12:50 AM
according to my input formatting code, combined with Knuth's algorithm X, as implemented in github.com/water-vapor/DancingLinksAlgorithmX , I've managed to determine that 3D T tetrominoes may tile a 6 by 6 by 2 area.
(this is news to me, because as of yet, I've struggled to find such a solution by intuitive brute forcing, however I could find a solution for 6 by 6 by 4.

1 hour later…
2:01 AM
@Mrcrg sry i fell asleep. yeah, you can make a 3-4 sentence summary of what you know (I am reading paper A and I saw this class of analytic functions holder at the boundary. I also saw it in paper B, its a banach subspace of C, but I want to learn more, for example D. I tried these notes/books (X,Y,Z) on Banach spaces of analytic functions but they don't have this space. Any good references?) and that should be enough.

2:14 AM
@CalvinKhor No problem, I already posted the question, thanks for the help!

@Asinomás I would have rejected as well but I feel reject and edit was abused. If amWhy's edit was suggested edit, I would have rejected that as well, on the basis of the mathjax typo introduced (now fixed).
@Mrcrg ok great :) any responses?
found the post, no response yet

3 hours later…
5:15 AM
please don't let's all talk at once.

5:29 AM

6:13 AM
$\begin{array}{l|l|l} & & X \\ \hline & & \\ \hline \ & & \end{array}$

\begin{array}{l|l|l} & & X \\ \hline & 0 & \\ \hline \ & & \end{array}

$$\begin{array}{l|l|l} & & X \\ \hline & 0 & \\ \hline \ & & X \end{array}$$

\begin{array}{l|l|l} & & X \\ \hline & 0 & 0 \\ \hline \ & & X \end{array}

$$\begin{array}{l|l|l} & & X \\ \hline X & 0 & 0 \\ \hline \ & & X \end{array}$$
is tic tac toe a win for p1 if the board is periodic?

I know what periodic means, but I don't know what it means with regards to the board being periodic.....lol

6:19 AM
what I mean is you win if you have 3 in a periodic line
so you can wrap around
eg $\begin{array}{l|l|l} & X & \\ \hline X & & \\ \hline \ & & X \end{array}$ would count

ah....Now I see.....well we can't really change the rules in the middle of the game can we?......lol....
that does make it more interesting though

hahaha im not suggesting that. just a random thought

doesn't the first move dictate the winner?

for periodic tic tac toe? idk I lost to myself lol
\tiny makes it use less space: $\tiny\begin{array}{l|l|l} & & X \\ \hline X & 0 & 0 \\ \hline \ & & X \end{array}$

6:37 AM
it bothers me when i get a downvote on a 9 yo answer.
i mean one without explanation

if you won't notice me here, copper, you'll notice me there.

just appeared from nowhere
i thought i found a nice convex question but they are asking silly things like the circumference of a convex set. that's for manifold people
why complicate life?
one of the references in my daughter's final year report had an author of the name s. shit. nothing wrong with that, of course, but just required a double take.
i always strive to raise the tone of conversation.

yeah, hah. i don't know what people are supposed to do, if the field publishes in english. change your name?
"so, my new first name is "some," i think this will boost my citation count"

my co-author is a ...

7:00 AM
@copper.hat i found a copy of "more exercises of B&V" with solutions floating on the internet, but it was before that question was added

@CalvinKhor Thanks. I am tempted to write to them and ask, I know Stephen from some time ago.
I can't see how you can use the dual to answer the question.

7:29 AM
One user who's been around on this site for a long time seems to have left this site now.

very cryptic

@Koro can you give us a clue?

7:47 AM
Here , In the 2nd diagram. U can notice resultant vector I.e AC is drawn with its tail joining tail of AB and head joining head of BC.

Define long @Koro

<10 years but >3 yrs
nope, they are still there. Only their account is suspended.

3 hours later…
10:38 AM
I cannot access ams.org and various links in the ams.org domain. Is it just me, or do other people have problems with that website, too?
Ok, so now I'm getting: "The site is down for maintenance and currently unavailable." A few minutes ago, I was just getting an error.

2 hours later…
1:09 PM
Hi! is there an easy way to simplify $c_i = a_i \cdot (b_0 + b_1 \dots + b_n) + b_i \cdot (a_0 + a_1 \dots a_n)$ where $i < n$.

2 hours later…
3:10 PM
@iyop45 Not that I can see, but we can simplify $\sum\limits_{i=0}^nc_i=2\sum\limits_{i=0}^na_i\sum\limits_{i=0}^nb_i$

Hey, easy question. If M and N are finite-dimensional vector spaces over D, E (division rings). How can i show that if End_D(M) ~= End_E(N) then M ~= N and D ~= E
?

3:42 PM
@iyop45 On the right, do you intend to have the product of $a_1,\dots,a_n$ or the sum?

4:04 PM
Can anything be said about $\lim_{b\to \infty} b^{1/2} B(b/2+1,b/2+1)$?
where B is beta function.

I know negative amounts about beta functions.

I was trying to figure out if the integral $\int_0^1 x^{b/2}(1-x)^{b/2} b^{1/2} dx$ is bounded below by $1$ for all large b.

Mathematica tells me the values are tiny ... tending quite quickly to $0$.
So, NO.
I think that's actually pretty clear.

4:20 PM
Source: trying to bound below $I(b)=\int_0^1 x^b b^x$ by 1 for large b.
$I(b)^2=\int_0^1 x^b b^x dx \int_0^1 (1-x)^b b^{1-x} dx\ge (\int_0^1 x^{b/2}(1-x)^{b/2} b^{1/b} dx)^2\implies I(b)\ge \int_0^1 x^{b/2}(1-x)^{b/2} b^{1/b} dx$
@TedShifrin oh no :(

if anything, $I(b)$ seems bounded -above- by 1

Look. $x^b(1-x)^b$ is maximized at $x=1/2$. So look at $b(1/2)^{2b}$.

isn't the maximum of x^(b/2) (1-x)^(b/2) b^(1/2) on [0,1] achieved at 1/2, where it is b^{1/2}/2^b?

The first inequality is due to Cauchy Schwarz.

or what ted said

4:22 PM
I think I'm suing leslie for copycat publication.

@Semiclassical bounded above case is easy. It's the bound from below which is tricky.

it's not tricky, it's false

well, i was copycatting the b/2's in koro's problem. i don't know about your b's.

if it was bounded above and below by 1 for large x, it'd have to converge to 1 at finite x

@Semiclassical it's true :(

4:23 PM
what are you actually trying to show?

koro what pit of hell did you pull this problem out of to begin with? i'm always curious. you draw from an eclectic group of sources.

I changed $b$ to $2b$ to get rid of the fractions. I figured that if $b\to\infty$, then $2b\to\infty$. Perhaps that was glib of me.

Infact, $\lim_{b\to \infty} I(b)=1$

then say so

3

I'm trying to find the value of $$\lim_{a\to \infty}\int_0^1 a^x x^a \,dx$$ My attempt: Let $\epsilon >0$ be given. $x\mapsto a^{x}$ is continuous at $1$ so there is a $d_a\in ( 0,1)$ such that $|a^{x} -a|< \epsilon$ for all $x\in [d_a,1]$. WLOG, let $d_a<1/2$. $|\int _{0}^{1} x^{a} a^{x} \... 4:24 PM robjohn and his mathrm{d}'s. Robjohn answered it but I find it complicated (in the sense that, the approach won't occur to me if I saw this question the first time). Sadly, robjohn is far from the only one who so indulges. if you ask WA for the antiderivative, you get some gamma function shenanigans. that gives the following plot: wolframalpha.com/… So I was trying to find lower bound differently than Rob's. what do you notice visually about that function 4:27 PM what is the go to book to find existence uniqueness results for first order ODEs to me it visually looks like it's strictly increasing @Monty I assume you don't mean systems. The classic serious book I studied from in college was Birkhoff-Rota. @Semiclassical it seems so. @TedShifrin what: complicated answers or mathrm{d}?$\mathrm d$, of course. Actually, for this sort of problem, at first glance, your approach seems the reasonable approach to take. 4:32 PM but my way failed as that limit value is 0 and not 1. @Koro Note where most of the weight of$x^a$lives for large$a$close to 1. compute where$1-\epsilon$of the integral of$x^a$is Then look at what$a^x$looks like there$\int_{\lambda}^1x^a\,\mathrm{d}x=\frac1{a+1}\left(1-\lambda^{a+1}\right)=\frac{1-\epsilon}{a+1}$I'll be back shortly. threats will get you nowhere. 4:43 PM Don't be so sure. 5:17 PM @TedShifrin no sorry I do mean systems of ODE Interested in the flows$\partial_t X(t)= V(X(t))$with$X(0)=\text{id}$of a vector field$V:\mathbb{R}^d\to \mathbb{R}^d$Hirsch/Smale is a great book. ODEs, dynamical systems.$x^a>1-\epsilon$for$x\in [\lambda,1]$so$I(a)\ge \int_\lambda^1 x^a a^x\ge (1-\epsilon) a^\lambda \int_\lambda^1 dx$6:04 PM The latest xkcd is mildly amusing. hah, i like it Burn the pulse and BP. I thought he was going to do an almost-anagram and interchange causality and casualty. Has munchkin called her mother or her teacher any more bad names? the older i get the more worthless those forms are. people are living forever and they all get diagnosed with stuff. it's not gonna help you treat me. If it weren't for three amazing surgeons, I'd surely be long dead. her mother was called a "bad mom" this morning for giving her two saltines. dad had given her three at dinner yesterday, when mom wasn't there. "you weren't there" also came up. 6:07 PM Saltines for breakfast? Ugh. saltines as an after-breakfast snack. On the way to school? no, at the table, like a civilized person. Civilized persons eat saltines right after breakfast? breakfast is usually a toasted waffle with a light coating of almond butter and some fruit and a glass of water. it's just easier to give her the saltines. she had a pretty bad meltdown last night. my wife had to go to the doctor and they took forever to see her. so i'd stupidly told her that mom would be home before bedtime and that didn't happen. not fixing the semi ambiguous hers although i would do that if i were billing for this in a legal document. 6:12 PM Yikes, I hope "her" is doing OK. fine now, but getting old sucks. @Koro I have added a second approach. 6:24 PM @leslietownes You folks are almost 30 years behind me. So please rephrase. getting incredibly young sucks. getting less young sucks? i'm 20. my wife is 18. Let's not regress too much further. i just realized that doesn't work out too well with my daughter's age. i'll recompute. i'm 25. that's about right. 6:28 PM Exponential decay? i'm aging like LaTeX version numbers. or is it TeX version numbers. 6:56 PM I think they're both frozen i thought they were designed to converge. like not 2.1, 2.2, but 2, 2.1, 2.18, 2.182, etc. anyway, that's what i'm doing. 7:08 PM @TedShifrin Hey, yea it is meant to be a summation on both sides. @TedShifrin Thanks a lot for the help with the triple integral yesterday. I understand now. Wow, @mechanodroid. I hardly helped. It's a pain in the butt to write out, but I think it does work (with four pieces) in spherical coordinates. 2 hours later… 9:04 PM @TedShifrin So the integral is not using the form that was named grotesque? @leslietownes TeX is converging to pi Looks like I ghosted Ted 9:22 PM @Koro I hope that the second answer is more understandable. @Ted welcome back. 9:40 PM I just indicated in chat that one should set it up with the vertex at the origin and use spherical coordinates. It's not pretty, but I thought about it enough to decide it's eminently doable. Perhaps grotesque, too, but no pretense of surface integrals. A doctor wrote to me to ask for differential geometric help with something called tortuosity in vessels inside the eye. They have some weird software that numerically computes it without divulging any of the "ingredients." It integrates$\kappa^2\,ds$and divides by the length. He gets numbers and wants to compare different eyes. I told him that mathematically, even with some constraints, it's just an ill-posed question. How can software compute that thing without being able to tell you the length itself? 9:59 PM @robjohn Can you tell this person that it is beyond bad form to delete questions once you've been told the answer? Granted, I didn't post an answer, but the question and comments might be helpful to others. @TedShifrin Ooh, that is BAD. 10:13 PM @TedShifrin done (and undeleted) 10:26 PM Thanks! 1 hour later… 11:43 PM Why is the notation$\zeta_n:=e^{2\pi i/n}$not standard? @AkivaWeinberger because it is a non-canonical choice of a primitive$n\$-th root of unity

one thought, random, is that it singles out a
oh, leaky got there
i remember using it in some analysis contexts where people are bothered less by arbitrary choices, but even there you're sometimes expressing something relating to all roots of a polynomial, not just one of them, and singling out any particular one of them for special indexing feels weird
i certainly prefer that notation being adopted once to seeing long expressions with exp(2pi i k/n) in them, which they love to do in some parts of signals processing