural numbers correctly.
uhhh @Mick 1+1 is not 1
$0\in\Bbb N$ is possibly the best troll ever
it's also correct IMO
we have $\Bbb Z^+$ for other purposes!
@Mike I honestly didn't think about the pointless case that hard
it makes sense order-theoretically
00:11
I'm more interested in the problem itself, is there an up to date theory on this?
@MickLH of course there is, it's a special case of FLT
have you never heard of fermat's last theorem
I don't even know what group of ideas number theory is exactly limited to
The case $n=3$ was actually proven by Euler
Roughly speaking, number theory studies $\Bbb Z$ and its arithmetic
but it uses tools from all over the place
the best way to prove that case is with $\Bbb Z[\omega]$ where $\omega$ is a complex cube root of unity
00:13
I basically do calculus because I love it, I like to stay in $\Bbb R$ but $\Bbb C$ is fun too sometimes
you can do calculus over $\Bbb C$ as well
A new low point. A plug-in-the-number task, probably with a mistyped constant, no indication where the problem for the asker is. I am becoming a grumpy old man, but there I should be even grumpier.
you can do calculus over $\Bbb Q$
heh I said "like" because I prefer it, not because I artificially limit my work
$\Bbb R$ just feels like the "real" numbers to me
hehe, I mean, the real real numbers :P
$\Bbb R$ is too weird to be real
3
00:16
@ccorn And of course, the answer handed on a silver plate.
@DanielFischer And that answer initially had three errors or so
@ccorn Let me quote Pedro: "Le SIGH".
Les sighs?
Oui, t'as raison, @Ted. More than one required.
@Ted That's too adult for this channel. The kids would get upset.
00:19
Évidemment!
You a kid now? @Mike
@Mike I hope it is not diabetes. Or anemia.
Yikes. Who? @Pedro
@TedShifrin Mikes says he's always tired.
Guys. How can I get better with hyper geometric series? I don't have a bunch of money to spend on books whort zhe fphouck dhoo I deouw?
Seriously, @Mike, get a physical.
00:23
b
@seaturtles I think that if you add "bounded",
Could also be thyroid out of whack @Mike
@MickLH: There are books on the internet...
Hey @Ted
@FernandoMartin So far I've collected the bateman manuscripts but basically everyone wants a bunch of money
I'm ending up just learning my way through by deriving it myself on Macsyma & Mathematica, which is kinda awkward you know
Isn't there a library in your university?
Lol I can barely get myself to do enough for-money work to keep food and a roof going
I'm not in school or anything like that
00:31
Then pirate the book
I don't know if it's frowned upon on this channel
trying! haha
but I can't deny I'm a pirate
@PedroTamaroff: what are the preffixes of books you can't take out of the library?
E1 and E2?
@FernandoMartin Yeah.
And I think some EX stuff too.
E1 means "Ejemplar 1". So it is basically the good one.
Meh
I wanted to take out an E2 one
My Atiyah Macdonal copy is totally wrecked.
00:34
I can't believe there's only one AM
There's only one?
ikr
@FernandoMartin Dang it.
I think this trick should be more known.
00:38
which trick?
I feel bad for using it so much.
Well, consider the nondecreasing sequences of natural numbers. Then sending $$n_1\leqslant n_2\leqslant \cdots\to (n_1,n_2-n_1,\ldots)$$ gives a surjection to $\Bbb N^{\Bbb N}$.
So that thing has cardinality $\mathfrak c$.
Similarly, consider the nonincreasing sequences.
Fix the tex please
you have one extra $
Thanks
Then if $k$ is where $k$ is the minimum $k$ where the sequence starts repeating, we have a bijection to $\bigcup_{n\geqslant 1} \Bbb N^n$ in a similar fashion.
That's a nice trick
That is, send $n_1\geqslant \cdots$ to $(n_1,n_1-n_2,\ldots,n_{k-1}-n_k)$.
And finally, consider the sequences of rationals that converge to $0$.
Then we get a convergent sequence to $r$ by $x_n+r$, and if $x_n$ converges we get a convergent to $0$ by $x_1,x_2-x_1,x_3-x_2,\ldots$.
=)
Or if you want, the Cauchy ones.
So it is $\mathfrak c$.
00:47
Also you get an injection from $\{-1,1\}^\Bbb N$ via $(x_n)\mapsto (x_n/n)$
@FernandoMartin Ah, cool. =D
If $f(z)$ is an analytic self-map of $\mathbb{D}$, it does not follow that $f$ fixes $\partial \mathbb{D}$, right?
is it bijective?
I don't believe that is given.
Well, I was following a dead end anyway. Hah. Schwarz Lemma OP.
no: constant maps
if you don't allow constant maps: $z \mapsto z/2$
00:53
yeah.
Thanks @Mike
Gosh I am slow. I am taking this graduate complex analysis course at the moment. I'm a senior in undergrad.
Those were such simple examples, Mike. They should have occurred to me :(
@FernandoMartin I like that better.
that sounds faster than most :D
btw - bijective holomorphic maps have holomorphic inverses.
@Pedro: I didn't understand your argument
It was poop.
so if it was bijective, it would fix the boundary I
I guess the fixed boundary is easier than the result I have
00:56
Wrong.
no, the fixed boundary follows from the fact that $f$ is a homeomorphism
there's an obvious uncountable set that obviously injects in the bounded natural sequences
Btw @Mike. I was trying to show that if $f$ is a non-constant analytic self map of $\mathbb{D}$ then it has at most one zero in $\mathbb{D}$. The answer was to move the fixed point to zero and apply Schwarz Lemma (At least, I think so.)
@FernandoMartin yes.
I don't even know what that Lenma is!
01:02
Err, not non-constant. Non identity
Can you tell me @Prototank ?
The Schwarz lemma is really cool
is that the reflection theorem?
that's the main Schwarz thingie I know and I don't see how to apply it
Let $f$ is analytic for $|z|<1$. Suppose $|f(z)|\leq 1 for all |z|<1$, and $f(0)=0$. Then $|f(z)|\leq|z|$ for $|z|<1$. Also, if we have equality in this case for z's that aren't zero, then $f$ is a rotation.
If we have equality, then $f$ is a rotation and surely has at most one zero. Otherwise, for all other $z$, $|f(z)|<|z|$ and hence they cannot be fixed.
that's cool
01:07
we are almost talking about the reflection principle
if that is what you were talking about
don't know about it yet
@Prototank why can you use that though? just because you know your map has a zero doesn't mean $f(0)=0$
yah
@FernandoMartin Tomorrow will be fun.
Topology stuff in CA right?
You know all that stuff already @Pedro
(IE I don't see how you move your fixed point to 0)
@FernandoMartin I mean Comm Alg.
01:10
Ohh
There are some details I need to work out. Suppose $z_1$ is some fixed point. Then let $\phi$ be the L.F.T. that takes $z_1\to 0$. Then $f(\phi(z_1))=0$.
I'm really oblivious to how topology gets in the picture here
I get that you use the Zariski topology
err, not quite
@FernandoMartin Jacobson's book has some explanations, I think.
01:11
but I still don't know why giving it a topological structure makes it easier to study it
Also MSE has good explanations.
@Mike, oh no I don't know what I am doing.
haha.
me neither man
If $\phi$ is the conformal self map L.F.T. that takes $z_1 \to 0$ then the map $\phi \circ f \circ \phi^{-1}$ is also a self map. It is analytic. It fixes zero. The lemma applies.... but what does that mean?
In the context of a eucalidean metric space over ℤ, the intervals [1,2] and [0.5,2.5] actually mean different things, right?
01:21
uh
not as far as I can tell @Greg
It means that $\phi \circ f \circ \phi^{-1}$ has at most one zero. Does this imply that $f$ has at most one zero?
@Prototank I bet it does
hmmm
I mean a linear fractional transformation is totes linear. It won't add zeros by composing... but that is just my intuition
Chase down those zeroes, I think
I bet it'll come out by putting pen to paper.
Enfin! A good question. Alas, numerators of the Bernoulli numbers are not easy, but a definitive answer might be possible nevertheless.
01:46
why are you apologizing to asaf?
@Pedro: when do you think you'll be around Ciudad?
02:29
@PedroTamaroff An outline of a nice way to prove that the complement of a saturated multiplicative set is a union of prime ideals is as follows:
(1) Given a multiplicative set $S$, define it's saturation $\hat S=\{x\in A: \exists y, xy\in S\}$. Show that $\hat S$ is saturated.
(2) Saturations relate to localization as follows; if $S^{-1}A=T^{-1}A$, then $T\subseteq\hat S$.
(3) The prime ideals of a localization $S^{-1}A$ are of the form $\mathfrak pS^{-1}A$ where $\mathfrak p$ is disjoint from $S$.
man
just changing my system clock doesn't let me see unicoins
@FernandoMartin All day long! =D
@Mike wtf are unicoins
I even cleared my cache.
@KarlKronenfeld I'm not sure they exist
the name sounds like bs
02:35
but when will you arrive?
@Pedro
Around 9 a.m.
Oh cool
I may have to help a friend with MATLAB at some point, so maybe we could study near the labs
I have no clue how to get there though, haha
I went there once and it's maze-like
@KarlKronenfeld Suppose $S$ is saturated. Note that if $x\notin S$, then $(x)\subseteq R\smallsetminus S$; in particular $S\cap (x)=\varnothing$ so there is a prime ideal $\mathfrak p_x$ containing $(x)$ that doesn't meet $S$. It follows $R\smallsetminus S=\bigcup\mathfrak p_x$. Guess it is similar to you idea.
@FernandoMartin Computer labs?
Yup
The architect that designed our university's building is an evil genius
> The April Fools "unicoins" prank activates based on local time, not server time Some mod on SO
@PedroTamaroff no
02:38
@KarlKronenfeld It is wrong?
I was just commenting on whether it's similar to my idea.
@KarlKronenfeld I changed my system clock and no unicoins.
@PedroTamaroff It's right though
@KarlKronenfeld Oh, OK. Phew.
@Mike Ah, I see.
02:40
I'm pissed, @KarlKronenfeld
@KarlKronenfeld I told my system I moved to Casablanca
I now have 75 unicoins.
Aha, that local
anyway, gotta go
They're... letting you buy them
ahahaha
"This padlock icon ensures that all your details are transmitted securely"
@Mike How do I do that?
@PedroTamaroff Wait 17 minutes
"Voting animations
Happy unicorn animations every time you vote. You know you want it."
um
yes please
02:44
16.
@TedShifrin You're back.
15
I am?
Hi @TedShifrin
I hope you're excited for unicoins tomorrow.
LOL ... I have no unearthly idea.
These take too long to mine.
02:47
I deny having anything to do with unic*s.
Shorter to your?
oh my god
the voting animation was worth my 99 unicoins
Resigns
"No downvotes, please
Remove the downvote arrows from all your questions and answers."
I know swome users that would like this
Hell no
Like in our culture every kid deserves an award and every student deserves an A
02:51
could anybody help me?
0
Q: KdV equation and classical linear wave equation

Poli TolstovLike we know, the standard form of KdV equation is (1) $u_{t}-6uu_{x}+u_{xxx}=0$, where this equation describes a solitary wave propagation and $u=u(x,t)$. On the other hand, we know the classical wave equation (2) $\frac{{\partial}^{2}u}{{\partial}t^{2}}-\frac{{\partial}^{2}u}{{\parti...

@TedShifrin "Guaranteed answer
No more need to wait: We guarantee you an answer on all of your questions."
Only 50 unicoins
there is nobody at Physics chat now
fix $x$ and watch the wave's amplitude vary with $t$ @PoliTolstov ... But surely you know this
@PoliTolstov As Ted said: Fix $x$... this means: Cauchy-Eulerian (spatial) velocity, not (Lagrangian) particle velocity
$u_t$ is the spatial velocity then
Yields the floor to @corn
But group velocity is one of the coolest things I've ever seen ...
03:00
what do you mean with "cauchy-eulerian (spatial) velocity"??? The only Cauchy - euler I know is an differential equation of non-constant coefficients
@TedShifrin I learned today that FTA and 1D Nullstellensatz are equivalent. Cool stuff.
@ccorn
what do you mean with "cauchy-eulerian (spatial) velocity"??? The only Cauchy - euler I know is an differential equation of non-constant coefficients
@PoliTolstov This mostly means that the $x$ in this case is a spatial coordinate. Fixing $x$ means staring at a fixed position and watching the stream through that point (or, in this case, the height $u(x,t)$ of the flood at $x$).
Then $u_t$ is the rate with which $u(x,t)$ changes at that fixed $x$
@Mike, over $\Bbb C$
@TedShifrin Right.
When I say FTA I mean over $\Bbb C$
03:08
I meant the Nullstellensatz
yes @ccorn, i understand... but what about $u_{tt}$?
$u_{tt}$: Again, the rate at which $u_t$ changes. In that sense, the acceleration of $u(x,t)$ for fixed $x$. Note that this is not a particle acceleration, just some visual acceleration of the observed height $u(x,t)$
@TedShifrin Ah, right, I thought my restriction to $\Bbb C$ was implied :P
oh, i see
You can never be too careful @Mike
03:12
@TedShifrin Well, I could stop doing math.
Hey all
@Mike i don't see no unicoins
@PedroTamaroff refresh
You wouldn't like that @Mike
03:13
@Mike refresh what'
@PoliTolstov The physics folks may well give illuminating answers, just a matter of hours :-)
@PedroTamaroff the main site!
@TedShifrin I agree - I think that would be being too careful
@ccorn ok, but thank you. it was very instructive
@PedroTamaroff THANK YOU
03:17
@PoliTolstov Hope to have helped. In any case, I'm quite confident that the physics SE question will receive feedback
@PedroTamaroff how do i work this
@Mike it's not downloading for me
sigh
@ccorn it was helpful... thanks
@Mike does the thing really use up reputation?
@PedroTamaroff I haven't tried it and don't plan to :D
It probably won't let you even use your rep - it won't let me buy coins with my visa, after all
03:28
Yas
There's a user who seems to be unhappy with me; he sent me 26 emails in 5 minutes.
19 of them consist of the subject line "
Re: Who are you I killed Obama you this bottom frogs"
that's not very nice of him
Man, I answer a question and then Matt Emerton answers it after me.
That's not even fair.
03:45
@PoliTolstov You are welcome.
04:04
Oh those comments :-) :-) :-)
Suffering is good for the soul, I suppose.
Neat: All the comments I see are now in teal comic sans.
@T.Bongers Is that readable at so small a font size?
@ccorn ctrl-+ is my friend :)
 
4 hours later…
08:04
Unicoins.
08:24
helllllllllllllooooooooooo
r9m
r9m
@usukidoll hai :D
help me D:
r9m
r9m
??
0
Q: Building a partial injective relation

usukidollQuestion : A Partial Injective Relation from $A \rightarrow B$ is maximal if its graph of an injection function from $A$ to $B$ or the graph of an injection function from $B$ to $A$. Example: $A =[a,b,c]$ and $B=[1,2,3,4]$. Build a partial injection relation. Are there 3-4 elements or a lot o...

r9m
r9m
@usukidoll this is too complicated for me .. :'( .. sorry
08:31
do you know someone who does this?
gawd I just got out of spring break and my math prof is being an evil b9999
midterm next week wtf
r9m
r9m
nope .. i dunno
D:
gawd I hate this crap
modular arthimetic would be the easiest
then the relations....posets but cardinality is driving me nuts
D:
arghhhhh my brain hurts just looking at this
spring break was amazing but damn it fuxx at the surprise midterm announcement gawdddddddddddddd why bangs head
I'm gonna study during my english class tomorrow...I got some scratch paper ..
r9m
r9m
@usukidoll so does mine ..
:(]
bangs head
:(

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