hi

Hi :-)

@TeresaLisbon Your argument (for the convex geometry problem) looks good to me. Thank you!

@epsilon-emperor Oh , nice to know. Thanks!

@NazmulHasanShipon Sure, let me know what you got, and I am happy I managed to help you out.

Just curious, what are you studying nowadays?

Looking at spaces which satisfy a certain inequality common for PDEs, called the Harnack inequality.

The Harnack inequality is a controlling inequality for harmonic functions (Laplacian equals zero) in a ball. It basically states that a harmonic function cannot vary too much inside a ball, if it's harmonic in a slightly bigger ball around this ball.

We are trying to find equivalent conditions on a space that ensure that the HI holds there.

Remarkably, this has found a connection with probability, where you can use probabilistic techniques to show that HI on a large number of spaces is equal to "volume doubling" and an "isoperimetric inequality".

hello! can someone please answer my question in CURED?

@vitamind Done!

So probability comes in out of nowhere into PDE, @epsilon-emperor. Not the first way for it to enter, and probably not the last way. But it's beautiful, because there are times when I cannot believe what I read, and how far we've come in the whole theory. It's a magnificent field.

@TeresaLisbon Thanks!

Sure!

hey Teresa Lisbon, hope you are doing well! I have a quick question for you if you have time: is it true that $!(s \leq M) \rightarrow (s > M)$? say, if I know $B$ is an upper bound but $M$ is not an upper bound, and I want to show that there is an $s$ in the sequence greater than $M$

woops, it's meant to be negation

didn't have time to edit, meant $\neg$ instead of $!$

@shintuku Yes, why not? It makes sense to say so. [Any upper bound (probably of some set)] will be bigger than [any number which is not an upper bound of that set].

And that statement you mention is true.

"For feedback/discussion/requests of Close/Undelete/Reopened/Edit/Delete for questions and answers on Math SE" Shouldn't it be 'reopen' instead of 'reopened'?

So , if there's no $s$ in the sequence that is greater than $M$ , then $M$ must be an upper bound @shintuku.

ahhh, that statement is easier to understand than mine

thanks a lot!

@Wolgwang Good point, we can mention it there.

@shintuku Welcome!

@TeresaLisbon Done :)

@Wolgwang I saw it! I think I can edit it, but I'll leave it to the rest to decide whether to do it or not.

Yes.

@Wolgwang Oh I forgot to talk about your question? But replacing every + with a \vee and . with a \wedge should do your job. Actually, there's no problem I see with an expression like 1 \wedge v or something , but to eliminate the 1, you just need to use the fact that 1.a = a for all a and 1+b = 1 for all b. If you use that, you get rid of the 1s, and then you can put the \wedge and \vee in place.

It's just a change of notation, so it can always be done.

There's also another thing the author missed, which could have simplified a step.

Indeed, we have $X+XY+XZ+YZ$. Instead of going with just removing X from X+XY, (s)he could have written X(1+Y+Z)+YZ and used the laws 1+A =A and 1.B = B to directly get X+YZ.

@TeresaLisbon She

Thanks :-)

@TeresaLisbon Have you seen 1 written with wedges and vees in some textbook?

@TeresaLisbon Sounds really cool! Motivates me for my Probability Theory and PDE courses next year.

@Wolgwang Yes : I don't think it's out of place in the textbook I remember reading when I was in school and learning Boolean algebra. It may be a case of different countries, different notations (because I used an American authored textbook, if I recall).

@epsilon-emperor Yes, this will really be of great motivation.

Also sidenote: Are you in Karnataka LOL? I deduced that from some old chats here.

@epsilon-emperor That is true.

@TeresaLisbon Interesting. I'm an undergrad in India too. Although my username may suggest I'm from Greece (epsilon).

Oh, I see. I didn't know that.

Whatever it is, MSE is the best place for Indian undergraduation students. It's the best community I can think of which will instantly raise your profile.

I really think it benefitted me. I'd love it if I can give back to others and to the site. I think I'm doing this right now.

Got to go, so see you in some time!

@TeresaLisbon Raise profile in the sense?

@TeresaLisbon See you!

@epsilon-emperor She wanted to say this is the best place where you can become famous and well respected for your maths (or any subject) if you post maths wonderful answers.

Makes sense, thanks!

@epsilon-emperor Btw if you want an example although I know you have understood you type in google amWhy. The first three results will be of one of our users of math stackexchange - amWhy. Also if you would try your name mentioned in your profile your profile will come as first two results and one can know the good quality post you wrote.

@JitendraSingh Yes, very cool! I'll probably change it to my name sometime