« first day (3189 days earlier)      last day (468 days later) » 

12:06 AM
Is anyone even here?
2 hours later…
2:05 AM
Can I have some feedback?
2:27 AM
Hey, just looking for feedback
@thank you @MartinSleziak
@MartinSleziak but it's actually more general in that I was looking for feedback on all my questions
2:46 AM
@Ultradark Feedback on what?
Q: Generalising the Dirichlet L-Series

TheSimpliFireAn MSE user and myself were discussing the Dirichlet L-function from one of their posts when we thought of a generalisation to the Riemann Hypothesis. The Dirichlet L-series is defined by $$L(s,\chi)=\sum_{n=1}^\infty\frac{\chi(n)}{n^s}=\sum_{n=1}^\infty\frac{\Re\chi(n)+i\Im\chi(n)}{n^s}$$ and $...

Feedback as in you want answers, or want to improve the question?
3:41 AM
There are only countably many elements in $\Bbb F_2[x]$, because there are finitely many points of degree n, and hence $\Bbb F_2[x]$ is countable union of finite sets. Is this correct?
@user76284 improve the question or answers
@Silent Yes.
4:11 AM
do any traveling-wave solutions of the diffusion equation $a^2u_{xx}=u_t$ exist?
5:03 AM
math.stackexchange.com/a/3202098/592227 What am I supposed to conclude about $g$? If I know something particular about the z_n (like maybe z_n = 1/n), I could use the identity theorem to explain that g(z) = z. but I don't think that is the intended strategy here?
6 hours later…
10:45 AM
@loch hi
11:13 AM
Can a random variable be used to model a quantity that changes over time? I know of differential equations that are used to model such, but can random variables model such behaviour?
11:26 AM
Hi! If $\mu \models T \cup \{\sigma\}$, then does that follow that $\mu \models T$ ?
@LeylaAlkan how is $\mu \vDash T \cup \{\sigma\}$ defined?
If $\mu \models T$, we say $\mu$ is a model of $ T$
Actually that is a curly sort of the letter M, I dont know why I used $\mu$ instead hahaha
11:50 AM
So, there are four finite rings of order $pq$ where $p$ and $q$ are prime not equal to each other. Does anyone know what the multiplicative rules for these four rings look like? Can they be expressed easily?
I know one is trivial multiplication, but that's trivial.
Well, actually, trivial multiplication accounts for two rings. One where the underlying group is $\mathbb{Z}/pq\mathbb{Z}$ and one where the underlying group is $\mathbb{Z}/p\mathbb{Z}\oplus\mathbb{Z}/q\mathbb{Z}$
@leakynun hi
12:12 PM
@loch what did you fill in for the point of contact?
Let $H$ be a subgroup of $G$.Then $aH=bH$ iff $a^{-1}b\in H$, but we can't say $aH=bH$ iff $ba^{-1}\in H$ (where $a,b\in G$), right?
I believe that might require $H$ to be normal.
Because you want $Ha^{-1}=a^{-1}H$
Which, unless I'm remembering something very incorrectly, only happens when $H$ is normal.
@Rithaniel How did you get this?
It is a memory from a lecture. I can acquire a more hard reference, if you'd like.
12:27 PM
no :)
Fair enough.
12:44 PM
@LeakyNun what is this for? visa?
1:06 PM
@loch right
I forgot. Have you received your i20?
1:17 PM
what is i20? is it DS-2019?
hmm no it's a form that MIT mails to you
1:30 PM
yeah it's DS-2019
in any case it's pending
uh anyway
I think once you receive the form there'll be a name on the form which you can put as your point of contact
1:47 PM
@LeakyNun @loch Guys I have few calculus questions..are you free?
How do I solve question 18?
2:06 PM
Use the inequality to show $f$ is continuous
@loch I was wondering do you know what you should have in order to do postdoc in big name universities ?
MIT, Princeton, Harvard, etc.
2:18 PM
@Thorgott How do I do that?
It's uniformly continuous even. The definition of uniform continuity is that for every $\varepsilon>0$, there is a $\delta>0$, such that for all $x,y\in\mathbb{R}$, we have $|f(x)-f(y)|<\varepsilon$ whenever $|x-y|<\delta$. You know that $|f(x)-f(y)|\le|x-y|^3$. What $\delta$ suffices?
why is e^(1/e) the greatest real number that can be written in the form of x^(1/x)?
any help?
@Thorgott Could you be a bit more elaborate, I didnt understand anything!
2:38 PM
Well, what is your definition of continuity?
@Thorgott small change in input results in small change in output
Did you not formalize that notion?
@Mathphile do the first derivative and equate it to zero.you get x=e..find second derivative and substiutue 'e' answer turns out to be negative...so x=e gives you maximum value of x^1/x.....I dont think I answered your question..but ..did you find it useful
@Thorgott Nope,I only this much and I know how to deal with continuity at a point(L.HL=R.H.L=f(c))
@Thorgott I just read wikipedia page on coninuous function...and I now I understand your statement..please continue your explanation
Well, $|f(x)-f(y)|\le|x-y|^3$ gives you a direct bound on how the change in output changes with a change in input. Formally, you now take an arbitrary $\varepsilon>0$ and want to find a $\delta>0$, such that $|x-y|<\delta$ implies $|f(x)-f(y)|<\varepsilon$. What choice of $\delta$ works out in light of the above inequality?
2:56 PM
Hello iditos
@Thorgott Idk
3:12 PM
In hindsight, the four rings or order $pq$ (where $p$ and $q$ are unequal primes) were less interesting than I expected them to be.
I have a little linear algebra question here,
if I have understood my task correctly.
Any help is appreciated.

Let $\{f_i | f \in \mathbb{N} \}$.
$f_i \in Map(\mathbb{N}, \mathbb{Q})$ is defined as:

$f_i(n)= -n \quad\forall \quad n \geq i$

$ f_i(n)=0 \quad\forall \quad n < i$

Question: Is that linearly independent?

I think, that it is lineare dependent.

for $a \cdot f_i(n) + b \cdot f_j(n) = 0 \equiv $a \cdot 0 + b \cdot 0 = 0$ if $n < i < j \quad i,j,n \in N $
And thus not always follows that for a \cdot f_i(n) + b \cdot f_j(n) = 0 $ there must be $a=0, b=0$.
3:57 PM
@Thorgott ???
4:21 PM
Does anyone know where I can find a proof that $||f||_\infty = \sup \{M > 0 \mid m(\{x \in E \mid |f(x)| > M \}) > 0 \}$?
4:49 PM
if vectors $\mathbb{u},\mathbb{v},\mathbb{w}$ are linearly independent, then can we write $$\mathbb{u}\times(\mathbb{v}\times\mathbb{w})=a\mathbb{v}+b\mathbb{w}$$ with some scalars $a,b$?
i think that such scalars do exist, because $\mathbb{v}\times\mathbb{w}$ is perpendicular to the plane that is spanned by $\mathbb{v}$ and $\mathbb{w}$ and thus $\mathbb{u}\times(\mathbb{v}\times\mathbb{w})$ must lie in that plane
also as all three vectors are linearly independent then their cross products cannot be equal to the zero vector
are my arguments correct? or am i missing something?
5:03 PM
@Rithaniel that happens a lot lol (to all of us)
Often times something that initially seems interesting turns out to be totally uninteresting
hi @lurkingnun
5:40 PM
For every irreducible cubic polynomial $f(X) ∈ \Bbb Q[X]$, there exists a subfield $F$ of $\Bbb C$ such that $F \not\subset \Bbb R$ and $F \cong \Bbb Q[X]/f(X)$. How to see that this statement is false?
@Silent because of totally real cubic fields
it coincides with the casus irreducibilis
looking into it
an example should be $\cos\left(\dfrac\pi9\right)$, $\cos\left(\dfrac{4\pi}9\right)$, $\cos\left(\dfrac{7\pi}9\right)$ being conjugates
conjugates? in real line?
they are solutions of $\cos(3\arccos x) = \cos\left(\dfrac\pi3\right)$
which is a cubic equation
(the left hand side is a Tschebyschev polynomial)
5:48 PM
thank you. trying to understand this...
everything links together so beautifully
do you see why the equation is cubic?
@Semiclassical hey
do you want a joke?
vector calculus in spherical coordinates
5:49 PM
how do you even remember the divergence
Yeah, it’s a bit goofy.
and like the basis vectors aren't really independent?
are they
why does the divergence and the gradient look different?
There’s a reason why the back cover (or is it front? I forget) of Griffiths Electrodynamics book—which is the standard upper-undergrad level text—includes all the formulas and conversions between Cartesian/cylindrical/spherical coordinates in vector calc
Because while they expect you to know how to use those formulae, they don’t expect you to memorize them
First page is what appears on the inside of the front cover and on the first side of the first first page
oh no
the radial component of the laplacian is different in 2D and 3D...
5:55 PM
Second page in the pdf is what appears on the back side of the last page in Griffiths and on the inside of the back cover
Believe me, you use those front/back cover formulae a lot if you do E&M
let's say i want to derive them
yeah, that’s not fun
I mean, the expressions in Cartesian coordinates are easy enough
And in principle it’s just a matter of using the multivariable chain rule
oh hey @ÉricoMeloSilva
But it’s pretty miserable
i aint here
5:59 PM
is there really no easy way to derive them
like the intermediate steps are really long
I -think- you can derive them a bit more nicely if you work in terms of differential forms
terms happen to cancel out each other
I'm not even sure how 1-forms correspond to vector fields in polar coordinates
@leaky you can derive the coordinate rep of the laplacian really easy for any coordinates if u know the representation of the metric
$\langle \nabla u, \nabla \varphi \rangle = g^{ij} \frac{\partial u}{partial x^{i}} \frac{\partial \varphi}{\partial x^{j}}$ where $\varphi$ is a test function, just integrate by parts and u get the coordinate rep of the laplacian for a general metric $g^{ij}$ on any riemannian manifold
6:03 PM
note u get divergence on general riemannian mflds the exact same way
note this is in fact, a very easy computation
would you mind demonstrating?
$\int_{U} \langle \nabla u, \nabla \varphi \rangle dV = \int_{U} g^{ij} \frac{\partial u}{\partial x^{i}} \frac{\partial \varphi}{\partial x^{j}} \sqrt{\det g} dx$
$U$ a coordinate chart, now ibp off $\varphi$, and you get it immediately
namely: $-\int_{U} \frac{1}{\sqrt{ \det g}}\frac{\partial}{\partial x^{j}} \left( g^{ij} \sqrt{\det g}\frac{\partial u}{\partial x^{i}} \right) \varphi dV$
so $\Delta u = \frac{1}{\sqrt{ \det g}}\frac{\partial}{\partial x^{j}} \left( g^{ij} \sqrt{\det g}\frac{\partial u}{\partial x^{i}} \right)$ and you're done, if you know $g_{ij}$ you can find the coordinate exp for whatever u want.
Spherical/cylindrical/Cartesian are all orthogonal coordinates systems, so the metric tensor is diagonal in that case
this is what i do on my e&m exams cause i never remember these expressions lol
6:15 PM
Hence why the Laplacian in those cases doesn’t contain any mixed derivatives
@ÉricoMeloSilva in ours you’d be given the formula sheets
since remembering those is a colossal arse
idk why we dont get them in my 2 advanced e&m classes, once you figure out what spherical integral you're supposed to do, it's not even physics anymore really
does analytic continuation usually start with a power series
no, cf. zeta
I mean, there are some cases which you should be able to know either off the top of your head or be able to derive from memory
6:19 PM
@Semiclassical the curl in spherical coordinates is not one of them
Oh god no
all the examples i've seen start with a power series
I just gave you one that doesn't
I have seen that
What I have in mind is the radial part of the spherical/cylindrical Laplacian
That’s not so bad iirc
6:22 PM
can we derive divergence from $\lim_{R\to0} \frac1{|R|} \int_{\partial R} F \cdot dS$
But that’s an exception.
@LeakyNun this limit is really annoying to calculate in bad coordinates so it's not more useful than just giving a definition and physical interpretation
but sure you can figure out nice coordinate expression for div from this
@LeakyNun I asked because I'm analytically continuing $$ \sum_{k=0}^\infty \frac{(\frac{1}{\log(s)})^k}{k!} $$
that's just $\exp(1/\log(s))$
6:42 PM
looking for the largest domain for which it can be analytically continued
Idk I could just be sc
$3cos x + 3 = 5$
$cos^{-1}(x) = \frac{2}{3}$
$x = 48º$
How do I know that its also $312º$?
@JBis Do you know how to relate cosine and sine to the unit circle?
cosine is adjacent and sine is opposite
That’s a statement about right triangles. It’s useful when doing acute angles, but 312 degrees is not acute
Then no, not sure.
6:54 PM
How do you make sense of sine/cosine when doing larger angles?
For instance, cos(120 degrees) =-1/2
Do you know how to verify that?
One sec...
A/H = -1/2
Ok. How did you check that?
A (-1) / H (2)
Right. Try doing that for both 48 degrees and 312 degrees in the same diagram
Note that 312 = 360 - 48
7:05 PM
Does that make cos(48 deg)=cos(312 deg) obvious?
So just the x +/- 360º where 0º ≤ x ≤ 360º to find the solution set?
Pretty much. In formula terms: cos(360 - x) = cos(-x) = cos(x)
ah. Thanks for help!
You’d of course have other answers, eg x = -48 degrees. But the two you have are the only ones between 0 and 360
7:36 PM
can someone prove/disprove this statement:
For $n \gt 5$,$ n$ positive integers in A.P. can never give a palindromic product.
Hello everyone!!
Does someone know the classical definition of the expression $^2t$?
For example what does represent the following expression?: $1+t^2+{^2t}$
@manooooh this is tetration
$^2t= t^t$
@Mathphile thanks <3
8:22 PM
Hey guys, suppose A & B are functions, and their difference i.e.; (A-B) diverges... then what that means? (new to concept of convergence/divergence)
@Shivanshu: You need to explain more. "diverges" as what happens?
Sure.. I meant 'diverges' as in like sum of a GP to infinite terms when |r|>1
Better give the precise question.
You have two infinite series, and their difference is a divergent series?
Where are functions coming from?
@TedShifrin hi
hi @Leaky
8:26 PM
@TedShifrin do you have any tips for, eh, vector calculus in other coordinates?
Too vague a question.
You talking about computing grad, div, curl in different coordinates?
Differential forms for the win.
Otherwise it's just tedious chain rule.
Yes Ted. Two infinite series whose difference diverges... Consider them any typical/generic functions
How are they functions, @Shivanshu? Just give the actual question.
8:28 PM
@TedShifrin how do 1-forms correspond to vector fields in other coordinates?
let's say spherical coordinates
I'm going to post a handout-exercise I gave my multivariable math class.
@Ted i mentioned to him above that for div and the laplaciano u can just do a one line integration by parts against a test function to get the coordinate formula, so it’s only ever as hard as computing the metric
curl tho i think is easier to derive by doing formsy cuz curl is so weird
8:31 PM
Yeah, but you might as well understand the dual orthonormal coframe :P
@Ted there is no such actual question, was reading about convergence/divergence.. so thought if A-B situation.
@Shivanshu: Convergence/divergence of NUMERICAL series?
Yes @TedShifrin
can anyone prove or disprove that :
"The set of elementary functions which do not have elementary integrals is bigger than set of elementary functions which have elementary integrals"?
understanding diff forms perspective is always good and useful imo
8:32 PM
So if the individual series converge, their difference converges. If one converges and other diverges, then the difference diverges. If they both diverge, then anything can happen. You should write down specific examples.
@Mathphile: I can't prove anything, but the set of elementary functions which have elementary integrals is a set of measure 0. :P
Thanks @TedShifrin :)
@TedShifrin what is the strange star thing?
oh no
hodge star
Hodge star. $\star dx = dy$, $\star dy = -dx$ in $\Bbb R^2$.
* is overused.
Sure thing.
The point is that you need an orthonormal basis for the $1$-forms on your manifold. So you need to diagonalize the Riemannian metric, effectively.
8:41 PM
@Mathphile Well there's a bijection between the set of all elementary functions, and the set of all functions with elementary antiderivatived
namely: taking the derivative
(Well, not quite a bijection, since you can add constants)
(There's probably a way to fix that)
Hi DogAteMy.
Evening all
Hi @ÍgjøgnumMeg
How's it going? :)
Decently, and you?
8:47 PM
Not bad, I hear back from my scholarship application next week
We wish you all the best :)
@TedShifrin I prefer... chen lu!
Does someone know why there is a vertical line in $\mapsto$?
9:05 PM
to distinguish it from \to : $\to$, I guess
Because it indicates that it's where a single element is mapping. I write $\rightsquigarrow$, myself.
@Semiclassical @TedShifrin thanks guys! But there is an historical reason for that math symbol? I mean, why a vertical line?
i do wonder how far back that goes
Yes lol, it seems impossible to answer!
But your answers seem OK to me
Don't know how else you'd distinguish it from $\to$.
9:10 PM
@TedShifrin using $\rightsquigarrow$ for example :P
Well, that's harder to typeset (before LaTeX) :P
9:24 PM
Good news everyone! :)
Is anyone familiar with the convolution operation in the context of signal processing?
More specifically, how would you read this notation $(f \circledast g)(x)$?
It is the convolution of function $f$ with function $g$, but why $(\dots)(x)$?
Maybe it just indicates that the result of the convolution of $f$ and $g$ is a function of $x$
I think that's the meaning
Bad news, nobody! :)
I could have avoided asking this to you, given that I came up with an explanation alone (maybe because I asked to you and nobody answered after 1 minute)
1 hour later…
10:43 PM
@nbro Yes, that's just evaluating the convolution at a point $x$.
@user76284 Is that the evaluation of the convolution or just the definition of the convolution?
11:30 PM
@Semiclassical Are you any savvy at numerical analysis?:P

« first day (3189 days earlier)      last day (468 days later) »