Mathematics - 2024-06-22

copper.hat

12:09 AM

engineers can be prcise tooo

If $\{x\}$ denotes the fractional part of $x$, is $\{n\{nx\}\}$ dense in $(0,1)$ for ae. $x$?

user 85795

isn't "shut up and calculate" a call to application?

albeit, an authoritarian one

"Get the numbers out!"

Jakobian

12:34 AM

@copper.hat how about we first note that $$(n+h)\{(n+h)x\}-n\{nx\} = h\{(n+h)x\}+n\{hx\}$$

I think this might be equidistributed for x irrational

One can certainly apply Weyl criteria here (or maybe not)

Then it would be that $n\{nx\}$ is equidistributed too

I need to go to sleep but good luck

user 85795

cya

copper.hat

@Jakobian Thanks, will take a look!

Jakobian

1:39 AM

We can add any integer, so the sequence of differences is h(2n+h)x

Up to fractional part

Since nx is equidistributed, so is 2nx, hence 2nx+hx = (2n+h)x, hence h(2n+h)x

Alternatively {n{nx}} = {n^2x} which is known to be equidistributed for irrational x

Okay I just didn't realize it was n^2x

Xander Henderson

2:01 AM

@user85795 not necessarily, no.

I've met a TON of grad students who want to talk about high level stuff, but who never actually do any computations, and can't give basic examples off the cuff.

They often try to look for clever tools and big hammers to solve problems when what they should really do is just shut up and calculate.

"I'm trying to show that this series is bounded. I think that if I take the Fourier transform, I can apply this theorem from [X], which will let me view the series as a special Dirichlet series. Then a theorem from [Y] gives an upper bound, as long as this parameter is small enough."

"Okay... Cool. But did you just try a basic computation?"

"Uh... No?"

copper.hat

4:31 AM

@XanderHenderson i have found this true at many levels higher than grad students...

leslie townes

5:37 AM

i had a friend in grad school who was the opposite of that, haha, he seemingly never learned anything and would approach every problem like a blank slate

well not that he never learned anything, but he never tried to use anything that he learned

i was discussing someone with his officemate one day and he was there and not really paying attention, and then suddenly he interrupted and asked if we did something using some integral trick, and i realized he had been trying to evaluate an integral we had mentioned using, like, "calc 1" techniques and no other information about what it represented

and as admirable as that is as a sanity check or a tool in the toolbox, if that is literally the only kind of stuff you ever try, you wind up being pretty useless

he did finish a phd on time (more or less) but probably has not read any math since

Random Person

5:53 AM

Hey mods! Please go through this post on MSE regarding a Math SE user suspension: meta.stackexchange.com/q/400851 I suppose you people have more context than we do and can answer the user's queries.

gparyani

6:05 AM

@RandomPerson FYI there's a Math mods' office room. Might want to post there.

The deleted chat message referenced in the global meta post you linked is posted in that room.

Jakobian

7:55 AM

Exactly

Eggsactly even

@XanderHenderson me

Well I'm not a grad student but I classify as something close to one

But I'll admit, I am lazy. Who wants to do all those computations

user 85795

8:14 AM

10

Q: Who was the first to say "Shut up and calculate!"?

The best thing I could find on the internet was this apparently forgotten article from 2004: N. David Mermin, Could Feynman have said this?, Physics Today 57 (5), 2004.

psie

9:34 AM

I have seen in a couple of places, that $${\rm E}[\psi(S,T)]=\int_\mathbb{R}\int_\mathbb{R} \psi(s,t)\,P_S(\mathrm ds\mid T=t)P_T(\mathrm dt),\tag1$$where $\psi$ is a measurable function and $P_S$ is a regular conditional probability, and $P_T$ the law of $T$. From the law of the unconscious statistician (LOTUS), I'd expect $(1)$ to read $${\rm E}[\psi(S,T)]=\int_\mathbb{R}\int_\mathbb{R} \psi(s,t)P_{S,T}(\mathrm{d}s\mathrm{d}t).$$

I'm not sure I'm using correct notation for the joint law of $(S,T)$, but is it true that $$P_{S,T}\left(\mathrm{d}s\mathrm{d}t\right)=P_S(\mathrm{d}s\mid T=t)P_T(\mathrm{d}t)?$$

Maybe this is a definition somewhere (which I haven't seen) that the regular conditional probability can be expressed in terms of the joint law and the law of one of the random variables involved.

nickbros123

11:10 AM

lets say, hypothetically, I finish Hoffman Kunze Linear Algebra (give or take a few months from here), where can I apply that knowledge somewhat immediately?

Xander Henderson

@copper.hat Indeed, but I've found it particularly annoying and prevalent among grad students.

user 85795

11:52 AM

did you find it more prevalent among middle schoolers or high schoolers to try to mathematically run before they could walk

psie

12:23 PM

@psie something like this probably $$\int_A P_{Y\mid X}(B\mid x)\,P_X(\mathrm dx)=P(\{X\in A\}\cap\{Y\in B\})=P_{X,Y}((X,Y)\in A\times B).$$

where $P_{Y\mid X}(B\mid x)$ is the regular conditional distribution of $Y$ given $X$.

Take what I say with a grain of salt, I'm probably speaking gibberish...

psie

12:58 PM

Let's start out with $(\Omega,\mathcal{F},P)$ and $(\mathbb R,\mathcal{B}(\mathbb R))$. Let $X,Y$ be maps from the former to the latter. If I'm not mistaken, $P_{X,Y}$ is the pushforward of $P$ under $Z=(X,Y)$, so that $$Z_\ast P(A)=P\left (Z^{-1}(A)\right )=P((X,Y)\in A).$$ Thus, above in my reply to myself, I probably should have written $P(\{X\in A\}\cap\{Y\in B\})=P_{X,Y}( A\times B)$, as $P_{X,Y}$ is a measure on $\mathcal{B}(\mathbb R^2)$.

EE18

2:46 PM

Question for folks:

My text writes that the well-definedness of the floor function $\lfloor x \rfloor = \max \{k \in \Bbb Z \mid k \leq x \}$ is a consequence of the WOP

I don't see this at all. Wouldn't it be a consequence in fact of the completeness of $\Bbb R$ and then showing that the sup is in fact an integer in that set which is max?

Xander Henderson

And how, exactly, do you know that $\mathbb{R}$ is complete?

EE18

ya exactly, sorry

Well Ordering Principle

That the naturals are well-ordered by the usual ordering

Thorgott

every bounded above set of integers has a maximal element

this is equivalent to the well-ordering principle

Xander Henderson

Oh, wait... completeness of the reals is irrelevant here.

I read your argument without thinking.

Thorgott

I guess you also need to call upon the Archimedean principle to know this set is in fact bounded above

Xander Henderson

2:54 PM

You have a set of integers, bounded above by $x$. So the well-ordering of bounded sets of integers implies...?

EE18

@Thorgott You're saying I need to use Archimedean property in order to get a natural which is larger than $x$? In order to then use the "every bounded above set of integers has a maximal element"?

Thorgott

yes, at least that's one way of doing it

Xander Henderson

@EE18 Yeah, if you don't already have the naturals as a subset of the reals, with respect to the ordering on the reals, you would need that. I don't think that it is necessarily necessary in the given context.

Thorgott

you can also define the floor in terms of the ceiling function and then the appeal to the well-ordering principle is more direct

Xander Henderson

Or note that if $A \subseteq \mathbb{N}$ has a least element by the WOP, then $-A = \{-x : x \in A\}$ has a greatest element.

EE18

2:59 PM

Ceiling function defined as $\min \{ k \in \Bbb Z \mid k > x\}$?

or maybe not strict

Xander Henderson

Weak inequality.

EE18

OK, thank you all. Will review what I know from Enderton :) and then write this up in my notes.

neander

3:37 PM

Hi! Do people have ideas about definitions of transverse intersections over non-closed fields?

Sahaj

4:41 PM

hi friends

EE18

4:56 PM

A question as I go through an argument regarding the positional number system

Letting $g$ be the base of the system for context, one has $\sum_{k=1}^\infty(g-1) g^{-k} = 1$

Now in some arbitrary expansion of $0 \leq r < 1$, suppose I have that there is some digit $x_j$ with $x_j < 1$. Is that enough to guarantee that $\sum_{k=1}^\infty x_k g^{-k} < 1$ (i.e. strictly) or do I need this for almost all $k$?

I know in general that having elements of a sequence be strictly less than the elements of another is not enough to guarantee the same strict inequality of their limits

But that is in general. What about this case?

Arguing with some handwaving, I have $$\sum_{k=1}^\infty x_k g^{-k} = \sum_{k=1}^j x_k g^{-k} + \sum_{j+1}^\infty x_k g^{-k} < \sum_{k=1}^j (g-1) g^{-k} + \sum_{j+1}^\infty x_k g^{-k} \leq \sum_{k=1}^j (g-1) g^{-k} + \sum_{j+1}^\infty (g-1) g^{-k} = \sum_{k=1}^\infty(g-1) g^{-k} = 1$$ which I think does the trick?

nickbros123

7:27 PM

@XanderHenderson quick question reg. our brief discussion a few days back, when a sequence in metric space converges to x, the set {$x_n$} has only one limit point $x$ ? is that correct to say?

Jakobian

@Sahaj hi

Xander Henderson

@nickbros123 Can you prove that statement?

Jakobian

@user85795 high schoolers

EE18

@nickbros123 The set or the sequence? (I believe) this matters

psie

It looks like I've encountered two definitions of conditional expectation. The first one is, for any $\mathcal G$-measurable $U$, $$\mathbb E[XU]=\mathbb E\big[\mathbb E[X\mid \mathcal G]U\big].$$ The second one is $$\mathbb E\big[\mathbb E[X\mid \mathcal G]\boldsymbol 1_G\big]=\mathbb E[X\boldsymbol 1_G],\quad\forall G\in\mathcal G.$$ Are these definition equivalent?

Xander Henderson

7:31 PM

@EE18 Can you prove it?

EE18

I can prove that (convergent) sequences have unique limits yes

And I cannot yet prove but do know that certain sequences have different limits under rearrangement

Ergo my question :)

Xander Henderson

@EE18 That wasn't what I was asking. You asserted that it is different if you consider the set, rather than the sequence. Can you prove that there is a difference?

(e.g. give an example)

EE18

To be fair I did caveat with "I believe" :)

Let me think if rearrangements of series constitute an example

psie

@psie clearly, the second one seems to be a special case of the first one, but this bothers me, since Wikipedia doesn't seem to include the first one.

nickbros123

@XanderHenderson im in a bus rn, but I think I have the rough sketch: I can choose n so that $d(q_n,p)<\frac{1}{n}$, for all n> some natural number; I consider another limit point $q$ so that $d(q,p)>0$; $d(q_n,q) \geq d(q,p)-d(q_n,p)$, and i think i can argue about bringing $d(q_n,p)$ small enough

EE18

7:34 PM

I retract my comment Xander. It doesn't matter

Xander Henderson

@user85795 Not sure who you are asking the question of, or the entirety of the context, but the habit I was complaining about is not "walking before running". Most graduate students get to graduate school having already demonstrated an ability to wok "walk" (i.e. to work through computations). They just often fail to do so, instead hoping to apply big hammers to solve problems simply.

@EE18 Точно.

nickbros123

yh I think I have the proof

Xander Henderson

@nickbros123 First off, I think that you are using $n$ to mean two different things there. Second, I don't understand what you are trying to do. I would suggest that you consider what happens if you suppose that there is a second limit point to the set---put that point and the limit of the sequence as the centers of disjoint balls. How many points must be in each ball?

Jakobian

@XanderHenderson Tochka.

EE18

Note I am only including the pictures for context, I doubt they're necessary to read

The footnote 4 at the end of the second picture says "One should also check that no series constructed by this algorithm satisfies the condition xk =g−1 for almost all k."

Jakobian

7:40 PM

@nickbros123 yes

But not in general

EE18

I am trying to do just that but am currently unable to derive a contradiction when I assume that some expansion does obey that condition for almost all k.

nickbros123

@XanderHenderson ok so argument is after some n0 all x_n fall into the ball around the limit, so if it has to happen, only finite points can fall outside this, hence contradiction? thats clean..

Xander Henderson

@nickbros123 That is, more or less, the contradiction I was driving you towards, yes.

Note that this is a property of metric spaces, and doesn't generalize to all topological spaces. The argument I have in mind essentially comes down to the fact that you can always separate points with disjoint open sets (is this Hausdorfiness? $T_3$? I can never remember which separation axiom is which...).

Hausdorfiness. Which seems to be $T_2$.

Separation axiom

In topology and related fields of mathematics, there are several restrictions that one often makes on the kinds of topological spaces that one wishes to consider. Some of these restrictions are given by the separation axioms. These are sometimes called Tychonoff separation axioms, after Andrey Tychonoff. The separation axioms are not fundamental axioms like those of set theory, but rather defining properties which may be specified to distinguish certain types of topological spaces. The separation axioms are denoted with the letter "T" after the German Trennungsaxiom ("separation axiom"), an...

nickbros123

I see

I was just working on Compact <==> limit point compact (for metric spaces)

Xander Henderson

That said, I don't deal with point-set topology all that much, so while I can show that Hausdorfiness is sufficient, I am disinclined to try to figure out whether or not it is necessary. But you are working in metric spaces, and metric spaces are nice, so I wouldn't worry about it.

nickbros123

7:50 PM

I wasnt going to worry about it :)

Jakobian

8:47 PM

@XanderHenderson it is

Consider two point indiscrete space

Limits are not unique in non-Hausdorff spaces so that's reasonable

Soumik Mukherjee

Jun 17 at 17:06, by Xander Henderson

For example, the dumb space $\{0,1\}$ with the topology $\tau = \{ \varnothing, \{0,1\}\}$ has the property that every sequence converges to both $0$ and $1$. This is not a terribly interesting space otherwise, but it gives the example.

Heh, Xander himself gave the same example :P

Jakobian

9:11 PM

Yes. I remember that

EE18

9:44 PM

Knowing that its finite by hypothesis, how would you formally show that $t(G) = \{g \in G \mid g \neq g^{-1} \}$ has even order? Intuitively we "pair" $g, g^{-1} \in t(G)$ but I am struggling to formalize this -- i.e. struggling to show a bijection to some even natural. Any hints?

Soumik Mukherjee

You don't need to explicitly show a bijection. The pairing logic is enough.

EE18

If I wanted to (as I do) formalize this I would need to though surely? Having even order means by definition having a bijection to some even natural right?

Thorgott

daily reminder that most proofs in maths don't go back to first definitions for everything

EE18

Agreed, but I should be able to do it on demand and in this case when I thought about it I couldn't :(

Jakobian

16 hours ago, by leslie townes

i had a friend in grad school who was the opposite of that, haha, he seemingly never learned anything and would approach every problem like a blank slate

EE18

10:02 PM

I don't think that's a fair characterization here tho. Me having the intuition for why something is true is quite some distance from knowing or proving that that thing is true. If intuition were sufficient then this math thing would be easy :)

Thorgott

so here is something to think about: a disjoint union of sets with even cardinality has even cardinailty

Jakobian

Depends on how good the intuition is. I don't prove most things

EE18

Ah I think I see now Thorgott

Ugh formalizing this is nightmarish though. I think I would need to recursively produce a subset $A \subseteq t(G)$ where for each $g \in A$ we have $g^{-1} \notin A$. $A$ is my "index set" in what follows. Then I'd have to show that $t(G) = \bigcup_{g \in A} \{g,g^{-1}\}$. Then each of these doubletons has cardinality 2 and I can use the result you allude to: finite unions of disjoint sets let us sum over the cardinalities

Won't dwell any longer. Thanks all!

Thorgott

10:37 PM

indexing here just comes down to choosing representatives for an equivalence relation

EE18

@Thorgott Agh that's a much cleaner way to do it. Thank you!

HighAsAKiteOnMath

11:32 PM

@mick

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