Mathematics - 2017-07-10 (page 4 of 6)

Akiva Weinberger

5:01 PM

Why do we want a 3D version?

(Was that in response to Secret or to me?)

Semiclassical

you, but not very clearly.

but I was misunderstanding what dyadic rationals were :P

Having actually looked up the definition, there's no need for anything 3D.

Secret

@AkivaWeinberger Indeed, we have discussed this uncountable chain solution of yours with steamyroot and alessandro about 2 months ago

Akiva Weinberger

Oh, so we have

…Wait, did I use the same exact image then?

Whoops

Secret

yup, you did

Akiva Weinberger

Now I'm wondering if you could do 3D Ford circles though

Ford spheres

Hrm arxiv.org/abs/1503.00813

Secret

5:09 PM

Speaking about set theory, recently due to trying to understand what admissible ordinals are, I end up reading this:

Model theory

In mathematics, model theory is the study of classes of mathematical structures (e.g. groups, fields, graphs, universes of set theory) from the perspective of mathematical logic. The objects of study are models of theories in a formal language. A set of sentences in a formal language is called a theory; a model of a theory is a structure (e.g. an interpretation) that satisfies the sentences of that theory. Model theory recognises and is intimately concerned with a duality: It examines semantical elements (meaning and truth) by means of syntactical elements (formulas and proofs) of a corresponding...

It's pretty cool, though, and I am suprised that the use of the word "theory" can be made somewhat analogous to the word "theory" in physics

Akiva Weinberger

Model theory (or whatever tiny bit of it I know) is great

Secret

The rough analogy is as follows:
Physics: Theory/models predict what happens in nature
Mathematics: A model is a structure that satisfy a theory, so in a sense, theory in mathamatics played the role of nature in physics, and structures/models in mathematics played the role of physical models in physics

Akiva Weinberger

I took a five day class on it at MathCamp

Maybe I just like it because the instructor was amazing

(Shoutout to the amazing Noah Schweber, who uses MSE by the way)

@Secret I gotta say, though, the phrase "the theory of dense linear orders" feels strange

(Fun fact, all countable models of the theory of dense linear orders without endpoints, aka DLO I think, are isomorphic)

(I think I've mentioned that before but not with that terminology)

Secret

countably infinite?

Akiva Weinberger

You can't have a finite dense linear order, so it doesn't matter

You can't have a finite linear order that has no endpoints either

An example of a dense linear order without endpoints is $\Bbb Q$

Secret

5:18 PM

ah right, because the necessary condition $\forall x,y, x < y \exists z : x < z <y$ will guarentee infinite number of elements anyway

Akiva Weinberger

Right, that's one statement of the theory

I forget whether "theory" is like the ordered pair of (alphabet, axioms) or if it's just the set of axioms

or if it's the set of all statements provable from the axioms

Semiclassical

I like questions which I can answer with 6-word hints, heh.

Akiva Weinberger

What question?

Secret

alphabets in a formal language can include operators and relations such as =, <

Akiva Weinberger

"Just do the frickin' problem, dumbo"

@Semiclassical

Semiclassical

5:21 PM

It's a linear algebra problem, with the hint being "Take the determinant of both sides"

I could probably even just have done "Consider the determinant."

Akiva Weinberger

@Semiclassical "…and shove it up your —!"

Semiclassical

(They were given a matrix and asked whether it could be written as $ABA^{-1}B^{-1}$. The matrix doesn't have determinant 1 or 0, so...)

Akiva Weinberger

Oh cool

Secret

Meanwhile, I still cannot remember the sufficient condition for a dense set. In general, I cannot recall anything I don't fully understood unless it sounds "interesting" enough (the converse is not true either due to memory fade)

Akiva Weinberger

Does the converse hold?

@Semiclassical It can't have determinant 0

Semiclassical

5:22 PM

...well, no, no it couldn't. derp

Akiva Weinberger

because $A^{-1}$ and $B^{-1}$ exist

Semiclassical

yeah.

This is why short hints are better: They prevent me from saying too much :P

Akiva Weinberger

"Do this, do det"

Semiclassical

"Do det to dis"

Though now I do wonder: Can every matrix of determinant 1 be factorized that way?

Just to be clear, what are the two directions?

Certainly $\det (ABA^{-1}B^{-1})=1$.

Akiva Weinberger

The converse is what you just said, sorry

Semiclassical

5:26 PM

mmkay

Akiva Weinberger

If every matrix of determinant 1 can be written in that form

Assume everything's diagonalizable maybe?

Semiclassical

"det=1 is a necessary condition, but is it sufficient?"

I feel like I've seen this question before.

Secret

It is if you are working with finite dimensional vector spaces

Akiva Weinberger

Does determinant even make sense otherwise

parvin

hi

is this correct?

Akiva Weinberger

5:28 PM

Hai

p(nx) = n . p(x)

Maybe

:))

In mathematics, the Fredholm determinant is a complex-valued function which generalizes the determinant of a matrix. It is defined for bounded operators on a Hilbert space which differ from the identity operator by a trace-class operator. The function is named after the mathematician Erik Ivar Fredholm. Fredholm determinants have had many applications in mathematical physics, the most celebrated example being Gábor Szegő's limit formula, proved in response to a question raised by Lars Onsager and C. N. Yang on the spontaneous magnetization of the Ising model. == Definition == Let H be a Hilbert...

Semiclassical

laptop battery is running low; once it dies, I'm going to get lunch.

Tobias Kildetoft

5:28 PM

@AkivaWeinberger Not generally, no (thought it might for bounded operators, I am not an operator theorist)

Akiva Weinberger

What's $p$? @parvin

Semiclassical

@Secret hissss

Tobias Kildetoft

Ahh, and that link seems to confirm my guess

Semiclassical

Fredholm determinants are something I should probably understand but I really don't

parvin

@AkivaWeinberger probability

Akiva Weinberger

5:29 PM

Ohh

Semiclassical

Mostly because I know a lot about Szego's limit formula and its extensions

Secret

NB, I don't really understood what I found except at the superficial level. To understand that I often need to do some heavy reading

Akiva Weinberger

I don't remember much probability. Is $x$ a random variable? @parvin

I know the expected value is linear, so E(nx)=nE(x) if n is a constant and x is a random variable

Tobias Kildetoft

@parvin If $p$ denotes a probability then that equality will hardly ever be true

Secret

2

Q: How to determine the trace and determinat of a differential operator?

How to determine the trace and determinant of the operator like $\Box$ or $\nabla^2$ etc. But first of all how to find the same for the simpler operator $\frac{d}{dx}$? I proceeded as follows. What basis functions should I choose- $\{e^{ikx}\}$ or $\{\delta(x-x^\prime)\}$? Since, the first basis ...

parvin

5:31 PM

what if it is the pr of independent events?

Tobias Kildetoft

@parvin That does not make sense on its own

Semiclassical

Where I've run into Fredholm determinant stuff is when one wants to consider the spectrum of a structured matrix as the dimensionality of said matrix goes to infinity

But the math gets tough quite fast

parvin

@AkivaWeinberger yes that's why I wanna know if it's true for p or not, for we have E[nx]=$\sum nx.p(nx) = n * \sum x*p(nx)

then what will happen to p(nx)?

Akiva Weinberger

@Typhon Say $p$ has rational coefficients, and $p(x)=0$. Find a polynomial $q$ with rational coefficients with $q(x^2)=0$.

(^Unrelated. Different $p$)

Secret

Meanwhile in chemistry: One of my calculations said my complex don't like benzene. The stable structure end up having the bezene lying 3A away from the Rhodium complex in question

Semiclassical

5:34 PM

Probabilities are not linear functions of their indices

Astyx

How do you pronounce the second "i" in sacrifice (the noun) ?

(#math.stackexchange)

Tobias Kildetoft

@Semiclassical Do people in chemistry actually shorten it A?

@Secret I mean (wrong ping)

parvin

what do you mean ? @Semiclassical

Secret

No, its actually Å

Tobias Kildetoft

@Secret Ahh, good

Akiva Weinberger

5:36 PM

@Astyx Like in "price"

parvin

what does it have to do with p(nx)=n(px) ? @Semiclassical

Secret

(I sometimes type A because I am too lazy to find the letter)

Astyx

@AkivaWeinberger Right, that's what I thought

Tobias Kildetoft

@Secret Just switch to a Nordic keyboard then :)

@parvin Because that is linearity

Akiva Weinberger

(which, in my accent, is slightly higher in the mouth than the vowel in "prize," but whatever) @Astyx

Semiclassical

5:36 PM

my point is that that condition is not something which is in general true

Secret

Anyway, this calculation suggests I can rule out arenes as ligands to these complexes

Astyx

@AkivaWeinberger And in no context can it be said the same way the first one is right ?

Akiva Weinberger

What, like sacriffiss?

No

Semiclassical

Indeed, I'm not sure off the top of my head whether any distribution works that way

Akiva Weinberger

It's the "hat" vowel, "fit" vowel (or maybe a schwa I guess), and the "price" vowel, in that order @Astyx

Astyx

5:38 PM

@AkivaWeinberger Cause one of my english examiner said it so, and it really confused me

Akiva Weinberger

I've never heard that

I mean, could be a regional thing

Tobias Kildetoft

@AkivaWeinberger Definitely not the same vowel as in "fit"

Astyx

She was possibly wrong, she was obviously french

Tobias Kildetoft

more like the vowel in "grit"

Akiva Weinberger

I pronounce those the same?

@Astyx Maybe she was confusing it with "edifice"

Tobias Kildetoft

5:40 PM

@AkivaWeinberger now there you have the same vowel as in "fit" both times

Astyx

@AkivaWeinberger That's a possibility

Tobias Kildetoft

@AkivaWeinberger @Astyx Obligatory xkcd xkcd.com/1816

Akiva Weinberger

@TobiasKildetoft OK, but I don't know how else you'd pronounce "grit"

Where are you from?

Tobias Kildetoft

@AkivaWeinberger Denmark

Astyx

My math oral exam this morning : let $f:\Bbb R^2\to \Bbb R^2$ be $1$-lipschitz for the euclidean norm. Prove that for all $x$, ${f^{\circ n}\over n}$ converges to a limit $l$ that does not depend on $x$.

You have 35 minutes

Tobias Kildetoft

5:45 PM

I don't think I would even be able to make the same vowel sound as in "fit" with the consonant sounds on "grit". It would come out as "griit"

Secret

There are oral exams in maths??

Astyx

"Epitome"'s "i" is pronounced the same way as "fit" right ?

Tobias Kildetoft

yeah

shai horowitz

hello

Astyx

@Secret Many

Hi

shai horowitz

5:48 PM

what does it mean for primes to have arithmetic sequences if arbitrary size? more specifically does this mean countably infinite sized arithmetic sequences?

Astyx

The answer to the second question is no

Tobias Kildetoft

@shaihorowitz I don't think anyone claims that the primes have arithmetic sequences. Do you mean primes in arithmetic sequences?

shai horowitz

yeah probably what i mean

"there exist arbitrarily long sequences of primes in arithmetic progression"

Akiva Weinberger

Say a big number

Astyx

4

Akiva Weinberger

5:50 PM

There exists an arithmetic sequence of primes of that length

Astyx

Wooow :o

shai horowitz

@Astyx why doesnt this imply a countably infinite progression

Akiva Weinberger

@shaihorowitz Consider this sequence

1,(skip),3,4,(skip),6,7,8,(skip),10,11,12,13,(skip),etc.

(fixed)

So, like, 1,3,4,6,7,8,10,11,12,13,…

That contains consecutive sequences of arbitrary length but no infinitely long consecutive sequence

I realize again I messed up because that doesn't guarantee against arithmetic sequences of common difference other than 1

Revenant

Hello guys

Akiva Weinberger

How about this: 1, 10, 11, 100, 101, 102, 1000, 1001, 1002, 1003, 10000, 10001, 10002, 10003, 10004, etc.

Revenant

6:00 PM

I have got a quick question

Akiva Weinberger

That should have arbitrary long arithmetic sequences but no infinitely long one ^

@shaihorowitz

Revenant

what would be the formula to calculate % of number 40 if 0 is 30 and 100 is 70

Akiva Weinberger

So if the input is 0, the output is 30; if the input is 100, the output is 70; and you want to know what input has an output of 40? @Revenant

(I think that's what you mean; this isn't percents)

shai horowitz

could you rephrase? @Revenant

Revenant

hmm I'm thinking of what would be best way to ask

between 30 - 70 what would 40 be in %

Akiva Weinberger

6:05 PM

I don't know what you mean by % here

Did my rephrasing of the question above make sense? ^

Revenant

hmm no, not really

I mean if min number would be 0 and max number would be 100 than given input:40 would be 40%

what if the min number would be 30 and max number would be 70, what would given input: 40 be in %?

Akiva Weinberger

So, like, 50 (which is exactly halfway between 30 and 70) would be 50% of the way?

Revenant

yes

Akiva Weinberger

So how far is it from 30 to 70

70-30=40 units, right?

Revenant

right

Akiva Weinberger

6:09 PM

Now, the number 40 (from the question) is 10 units away from 30

So that's 10 units out of a possible 40 units

GFauxPas

math.stackexchange.com/questions/2353840/why-is-enx-nex/… lol, the answer that doesnt address the OP's question got +4 and my answer got +0 :P

Akiva Weinberger

What percent is the fraction 10/40?

GFauxPas

Such is life

Revenant

ah that makes sense @AkivaWeinberger thank you

Akiva Weinberger

@Revenant So the answer is…?

Revenant

6:11 PM

25 @AkivaWeinberger :)

Akiva Weinberger

Yep.

And I think the right way to phrase it would be,

Revenant

thank you very much :)

Akiva Weinberger

40 is 25% of the way from 30 to 70

Astyx

@shaihorowitz I guess your question was answered (sorry, I was away)

Akiva Weinberger

I think "of the way between" works as well

This is vaguely relevant to the notion of a convex combination, I guess

(Only vaguely)

shai horowitz

6:14 PM

@Astyx no i'm still not sure i understand why it isnt implied to have a progression of countably infinite size

Akiva Weinberger

What about my example? @shaihorowitz

1, 10, 11, 100, 101, 102, 1000, 1001, 1002, 1003, 10000, 10001, 10002, 10003, 10004, etc.

You can see it has arithmetic progressions of arbitrary size

but no infinite progression

Tobias Kildetoft

@shaihorowitz Let's take a different thing: a sequence with arbitrarily long sequences of 1's but no infinite sequence of 1's.

Akiva Weinberger

Whatever the common difference is in your arithmetic progression, it'll eventually get swallowed by the huge ever-increasing jumps in the sequence

shai horowitz

i'm not sure how to imagine such a sequence

Akiva Weinberger

10110111011110111110…

Tobias Kildetoft

6:16 PM

@shaihorowitz just take 1011011101111011111...

Akiva Weinberger

Like that? @TobiasKildetoft

Astyx

sniped

Tobias Kildetoft

heh

shai horowitz

its the ... thats the problem why cant i claim that at spot omega there is a string of omega ones

Akiva Weinberger

There is no spot omega.

There's no "end".

Every single number in that string is a finite distance away from the start.

shai horowitz

6:22 PM

to make it clear, its the definition of aribitraily long that i'm getting stuck with i think

i take it to mean give me an integer and i can give you a larger integer, and im trying to reconcile omega as an integer that has the property that given any two integers with one is omega then omega is the larger integer

Imago

I am currently trying to develop an understanding about basic probability theory, in particular about the term of a probability distribution. Would there be someone, who would counter check my understanding?

Astyx

@AkivaWeinberger Overdramatic music

evil laughter

shai horowitz

@Imago i think you're good to go

Imago

ok

Tobias Kildetoft

@shaihorowitz arbitrarily long means that for any natural number there is a sequence with at least that length

and omega is not an integer

shai horowitz

6:29 PM

got to work on my ordinal axioms

parvin

I saw some teacher said E[X] is equal to "mean" , why is that? shouldn't it be equal to average?

shai horowitz

@parvin what is a mean

Astyx

Mean means the same thing as average

shai horowitz

most of the time

GFauxPas

theres also more than one type of mean

shai horowitz

6:36 PM

more then one type of average

Astyx

^

GFauxPas

but "average" refers to things other than "mean"

as well

Astyx

We usually say expectancy in this case

GFauxPas

or expeCted VaLue

except without hitting the caps lock key randomly

Astyx

ExPeCtEd vAlUe

GFauxPas

6:37 PM

lol

shai horowitz

integral from a to b of P(x)/(b-a)

Astyx

Not necessarilly, that's if you have continuous probabilities

Akiva Weinberger

@shaihorowitz If you're a nonnegative integer, you're an ordinal, but the reverse is false

Most ordinals are not integers. All integers are finite.

(Also, negative numbers are not ordinals, but that's irrelevant)

Astyx

Are there more ordinals that are not integers ?

Akiva Weinberger

You mean the reverse? @Astyx

"Finite ordinal" is synonymous with "nonnegative integer"

Tobias Kildetoft

6:42 PM

@Astyx More than what? There are so many ordinals that they don't even form a set

Astyx

This answers my question :)

Akiva Weinberger

Oh, I misunderstood the question

Astyx

It wasn't very well formulated

Akiva Weinberger

You can get sets of ordinals with cardinality greater than $\aleph_0$

There's a lot of ordinals

In any case, @shaihorowitz, ordinals are unrelated to the problem. Sequences are indexed by the natural numbers, not by the ordinals.

parvin

:/

@Astyx ow not always! just if you have "even" number (discrete) of events, thus you have avg = mean. but not always.

he said E[x] is avg (same definition) but why mean?

Akiva Weinberger

6:46 PM

@Astyx Given any set of ordinals, we can find another set of ordinals of higher cardinality.

(That's why they can't form a set, by the way)

Astyx

@AkivaWeinberger That's cool

Akiva Weinberger

(Well, the usual argument for nonsettiness goes a bit differently but whatever)

Astyx

I'll be back in a while, gotta fly

Tobias Kildetoft

@parvin How do you define the mean? I think you might be confusing it with the median here

Akiva Weinberger

♫ Fly away, fly so fast… ♫

(Making references to things none of you could possibly know about)

> Lemma $1$. Let $X$ be a continuum [a compact, connected, Hausdorff space]. If $F$ is a non-trivial closed subset of $X$, then for every component $C$ of $F$ we have that $\text{Bd}(F) \cap C$ is non-empty.

Is there a counterexample to the above for compact noncontinuum things?

Oh wait, $X=[0,1]\cup[2,3]$ works

Is there a counterexample for non-Hausdorff but still compact and connected things?

parvin

7:06 PM

you are right! i'm not so used to these definitions in English. checked it it was median u r right. thanks @TobiasKildetoft

Tobias Kildetoft

@parvin Yeah, there are a bunch of those terms that mean (no pun intended) similar things, and they are easy to get mixed up.