« first day (4494 days earlier)      last day (823 days later) » 

00:26
Before high school, I was a country music aficionado. I could recognize almost any country singer pretty quickly. Since high school, I’ve expanded my tastes and have lost track of who’s who in country.
@lorenzo There’s no reason to think this is correct. Why magically does the same subsequence converge for every coordinate?
@TedShifrin mh, I have to think more about this. Thanks for the help, as always. Bye
01:04
@TedShifrin Do you have any idea what this notation means? $$\frac{1}{\lambda^2}/\times 10^{-15}m^{-2}$$
@PM2Ring I agree about the impressive human being. A bit unfortunate her other characteristics tend to interfere with that observation.
when i grew up in ireland, country music was the #1 best seller, but very few i knew would admit to listening to it.
01:34
@Ajay What do mean? Units of l/m$^2$?
02:11
G'night everyone
 
2 hours later…
04:11
Nevermind, I got an answer.
05:10
Suppose $f:\Bbb D\to\Bbb C$ is a bounded analytic function and extends continuously to $\partial\Bbb D -\{1\}$. Then if $|f|\leq 1$ on $\partial\Bbb D -\{1\}$ then $|f|\leq 1$ on $\Bbb D$. How can I show this?
05:25
Is it correct to say that any set A is contained in the union of itself? As in, the set A is in the union of the elements of the set {A}?
No braces. Yes, otherwise, but the first sentence is wrong.
@TedShifrin Right, thanks! I was asking this because then I thought we could prove that any open cover has a finite subcover.
05:44
why would someone insult an interval by calling it degenerate
@Seeker The open cover rarely contains the whole space as one of the sets.
06:08
@TedShifrin Oh right.
06:18
Can anyone please explain what I am missing in the following:- The statement is -"Topological X is a Baire space". If I prove this for compact Hausdorff space X, then I it automatically proves the statement for complete metric space X.
nvm, I figured it out.
06:52
wtf is Generalised stokes theorem
07:28
I don't think supE is upper bound deduce that $supE-e_0>=a$ for all a in E.
$supE>=a or supE+e_0$ makes sense
but I think supE is upper bound implies $supE-e_0>=a$ is true but antecedent doesn't deduce to that conclusion
I think it follows from this "Then there is an ε0 > 0 such that no
element of E lies between s0 := sup E − ε0 and sup E"
0
Q: Prove $\{x \mid d(f_n(x), f_m(x)) \le \varepsilon \text{ for all } n,m \ge N \}$ is closed

user200918In the proof of Theorem 48.5. in Munkres' Topology it is written : I can understand the intersection of closed sets is closed, but it is closed for a single pair $n,m$ in the first place? I mean, why "the set of those $x$ for which $d(f_n(x), f_m(x)) \le \varepsilon$ is closed in $X$ by conti...

Can anyone please explain why $A_N(\epsilon)$ is an intersection of two closed sets?
I understand why $A_N(\epsilon)$ must be a closed set.
@NotTfue what is your question exactly?
@Koro Just want to know that how does supE being upperbound implies $supE-\epsilon_0\ge a$ for all a in E.
Logically it is correct but how does $supE-\epsilon_0\ge a$ for all a in E follow from supE being upper bound.
07:45
sup E is an upper bound by definition. It does not imply $sup E-\epsilon_0\ge a$ for all a in E.
@Koro so book is wrong?
it follows from "suppose the theorem is false."
The book is correct. Think about the first sentence of the proof.
great example of a needless proof by contradiction, anyway. blegh.
Speaking of contradiction proofs.
07:47
@leslietownes yes due to contradiction
How can we use it to prove the mvt?
the proof that root 2 is irrational
oh i was confused by "it" my bad
@Koro what do you mean? if C is closed, then C = C intersect C is an intersection of two closed sets.
that's one interpretation but I think that was not intended by Munkres.
where does munkres say it's the intersection of two closed sets?
07:49
C= C_m intersect C_n, C_m coming somehow from f_m, C_n; from f_n. I just don't see how.
it's naturally presented as an intersection of a countable family of closed sets.
an arbitrary intersection of closed sets is closed, so there's no reason to focus on two, or how many.
in the last line in the image- A_N(\epsilon) is the intersection of these sets for all m,n >= N.
Not sure what 'these sets' are.
the sets $C_{m,n} = \{x: d(f_n(x), f_m(x)) \leq \epsilon\}$.
$A_N(\epsilon) = \bigcap_{m,n \geq N} C_{m,n}$.
each $C_{m,n}$ is closed, so $A_N$ is too.
Thanks a lot. :-).
Sorry for causing you the trouble of typing out mathjax. :P
haha. once i need to do \bigcap, i might as well.
or ted yells at me.
07:53
3
Q: Contour integration $ \int_0^\infty \frac{\log x}{1-x^{8}}\,dx\, $

AgonyI am tasked to compute the integral $$ \int_0^\infty \frac{\log(x)}{1-x^{8}}\,dx\, $$ using contour integration. I've seen some approaches to complex logarithms, but never with such a function in the denominator. I can't seem to figure out which contour fits this problem best. If I choose a semic...

This is a similar question I asked yesterday. The appearance of $\int{1\over 1+x^b}$ is not a weird thing.
@Ajay the map $f(x)=x^2-2 $ is irreducible in $\mathbb Z$ by Eisenstein's criteria hence irreducible in $\mathbb Q$ by Gauss' lemma. So \sqrt 2 is not rational.
Leslie, is this correct?
And the question I just asked a few hours ago, I can consider $f_{\epsilon}(z) = (1-z)^{\epsilon}f(z)$ which is conti on $\overline{\Bbb D}$ so MMP gives the result letting $\epsilon\to 0$.
koro: sure? for polynomials of low degree (which factor only if they have a root) the rational root theorem is a simpler approach.
@Koro I mean't using the proof of irrationality of root 2 to prove the intermediate value theorem.
The hint of that problem is to use some subharmonic function related to $f$ and use maximum principle but... not sure what it's talking about.
07:59
not mvt, my bad.
I remember reading it somewhere but I can't find it. Nor do I know how to do it. I think it might have something to do with least upper bounds?
yes, rational root theorem :-).
Algebra is amazing :-).
algebro
Hi algebro @shintuku!
am not algebro until 19th of december
hello
 
2 hours later…
09:50
residue calculus is so annoying
10:27
0
Q: Calculation of $\int_{-\infty}^\infty {\cos x\over a^2-x^2}\ dx$ using residue theorem

one potato two potato Compute the integral using Residue theorem: $$\int_{-\infty}^\infty {\cos x\over a^2-x^2}\ dx\quad a>0.$$ My attempt: Choose the contour $$\Gamma = \{Re^{i\theta}:0\leq\theta\leq\pi\}\cup [-R,-a-\epsilon]\cup\{-a+\epsilon e^{i\theta}:0\leq\theta\leq\pi\}\cup[-a+\epsilon,a-\epsilon]\cup\{a+\epsi...

I don't even know what I typed.
11:07
indian food where mouth says yes but arse don't agree
this is the first thing I saw when entering this room
$\operatorname{Aut}(\Bbb C) = \{az+b\mid a,b\in\Bbb C, a\neq 0\}$: Let $F\in Aut(\Bbb C)$ then $f(z) = \sum_{n=0}^\infty a_nz^n$ on $\Bbb C$. Since $f$ is conformal equivalence, $f$ is injective so $f'\neq 0$ on $\Bbb C$. Hence, $f(z) = a_0+a_1z$, $a_1\neq 0$.
Does it make sense?
It's not clear why $a_k = 0$ for $k\geq 2$
from what you wrote
Oh right. Thanks.
also are those supposed to be holomorphic automorphisms or something?
lastly you wrote $F$ instead of $f$ which is a minor mistake
11:27
$Aut(\Bbb C)$ is a group of biholomorphic functions.
 
2 hours later…
13:32
Hi folks! For using the method of moments, do I need to assume that the parameter space is open?
13:45
Is a moderator online?
@Goku @ one of them.
Try Xander or Robjohn
A question needs to be closed. The person who was cheating on USAMTS is back with an alt account
And he posted one of the problems again
Send the link here.
Actually no, post it on cured
the chat room
@Ajay done. Check it
Does the function ${(-e^t+te^t+1)t^z\over (e^t-1)^2}$ integrable on $[0,1]$ for $Re(z)>-1$?
13:54
@onepotatotwopotato
Can you help to close this question
-1
Q: Equation of line EF (coordinates)

Thi nkLet $r,s$ be positive integers, and let $A=(0,0)$, $B=(1,0)$, $C=(r,s)$, and $D=(r+1,s)$. Find the coordinates of a point $P$ in terms of $r,s$, such that if $E$ is any point on $AB$, and $F$ is the unique point other than $E$ that lies on the circumcircles of triangles $BCE$ and $ADE$, then $P$ ...

The OP is cheating on a math contest.
Man we really need mod presence at all times
when people volunteer all of their time and effort for free, having their presence here at all times is a bit too much to ask
9
I flagged the question. Moderators would check.
That is true, moderators have to deal with some real shit.
@onepotatotwopotato ty
14:20
I left a little note, seems as if there are multiple users.
14:48
@Ajay what math contest is this?
@user4539917 True.
USAMTS
How do you guy know about these math competetions. My teacher never informed me about this stuff except IMO.
And why is competetion question posted publicly?
Are there competetion like this for undergraduate?
15:46
Go to aops
They have loads of math comps
see which ones are available in your region.
16:19
@Goku As I said the other day,
yesterday, by PM 2Ring
Mods prefer stuff like this to be done via the flag system (rather than Chat), since that makes it easier for them to keep track of it.
Chat is great when you want to discuss things. But when you need a mod to do something you need to use the flag system. That's what it's for!
FWIW, you can see the full list of moderators (for all sites) here: stackexchange.com/about/moderators
Ah, here's the site-specific page: math.stackexchange.com/users?tab=moderators
17:03
$$\frac{\sqrt{|x|}}{x}=\frac{(x^2)^{1/4}}{x}=\frac{\sqrt{x}}{x}=\frac{1}{\sqrt{x}}.$$ By simplifying as I do above, am I making any assumptions on $x$?
I am uncertain about the second equality; when does $(x^2)^{1/4}=\sqrt{x}$ hold?
@PM2Ring: I have improved the ticks and labels on the axes, and changed the connected data to points. Here is a slightly better chart.
@schn You are assuming that $x\ge0$.
17:57
@robjohn I agree that the new version looks a bit nicer. I'd probably draw a Bézier spline through the points. OTOH, those dots make it easier to read off the time to the minute.
That was my thought on the points.
I'm currently working on code for What is the velocity of the ISS relative to the Earth's surface?. JPL Horizons can give me the velocity vector relative to a fixed point on the Earth's surface, but not relative to the ground point currently below the satellite. So I'm going to approximate, using the satellite's declination as the geocentric latitude of the ground point.
 
2 hours later…
20:17
Chebyshev's function $\theta(x)=\sum_{p \le x} \log p$
some say it's more natural than the prime counting function
are there any other natural functions to analyze the distribution of primes of th form $f(x)=\sum_{p \le x} g(p)$ for some $g(p) \ne \log p$?
21:18
@TedShifrin Following your advice I have continued working on Exercise 2.2.15 and I understood what my mistake was (mainly by thinking about the concrete example $S\subset\mathbb{R}^2$).
My proof is as follows (the notation is admittedly a bit cumbersome).
The case $n=1$ has been established in Exercise 2.2.14(a).
Suppose now that the thesis is valid for some $n>1$; we claim that it is true also for $n+1$. Let $\{\mathbf{x}_k\}_{k\in\mathbb{N}}$ be a sequence of points in $S$.
We use the notation $\mathbf{x}_{k}=\begin{bmatrix}\mathbf{x}_{k,n}\\ x_{k,n+1}\end{bmatrix}\in\mathbb{R}^{n+1}$.
22:00
@lorenzo You could save notational stuff by doing the induction step with the notation of just a sequence (rather than subsequence), but now you get the right idea :)
good, thanks
 
2 hours later…
23:49
The set of all vectors orthogonal to a subspace W is just the subspace's orthogonal complement and is equal to the kernel of the basis of W, correct?

« first day (4494 days earlier)      last day (823 days later) »