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12:00 AM
@XanderHenderson That's right, you lived in iowa, as well!
@amWhy Yup.
luck o' the iowish.
But Holbrook is right on I40, @Xander. Therefore less in the middle of nowhere, by definition.
But @leslie no longer speaks standard English. That's what leaving Iowa or some parts of WI does to people: their speech gets distorted. I imagine robjohn also believes Don and Dawn are homonyms?
12:02 AM
@TedShifrin Yes, but it takes more time to get anywhere interesting from Holbrook than it does to get from Bisbee to anywhere interesting.
Actually, having just checked the map, Bisbee to Tucson is almost the same duration as Holbrook to Flagstaff.
@amWhy yes, but it's not PC to mention it ;-)
@robjohn Hah! Okay!
@robjohn Don and Dawn in particular, goes back to when my family spent Christmas in Claremont with my aunt and uncle and cousins. We went out to a place serving "family style dinners" (not a family diner) and among us cousins, we were surprised how we each spoke. Don is one of the cousins, who the other cousins called Dawn!
@amWhy would that be Dawn and Down being homonyms?
depending on region
My relatives in Tennessee would pronounce those the same
@robjohn DON is not Down, nor DAWN. DON, to say correctly in Spanish, would be Dan, ahhhhhhhhh Dahhn. Dawn is like D-awe-n. Down rhymes with Town, and Brown,
when people say this kind of stuff they sound crazy. don isn't don, it's pronounced don. whereas DON is pronounced DON.
we agree about down.
12:14 AM
@robjohn I can believe that. I am accent-free!
this reminds me a little bit of my daughter's "lowercase three"
@leslietownes ;P My tutoring has failed you, I guess!
the issue is that a lot of the transliterations read the same too. 'awe' and 'ahh' are exactly the same. they don't help me to differentiate don and don (which are the same)
@leslietownes is that a small three, closer to two than a big three?
anyway, lowercase three is your pinky, ring, and middle finger.
as distinguished from three, which is index, middle, ring or thumb, index, middle.
12:16 AM
don is don. but it isn't *dawn!!
@leslietownes ah
@robjohn Indeed!
@amWhy it is around here
in law school i got a lot of guff from people for saying 'defendant' in a way that i was told sounded like 'defindant'. guilty as charged, it's the same sound. the fin in a fish is the same as the fen in fen-phen.
@leslietownes No thumb needed for three!
12:19 AM
this is how you get identified as a spy in europe, amwhy.
@robjohn Well I can either lament lost causes, or lament lost causes.
@leslietownes I'll have to remember that next time traveling there!
well, at least wait until we're at war with something in europe. shouldn't be too long
@leslietownes Yikes, indeed. I assume Ukraine is in Europe...
i'm not sure that something like that exists, but for now it might be pretending to be.
@leslietownes I understand.
@leslietownes If it wants to be part of Europe, I wouldn't say it's pretending, at least Western Ukraine. But I do not know what's all at stake. I just know a two hour conversation between Biden and Putin took place today precisely about the Ukraine.
12:30 AM
i'll report this to putin, he will be pleased.
@leslietownes Sorry if I was mistaken. Tell him I'm sorry but I don't follow Fox News.
it's all going in the report.
1:23 AM
@TedShifrin imgur.com/vHe7vGb
its dark & rainy at the moment...
re putin keep your friends close and your enemies closer...
looks great.
the google maps image has uncharacteristically sunny weather. this is the authentic experience.
hey, what can i say, the irish weather touch
we had some irish weather earlier in the week but today it was pretty sunny.
Takes me back 60 years, @copper. Thanks!
1:45 AM
2:34 AM
Not sure how to go about showing this. $f(x) = \frac{1}{1-2x} + \frac{1}{1-5x}$ looks like the sum to infinity of two geometric series. For $f(x) - 7xf(x)+10x^2f(x)$, I've got that this equals to $\sum_{n=0}^{\infty} a_n x^n(1-7x+10x^2)$. But not sure how to proceed. Would appreciate some hints
oh, you wanna do it with the series instead.
$7x f(x) = \sum_{n=0}^{\infty} 7 a_n x^{n+1} = \sum_{n=1}^{\infty} 7 a_{n-1} x^n = 7 a_0 x + \sum_{n=2}^{\infty} 7 a_{n-1} x^n$, and $10x^2 f(x) = \sum_{n=0}^{\infty} 10 a_n x^{n+2} = \sum_{n=2}^{\infty} 10 a_{n-2} x^n$ so if you collect the x^0 and x^1 terms separately from the sums, $f(x) - 7x f(x) + 10x^2 f(x) = a_0 + a_1 x - 7 a_0 x + \sum_{n=2}^{\infty} (a_n - 7 a_{n-1} + 10 a_{n-2}) x^n$. maybe some typos there.
and because of that recurstion the series goes away.
this gives you $(1 - 7x + 10x^2) f(x) = L(x)$ for some linear polynomial $L(x)$ and presumably algebra shows you that $L(x)/(1 - 7x+10x^2)$ is what it's supposed to be. that would be a partial fraction decomposition.
look at my chatjax, everybody. this is rare stuff from me.
2:50 AM
Don’t worry. You forgot plenty of $$s.
@leslietownes Thanks! How did you see through it so quickly?
experience. if you think of a power series as just a sequence of coefficients, multiplication by x is a shift. (a,b,c,...) goes to (0,a,b,c...) etc. and if you equate linear combinations of those you get linear recurrence relations in the coefficients.
3:19 AM
Is anyone familiar with the definition of continuity when it is written as 'a function $f$ is continuous on $X$ iff $f^{-1}(U)$ is open for each open set $U$'? $f^{-1}(U)$ is the functions inverse image $\{x\in X|f(x)\in U\}$, not the inverse function.
This definition is meant to be a generalisation of the $\epsilon$-$\delta$ definition, but for the life of me, I can't figure out if the function, $f(x)=1$ if $x>0$ and $f(x)=0$ if $x\leq 0$, is continuous under the new definition or not. It is definitely not continuous under the epsilon delta definition, but for the new definition there are no open sets in the range of $f$ to look at.
look at $f^{-1} ((-\infty, \frac{1}{2}))$
you get to pick the open sets in the range to look at. it does require some mental rearrangement to think in terms of this definition. some books prefer one view to the other, to the point of not quite acknowledging that the other exists.
@leslietownes That would return $(-\infty, 0]$. Which is neither open nor closed... hmm, thank you for the help.
@user400188 Well, it's closed. (Don't be tricked by the left paren; a mistake I've made quite a lot.) But you are right that it's not open.
I learned my lesson. My last problem set of the semester is due Monday and I'm starting it tonight like a disciplined student.
note that the same kind of idea can work with any f that has a jump discontinuity at a point (in the usual sense of the term).
3:24 AM
@Fargle A set is closed iff its compliment is open. The compliment of that is $(0,\infty)$, so I guess yeah it is closed.
Thanks too @Fargle
Indeed! No problem.
I am useful for pedantry and pedantry only. ;)
That's exactly the stuff I like
It has its place, to be sure.
might help to work that out in detail. e.g. if $\lim_{x \to a^{-}} f(x)$ is less than, or greater than, $f(a)$, how do you grab open sets that will show the discontinuity in the open set sense
I haven't covered limits yet in the context of point set topology, which is where I am learning all of these definitions from @leslietownes.
3:27 AM
we may be a few days ahead of that, then. :)
huh, now I realised why I was confused. The full definition was 'A function $f:(X,T_X)\rightarrow(Y,T_Y)$ is continuous on $X$ iff $f^{-1}$ is open for each $U\subset Y$'.
I was thinking of $Y$ as the target set, which would be $\{0,1\}$ in my example, but then since every $U$ needs to be taken out of that, I wasn't able to get that larger subset of Leslie's $(\infty,\frac{1}{2})$. Come to think of it, is it even compatible with the definition?
@UnderMathUate Good boy.
@user400188 No, $Y$ is $\Bbb R$. The image of the function happens to lie in $\{0,1\}$.
You never carefully wrote the domain and range of $f$.
3:42 AM
In the definition given, the domain and range are not specified. Unless you are referring to my example in which case yes I didn't write out the domain and range.
Yes, I'm referring to that.
That's part of your confusion, I think.
Hmm, ok. Well thanks for the clarification, but it does make me wonder. If the range was $\{0,1\}$, would the function be continuous?
I am trying to code multivariate polynomial division, and I wonder if, by following the long division algorithm, is it possible to get terms with repeated monomials in the quotient, or terms out of the chosen monomial ordering.
Depends. What topology are you picturing on $\{0, 1\}$?
(There's one natural answer here.)
Ah, that's right, I hadn't even considered it. The trivial one which is just $\{0,1\}$ and the empty set?
3:44 AM
@TedShifrin :D
Depends on the topology on $\{0,1\}$ (and on the topology of the domain).
I was thinking in this case of the subspace topology. Though I guess I'll get in trouble for using the word "natural"...
I suppose the discrete topology works too, either of them would make the problem the simplest. I don't know what a/the subspace topology is. The discrete topology to me is all the subsets of $\{0,1\}$.
@TedShifrin Let's say the domain is just R, with the metric topology on it.
The subspace topology on a subset $Y \subset Z$ is the topology whose open sets are exactly $U \cap Y$ where $U$ is open in $Z$.
In the present case ($\{0, 1\} \subset \Bbb R$), the subspace topology is one of the ones we've named (discrete or indiscrete). But which?
@Fargle: Have you been in touch with Ken K.?
3:48 AM
Rephrasing my definition, a set $V$ is open in $Y$ in the subspace topology if you can find an set $U$ which is open in $Z$ such that $V = U \cap Y$.
I haven't, not in a long time. I should email him again.
Congratulate him on being a daddy.
Oh, wow!
He did teach me all this stuff I'm talking about now. I haven't thought about my years at UTK in a long time.
I serve a porpoise here.
I feel kinda bad that I wound up dropping his Top 2 class. Though I might have felt worse staying in, considering that was my, uh, catastrophe semester.
I'm sure he holds no grudges.
3:51 AM
I talked to him about it; my impression is that he doesn't. He warned the class early on that it was, for reasons at least partly outside his control (as I remember), going to turn into quals prep.
Oh, intro alg top?
Not sure the nature of it. I know it was going to be Munkres part 2 before that, so roughly, yes, but I don't know about after because I dropped very early.
@Fargle The discrete topology (I used the first definition not the rephrasing).
Well, the last third of Munkres is fundamental group and covering spaces.
Right @user400188
Yeah, that's the part we were doing (I think that's part 2, anyway).
And yes!
3:55 AM
Anyhow, you may tell him I instructed you to email hi.
I lasted through that material for a little while, and then his announcement + my dropping, which came just a bit before all the cool theorems (FTA, Borsuk-Ulam, etc) and Seifert-Van Kampen, IIRC.
Or whatever the theorem is called. It has been too long since I looked at AT.
@Fargle Elements of the topology are called 'open sets' of the space. So $\{0\}$ is an open set in this context. Since $f^{-1}(\{0\})$ is $(-\infty, 0]$, the function will not be continuous.
Diff top is still more cool. :)
I agree.
(With both. ;) )
@Fargle Oh cool, so I didn't make a mistake?
Or was that in reply to something else?
3:59 AM
@user400188 You got it right.
phew... I have been studying everything without examples, so I am relieved that I managed to get that right.
4:45 AM
@TedShifrin I told him exactly that.
And wrote a practical novella. Why are all my emails to professors like that?
"A figure consists of semi-circles." Does this mean that the figure consists only of semi-circles?
possibly. is there more context?
> There are three paths from $\mathrm{A}$ to $\mathrm{B}$ each consists of one or more semi-circles of unknown radii. The paths $\mathrm{AXB}, \mathrm{AYB}$, $A Z B$ are called I, II and III respectively.
The middle figure in AZB doesn't look like a semi circle :-/
given that context i'd assume they are all supposed to be semicircles and the picture is not to scale
1 hour later…
6:06 AM
hmm, my mse swag had been delayed with no explanation.
maybe it's some kind of subpoena or other legal document
and they want to catch you unawares
my alter ego will deal with the sdt
Is it possible to calculate a primitive, even if it does not exist? Let me explain: I want to calculate $\int \frac{1}{\sqrt{1-u^{2}}}\sqrt[4]{\ln u}{\rm d}u$ but it seems that over $[0;1]$ this integral does not exist. Still, is it possible to calculate it?
6:13 AM
are you asking if it is integrable over $[0,1]$?
How to type $\int_a^b \sqrt{1+(f'(x))^2 } \mathrm dx$ in MS word?
why would you want to do such a thing?
wolg there is an 'equation editor', and god help you
@leslietownes Yeah I am using that only :-(
it is easier to bang your head repeatedly against the wall
6:15 AM
you'll want to do it through menus, for a loto f stuff, so if you have that off by default it will be confusing
@copper.hat I came across a question where they asked to calculate that indefinite integral. However, it seems that this integral does not really exist.
you insert an int_{blank}^{blank} blank symbol and then can fill in the blanks
6:52 AM
Ohk that works. :-)
don't bother with that mathrm foolishness on the d
that looks goofy even in latex
7:52 AM
@copper.hat they giving out swag again?
@leslietownes it looks right
@Alex the integral exists; it may not have an elementary closed form.
@Wolgwang none of the pieces of AZB look like semi-circles.
@Alex $\sqrt[4]{\log(u)}$ is complex on $[0,1]$. $\sqrt[4]{\log(1/u)}=\sqrt[4]{-\log(u)}$ is real.
@Wolgwang: did you get my late explanation?
$\sum_{k=1}^m a^{\gcd(k,m)}\equiv 0\bmod m$ for any $a,m\in\Bbb N$.
8:08 AM
@robjohn Yes, Thanks :)
1 hour later…
9:26 AM
I tried to show if $\int_0^\infty (\cos^2x)^{x^4}\,dx$ converges but nothing seemed to work. @robjohn
Any hints on that please?
I tried considering $exp (x^4 \log (\cos^2x)$ and then $x^4 \log (\cos^2x)\le x^4 e^{-x}$ and integrating both sides gives RHS as $\Gamma {5}$
But then I realized that LHS is negative and I couldn't conclude convergence of the integral. And even if the convergence could be concluded still convergence of $\int exp (x^4 \log (\cos^2x)$ could not be established, I think.
Background: this integral came up while discussing one day: if there exists a continuous function $f$ such that that $\int_0^\infty f(x)\, dx$ exists but $\lim_{x\to \infty} f(x)$ does not exist.
2 hours later…
12:01 PM
@Koro there are much easier ways to construct such functions
Take the function that looks like a triangle with base $1/2^n$ and height 1, then 0, on [n, n+1]
12:17 PM
Astyx: I know that. But my question is strictly in reference to that integral.
12:29 PM
I keep seeing big lists of invariants for Klein geometries, but how does one actually find those invariants?
Is there a systematic method or do you just wing it
1:00 PM
Hi, i am getting a very serious problem, because suddenly i cant comment, neither answer nor ask a question, also i cant bote and i got more than 500 in reputation, i dont know if it is compatibility issues on my phone, any suggestion?
1:25 PM
How can we directly assume $x_k<1/2$ ?
and bounded below by 1/2?
why do mathematicians say formal when they mean informal?
@Rover Your handwriting might actually be worse than mine. :/
@jay example?
a formal derivation blah blah blah
where they go on to use IBP (ignoring boundary terms) and swap limits with integrals etc
is always said to be formal computations
or a formal proof
when really its more informal
@XanderHenderson well, It's not actually mine.
But, is it understandable ..?
1:46 PM
A formal argument is one that is done in accordance with the rules. It fits the common English definition.
well in reality they are done by sweeping the details under the rug
Hii , I want to know what are the uses of mathematics in engineering , physics , economics etc.Please share your points on it.
2:01 PM
they are all applied math
@S.M.T The presupposes that mathematics is a whore, who only gains value by satisfying the needs of other disciplines. In reality, mathematics is a beautiful queen, who serves her own purposes, and occasionally deigns to grant a favor to the dirty sciences.
@XanderHenderson I like this characterization.
2:34 PM
@XanderHenderson Hmm. That’s deep & beautiful
1 hour later…
3:54 PM
jay: yeah, that use of "formal" means "following an expected or established form" and not formal in the sense of "formal proof" in a proof writing context, or formal reasoning. and in context is a signal that what follows probably isn't rigorous at all
very common e.g. to write the so-and-so transform is 'formally' given by some expression, when it maybe the expression only makes sense for a subset of the domain it is being considered on
i just skimmed one of my old papers where 'formally' i deduced f(z) = Re g(z) using a comparison of power series that diverged for a lot of z i wanted to use that equality on, where the real proof would have detoured through some integrals and limits
but nobody's trying to read that crap
hi guys
so is the column space of an m x n matrix basically just the image under its corresponding linear map?
4:11 PM
now im curious what the intuition behind the row space
dont spoil it!!
it doesn't relate as directly to the transformation. it relates to the image of the dual map.
you can kinda ignore it
:) they did say -don't spoil it and you didn't spoil it.
idk what the hell a dual map is
if you want to introduce an inner product you can think of it as the orthocomplement of the null space.
again, that's not quite as directly tied to the transformation as its image is.
i am a huge fan of matrices but unless you have to calculate something i think it is better to think in terms of properties of the maps themselves. matrix entries, whether rows or columns, aren't themselves very interesting.
sometimes even if you are calculating something.
4:22 PM
@leslietownes This. "Formal" is about "form".
see every engineering treatment of every integral transform ever
for nonsingular matrices i assume the nullspace just consists of the 0 vector?
yeah. this is sometimes or maybe even usually the definition of 'nonsingular' (note that it makes sense for rectangular matrices that are not square).
although some books define 'nonsingular' only for square matrices by matrix algebra (meaning: one of a long list of properties equivalent to invertibility) in which case this is maybe a theorem and not a definition.
yea my book defines nonsingular/singular for only square matrices
4:37 PM
After a couple of months, I finally figured out how to use chatjax.
Now the "$" and "\" makes sense.
welcome to the party. :)
@leslietownes which paper is that of yours
it's a secret
4:57 PM
Hi 👋
@robjohn Nice, thanks
If $x_{1}\in (0,1)$ and $x_{n+1}=x_{n}-x_{n}^{2}$ so $\sum
If $x_{1}\in (0,1)$ and $x_{n+1}=x_{n}-x_{n}^{2}$ so $\sum_{n\in \mathbb{N}}a_{n}$ is divergent series?
x_n rather than a_n?
4 hours ago, by Rover
user image
Is it readable ?
yes. it might help to crop the image, and some people here hate images, so be prepared for that
also expect mockery of your penmanship (which is better than mine)
@leslietownes i know i know :-))
@leslietownes Is it better to write and post in a good handwriting?
Here it’s better to use Chat Jax.
5:07 PM
@leslietownes It's a typo. This is the correction: if $x_{1}\in (0,1)$ and $x_{n+1}=x_{n}-x_{n}^{2}$ so $\sum x_{n}$ is divergent series? I think we can prove that $a_n \not\to 0$ as $n\to \infty$ so we can conclude that $\sum x_{n}$ is a divergent series. But I'm no sure as to work here.
@Alex It would be nice to learn how to use the comment links. See how this message is linked to your last message? I did that using the arrow at the far right of the message:
@Alex if $x\in (0,1)$ then $x-x^2=x(1-x)\in (0,1/4)$. if you go one step further the interval shrinks further, and so forth. so i'm not sure on what basis you're concluding that $x_n$ isn't converging to zero
@Alex: then we know to what you are replying. This is good in a busy room and when replying to something that was a while ago.
@TedShifrin yep okay.
@Koro $$\int_{\left(k-\frac12\right)\pi}^{\left(k+\frac12\right)\pi}\cos^{2n}(x)\,\mathrm{d}x=\frac{\binom{2n}{n}}{4^n}\pi\sim\sqrt{\frac\pi{n}}$$
since $n\sim k^4$...
5:15 PM
Oh my God! I think that's Wallis formula.
howdy @robjohn
Hey @Ted. what's up?
(i) Let $x_1$=1 and $x_(n+1) =(x_n+1)/3$ where n€N. Then prove that ($x_n$) is congruent , lim n-->infinity ($x_n$) = 1/2 .
@Rover convergent
Your proof was fine, except for one thing. You never proved the limit was actually $1/2$.
@robjohn I understand @robjohn thank you. I think that I know that now.
5:21 PM
@Alex If $f(x)=x(1-x)$ then the sequence can be recursively defined as $x_{n+1}=f(x_n)$. Now if $x_1\in (0,1)$, it follows that $(x_n)$ is a sequence of positive terms and decreases. Therefore $\lim x_n$ exists and can be shown to be equal to $0$.
@Koro so, it is not helpful to see that $a_n\not\to 0$ as $n\to \infty$ because it's not work?
why do you think that $x_n$ doesn't converge to $0$?
@TedShifrin actually it's not mine. I was having problem understanding why assume $x_k>1/2$ ?
You want to prove that for all $k$ by induction.
It's an intelligent guess once you argue that the limit must be $1/2$ if it exists.
@robjohn thanks. I'll try with this now :)
5:27 PM
@robjohn @Koro What does one get from looking at Taylor polynomials near multiples of $\pi$?
@Koro Sorry and I just thought I could use the convergence theorem. I should have thought a bit more about the limit of the terms. So how could you prove that the series diverges?
Ted, I'll expand $(\cos^2x)^{x^4}$ around $\pi$ and see. I think that will also, alternatively, let me see convergence of the integral.
Well, around $0$ is just the same.
@TedShifrin Hmm ok. It's just a guess and we can assume anything else like $x_k> 1/3$
@Alex Ah, I have also used monotone convergence theorem -$(x_n)$ is decreasing and bounded hence must converge. About convergence: I think it's easy to see that $\sum x_n^2$ converges. I'll think about convergence of $\sum x_n$ though.
5:33 PM
Well, see if the induction proof works with $1/3$, @Rover.
@Koro I meant for $\cos^2 x$. Of course, the shift in $x$ will be crucial. So we need to do $x\approx k\pi$.
@Koro take a look at Theorem 1 from this answer and equation $(9)$ from this answer
@robjohn Stationary phase is one of the favorite things I learned teaching the applied math year-long course back in 86-87.
@TedShifrin a very useful tool
It is used all over Stein's books
@TedShifrin were you being punished for something? ;-)
5:38 PM
No, no, I wanted to teach it. I used Strang's brand new book, and it made a fabulous course, although I had to fill in all sorts of proofs and application derivations ... and make up a lot of better homework.
I actually don't think stationary phase was in that book, but a lot of good stuff is in Bender/Orszag (a course I should have taken at MIT, but I was too busy doing grad pure courses).
@robjohn okay, I will. Thanks. :)
@Koro So what happens at $k\pi$?
@robjohn The typical applied course was taught out of Kreyszig or some such and was just a compendium of unrelated stuff. Strang's book was very wonderfully thematic.
@TedShifrin $\cos^2x=\frac {1+\cos 2x}2=\frac 12 +\frac 12 (1-(2x)^2/2+\frac {(2x)^4}{4!}+O(x^6))$
5:44 PM
it's not quite the same thing but it's not too far from WKB if i remember right
at $k\pi$, the function $(\cos^2x)^{x^4}$ is 1. But apart from that I don't see anything else. :(
I wanted the expansion at $k\pi$ to see how the integral behaves in a neighborhood.
@TedShifrin I never looked at it. I was teaching at UCLA 86-88, so it would have been nice to have a good book to teach from, but I was not in any position to create a course.
@Koro thanks, can we use the monotone convergence theorem?
When I taught the course, many people were doing a year-long course out of Haberman's PDE book. That seemed quite inappropriate to me, @robjohn.
5:51 PM
@Alex I used that to conclude that $\lim x_n$ exists. Then note that $x_n-x_{n+1}=x_n^2$ so telescoping it will give you insight into convergence of $\sum x_n^2$
Ted, you taught functional analysis also?
@Koro I think my series idea is doomed. Forget about it :)
Functional analysis, no. This was a year-long course in applied mathematics, so various things from linear algebra, ODE, PDE, transforms, complex variables, asymptotics, ...
@TedShifrin Well, $\cos^{2n}(x)\sim\left(1-\frac{x^2}2\right)^{2n}\sim e^{-nx^2}$
That integrated over the interval gives $\sqrt{\frac\pi{n}}$
so a series can be used
Ah, sure. My original ill-founded idea was to see the growth rate of an integral in a tiny interval around $k\pi$ by Taylor. That doesn't go anywhere.
5:56 PM
Without scrolling till author's name, I just know these days if an answer has been written by robjohn :)
@TedShifrin That is sort of what I was using with the asymptotic estimate above, so I don't think it is that doomed.
Integrating over the interval $[k\pi,(k+1)\pi]$ makes it a lot easier than doing separate uniform estimates when the function is $<1-\epsilon$.
@Koro Yes, robjohn has a certain style :P ... I don't know if you'll recognize me so easily ... but you don't look at many questions I answer anyhow :D
Ted, yeah robjohn has a style :). I have not yet started revising my multivariable calculus so I'll come across yours too soon :)
I recognize user fleablood's answers too without scrolling till author's name :)
I have by far the greatest number of answers in differential geometry and differential topology. For multivariable calculus there are some obvious people who try to answer every question.
See, now those are the subjects that I haven't studied yet :(
6:04 PM
@robjohn They seem to have again reformatted the answers page.
@Koro My third subject is linear algebra :P
@TedShifrin yes, they fixed one of the bugs that I reported, but another still remains. The scores for your answers should be correct, but the acceptance state is the acceptance state of the question, still.
Any hint for prove that $\sum x_{n}$ is a divergent series? :-(
@robjohn Are you sure? The few of mine I've looked at seem correct.
They got a crapstorm of negative criticism yesterday
@TedShifrin let me see, it might have been fixed
I bet they did.
6:12 PM
@TedShifrin you might just have gotten the accepted answers on those questions
I see answers as accepted that are not. The question has an accepted answer, but it's not mine.
Let me look some more.
If you look at my activity page, the first answer is shown as accepted, but it's not.
Yes, on my own page, it indicates if there is an accepted answer and then it tells me below that if my answer is accepted.
See this if you can.
Ah, that's your Answer page. I am looking at the Summary page
that is what usually comes up when you look at someone's info
Oh, so confuzling.
6:19 PM
And this confusion makes it hard to describe these problems to the developers
Well, the five that show up on my summary page are correct. I had to click to get more answers to see more, and then I ended up where you don't want me :P
@TedShifrin I checked and that is because of those 5 questions you've answered, you've gotten all of the accepted answers.
So you are seeing the fact that the question has an accepted answer, but it looks as if it is saying that you have the accepted answer.
they are the same in your case, but if you look at my profile, I have answered some questions with accepted answers that are not mine.
so it shows some of my answers as accepted, when they are not
Yes, I understand the issue.
ok, I'll stop producing hot air
No, no. I'm not criticizing. What is interesting, I suppose, is that we only see the top five. To see more, it takes us to the page with more complete information.
6:32 PM
@TedShifrin and that page has a lot more empty space than it used to. It requires a lot more scrolling.
The answers page? Yeah, way more stuff displayed than before.
And I hate that to see more detailed information, we cannot just click on the "Reputation" title, which is simply text, but we have to scroll back to the top and click the "Reputation" button.
@TedShifrin more information, framed by a lot of empty space
What this conversation teaches me is that I spend way less time on these pages than lots of people do :P
@robjohn I don't know why I am getting a second round of swag. In any event, it has been delayed by fedex.
i click on the differential geometry tag and i get a picture of a cesspit, is this happening for anyone else?
6:35 PM
smacks leslie 4 times
@copper.hat Ah, the bottle-neck is FedEx, not SE
Indeed :-)
@leslietownes I just see black... when I stare into the abyss, the abyss says, "welcome".
@robjohn Sometimes not much of the title is shown.
@RandomVariable when there is MathJax? yes I have noticed this.
6:40 PM
Hi everyone!
If S is a subset of $\mathbb R$ such that $S'\ne \mathbb R$, here S' denotes set of all limit points of S in $\mathbb R$. Is there an example of S such that $S'\ne \mathbb R$ but still S is dense in $\mathbb R$?
koro, no? this is a definition chase.
if y is in R \ S' there is a punctured interval around y containing no points of S, and hence subintervals of R containing no points of S, so S is not dense.
I have a sample of size d from each of n normal populations with common
unknown variance bur possibly different unknown means $X_{ij} \sim \mathcal{N}(\mu_i, \sigma^2)$

Does it mean my data looks like this:?
@TedShifrin I'm beginning to see a pattern of the smacking @leslie. Should I be concerned? ;D
it's a natural phenomenon. like the sun rising every day.
6:49 PM
@Koro ok, I think it works: since $x_{n}\to 0$ so $\frac{1}{x_{n+1}}-\frac{1}{x_{n}}=\frac{x_{n}}{x_{n+1}}\sim 1$ because $x_{n}-x_{n+1}=x_{n}^{2}$. Now, by Cesaro we have $x_{n}\sim \frac{1}{n}$. Hence $\sum x_{n}$ is a divergent series.
hi @TedShifrin
@leslietownes I think the question also implicit here is Why do your provoke @Ted! ;P ;D
amwhy: this is a bit like asking why the sun rises.
Leslie: Yes of course. Thanks a lot. Instead of R, if I take any metric space then also the result holds :)
@leslietownes Haha! Love it!
6:54 PM
koro: is it? what about the discrete metric.
at least, the above argument implicitly uses the fact that a punctured interval/ball in R contains nonempty open intervals/balls as subsets.
so the argument might not work in general even if the result holds.
not to be that guy, of course. you never want to be the 'what about the discrete topology' guy
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