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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. $x$?
isn't "shut up and calculate" a call to application?
albeit, an authoritarian one
"Get the numbers out!"
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
@Jakobian Thanks, will take a look!
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
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?"
 
2 hours later…
4:31 AM
@XanderHenderson i have found this true at many levels higher than grad students...
 
1 hour later…
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
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.
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.
 
2 hours later…
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
8:14 AM
10
Q: Who was the first to say "Shut up and calculate!"?

user 85795The 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.

 
1 hour later…
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.
 
1 hour later…
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?
@copper.hat Indeed, but I've found it particularly annoying and prevalent among grad students.
11:52 AM
did you find it more prevalent among middle schoolers or high schoolers to try to mathematically run before they could walk
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...
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 $\mathcal{B}(\mathbb R^2)$.
 
2 hours later…
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?
And how, exactly, do you know that $\mathbb{R}$ is complete?
ya exactly, sorry
Well Ordering Principle
That the naturals are well-ordered by the usual ordering
every bounded above set of integers has a maximal element
this is equivalent to the well-ordering principle
Oh, wait... completeness of the reals is irrelevant here.
I read your argument without thinking.
I guess you also need to call upon the Archimedean principle to know this set is in fact bounded above
2:54 PM
You have a set of integers, bounded above by $x$. So the well-ordering of bounded sets of integers implies...?
@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"?
yes, at least that's one way of doing it
@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.
you can also define the floor in terms of the ceiling function and then the appeal to the well-ordering principle is more direct
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.
2:59 PM
Ceiling function defined as $\min \{ k \in \Bbb Z \mid k > x\}$?
or maybe not strict
Weak inequality.
OK, thank you all. Will review what I know from Enderton :) and then write this up in my notes.
3:37 PM
Hi! Do people have ideas about definitions of transverse intersections over non-closed fields?
 
1 hour later…
4:41 PM
hi friends
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?
 
2 hours later…
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?
@Sahaj hi
@nickbros123 Can you prove that statement?
@user85795 high schoolers
@nickbros123 The set or the sequence? (I believe) this matters
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?
7:31 PM
@EE18 Can you prove it?
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 :)
@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)
To be fair I did caveat with "I believe" :)
Let me think if rearrangements of series constitute an example
@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.
@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
7:34 PM
I retract my comment Xander. It doesn't matter
@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.
yh I think I have the proof
@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?
@XanderHenderson Tochka.
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."
7:40 PM
@nickbros123 yes
But not in general
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.
@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..
@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 $T_2$.
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...
I see
I was just working on Compact <==> limit point compact (for metric spaces)
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.
7:50 PM
I wasnt going to worry about it :)
8:47 PM
@XanderHenderson it is
Consider two point indiscrete space
Limits are not unique in non-Hausdorff spaces so that's reasonable
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
9:11 PM
Yes. I remember that
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?
You don't need to explicitly show a bijection. The pairing logic is enough.
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?
daily reminder that most proofs in maths don't go back to first definitions for everything
Agreed, but I should be able to do it on demand and in this case when I thought about it I couldn't :(
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
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 :)
so here is something to think about: a disjoint union of sets with even cardinality has even cardinailty
Depends on how good the intuition is. I don't prove most things
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!
10:37 PM
indexing here just comes down to choosing representatives for an equivalence relation
@Thorgott Agh that's a much cleaner way to do it. Thank you!
11:32 PM
@mick

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