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15:14
I have a pretty stupid question. But I have the series from 1 to infinity. $$\cos^2(\frac{\pi}{n})$$ How exactly do I take the limit of this? I know that it's supposed to diverge. But I have no idea how to write it
15:32
@Link you mean what is $\lim\limits_{n\to\infty} \cos^2\left(\dfrac{\pi}{n}\right)$ ?
Hi @AlessandroCodenotti
actually $\cos (x)$ involves all the even powers of $x$
here in denominator
and $1/n^p$
diverges for $p>1$
can any one tell me is this the correct logic or approach?
@MikeMiller can you help me in above question
actually related to complex analysis I know in MSE Daniel Fischer @user123733
You may ping him and he may respond if he will be available!
@Fawad
he meant series i guess
15:41
Alright
0
Q: $[\mathbb{Q}(\sqrt{2})(\sqrt{6}) : \mathbb{Q}(\sqrt{3})]$ = ?.

BAYMAXIn order to find the value I took $F =\mathbb{Q}(\sqrt{2})$. So,$F(\sqrt{6}) = \{a + b\sqrt{6} | a,b \in \mathbb{Q}(\sqrt{2})\}$ So expanding the elements $a$ and $b$ as elements of $\mathbb{Q}(\sqrt{2})$, $= (a_{1} + b_{1}\sqrt{2}) +(a_{2}+b_{2}\sqrt{2})\sqrt{6}$ Solving a bit more gives me ...

Hey everybody!
hololulu@Daminark
How's it going?
@Daminark Hey dami
15:46
In mathematics, the method of equating the coefficients is a way of solving a functional equation of two expressions such as polynomials for a number of unknown parameters. It relies on the fact that two expressions are identical precisely when corresponding coefficients are equal for each different type of term. The method is used to bring formulas into a desired form. == Example in real fractions == Suppose we want to apply partial fraction decomposition to the expression: 1 x ( x ...
Yeah nice, how about you?
It didn't say why it works every time
Doing pretty well. Just gotta finish the last bit of a paper for TAPS, as well as reading for it, and then I can dedicate wholesale to analysis
@Fawad was that for me?
@BAYMAX can you explain?
15:48
oh no I meant were you correcting my solution I told above regarding the cos^2 thing ?sorry@Fawad
ignore. not an interesting/useful example.
@Fawad As far as I understand, they all derive from a fundamental linear independence of the terms.
It always comes down to: There's no nontrivial polynomial in $x$ which vanishes for all real $x$.
@Fawad Yes
Or in other words, a polynomial of degree n where n is non-zero has n roots.
15:51
That's what I mean, but I'm not sure how to like... go from there?
Like I have what you wrote, but I'm not exactly sure what to write next? I can't just write not zero and then say it diverges, I think
So suppose you have two polynomials. These can only be equal for all $x$ if their difference is zero. But such a difference would itself be a poolynomial in $x$.
each term in the cosine diverges ,then the series diverges , is there more to this ?@link@Semiclassical@Fawad
@Semiclassical but at same time we can put values of $x$ and get equations for coefficients. And solving those equations gives same result.
@BAYMAX I know that it diverges, I just don't know what to write, basically? How do I show that it's diverging by taking the limit?
Sure. If it's valid for all $x$, then it's valid for specific $x$.
15:55
ok
As for the divergence, it should follow from the fact that $a_n\to 1$ not $0$ as $n\to\infty$.
bullet
Hi guys
15:56
But I don't have the mental energy right now to give an actual proof of that.
how are ya? @Tim
Also, question on the screenshot: We've established that $g’(y)=0$, but didn’t we need that $g(y)=0$?
@Link $\lim\limits_{n\to\infty} \dfrac 1n =0$ so $\lim\limits_{n\to\infty}\cos^2\left(\dfrac{\pi}{n}\right)=\cos^2(0\pi)=1$
btw, $L(y)=\int_1^y\frac{1}{t}dt$
Hey @Sha!
@ShaVuklia if $g(1)$ is $0$ and $g'(y)=0$ for all $y$ what is $g$?
15:58
It might also be smart to use the identity $\cos^2 x=\frac12(1+\cos x)$.
Hiii Dami
oh right! @AlessandroCodenotti I missed that part
thanks!
@ShaVuklia Pretty good, wbu?
@Tim well, pretty good also, mainly because I just found out I did really well on a test that I had been preparing for during Easter:P (I even cancelled a family event to prepare for this test XD)
sorry, I don't want to give myself a pat on the shoulder :P or actually I just did XD ugh, never mind!
@ShaVuklia *pats shoulder *
haha, *accepts the pat *
16:04
How's it going?
well I'm doing well! you?
Busy but well, thanks!
And lol congrats on everything having worked out on the test!
haha thank you!
life is always busy as a student, isn't it:d
Hey @Ted!
And yeah definitely
Hi Demonark.
@Alessandro: Did it go well?
16:10
Hi @Ted!
Hi @Sha!
How's life?
A finite extension is algebraic
Let $K$ be a finite extension of $F$
Let $[K:F ] = n$
Let $a \in K$
Then why is $1,a,a^2,a^3,...,a^n$ are linearly dependent?
@user123733: The answer given in the problem is correct. Note that $z-z_1 = t(z_2-z_1)$, where $t$ is real. So the conjugates are related by the factor of $\bar t = t$. So the entries of the second row of the determinant are both $t$ times the entries on the first row. If you have $\left|\begin{matrix} a & b \\ ta & tb \end{matrix}\right|$, you get $0$.
@BAYMAX: What does $[K:F]$ mean?
Dimension of vectorspace $K$ over $F$@TedShifrin
?
16:14
And if you have $n+1$ vectors in an $n$-dimensional vector space?
hololulu
They're all orthogonal, clearly
they must be linearly dependent]=
smacks Demonark for rudeness
hololulu
16:15
ow
Hey @Adeek!
hi @Daminark @TedShifrin
hi Karim
You know @TedShifrin I will spend this summer as hermit at home focusing on research and I will have TA duties which I will do as hermit at home
I want to also solve some books
I'm not a big fan of hermits, but whatever ...
why ? I can finish a lot of things
16:20
hi @ted
I think it's important to interact with people ...
Hi @Semiclassic
Truth to that.
I interact with you guys hehe
MSE is a nice place to interact
3
Well, you do that too much.
16:21
haha
Well, then you won't be a total hermit!
Even though I'm overall an introvert, I know that I start going batty if I don't interact with people enough.
Plus it helps develop important life- and professional-life skills.
16:22
Which is often why the TA stuff can be kind of a problem: I interact with my students in the second half of the week, and in the first half I end up trying to work on their grading by myself.
Not a great combo.
oh
@Semi Did/do you TA for a smaller class?
I don't even want to try to count the thousands of hours I spent grading in my career.
Not this semester.
I have two sections (about 36 students total) and I meet with each of them twice a week for discussion + lab sections.
That's out of 150 students total for the entire course.
I only ever TA'd one time but there were a few of us grading since it was a slightly larger group, 30-40 students, so we often talked it out, which I found nice
Ah
16:24
I am honestly really f***ing tired of lab report grading.
LOL, @Semiclassic, as one of my old friends likes to say, "Wait 'til you're my age." :P
Plus I really haven't made time for a lot of research progress this semester.
So should I mention how many dozens of hours you spend here distracted by math problems when you could work on your own research?
Eek, yeah I imagine that's kind of annoying. At least we were grading linear algebra and combinatorics puzzles :P
You'll note that I said "haven't made time" not "haven't had time."
16:25
@Semiclassic: I duly noted. I'm being blunt.
"For your own good."
Fairly noted.
Thing is, the main reason I end up being distracted by math problems here is because I"m avoiding working on the grading.
How 'bout two hours of grading and then 3 hours of work on your own stuff?
Or maybe in the other order.
That would work if I could follow through on it.
Hello. I am reading a proof and it seems to hinge on the following fact, which I am having trouble proving: if p^a divides p^bm, where a > b, then p^a divides m. Is this statement true? How does one prove it?
Neither order works too well for me
16:28
Not true, @user193319.
If I try to grade for two hours, I decide to take a break in between that ends up being a very long break
But inevitably it becomes hours of nervously putting off the grading and then being so burnt out that I can't do much else.
^
Hmm....So is the proof of problem 7 here math.wvu.edu/~hjlai/Teaching/Math541-641/I-4.pdf
fallacious?
That's why I suggested doing your own work first, in hindsight, @Semiclassic.
And if I do stuff of my own first, then I don't want to quit after those three hours
(well, unless I really just ended on a result or got stuck)
16:28
@user193319: But it is true that $p^{a-b}|m$.
(I'm assuming $p$ is prime.)
The other problem, though, is that my patterns in this semester have led to me not being in the same building as my advisor as much as usual.
I guess it doesn't matter whether $p$ is prime or not.
And he doesn't seem to have been around much period this semester.
So less face-time/ interaction means less impetus to do research.
@Semiclassic: I found it extremely helpful to schedule a regular meeting with my adviser (it was Friday mornings at 8 am). He was very generous to do that. But I had to have stuff to talk about and he had committed to being available to me at the time.
But then there's finally just the simple fact that I've been in grad school quite a while and a big part of me just wants to move on.
16:30
@TedShifrin Yes, it is prime. In the link I gave, how are they able to conclude that that p^{k+d-b} |p^km implies gcd(p,m) \neq 1?
It's what I just said, @user193319, up there ...
Cancel the $p^k$. Then you have $m$ divisible by some power of $p$.
I don't see nor really desire, at this point, an academic career.
I don't know what career I do desire, unfortunately, but it's hard to care about research progress when the goal no longer seems to make any sense.
@TedShifrin Ah! Thanks!
Fair enough, @Semiclassic. But you still want to finish your degree after all this work ... You have more options in the "real world" with the higher degree (and computer skills), most likely.
Yeah. A big question is which degree (masters or phd).
I've got enough coursework to leave with a masters at this point.
16:33
Hi, I am working on unweighted random graphs with $K$ vertices. When $K<\infty$, using the discrete $\sigma$-algebra looks like a good option. When $K=\infty$, however, there are infinitely many possible graphs. I wonder, does using a discrete $\sigma$-algebra somehow become a problem then?
Oh, I had assumed that was true a few years ago.
Yeah.
It's been true for a while.
The thing that I just don't know is what careers are really out there. I can well believe that there's a lot of options, but
Lots of people finish math Ph.D.s and don't pursue an academic career. There are various industry jobs. I'm sure the same is true (probably more so) in physics. Especially with all the biotech going on these days.
Of course, the Orange Cheeto wants to dismantle all of science. But this too shall pass.
Yeah. I think the right rubric for me at this point is probably 'applied math' in a fairly broad sense.
I think you should look into the variety of biotech things, Semiclassic.
16:35
I'll confess, I don't really know what that means.
Start with CT scans.
Ah, so imaging stuff?
Lots of fascinating applied math.
The joys of inverse problems.
Yup.
I think computation skills are highly important, too.
16:36
Right.
One project I should make for myself is to learn how to program in Python.
Waste an hour googling and browsing ...
Any programming experience you can boast, will open up a lot of jobs
I know how to do stuff in Mathematica and Matlab.
Ah, so I was coming somewhat close to taking medical physics, someone who actually took the class said it was really fun
Hi @Ted I don't think so, I'll probably write it again in June
16:37
EdX,Coursera are nice
in Python
courses
Oh, sorry, @Alessandro :(
The tricky thing with 'programming' is that I have no knowledge of 'development.'
Ask me about unix/linux and I'll just shrug.
The language itself doesn't matter so much. More the mindset and the ideas
It's not a big deal, it was just a partial exam
(and the willingness to learn new languages/environments, of course)
16:38
Right.
The one in June will be the actual exam
They too give you certification
@Semi You may enjoy Haskell or Racket
The nice thing about python is that it's all open-source.
vs. matlab or mathematica.
(I should probably say something re: machine learning but I really dun know about that)
you tried coursera,edx @Semiclassical
?
16:41
Haven't, no.
I've been able to make do with mathematica for research computations up to now.
@Alessandro: The European system is so different ... :)
they offer nice courses and certification too .
@TedShifrin and it varies a lot among European countries as well! We get a lot of chanches for exams in Italy
It sometimes even varies quite a bit within a country
16:51
Could someone tell me how it follows that $\log_e y=L(y)$? We defined $L(y)=\int_1^y\frac{1}{t}dt$.
Because $e$ is defined by $L(e)=1$.
yes, but is that all I need?
also, we defined $E(x)$ to be the inverse of $L(y)$
Precisely.
i just stated it btw, I don't know how it helps
oh!
$e^x=y \iff \log_e(y) = x$.
16:53
I just see now that they say $\log_e$ and not $\log_b$
Is $<3>$ a maximal ideal in $Z[i]$?
Good question, @BAYMAX. Figure it out.
Hi chat
hi @Astyx
Hi @Astyx.
16:54
I'm so used to using $\ln$ for natural log that $\log_e$ looks weird to me now.
or is $Z[i]$/$<3>$ a field
@Semi same!
I've never used $\log_e$
I always used $\log$ in "real" math classes, @Semiclassic, and $\ln$ in regular calculus and applied courses.
I tend to do that as well.
16:55
Only to chemists does $\log$ mean $\log_{10}$. :P
Though I guess I use $\ln$ in the former case as well at times.
haha true :P I used that for 3 years in high school, and now $\log$ means $\ln$
In maths classes I've taken we always always use $\ln$
I thought they used lg in France.
I donot think it is a maxiaml ideal
16:56
Fun fact, my students are doing nuclear physics stuff lately. So they've got half-life stuff e.g. $\log_2$.
checks Dieudonné
Lol
Why, @BAYMAX?
@TedShifrin That sounds like an anatomical consideration, and one which shouldn't be done in polite company :P
as $Z[i]$/$<3> = \{a + ib + <3> | a,b \in Z\}$
16:57
Hello, i want to prove that in a metric space if $A$ is compact , $B$ is closed such that $A\cap B=\emptyset$ then $d(A,B)>0$. i want to prove this by contradiction if i suppose that $d(A,B)=0$ how to find a contradiction ? please
@BAYMAX: You've given me a definition. So what.
then inverse of those elements need not exist in $Z$
that is coeeficients $c,d$ may not be integers if $c + id$
Prove that, @BAYMAX (and you mean in the quotient ring).
is an inverse of $a + ib + <3>$
Better to think about it a different way, @BAYMAX.
But you can try to do that if you want.
Is the ideal prime?
16:59
If <3> is a prime ideal then
$Z[i]$/<3> is an integral domain
Now I have to check that
Don't think about quotient rings.
Think about factoring.
someone have an idea about my question please
Heya tern! I hope you're feeling better.
ish
haven't been sick a whole month
Where is my another circle? desmos.com/calculator/bregegqbie
17:01
I'm finally on the mend ... my doctor finally gave me strong antibiotics (which I hated to have to do).
But I have a dental implant next week and couldn't still be sick :(
Get well soon @TedShifrin
yes <3> is a prime ideal
@Fawad: LOL ... it's way off the page. Look at where the center is.
@Vrouvrou pick two sequences of elements, one in $A$, the other in $B$ such that the distance between them goes to $0$ and find something absurd
Thanks, @BAYMAX. I'm feeling much better today.
Glad to hear it @Ted
17:03
@TedShifrin :P thanks
Thanks, @Astyx.
Strong antibiotics implies high sleep
I hate having to use antibiotics. The world uses way too much of them.
All the beasties have become immune to them.
@TedShifrin can you help me how to get like a face ?Here till I got desmos.com/calculator/i8li67jjpx
17:05
maximal ideal implies prime ideal but not converse
generally
In the ring $\Bbb Z[i]$ they are equivalent, @BAYMAX. Why?
@Fawad: Not me.
I know that its PID,UFD,ED
but still how?
can you tell please@TedShifrin
How do you prove that $\Bbb Z/\langle p\rangle$ is a field when $p$ is prime? Can you do the same thing here for $\Bbb Z[i]$? [This is one way.]
@BAYMAX: You should be able to prove that in a PID every prime ideal is maximal.
lemme
prove the first
17:11
@MikeM: The OP that asked that question about codimension 2 submanifolds apparently doesn't really know what Euler classes are. Sigh. And he was originally concerned about the ambient space being $\Bbb R^n$ rather than a general manifold.
$\mathbb{Z}/<p>$ is an integral domain
But why a field?
and since it is isomorphic to $Z_{p}$
it is finite
and finite integral domain is a field
so I have been studying Ioannis and Shreve's Stochastic Calculus book and within 2 hours I'm so tired
I don't recommend this argument. I recommend explicitly understanding how to construct the inverse of an element.
17:14
hi @TedShifrin
hi @SoumyoB.
like $a + bp . * =1$
$a \in Z$
@TedShifrin
If I ask you to find the multiplicative inverse of $6+\langle 11\rangle$ in $\Bbb Z/\langle 11\rangle$, how do you do it?
Well, that was too easy to guess. But, in general, what computation do you do?
so if I'm living in the US, and someone tells me "Delaware", without having any knowledge of the context how do I tell if they meant the US state or the city?
the naming is so ambiguous
I don't know a city, @SoumyoB, so I opt for state.
I don't think I know any city and state with the same name, other than New York.
17:19
@TedShifrin
https://en.wikipedia.org/wiki/Delaware_City,_Delaware
Delaware City
Kansas City
"City" is part of the name of the city!
OHHH
:)
And officially, New York is really New York City, NY.
But no one calls it that.
I see, so the ambiguities aren't nearly as big as I had thought
Right.
17:21
I am not getting that now,@TedShifrin ,I only thought of $(6 + <11>).() = 1$
@BAYMAX: Think Euclidean algorithm.
@TedShifrin do you also happen to know about the Fighting Blue Hens?
LOL, no, @SoumyoB.
@Astyx thank you, i apply the caracterisation of $\inf$ i found that $$\forall n>0, \exists (a_n)\subet A, (b_n)\subset B, d(a_n,b_n)\leq \frac{1}{n}$$ As $A$ is compact there exists a convergent subsequence $(a_{n_k})$ but what to do to $(b_n)$ ? please
whenever I search for Delaware this result always pops up
17:22
inverse of 17
will think about it !
Ok till then bye all.
@BAYMAX: Hint: What is the gcd of 6 and 11?
So don't you know that Euclid says $\langle 6,11\rangle = \langle 1\rangle$, i.e., $6m+11n=1$ for some $m,n\in\Bbb Z$?
Yes
Use that to finish.
Then think about your original question.
17:27
ok
will think,bye.
@Astyx are you there ?
@Ted in the R^n case this is the more well-known result that submanifolds have Seifert surfaces.
Good point, @MikeM, although not totally trivial to prove, either.
I guess Rolfsen is a good reference for that.
Nope, not at all.
Does he do the cohomological argument or the constructive one (which works for knots)?
17:30
I no longer have the book. But I remember a cohomology argument, I think, along with the geometric one.
Ryan Budney has a blog post somewhere that has the argument.
You might need co-orientability... I don't remember.
@TedShifrin there is city of Newark in Delaware as well as a city of Newark in New Jersey
Oh that happens a lot, @SoumyoB: repeated cities is common.
this reminds of me of proofs where I have to introduce so many variables I can't name them without using unsuitable names or using tildes and bars
Given a linear transformation $L\in\hom_{\mathbb R}(V,v)$, we define $L_\beta^\beta\in\mathbb K^{n\times n}$ as the unique matrix, such that
$$
\forall v\in V:\quad co_\beta(L(v))=L_\beta^\beta co_b(v).
$$
The proof is simple (i.e., the columns of $L_\beta^\beta$ are the coordinate vectors of the images of the basis vectors), however, how do we know a priori that such a matrix exists?
17:41
why would you use $v$ for a vector space if $V$ is already a vector space. how is $L_\beta^\beta$ a matrix; wouldn't you write $L_\alpha^\beta$ or something? what does $co_\beta$ mean? what is $b$?
$co_\beta$ is the coordinate mapping w.r.t. basis $\beta$ of $V$
so we assume the same basis for the domain and codomain, hence $L_\beta^\beta$
ohhh right, the $v$
sorry, that was a typo
it should say $L\in\hom_{\mathbb R}(V,V)$
@Vrouvrou what can you tell of $d(a, b_n)$ where $a$ is the limit of $a_n$ ?
=0
?
@Astyx
Not =
but $d(a_n, b_n)\leq \frac1n$
17:48
@arctictern I'll post it on the main site btw, so I can leave it for what it is for now :P
mind just went blank... $f(x+a)$ suggests a horizontal translation of $-a$ but why does the intersection for $y=ln(4-x)$ have an asymotope at $x=4$??
dont know why i started talking about intersections
but anyway... any help?
graph*
i figured it out
damn
so simple
18:06
Not really @Vrouvrou
18:33
[Function growth rate]
It seems for BEAF Rule 2 is a special case of Rule 3
Given we knew the following:
18:45
1. Same as hyperoperations $H_n(a,b)$. ("n-tation)"
$a\{1\}b=a^b$
$a\{n\}b=a\{n-1\}(a\{n\}b-1)$

2. Multiexpansion (hyperoperation on itself with the rank/index obtained b times)
$a\{\{1\}\}b=a\{(a\{()\}b-1)\}$
$a\{c\}^d b=a\{c-1\}^d (a\{\}^d (b-1))$, where $d $counts the number of curly brackets/expansions

3. Linear notation
$a\{c\}^d b=\{a,b,c,d\}$
Now consider:
Sorry, slight typo:

$a\{\{1\}\}b=a\{(a\{a\}b-1)\} 1$
$\{a,b,c,d\}=a\{c\}^d b= a \{c-1\}^d (a\{c\}^d (b-1))$

Now rewrite in linear notation:

$a \{c-1\}^d (a\{c\}^d (b-1)) \equiv \{a,a,\{a, b-1,c,d\},c-1,d\}$
Now set $c=1$ and compared with the index notation, we found:
$\{a,stuff,0,d\}=\{a,a,stuff,d-1\}$
Therefore rule 2 is a special case of rule 3
Will see whether this observation holds for all other cases, including the multidimension arrays
Simple art's latest question on the ordinal collapsing function arose my interest on understanding how fast the fixed points for recursive functions grow, in particular, in the transfinite domain
As discussed with him earlier, there are different "Tiers" (precise term to be found later) where the limit ordinals are located.
I am interested in a function that evaluates the first member of each Tier

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