Mathematics - 2020-11-25 (page 3 of 3)

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11:00 PM

It does remind me of something my discrete math prof told me once, as a funny reversal of the usual statement: "The continuous is sometimes a useful approximation of the discrete."

geocalc33

foliation of R^2 (take zero away) example?

Thorgott

I'm not sure what the boundaries of what's considered a "finite concept" are

Semiclassical

yeah

Shaun

That's a point I'm stuck on, too, @Thorgott.

manooooh

@Thorgott considering $x\leq y\iff x\not>y$, I see. What should be the correct phrase? I am thinking but nothing comes to my head

Thorgott

11:02 PM

I'd say the area of a bounded region in the plane or in space is a finite concept (in the sense that it should be a finite number), but good luck making sense of how to define this in any generality extending beyond like polyhedra without invoking infinity

manooooh

It's obvious that if $|A|<|B|$ then $|A|\neq|B|$, but here we have $|A|\leq|B|$...

Thorgott

@manooooh yeah, that simply isn't enough to conclude

manooooh

@Thorgott so?

Could I say this in words?: Note that $ V_t^* $ represents the set of all words of any length, therefore it is impossible that there is another set greater than it, since all the words (including the null word) are in $ V_t^* $.

Shaun

Right, I'm going to post the question soon. If it is heavily downvoted and/or closed, I can always delete it. Thank you for your thoughts.

Thorgott

so you need to make another argument

geocalc33

11:08 PM

a function was once thought of as a spot a point could be. Now a function is thought of as a set

mapping between two sets

crazy how mathematics has gone through so much change over the past few thousand years

shaun if you post it here you could get up to 16 upvotes?

mick

Q: Why does this weird iteration converge to the square root ??

Let $1 < x < 4$ , $a_1 = x$ and $b_1 = 0$. Now consider the (conditional) iterations if $a_n > b_n$ then $a_{n+1} = 4(a_n - b_n - 1)$ $b_{n+1} = 2(b_n + 2)$ else ( $a_n = b_n$ or $a_n < b_n$ ) $a_{n+1} = 4 a_n$ $b_{n+1} = 2 b_n$ Now consider $c_n = \frac{b_n}{2^n}$ Now define $f(x) = \lim_{n \to ...

any ideas ??

Shaun

Here goes nothing . . .

Q: Finite Concepts that Require Infinity.

Answers to this question, I hope, would address the objections one might have towards the (many different) foundations of infinity by examples of finite concepts - theorems, definitions, intuitions, perspectives, etc. - that require the infinite, whether that be in a proof, an important source of...

examples-counterexamples infinity big-list foundations

Semiclassical

hmm. it's always frustrating when someone's work is correct except for a minus sign error

especially when there was an opportunity to detect the problem later on and they missed it.

manooooh

@Thorgott yes, but how?

Thorgott

11:23 PM

i dont know

the statement you're trying to prove false doesn't even seem false to me, but I know nothing about languages, so I'm not gonna make an authoritative claim

Ted Shifrin

Howdy, @Semiclassic, stranger.

Thorgott

at the very least its definitely not "obviously" false

Fargle

Heya @Ted

Rithaniel

Happy Thanksgiving Eve

Thorgott

if you call something obvious, you should be able to prove it with immediacy

Rithaniel

11:25 PM

(Disclaimer: Thanksgiving Eve is not a thing)

manooooh

@Thorgott it isn't true that for any grammar, $|L(G)|=|V_t^*|$. Take for example $G=(\{S,A\},\{a,b\},P,S)$ where $P$ is $S\to aAa$, $A\to b$. Then $L(G)=\{aba\}$ but $V_t^*=\{a,b\}^*=\{\lambda,a,b,aa,ab,ba,bb,aaa,\ldots\}$ so $|L(G)|\neq|V_t^*|$

@Thorgott I can justify by an informal argument, but I would like to know how to prove it in a formal way

Thorgott

oh, I misinterpreted your original task

that example is everything you need

Ted Shifrin

Fargle!!

Thorgott

cause it proves that $|L(G)|\neq|V_t^{\ast}|$ for at least one grammar

Ted Shifrin

Thor ... any day can have an eve.

manooooh

11:31 PM

@Thorgott that makes sense. Sorry for my comings and goings

Fargle

"You have any plans for Tomorrow Eve?"

Semiclassical

@TedShifrin hey

been away, yeah

Rithaniel

Yeah, Thor, any day can have an eve

Semiclassical

@Thorgott on that note, i love the anecdote here math.stackexchange.com/a/151798/137524

"Perhaps another story will be illuminating: a professor of mine once made an assertion in lecture that I didn't quite see instantly. I asked him "is that obvious?" and he replied "yes." I asked him "is it obvious that that's obvious?" and, after a short pause, he replied "no.""

@TedShifrin it's a regular math-chat reunion

just in time for thanksgiving

Thorgott

yeah, love that anecdote

@Rithaniel I'm sorry!

Rithaniel

11:39 PM

How are things going on your side?

Ted Shifrin

Happy Thanksgiving, Semiclassic and Fargle.

Semiclassical

Same to you

Rithaniel

Heya Ted

Semiclassical

and a belated happy quasi-end-of-election-season

@Thorgott a play on words: that which is evident requires no evidence

Ted Shifrin

Hi Rithaniel.... and to you too.

Fargle

11:48 PM

Ditto, Ted. How're things?

Semiclassical

(though I guess you could also say that evidence is that which makes a conclusion evident)

Ted Shifrin

The world's a mess, Fargle, but I'm OK.

Fargle

Yeah. I know the feeling.

Semiclassical

hopefully less of a mess in a few months...hopefully

Rithaniel

The human mind is very good at identifying messes, to the point that we can find them even if they aren't there (This also gives rise to new messes, more often than not)

Ted Shifrin

11:51 PM

No, Semiclassic, those 70 million voters are still in this country.

Rithaniel

I have a strong feeling that a decent portion of those 70 million voted for their team, instead of thinking about whether they want to stay with that team

Ted Shifrin

The racists are emboldened ... Sorry, gotta disappear for now.

Rithaniel

I honestly look forward to a hindsight analysis of why people were so into Trump, maybe 20 years down the line when heads have hopefully cooled and all fact have hopefully come to light