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00:00 - 16:0016:00 - 00:00

12:02 AM
@user193319: Did you sort out the $\infty$ stuff?
 
12:19 AM
Yes, I did. Thanks for reminding me of the definition.
 
12:44 AM
Oh, cool. Great.
 
@quallenjäger Heidelberg. Why do you ask?
 
Im wondering because you say its 1 hour away from heidelberg
I live 1 hour away from heidelberg as well
 
ah. In which direction? A small town near Heilbronn, you probably don't know it
 
hi @quallenjäger
 
@TedShifrin Ted! What a surprise
Haven't see you for a while
 
12:49 AM
Hardly a surprise :) But you're awake in the middle of the night/morning.
I had visitors here for 4 days.
 
@MatheinBoulomenos I am from Worms
You know it?
@TedShifrin I see, well, I am still struggling with my thesis
putting the final polish together.
 
Well, yes, thesis struggle leads to sleeplessness.
 
By the way, one thing I need to be confirmed is that, suppose I have a Lie-series
Which is a formal series of Lie-brackets
 
I've nearly got 100 upvotes! math.stackexchange.com/q/728069/104041
 
what happens if I exponentiate it, does it give a Lie group structure?
 
12:51 AM
@quallenjäger ah, I've been in Worms before! you have a medieval market
 
@MatheinBoulomenos Haha yes, a lot of people coming here for that
The Lie series clearly forms a Lie algebra structure and we can use the BCH formula to define partial group strucutre.
Near the origin
What happens if we stick these neighbourhood together?
 
@quallenjäger: I don't totally understand your question. Series means sequence?
I don't know what you mean by sticking neighborhoods together. You only have things on a neighborhood of the origin.
 
$E\oplus [E,E]\oplus [E[E,E]]\oplus......$
Is some element from this, but infinite series.
 
Oh ... hell, I don't know.
 
The normal free-lie algebra are the Lie polynomials, which is finite series
So I am basically interested in the completion, which are the formal series. I don't understand why the Lie's third theorem breaks down in this case.
 
12:55 AM
Any time you exponentiate a Lie algebra, you get a Lie group. Isn't that thing a Lie algebra? I guess the infinite sums cause issues. But direct sum should mean you only take a finite sum ...
Oh, there may be some theorem about dealing with completions, but I don't know it.
 
Do you know any sources to look for that?
 
Nope. This would be a reasonable question to post on main.
 
Ok thanks for advice.
I have such a headache from the thesis.
 
You don't want to hear about my sleepless nights with my Ph.D. thesis many years ago.
 
I can imagine that.
MIT must be level harder
Also I have a huge problem with the language.
 
12:59 AM
Well, Ph.D. was Berkeley, but people have sleeplessness with Ph.D.'s everywhere.
 
I have a picture in the mind, I cannot describe in word unless I draw it.
 
LOL
 
sleepless nights have been quite productive for me in the past
 
What did you do at MIT actually if I may ask?
For me as well, but I noticed, my brain has aged too much since I worked in the night.
I forget a lot of things in the daily lifes.
Like I lost already 3 wallets 2 sets of keys and 1 mobile phone
lol the numbers are such nice sequence
 
I was an undergraduate at MIT and then a postdoc there (taught a few advanced courses and two 350-person multivariable calculus lectures).
 
1:03 AM
3.2.1
 
Yikes, that's not good. You're too absent-minded ...
 
@quallenjäger that's worrying
 
Das stimmt, @Mathein.
 
I used to think in the train and in the bus and everywhere, then it is easy that you are not really focused on such "small" things.
I can't count how many times I missed the bus stops.
 
I've never been absent-minded.
You should never ride a bicycle or drive a car when you think, @quallenjäger
 
1:05 AM
Haha I never do
I used to drive cars in germany.
 
Before you learned to think?
 
But now in London I dont really need cars or bicycle
 
Thank goodness. Or we'd be scraping you off the pavement.
 
haha true
 
@TedShifrin that's the reason why I like walking so much. Thinking while walking is not that dangerous (of course you shouldn't be completely absent-minded)
 
1:06 AM
But isn't it usual? A lot people around me happens the same.
 
@Mathein: I see people in the US distracted by texting on their phones and walking right into moving cars in traffic.
 
good morning all, wonderful day ahead.
 
OK, night, guys. I'm going to cook dinner and watch some tennis.
 
@TedShifrin I don't text while I walk. I try to observe my environment, but thinking while doing so
 
LOL, ok, @Mathein :P
it's evening here, @Nick, but OK :P
 
1:07 AM
@TedShifrin enjoy your dinner
 
Vielen Dank.
 
@quallenjäger losing that much strikes me as a bit unusual
 
@TedShifrin Guten appetit.
@MatheinBoulomenos It accumulates over the year.
 
I see. I've somehow managed to be a bit dreamy without losing things
 
Lucky you.
What is actually the character group of the Hopf algebra?
 
1:12 AM
@quallenjäger I have no idea
 
What is general a character group?
 
I know what the character group of a group is, it's just the group of homomorphism to $\Bbb C^\times$, but that probably doesn't help you with hopf algebras
 
What is $\Bbb C^x$?
 
non-zero complex numbers with multiplication as a group operation
I use $^\times$ for the unit group of a ring
 
I see.
What goes wrong with Hopf algebra?
 
1:19 AM
I don't think the set of homomorphisms between two hopf algebras typically forms a group
 
2:01 AM
I'm trying to establish a base case for an induction problem $\forall n \in \mathcal{N}, (n+1)\cdot(n+2)\cdots 2n = 2^n\cdot 1\cdot 3\cdot 5\cdots (2n-1)$. How would I interpret the case $n=0$? Would it just be $2\cdot0 = 2^0$ (which isn't true)?
 
2:42 AM
Sometimes zero is not included in the naturals.
Was the n in N bit in the problem statement, or did you frame it that way?
 
Problem.
 
@Li357
 
It's awkward because other questions explicitly mention $n\geq 1$
Which would kind of imply $n \in \mathcal N$ includes 0?
 
if it doesnt work for n=0 just prove it for where it holds
 
So then I guess $\mathcal N$ excludes 0 since it holds for 1.
 
2:46 AM
it depends on the book
some do some dont
most normal people just use $\mathbb Z_{+}$
 
Not from a book. Also, I heard that if you're in the US, it's usually that natural numbers include 0
 
depends on the book
thats most people use the notation i used to prevent confusion
 
Yeah, I'll just ask my prof
 
In my experience it can vary from one lecturer to another in the same set of courses.
 
there testing you on induction anyway its not important
 
2:48 AM
It's just preferences.
 
i include 0 personally
but later on in math you just define what your using or just use what makes sense
 
I'm neutral, maybe if I learn more maths/teach I'll have preference
 
sometimes its useful to have 0 sometimes its useful not to have depends on the subject
if you use algerbraic axioms better not to have the 0
analysis better to have it
 
In my analysis course didn't have N didn't contain $0$.
 
Okay, just shot my prof an email. Thanks
 
2:52 AM
But there was a long preface explaining how it doesn't matter etc
@Li357 Well, that escalated quickly lol.
 
@Ryan its better to have it but it can be done without it
 
Heh
 
@Ryan What about them?
 
 
1 hour later…
Zee
4:16 AM
Ya whatever
 
Does anyone have experience with sympy?
I'm trying to bridge a number of math concepts at once, and trying to do it with sympy
currently, I'm representing the products of finite k-algebras as vectors (or "lists" in python) of multivariate polynomials
 
If I have a sequence $S = f(c_1S_1 - c_2S_2)$ such that $f(x) = \begin{cases} x & x \ge 0 \\ \frac x 2 & x < 0 \end{cases}$, is there a term for this type of sequence?
$S_1$ and $S_2$ are other sequences.
Or at least is there a term for the $f$ function? I've seen it before used as a transfer function in a neural network, but if it has a name, that would help out.
 
4:58 AM
@TedShifrin For some reason, I always picture you as Brad Sherwood from Whose Line.
 
I think of him as more of a Colin
 
(fyi I decided to ask my question in the sympy gitter channel)
 
5:40 AM
how many are active now
 
 
1 hour later…
6:52 AM
None.
 
 
1 hour later…
8:05 AM
@TedShifrin should $\nabla f$ be a column vector (an element of the domain) or a row vector (i.e. an element of the dual of the domain)?
 
8:40 AM
New 3B1B drops tomorrow
Quaternions
(What's $1/(i+j)$?)
 
9:01 AM
that’s (-i-j)/2 right
 
9:26 AM
Hello guys.
 
The subset {5, 1-2i} of R-vector space C is linearly independent because 5a+b(1-2i)=0, so (5a+b)-2bi=0 holds only when a=b=0 because a and b must be reals, right?
 
@MaryStar Yes
 
I have to choose electives for my course and "advanced number theory" and "algebraic number theory" both are open. Could you suggest which should i take? I like number theory in general, but i am not sure which should i take. Advaned N.T comprises of fermat, wilson theorms, congruence theory, dirchlet prime number theorom etc... whereas Algebraic N.T consists of algebraic number fields, nonetherian rings,
dedekind domians, dirchlet unit theorom etc. I have some exposure to advanced N.T but almost none to Algebraic N.T. Which would be better, like, if i were to do Phd in N.T (say).
 
But if we have the subset {1, 2+i} of C-vector space C then this is linearly dependent because a und b can be complex, right? @TobiasKildetoft
 
right
@Shobhit Which Fermat and which Wilson?
 
9:33 AM
the usual ones
 
And what do they mean by congruence theory?
Those topics don't sound advanced at all, except for Dirichlet (which usually is a part of analytic number theory)
 
i agree. thats what i thought too
these topics are generally well known
why would they name it advanced N.T
but it is what it is
so i should take algebraic N.T?
i have no knowlede of the topics listed under this course
 
It does sound weird. I mean, Wilson's theorem is one I give as an exercise in the third week of Algebra, and Fermat is a special case of Euler which is covered in week 2.
 
yes, i did it in my olympiad prep in class 10th
 
Algebraic number theory is certainly much more advanced than those topics, so you would probably learn a lot more from that
 
9:38 AM
ok. Any prerequisites?
 
Depending on the way it is structured, probably the basics of algebra up to some Galois theory
and of course a solid grounding in linear algebra
 
How can we check if the operator L of the C-vector space C^2 defined by $L(z_1, z_2)=((1+i)z_1+(2-i)z_2, (2+i)z_1+3z_2)$ is hermitian? @TobiasKildetoft
 
@MaryStar First write up what that means
 
Do we have to check if $\langle L(z_1, z_2), (w_1, w_2)\rangle = \langle (z_1, z_2), L(w_1, w_2)\rangle$ ? @TobiasKildetoft
 
Well, is that what hermitian means?
 
9:46 AM
Isn't A hermitian if $\langle Ax, y\rangle = \langle x, Ay\rangle $ ? @TobiasKildetoft
 
@MaryStar You are the one with the exercise and the accompanying book of definitions
 
@LeakyNun Yeah
I was hoping someone would fall into the "trap" of$$\frac1{i+j}=\frac1{i+j}\cdot\frac{i-j}{i-j}= \frac{i-j}{i^2-j^2}=\frac{i-j}0$$
Probably a bit mean of me
 
ij=-ji
 
I love how nicely $(ai+bj+ck)^2$ turns out
 
@AkivaWeinberger lol you got me for a second
 
9:59 AM
I'm a bit confused by the definition of measurable. Why is it chosen to be such that it's measurable if one can remove parts of the set to make it arbitrarily small? How does that exclude bad sets?
The Cantor set isn't measurable, right? What is it with this that doesn't make it measurable according to this definition?
 
Cantor set is measurable
Measure 0
 
Okay. Hm... I don't get why it has measure zero.
 
Draw it, see if you can cover it with open intervals whose sum is arbitrarily small
Alternatively, look at the measure of its complement in $[0,1]$
 
Isn't the Cantor set of zero cardinality?
Given that $$C:=\bigcap_{n=1}^\infty C_n$$
 
@OskarTegby What? that is nonsense
 
10:13 AM
Okay. Sorry! I'm just trying to learn.
 
@OskarTegby cardinality does not change meaning from the usual one in measure theory
 
Cardinality is the number of points in it
The only set with zero cardinality is the empty set
 
I was just thinking that none of the sets have any points in common, and if you take the intersection you should get zero.
By axiom, yes.
 
What is $C_n$?
$C_n$ is the $n$th row in that picture, yes?^
 
Yeah!
Exactly.
 
10:16 AM
Note that $\frac13$ is in all of them (it's the endpoint of an interval in each of them)
 
Yes.
 
Note also that $\frac14$ is in all of them (though it's not the endpoint of any interval in $C_n$)
 
Ah! Silly me. Thus, it's not empty. I was looking for something that isn't measurable to see why the definition is a good one.
 
Have you heard of the Vitali set?
Any nonmeasurable set is extremely weird
In fact, you need something called the Axiom of Choice in order to prove that nonmeasurable sets exist at all
 
No.
Okay.
Maybe my approach for understanding the definition is a bad one.
It just doesn't seem that clear why being able to make the set arbitrarily small would be important here.
 
10:23 AM
What exactly is your definition of a measurable set?
 
"A subset $E$ of $\Bbb{R}^n$ is called Lebesgue measurable if for any $\epsilon>0$ there exists an open set $\mathcal{O}$ where $E\subset\mathcal{O}$ such that $m_\ast(\mathcal{O}-E)\leq\epsilon.$
Where $m_\ast(E)$ is the exterior measure of $E$.
 
Take $[0,1]$ as an example. Why is it measurable according to that definition?
 
Hello!!!
Let $W$ the subspace of $\mathbb{R}^3$ that is spanned by $w_1=(1,-1,0)$, $w_2=(1,1,2)$ and $p=(x,y,z)$ the projection of the vector $v=(-1,1,2)$ onto $W$. What is the number $2x+7y+3z$ equal to?

I got that

$$(x,y,z)=(-1,1,2)- \left( \frac{(-1,1,2) \cdot (1,-1,0)}{(1,-1,0) \cdot (1,-1,0)}(1,-1,0)+\frac{(-1,1,2) \cdot (1,1,2)}{(1,1,2) \cdot (1,1,2)}(1,1,2)\right)=\left( -\frac{2}{3},-\frac{2}{3}, \frac{2}{3}\right)$$

Then $2x+7y+3z=-4$. But this isn't a possible answer.... Have I done something wrong?
 
I don't think you want $(-1,1,2)-{}$ at the start there
That would give you the component of $v$ perpendicular to $W$ I think
@OskarTegby Basically, the exterior measure gives you an upper bound on what the measure should be. The criterion says it's also a lower bound.
(Namely, the limit of $m(\mathcal O)$ as $\epsilon\to0$)
 
11:00 AM
I see @AkivaWeinberger Thanks
 
11:53 AM
@AkivaWeinberger: Ahh! I get it now. I'll look at the Vitali set in a minute. Then I guess that it'll feel more reasonable.
 
@TobiasKildetoft this is how it is structured. As you can see no prerequisites are given, but, i know none of the things listed.
 
12:13 PM
hi
 
hi
 
12:25 PM
how are you?
 
@Shobhit another algebraic number theorist :D
 
fine thanks, how are you?
 
XD
have you studied algebraic N.T? @LEA
@lea
 
by myself, a bit
 
@LeakyNun
oh
 
12:27 PM
yea i get it
 
any prerequisites?
 
you might want to know some galois theory
(that includes field theory)
 
yeah, just googled
Thats a start, anything else?
commutative algebra?
 
linear algebra
I don't think general commutative algebra is very much needed, since our rings are nice
but you might want to look at Atiyah-Macdonald Ch. 5 (that's commutative algebra)
 
Commutative algebra is also open for us, so i was thinking of taking both of these.
 
12:34 PM
@Shobhit how much do you know?
 
about? the topics listed in algebraic N.T?
 
that, and the prerequisites
 
nothing and nothing, BUT, i realllly want to know more number theory >_<
 
do they offer elementary number theory?
 
not to us, there is a course called advanced number theory though, but i know almost all of the things listed under it, wait i'll post a picture
@LeakyNun idk why are they calling it advanced -_-
 
12:44 PM
that's really advanced, you know
 
are you serious?
only the last bits
 
not at all
 
oh ok XD
 
 
2 hours later…
2:16 PM
Can someone mention some book, or lecture note or link ... that defines $a^p$ (where $a>0$ and $p$ any real), and proves properties like $(a^p)^q=a^{pq}$ or $0\le a\le b\implies a^p\le b^p$?
I know that Rudin in PMA defines, and asks some properties as exercise, but he uses many more in his text.
 
hi, can someone help me with this:
How do I find : S=∑k^2 * (n-k)C(n-r) from k=1 to k=r
2
 
n-k choose n-r?
combinations?
 
2:59 PM
Hi @Mathei
 
Ciao @Alessandro
 
I don't quite understand what's a local uniformizing parameter when talking about completions of nonarchimedean fields
 
not sure about the added "local", but a uniformizing parameter (also called uniformizer) is just an irreducible/prime element in a DVR
note that in a DVR, there's only one irreducible element up to a unit
 
More specifically I'm looking at chapter 7, section "Completions in the nonarchimedean case" of Milne's ANT notes. He writes "Let $|\cdot|$ be a discrete nonarchimedean absolute value on $K$, and let $\pi$ be an element of $K$ with
largest value $<1$ (therefore $\pi$ generates the maximal ideal $\mathfrak m$ in the valuation ring $A$). Such a $\pi$ is called a local uniformizing parameter."
 
in terms of the valuation, it's just an element of valuation $1$ (if your valuation is normalized, otherwise minimal non-zero valuation)
okay, that seems like a fine definition, what are you unsure about? why it generates the maximal ideal of the valuation ring?
 
3:05 PM
Hmm it seems to be of maximal smaller than 1 valuation here
@MatheinBoulomenos I wasn't parsing the "largest value" correctly, but now it suddenly makes sense!
I'm still not sure why it generates $\mathfrak m$ though
 
forget that, I was thinking about the additive valuation, (i.e. $v(ab)=v(a)+v(b)$), while Milne uses "absolute values" aka "multiplicative valuation"
$\mathfrak{m}=\{x \in K \mid |x|<1\}$. Let $y \in \mathfrak{m}$, $y \neq 0$, then consider $y^{-1} \pi$
 
@MatheinBoulomenos Oh, yes, he also talks about additive valuations and the relationship between the two earlier, but he uses multiplicative ones
 
Help?
1
Q: Describing shapes independent of a co-ordinate system?

More AnonymousIdea Let's try to express objects in terms of their line elements and line element vector rather than the co-ordinate system. Let the object be a line in three dimensions: Line At the origin we have: $$ d \vec s|_{(0,0,0)} = dx \hat i + dy \hat j + dz \hat k $$ $$ d \vec s|_{(a,a,a)} = dx ...

 
@Alessandro sorry, wrong hint
you should consider $y\pi^{-1}$ instead
 
Ok, so I want to show that $|y\pi^{-1}|<1$
 
3:16 PM
start from $|\pi| \geq |y|$
 
ln that case I have $|y\pi^{-1}|=|y||\pi^{-1}|\leq|\pi||\pi^{-1}|=1$
In the other case I have $|\pi^{-1}|>|y^{-1}|$ and I still obtain $|y\pi^{-1}|<1$
 
but you always have $|\pi| \geq |y|$, you don't even need a case distinction
 
Oh, right, $\pi$ is chosen of maximal norm in $\mathfrak m$
 
Hello! I need help regarding a probability question
Find the probability that the square of any integer has last digit as 1,5
I dont know what will be total number of outcomes will it be all digits from 0 to 9 or will it be all probable digits which can be the last digit of a number
Please someone see this
 
@Alessandro si manga pasta con salsa di vino bianco nel Italia?
not that I'm worried about cooking authentically Italian
 
3:31 PM
The last digit of squared integer is just depending on the last digit of original integer.
 
@ChoMedit so the probability will be 1/2 right?
But the answer is given 3/10
 
How could you get that result?
 
@MatheinBoulomenos Mangia*, in Italia*
And no, I've never heard such a recipe
 
@ChoMedit favourable otcomes when the 1,5 is 2 and total outcomes 0,1,4,5,6,9 are 6
 
oh okay
thanks
 
3:35 PM
Oh, by the way, it seems that I found a dorm room for Bonn! (but just for the first term, this is still pretty good though, finding a flatshare should be much easier when I know the city, people and can actually visit the flats)
 
nice, congrats!
 
Did you make that pasta with goat cheese in the end?
 
@ChoMedit favourable otcomes when the 1,5 is 2 and total outcomes 0,1,4,5,6,9 are 6 so 1/3 and not 1/2
 
@Jasmine But your entire picking is occurred in 0, 1, 2, 3, ..., 9. You should count in 0...9.
 
no, I think I'll just eat the chèvre on bread and make the pasta in a less experimental way
 
3:36 PM
I like this more cautious approach
 
current plan is fresh linguine with chicken, carrots, zucchini, fresh tomatoes, marjoram and white wine sauce
 
We add wine to some dishes (especially when making a risotto) at the beginning of the cooking process but it's supposed to evaporate and just leave some flavour
 
oh I cooked risotto Milano before, I love that dish
 
You choose the case of "the last digit of squared integer is 1, 5" from the case of "integer of any last digit"
 
@ChoMedit i am comfused
 
3:40 PM
Think about this. $\{(0:0), (1:1), (2:4), (3:9), (4:6), (5:5), (6:6), (7:9), (8:4), (9:1)\}$
(original last digit : the last digit of the squared)
 
*Milanese
 
There are a lot of ways to make a risotto. I especially like the recipes using asparaguses
 
hmm, I should try that some day, sounds good
 
It does
 
Risotto with asparaguses and scamorza is great in my opinion
 
3:45 PM
@ChoMedit so you mean total number of outcomes will be 10 that is 0,1,4,9,6,5,6,9,4,1
It means you are counting 1,6,9,4 two times
I dont get this
Please someone reexplain
 
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