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00:29
Stupid question: when Serge Lang writes, for example: "Let L, L' be parallel lines ..." (in Basic Mathematics), how is "L'" pronounced?
I would say like the letter: "ell".
But L is already pronounced "ell". How do you pronounce L' to distinguish it from L?
Oh, I missed the prime.
"ell prime".
Ah, thanks. I didn't know that. (I only know how it is said in Dutch, but that doesn't help much.)
Yeah, in English, $[\text{thing}]'$ is pronounced "thing prime".
00:35
And there are different Unicode codes for prime and apostrophe.
@Tsundoku Sure, probably. But in typeset mathematics, both LaTeX and MathJax interpret f' as "eff prime", not "eff apostrophe".
And there really is no such thing as an apostrophe in mathematical expressions. It is always a prime.
Well, typing an apostrophe is easier than remembering how to enter a prime using Unicode references. Since I write maths on paper, I have had no need to worry about that.
@Tsundoku No one in mathematics is going to be able to tell the difference between an apostrophe and a prime in un-TeX'd mathematical writing. Or rather, if they do notice (since I am the kind of person who would notice), they won't care.
xander i would be so afraid of going into the latex SE and learning that f' is now deprecated and it's supposed to be something else.
@leslietownes ' is actually, under the hood, a macro for \prime.
00:46
with like a 50 paragraph explanation of how ' was tex from the 1980s and nobody in their right mind would have dreamt of doing it after 1992 when [some squiggle code] became available
I don't think I've ever asked here
Does anyone know who Cleo was?
@EE18 In what sense? There was a user on Math SE named Cleo, who answered a number of questions involving "difficult" integrals with cryptic and unenlightening answers.
If you are asking who the "real world" person behind that account is, don't. Doxxing is against the rules.
Fair enough, I was doing the latter
Well, don't.
01:12
yep
01:25
Welp, I proved twin primes (again) 🤓
Previous proofs all had errors, but not seeing an error in this one
Topological, Furstenberg-inspired, but totally different - argument by density of semiprimes
Uses two applications of Dirichlet's Theorem
@leslietownes
@Jakobian hi
@HighAsAKiteOnMath No, you didn't.
I know I know
But it's useful to find out where the error is
I'm meeting tomorrow with a professor to discuss the theorem I've proven
Hopefully its not totally worthless
01:43
I am really, really tempted to write a book on fractal geometry titled "No, That Isn't a Fractal, Dumbass."
@Jakobian No, dumbass.
02:00
LOL
 
2 hours later…
03:34
Whether or not something is a fractal is subjective
@HighAsAKiteOnMath No, not really.
Unless you're using the math def
Unless you consider the choice of definition to be subjective.
oh, the math def
Fraction vs fractal.
03:51
Okay, topological proof of twin prime infinitude is up
I fixed all the errors I found
Leslie this is an example of one of those questions we were discussing where the limit's not given to you, and I'm failing miserably in determining what it might be
I played with a few examples on Desmos and it seems to go somewhere near $a^{1/3}$ I think? But not quite sure
04:08
so the algebraic equation you get here is L = a/(1 + L) (which is equivalent to a quadratic in L that generally has two roots, but from context you know that L will be nonnegative if the limit exists, which in this instance is enough to tell you which root)
@leslietownes I know that because I started with a positive number and I'm just doing field operations on positive numbers, so the limit (if it exists) can be at least 0?
if your boy enderton were here, maybe he and you could spend half of a page proving that the unique sequence x_n that exists satisfying [and therefore defined by] that recursion is positive for all n. but yes, that's where you'd end up
you'd end up with L = a/(1 + L) and L >= 0 like the cs majors :)
Naturally :)
Now one last question if possible. How do we establish the limit exists? I can see its bounded but not obvious that it's eventually increasing or decreasing?
Maybe it should be
i would not expect a question like this to be 'obvious,' i would expect it to be a somewhat challenging exercise in (a) finding out exactly what you're wondering about, i.e. is this thing going to be eventually monotone and if so which way, and (2) finding the right thing to induct on so you can prove that and then also the limit
because life is hard i'm guessing that it's going to eventually be monotone increasing, because if it were eventually monotone decreasing, the proof that it is bounded below is too easy (0 being the obvious lower bound)
but that's meta-gaming the homework exercise and not actually thinking about anything, let alone following any kind of general program
Surely the upper bound is easy too? Namely $a$?
an upper bound*
04:19
i would not let any sequence of exercises like these fool you into thinking that you can do this with x_{n+1} = f(x_n, a) for any 'simple' f you can write down (see e.g. the logistical map)
okay, see i wasn't thinking
but that is the part of the problem that can, in general, get very hard, even if L = f(L,a) is easy to solve algebraically
The part being establishing that the limit exists?
the (1) and (2) above
also, note that a is not necessarily an upper bound for the whole sequence, although max(x_0, a) might be :)
@leslietownes That is well taken. That's the level of formality my guy Enderton likes to see
which also illustrates a theme, which is that even if the general behavior is easy to analyze (e.g. 'eventually monotone'), it is very common for there to be some static at the beginning arising from a choice of initial condition
Got it. OK, will get after it here...hopefully successfully before bed. I did want to ask one thing about the next exercise if i can. When I apply the method we discussed of solving $L = f(L,L)$ I get $L = L$ which gives no information. Is this problem not amenable to the aforementioned strategy then?
04:22
and as the logistical map illustrates if the general behavior is not easy to analyze you could expect it also to be reflected in the choice of initial condition really mattering
EE18: indeed, you will have to be more clever
Disaster, asking me for cleverness. OK, time to work. Thanks as always for the enlightening discussion and help
04:39
i think from now on instead of your boy enderton his name will be 'herb'
as in, "i think you'll have to get out the herb for this one"
Not the most famous American named Herb in my books im afraid
Without telling me how to show it, are we sure this actually is eventually monotonic? All the examples I put into Desmos seem to show oscillation?
Like if I "knew" it was Cauchy I could use that easily but I can't use that yet
no, i'm not sure at all, i haven't thought about it
but it must have simple behavior if it's an exercise in a book like this
so if "oscillation" means what i think it does, that's not too bad as behavior goes
e.g. if the even terms are eventually monotone and the odd terms are eventually monotone, something like that
if there's one word it might help to banish from your vocabulary, even in informal chat, it might be "oscillating" (nothing wrong with it here but when you teach real analysis in a classroom people start putitng it into their solutions to exercises and its just a nightmare of ambiguity)
i guess you replace step (1)'s "find out if this thing is going to be eventually monotone or what" with "find out whatever the simple behavior is", assuming that it's a solvable textbook exercise it will have simple behavior
i was definitely not trying to certify for you that "for some galaxy brain reason that requires no work, this is going to be monotone and the exercise is to get you to see that galaxy brain reason"
nothing like that exists
step (1) generally sucks see e.g. the logistical map
i wonder if there's a book that just tosses x_{n+1} = r x_n (1 - x_n) as like #3 on a list of 5 otherwise standard exercises
04:57
Just thinking about your comment for this exercise. Need to do the legwork of formalizing that into the conclusion that the sequence itself converges but hopefully I can
Lol isn't there some textbook that has Riemann Hypothesis in it?
Tongue in cheek no doubt but still...imagine self-studying...
god, haha, it wouldn't surprise me
humor in the form of exercises in a math textbook is definitely 'a choice'
presumably in any book where you could even formulate the RH it would be hopefully be an obvious joke by the time you got to it
i guess the sadistic thing would be to put something non obviously equivalent to RH, or at least implying RH, into the exercises
05:13
WLOG let $x_1 \geq x_0$ (I think $x_0 > x_1$ is treated in a complementary way). Then I can show by induction that for all $n$, $x_{n+2} \geq x_n$ and $x_{n+3} \geq x_{n+1}$. Thus the "even" and "odd" subsequences are monotonic (and bounded as discussed earlier) and so converge to $L_e,L_o$. Further, I can show by induction that the odd subsequence elements are greater than the even subsequence elements. If $L_e \neq L_o$ then we must have $L_o > L_e$ ()$L_o < L_e$ impossible because the odd
elements greater than even elements.
Now take $\epsilon = (L_o - L_e)/2$. I suspect I'll manage to get some contradiction. Does this seem like the right strategy or is there a cleverer way?
you can probably write down a recursion relation satisfied by either the even or the odd subsequence and use the limit algebra trick on that to find what those respective "L's" have to be (and thus prove they are the same without epsilonology)
arguable hint wolframalpha.com/…
There's that same L
Much cleaner, thanks very much Leslie!
Not for a couple hundred pages it seems
well some sequence version of it (it can be stated in absurd generality) is sometimes helpful here although even in simple examples there can be some work involved in identfying a domain on which the function you'd want to apply it to actually is a strict contraction in the sense of that page
if it hasn't come up maybe don't bother
05:23
Fair enough to that
Incidentally
3
Q: Limit of a sequence defined by recursive relation : $ a_n = \sqrt{a_{n-1}a_{n-2}}$

Parth ThakkarWe're given a sequence defined by the recursive relation: $$a_n = \sqrt{a_{n-1}a_{n-2}}$$ $a_1$ and $a_2$ are positive constants. We have to show the following: The sequences $\{ b_n \} = \{ a_{2n-1} \}$ and $\{ c_n \} = \{ a_{2n} \} $ are monotonic, and if one is increasing, the other is decr...

Clever :/
yeah, a lower tech idea would be to note that if a_1 = a and a_2 = b then fairly obviously a_n is going to be of the form a^(x_n) b^(y_n) and maybe the sequences x_n and y_n are possible to expressly identify, or at least analyze individually (e.g. because they maybe they individually satisfy simple recursions)
that's where my mind would have gone
i think if you bashed your head against it for an hour you would at least discover the formula for the answer, which might give you further ideas about how to prove it in some slick way like the answer
like in that accepted answer i mean (realized i used 'the answer' to also mean the value of the limit in terms of a, b)
9
Q: Prove that this sequence is eventually one

GustavoLet $ \ \mathbb{N} = \{ 0,1,2,3,4,...\} \, $, $ \ O = \{ n \in \mathbb{N} : n \text{ is odd} \} \ $ and $ \ T: O \to O \ $ be such that, for all $ \ n \in \mathbb{N} \, $, \begin{align*} T(8n+1) & = 6n+1 \\ T(8n+3) & = 12n+5 \\ T(8n+5) & = 2n+1 \\ T(8n+7) & = 12n+11 \end{align*} For each $ \ n \i...

EE18 i heard u like sequences so i put 4 sequences in your sequence
 
4 hours later…
09:20
hi
 
1 hour later…
10:28
There is a theorem in Rudin (2.37) which is "If E is an infinite subset of a compact set K, then E has a limit point in K".
The proof in rudin uses some kind of contradiction argument.
However, I would like to see a proof where we explicitly find that limit point.
Something like how we can find the explicit limit in the proof of bolzano-weierstrass theorem.
Can someone show me a proof like that?
10:58
In mathematics, a topological space X {\displaystyle X} is said to be limit point compact or weakly countably compact if every infinite subset of X {\displaystyle X} has a limit point in X . {\displaystyle X.} This property generalizes a property of compact spaces. In a metric space, limit point compactness, compactness, and sequential compactness are all equivalent. For general topological spaces, however, these three notions of compactness are not equivalent...
The problem is that in Bolzano-Weierstrass theorem there was a countable sequence of points, whereas here we are considering a set $E$ which can be arbitrary in size
So we would first have to be able to find a countable subset which wouldn't be discrete, which most likely requires some axiom of choice, and only then try to argue like in the theorem
11:58
12:33
Humble.
13:12
Is there some notation to denote an integral over a region defined by a position vector $\mathbf{r}$, for example $\int_{something}$
13:31
He said that if I were to write it up it'd be enough for a publication
And that he never seen a theorem like this
@Jakobian what is "it"?
And who is "he"
@AlessandroCodenotti theorem I've proved and the topology professor I met with today
13:49
Nice!
What kind of terrible spaces is the theorem about?
lichess.org/jAmGgyYF anyone want to play this?
Metrizable ones
14:11
@Jakobian nice
Yep. I know. What kind of general topologist am I, to work with those awful spaces
14:36
I only work with Polish spaces (if I can)
14:56
Hi, I need an expert for my question on NFA: math.stackexchange.com/q/4928500/390226. Cannot figure this out :(
@linear_combinatori_probabi sadly there's no people knowledgable about automatas here
If I really tried I could probably answer you in under a week
But I'd rather learn what I am learning right now
15:13
@Jakobian what are you learning?
hope you won't get shocked by the ping like I did, as always.
I have it muted + I'm on phone
@linear_combinatori_probabi topology of various kinds. But I've studed semigroups and universal algebra before
15:33
@RyderRude what's your rating?
 
1 hour later…
16:47
@RyderRude No. I'm not Magnus Carlsen :-D
17:18
by the way you can mute notifications in upper right corner, left of the button that reads "all rooms" @linear_combinatori_probabi
this way you won't hear when someone messages you here, so you won't get irritated each time by the annoying sound
 
1 hour later…
18:18
-5
Q: Does Paul Erdős number have any value or importance?

SaneSometimes I come across some mathematicians' CVs, in which they report Paul Erdős number. I was wondering whether in mathematics communities this number is taken seriously, and to what extent? Ref: Erdős Number

The answer to the question is "Yes". Whenever two mathematicians want to know who is the more viral, sexually potent person, they compare Erdos numbers. Lower wins.
18:33
@XanderHenderson I have answered this question with "no". I am pretty sure that the vast majority of the mathematicians does not care about this.
@XanderHenderson my Erdos number is 69
@Peter That's probably just because your Erdos number is very high (infinite, even?), indicating a lack of raw sexual energy. :P
I would also argue that most mathematicians do care, at least to the level of knowing their own Erdos number. It is a fun bit of trivia, and an interesting shibboleth of the mathematical community.
it is closer to listing a favorite hobby on a CV than it is to listing anything that people might "actually" care about
18:51
For a finite collection of Bernoulli random variables the expectation of their product is positive if and only if their covariance matrix is nonsingular
i guess if you recently attained erdos 1, almost anybody (inside or outside of the mathematics community) would be interested in how
does that proposition sound familiar to anyone?
that would be one hell of a conversation starter
can't wait to co-author with Erdös AI
Suppose I have two spaces: a pair of mutually tangent circles and an
ellipse with a diametrical segment. I want to show that neither is a deformation retract of other. So if I show that neither is homeomorphic to a subset of other space then that will do right? the first space has a point that locally disconnects 4 arcs but the 2nd space has no such points. Also the 2nd space has a point that locally disconnects 3 arcs but the first space has no such points. So neither is homeomorphic to a subset of the other spaces. Is this approach fine?
19:06
I'm not sure what an ellipse with a "diametrical segment" is supposed to be. The shape of a $\theta$?
ok, so the question does not make sense as far as I'm concerned
it does not make sense to ask if A is a deformation retract of B if A is not a subspace of B
'I want to show neither is homeomorphic to a deformation retract of other', in the first line
sorry i missed homeomorphic
ah ok
I think your argument is incomplete though
the subspaces in question are not open, so you can't immediately transfer local disconnection properties
@Derivative covariance matrix is singular iff $X_1, ..., X_n$ are linearly dependent as stated here
though I'm not sure what it has to do with products yet
$X_1, ..., X_n$ here are non-negative so we're saying this is equivalent to $E[X_1...X_n] = 0$
19:18
@XanderHenderson I am not surprised about this comment , neither about that you apparently find such comments funny. I do not care about it anymore , I anyway expect you will delete this reply , as you did it with a critical comment in CURED which I left because of being attacked without a reason. But apparently , everything is censored here except someone "powerful" says it. Your "conclusion" is either ridicoulous or a bad joke.
19:34
Did you quit contributing in CURED?
@SineoftheTime Yes.
Me too
A long time ago
I also use chatrooms quite rarely. Why ? See my above comment.
this chat is like social media for me
Does anyone have Spivak's CoM's book? I have a question about a last step in a somewhat lengthy proof, which I don't want to post all the details of. He proves that if all first partial derivatives of $f:\mathbb R^n\to\mathbb R$ exist in an open set containing $a$ and are continuous at $a$, then $Df(a)$ exists.
Basically, to show $Df(a)$ exists, he computes \begin{align*}\lim_{h\to0}\frac{\left|f(a+h)-f(a)-\sum_1^n D_if(a)\cdot h^i\right|}{|h|}&=\ldots\\ &\leq\lim_{h\to0}\sum_1^n \left|D_i(c_i)-D_if(a)\right|\\ &=0,\end{align*}where $c_i=(a^1+h^1,\ldots,a^{i-1}+h^{i-1},b_i,a^{i+1},\ldots,a^n)$ and $b_i$ a number between $a^i$ and $a^i+h^i$.
He says the last equality in the aligned portion follows from continuity at $a$, which I don't understand. It's not clear to me how $D_if(c_i)$ depends on $h$. It looks like not all components of $c_i$ depend on components of $h$, hence I am confused about $h\to0$. I'm used to limit expressions where we have a function depending on a variable and then the limit is with respect to this variable.
19:41
@psie I don't have Spivak, but yes this is a standard result
@Jakobian theorem is false!
I managed to produce a counterexample by considering a random variable uniformly distributed in a unit-length circle and X_i=1 if you land in an interval of length 4/9 and the three intervals are symmetrically distributed in the circle with intersections of length 1/9
in this case n=3
I'm trying to generalize the Lovasz Local Lemma to the case where you have lots of weakly correlated random variables
But I haven't managed to figure out what the correct statement is yet
Afterwards it occurred to me that the autocorrelation matrix might be more appropriate than covariance matrix
@psie $D_i f(c_i)$ doesn't depend on $h$. I think this actually gets cleared when you do the computation on an abstract manifold because then $h$ is an element of the tangent space $T_{c_i}M$. If you don't understand what that means don't worry
ah wait sorry your $c_i$ depends on $h$
yeah, it depends on $h$ in some strange way (i.e. not all components of $h$), this is what's confusing me
I guess one could just write $c_i=(a^1+h^1,\ldots,a^{i-1}+h^{i-1},b_i(h^i),a^{i+1}+0h^{i+1},\ldots,a^n+0h^{n})$ to be very explicit about the dependence on $h$ (here $b_i$ depends on $h^i$)
okay, write $Df(a)=\sum_i D_i f(a) dx_i$ and by hypothesis you have that the partial derivatives are continuous at $a$
then your result follows
the hypothesis that your partial derivatives are continuous is telling you that $D_i f(x)$ is a continuous function of $x$ as $x\to a$ for every $i$
@psie you're not making it more explicit in this way
19:56
then you want to prove that a function whose domain has several variables is continuous if and only if it is continuous in each variable
I hope that's clear
@Derivative ok, what is $dx_i$ here? :)
Its a linear function from $\mathbb R^n$ to $\mathbb R$ that takes $x_i$ to $1$ and $x_j$ to 0 for $j\neq i$
@Thorgott Can you please elaborate? do you mean that I need to write that the sets are given subspace topology from $\Bbb{R}^2$ and only then proceed with the same argument?
20:13
no, I'm saying the argument is not complete
just cause the second space does not have any point locally disconnecting it into 4 pieces does not generally mean that a subspace of this space couldn't have such a point
@SoumikMukherjee ^
:^)
20:34
:^)
^ my face when :^)
:(
Y'all have big noses...
😎
🤥😷
This is my face: 🍕
❌🍍🍕
20:50
^💯✅
✔🍍🍕😋

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