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12:15 AM
Perhaps the simplest true-but-unprovable sentence in PA would be Conf(some theory that is bi-interpretable with PA but better handles recursion)
 
1:11 AM
Hm: Arithmetic can be defined by strings
 
In a finite field of size $q$, how many $n$-degree polynomials pass through $k$ distinct points?
Obviously for $k$ = $n+1$ the answer is 1.
 
Like if we think of a number as a string such as "oooooo" (with as many 'o's as the number we want to represent), then we can multiply two strings by "substitute the second string into every instance of the letter 'o' in the first string"
i.e. to multiply 3 and 5, substitute "ooo" for every 'o' in "ooooo"
This feels to me like a more natural setting in which to prove Gödel's incompleteness theorem
because it's clear how sentences can talk about sentences
In PA, you have to encode sentences as numbers. Here you just have to turn sentences into strings of symbols, which they kind of already are
so I guess you just need to put quotation marks around them
(Annoyingly, the quotation mark would have to be one of our symbols… I guess you could fix that with escape characters)
@user76284 Possibly infinite? Like in $\Bbb Z_p$, the polynomial $x^p+x$ is zero everywhere
so you could add it to any polynomial to get another one that passes through the same points
This relies on you considering $x^p+x$ to be a different polynomial from $0$, even though they're equal as functions
(Usually we define polynomials on finite fields in such a way that those are different)
 
DogAteMy, you mean $x^p-x$.
In a field of degree $q=p^n$, the polynomial is $x^q-x$.
 
Yes, thanks
 
hello
@TedShifrin Hi Ted!
 
1:22 AM
Heya Stan.
Do I owe you an email?
 
No, I am stuck and have been pondering how to phrase what I'm stuck on.
 
I was waiting for you for you to fix the double-perp, etc., problems.
 
yeah that's the one
 
Oh, stuck whereupon?
 
Firstly, I'm confused about what an infinite dimensional vector space is.
So we can like abstractly define vectors with those axioms of like closure under scalar multiplication and so forth
 
1:23 AM
Well, you don't need to worry about that now. But the set of all real sequences or all real polynomials are examples.
 
Do those axioms include infinite dimensional ones
 
All continuous functions, differentiable functions, etc.
Yup.
And very important in differential equations, etc.
But don't worry about infinite dimensions. You'll get to discussion of that in Chapter 3.
 
I was just warning you that your proof couldn't rely on finite-dimensionality.
 
Dunno if this is a silly question: is it "obvious" in some sense that $\sum_{n \in \Bbb Z}e^{\pi i n^2 z}$ is holomorphic on the upper half plane?
 
1:25 AM
So you just want to show $V\subset (V^\perp)^\perp$.
@ÍgjøgnumMeg: Can you show uniform convergence on compact subsets?
Any finite sum is obviously holomorphic.
What does the upper half plane limitation do to that?
Announcement: I need to leave in about 5 minutes.
 
@Ted I'll have a think, thanks for the tip :)
I tried just hitting it with Morera's theorem
 
@TedShifrin so you named several different examples of infinite dimensional vector spaces and I have very little exposure to most of those, if any. Will Chapter 3 give me any exposure?
 
Nah. I don't think that helps.
 
@TedShifrin I will then proceed on problem 17 and try again.
 
or at least, I tried hitting $\sum_{n=1}^\infty e^{\pi i n^2 z}$ with that, but that suffices, but unfortunately the line integral I get is too hard
lol
 
1:28 AM
Yes, @Stan. Section 3.6.
@ÍgjøgnumMeg: Upper half plane is crucial.
 
What's an example of a completely multiplicative function s.t. $f(ab)=f(a)f(b)$ for $a,b \in \Bbb R$
 
@Ted okay :D Thank you for the tips, I'll go and bang my head against the page
 
LOL ... here's a hint: Think about $|f_n|$. @ÍgjøgnumMeg.
 
Those guys are all bounded right?
I mean.. $f_n$ are all bounded because of the upper half plane condition I guess
 
Yeah. Can you control the bounds?
 
1:30 AM
err
 
@TedShifrin excellent. ok I will email you soon. I just saw Schumann's 3rd symphony performed. Highly recommend it! youtu.be/OfR8d3aJKEs
 
I'm not sure if I know what you mean by "controlling" the bounds
 
Remember you want compact subsets of the (open) half-plane.
OK, thanks, @Stan. Unless you time-traveled, that was not the performance you saw.
@ÍgjøgnumMeg: Something like Weierstrass M-test?
 
Oh I've shown that the $f_n$ satisfy the relevant conditions for the Weierstraß $M$-Test, so I've got that $\sum f_n$ converges abs. uniformly on $\Bbb H$
because the sum over the bounds on $f_n$ is finite by some ratio test argument
 
@TedShifrin hahahahaha omg funniest math professor I know. yes, Bernstein was not conducting I must admit. Ricardo Muti. ok off to study stats and linear algebra. à bientôt!
 
1:38 AM
Uniformly on COMPACT subsets.
Bye all.
 
:) Bye @Ted thanks for the help, COMPACT
 
2:03 AM
Hate to admit it, but I'm really enjoying doing lots of complex analysis atm
 
@ÍgjøgnumMeg number theory question: Does $\sum_{n=1}^\infty \frac{1}{\phi(n)^s}$ have a euler product for any multiplicative function $\phi(n)?$
 
I don't know, it certainly does for the identity ;)
but you can for example look at $\sum_{n=1}^\infty \frac{\chi(n)}{n^s}$ which has an Euler product for real part of $s$ greater than $1$
and $\chi$ a Dirichlet character
which are completely multiplicative
 
okay here's the sequence $a=\{2,3,4,4,5,2,6,6,4,5,2,10,2,...\}$
which is the Number of values of k such that phi(k) = n, where n runs through the values (A002202) taken by phi.
 
Hi all
 
(this is for the euler phi function)
 
2:18 AM
5 hours ago, by Akiva Weinberger
Peano Arithmetic (PA) is a formal proof system. Statements and proofs must follow a very strict syntax.

Step 1: Encode sentences in PA as numbers (regardless of if they have a free variable or not)

Step 2: Encode proofs in PA as numbers

Step 3: If x encodes a sentence F with a free variable and y is a number, let sub(x,y) be the number encoding F(y) (i.e. y substituted into the free variable of F)

Step 4: Let P(x) be the sentence "the sentence encoded by x (has no free variables and) has no proof."
Brief, rushed explanation of the Second Incompleteness Theorem: What happens when we try to write the above proof in the language of PA? (Never enough recursion!) We run into a problem between steps 6 and 7, because we need to assume that PA is consistent. If PA proves the sentence "PA is consistent" (aka "PA doesn't prove a falsehood" aka "there is no number encoding a proof of the sentence 0=1"), then it can go from step 6 to step 7, and prove that P(sub(n,n)) is true. But we already know that P(sub(n,n)) is unprovable, so contradiction. The only resolution is that PA doesn't prove that "
 
@Ultradark I would expect this to be a super difficult question to answer tbh
 
(Er - the quote there got cut off at the end)
 
@TedShifrin glad to find you still active here Prof, really benefited from your lectures and messages here, ty
 
@ÍgjøgnumMeg but i just have to determine if that sequence is completely multiplicative right?
this one $2, 3, 4, 4, 5, 2, 6, 6, 4, 5, 2, 10, 2, 2, 7, 8, 9, 4, 3, 2, 11, 2, 2, 3, 2, 9, 8,$
oh but if it's muliplicative but not completely multiplicative or it's not even multiplicative I can see how it would be hard
 
Establishing the existence and convergence etc. of an Euler product is hard at the best of times lol
 
2:51 AM
how on earth do you read this
@ÍgjøgnumMeg kannst du es lesen?
 
@Leaky hahaha nein aber ich hab's schonmal gesehen
Über die Anzahl der Primzahlen unter einer gegebenen Grenze
I think
oder größe
@Leaky I imagine your reaction to this paper is the reaction of most tutors to students' homework solutions
 
lol
 
@Secret The Second Incompleteness Theorem is what happens when you try to write the proof of the First Incompleteness Theorem in the language of PA
(Why do we know that the unprovable Gödel sentence is true? Because we know that PA is consistent. If PA knew that PA is consistent, then it'd know that the Gödel sentence is true, too. But it can't, because that's what it means for the Gödel sentence to be unprovable.)
 
3:21 AM
(will get back on you on the above shortly)
In numerical analysis, some methods for the numerical solution of ordinary differential equations (including the special case of numerical integration) use an adaptive stepsize in order to control the errors of the method and to ensure stability properties such as A-stability. Using an adaptive stepsize is of particular importance when there is a large variation in the size of the derivative. For example, when modeling the motion of a satellite about the earth as a standard Kepler orbit, a fixed time-stepping method such as the Euler method may be sufficient. However things are more difficult...
Thoughts on numerical integration schemes:
Take $y=\frac{1}{f(x)}$ where $f(x)=0$. Suppose your task is to integrate it:
If you do it the normal way without the correct step size, you either end up integrating over the singularities, or you accumulate a lot of numerical errors due to mismatch stepwise with function magnitude
Adaptive step size can help to solve this problem by matching the two magnitudes together so that each derivative and hence the approximated area is A stable
Combining this with trapezoidal integration or mid point integration, and with sufficiently matched step size, you can minimise both truncation error and numerical instability
Might be interesting to investigate later on how local curvature influence the truncation error of the adapted step size
and also to investigate whether given f(x), a step size of 1/f(x) is always ideal as they multiplied to numerically one
 
PA proves that PA doesn't prove Conf(PA)
T doesn't prove Conf(T) if and only if T is consistent
Therefore, "PA doesn't prove Conf(PA)" is equivalent to "PA is consistent"
Therefore, PA proves that PA is consistent
Problem?
 
Resolution: I think the faulty statement is the first one?
PA proves that either PA doesn't prove Conf(PA) or PA is inconsistent
In other words, PA proves that if Conf(PA) then PA doesn't prove Conf(PA)
 
@Ultradark cool :)
 
3:36 AM
PA is inconsistent iff PA proves that PA is consistent. The previous sentence is a theorem of PA
Math proves that [ math is inconsistent if and only if math proves that math is consistent ]
I realize I'm saying the same thing repeatedly but this is kinda wild
 
3:52 AM
is it paranoid to think that they overhauled the reputation points system just because I was approaching my favourite number?
it is a little yes
 
Is conf(T) the same as con(T), that there is a contradiction in T
 
I can't spell. I meant Con(PA), which says that PA is consistent
(aka the opposite of what you just said)
In my rushedness, I think I glossed over a key missing ingredient. It turns out that, for every number n, if PA proves "the statement encoded by n has a proof", then PA proves the statement encoded by n. You prove this by structural induction
Wait
That's not true
Argh
OH
It's the reverse
Or is it
Hold on
For every n, if PA proves the statement encoded by n, then PA proves "the statement encoded by n has a proof"
That seems fine. Question is, is that the missing ingredient
OK so G is our Gödel sentence
G is equivalent to "There is no proof of g" where g encodes G
Thus PA proves: [ "There is a proof of g" if and only if "There is a proof of 'there is no proof of g'" ]
Wait no I'm confused
 
No evidence of existence is not the same as evidence of nonexistence. Likewise, "does not prove x" does not implies "proves not x"
So at least intuitively, if PA failed to prove it is consistent, then PA is consistent is not a contradiction
 
4:08 AM
PA proves "P(g) iff P(~P(g))"
Thus PA proves "P(P(g)) iff P(~P(g))"
Thus there exists a number k such that PA proves "P(k) iff P(~k)"
 
@anakhro ah i see that... It seems like a pretty good geometrical picture of one-forms though
 
Oh
OK, here is where I bring Conf(PA) in
Conf(PA) is the same as "~P(k and ~k)"
Thus it implies "~(P(k) and P(~k))"
(I've used the fact that for all a,b, PA proves that P(a and b) implies P(a) and P(b))
(No - I've used the converse of that)
Oh and also we need the law of the excluded middle
K or ~K
 
4:32 AM
this is ok right? $n=a^2+b^2 \land a^2 \equiv 1 \pmod 4 \land b^2 \equiv 1 \pmod 4 \Rightarrow n \equiv 2 \pmod 4$
 
ok ill do the other bit tomorrow then if I have to ask I should be in bed
 
4:45 AM
How do I calculate the surface integral of a surface which is such that a part of it has the same projection on a plane as the complete surface itself?
*over a surface
 
5:06 AM
@Secret OK so if P is "proves", and wherever F is a sentence I write 'F' for its Gödel numbering, then:
P('A and B') is a theorem of iff P('A') and P('B') both are
If P('A') or P('B') are theorems of PA then so is P('A or B') (but not necessarily the reverse!)
If P('A') and P('A implies B') are theorems of PA then so is P('B')
If PA proves P('not A') then PA does not prove P('A'). However, that is not a theorem of PA
And the reverse is not true: If PA does not prove P('A') then it does not necessarily mean PA proves P('not A')
For all A, Conf(PA) is equivalent to "P('not A') implies not P('A')"
 
6:10 AM
Morning @Alessandro
 
6:28 AM
There are only three types of islands: shipping hazards, resources, and potential markets
 
@Rrjrjtlokrthjji Um, thanks. Who were you before? :P
 
It's been three years
Apparently reputations are being recalculated
and I suddenly have 16,380 reputation
I think they changed an upvote on a question from +5 to +10?
Yup
Sara Chipps on November 13, 2019

In my very first blog post, I wrote about what a personal experience taught me about the Stack Overflow community. I said we were going to step back and re-evaluate how we deliver feedback, how we can improve content quality, and how we can reduce friction between people. I said that our goal is to have the question asking process be painless and beneficial for new people and Stack Overflow veterans alike.    

During this re-evaluation period, we noticed something in our reputation reward system. We give anyone who receives an upvote on an answer ten additional reputation points, but only give five reputation points to people who receive an upvote on a question.  …

 
@Ted I got the holomorphy with your tip! Thanks :)
although I didn't use the M-Test because the sequence of functions in question were themselves series
 
6:57 AM
Talk about it with my advisors and professors? um strangers on the internet is the same thing
pff
 
@AkivaWeinberger That makes a lot more sense because not all of them are iff statements. However, my logic is way too low to justify whether my judgement makes sense. I have also posted that question in the logic room for user21820 to have a look. He is much better at logic than me
 
 
2 hours later…
9:04 AM
Hi @ÍgjøgnumMeg
 
9:27 AM
Hi everyone, im getting that the absolute values is less than $Mh^2$ instead of $\dfrac{1}{2}Mh^2$
can someone help?
 
9:40 AM
hey guys good morning, I am studying the Skorkhod's theorem and I have a doubt, the statement starts by saying "Let be $P_n$,$n\in\mathbb N$ a sequence of probability measures defined on a metric space $(S,\mathcal S)$. I might be wrong, but by definition $(S,\mathcal S)$ is a measurable space, it can happen that $S$ is metric, but this statements seems off to me.
does the author intend to say that $(S,\mathcal S)$ is a measurable space and $S$ must be metric?
 
 
2 hours later…
12:07 PM
Is a vector function like f(x,y) = P(x,y)i + Q(x,y)j representing a curve in the xy plane or a vector field in the xy plane or it can be both based on interpretation?
 
12:54 PM
The latter. The former would only have one input
 
@AkivaWeinberger Hi Akiva, thanks, why do you say the former would only have one input? I'm not sure what that means.
for example x=1,y=1 is one input, and x=1,y=2 is another input and etc.
 
I mean f(t)=P(t)i+Q(t)j
x and y are two inputs
 
That's right... thanks.
 
1:23 PM
@Alessandro you'll be interested to know I spent the entire night enjoying doing analysis
 
What kind of analysis?
 
complex analysis lol
for modular forms
I'd say I'm probably enjoying that course more than I'm enjoying algebraic number theory, at least at the moment
 
$\sum^\infty_{n=1}\frac1{\phi(n)^s}=\prod_{\text{prime }p}\left(1+\frac{(1-1/p_i)^{-s}}{p_i^s-1}\right)$ Can anyone help me find a functional equation?
 
 
2 hours later…
3:11 PM
if X and Y are uniform random variables on (0,1), what is the integral to compute P(XY < t)?
 
3:22 PM
$\displaystyle \int \int_{xy < t} f_{X,Y}(x,y) \ \mathrm d(x,y)$
 
3:51 PM
Another way to write the Gödel sentence
P(“∃y,Sub(\“∃y,Sub(x,x,y)∧P(y)\”,\“∃y,Sub(x,x,y)∧P(y)\”,y)∧P(y)”)
Here, Sub(x,n,y) means that if you substitute the number n into the formula coded by x, you get the formula coded by y
and if F is a sentence then “F” is the number encoding it
 
One more vote to reopen this question, please.
It's a result of classification of groups of order $\leqslant 50$
 
P(「∃y,Sub(「∃y,Sub(x,x,y)∧P(y)」,「∃y,Sub(x,x,y)∧P(y)」,y)∧P(y)」)
 
It's not a question o.O
 
It was converted to a question.
 
If the question is the first sentence of the post, isn't the answer just the rest of the post?
 
3:57 PM
@AkivaWeinberger Just partially. Some results may need verify, and classifiations of groups of order 16,32,36,48 are not included.
 
@Andrews I would recommend editing something like the above into your comment (click to enlarge)
just to make it clear
 
Doesn anybody have a simple (not invoking compactifications) example of a nonmetrizable space with a dense metrizable subspace?
 
@AlessandroCodenotti Sierpiński space ($\{0,1\}$, with open sets $\emptyset,\{1\},\{0,1\}$). It is not Hausdorff so not metrizable, but the subset $\{1\}$ is dense and trivially metrizable
 
Oh sure, I forgot about finite spaces :P
 
I think I can turn that into a nonfinite example
er - "in"finite
Consider $\{0\}\cup S$ where $S$ is any metrizable space and $\{0\}$ is disjoint from it. The open sets are the open sets of $S$, and the open sets of $S$ union $\{0\}$
Wait
That might be metrizable unless you include other conditions
 
4:07 PM
@AkivaWeinberger Hmm I'd be more interested in Hausdorff examples since just by looking at $X\to\beta X$ you get normal examples easily
 
If $S=[0,1]$ then this is Hausdorff and not metrizable, I think
Wait
No hold on this is just $\{0\}$ disjoint union $[0,1]$, aka it's homeomorphic to $[0,1]\cup\{2\}\in\Bbb R$
@MatsGranvik Humanity is the original singularity
 
Never heard that.
The universe created you in order to think faster. (Which was what I deleted above.)
To think locally faster.
 
Earth accidentally created humanity, and we almost immediately invaded the entire planet, changing entire biomes and consuming resources almost without limit
 
4:25 PM
Abstract capitalism: A system that over time, gone into a runaway feedback loop where it expands without limit, divide space infinitely, compress time infinitely, and in the process, drain any resources in its way
and no major improvement in quality results
If you think about this for a second, there are basically 4 singularities in this one object
1. $V \to dV$
2. $t \to dt$
3. $V \to \infty$
4. $\frac{d^n}{dt^n} \to \infty, n \in \Bbb{N}$
 
@AkivaWeinberger Thanks, I've edited it.
@AkivaWeinberger It's now reopened. Thank you very much!
 
4:49 PM
Hello
I don't have a good and deep math-language ... But i'm looking for help :

https://math.stackexchange.com/questions/3436751/game-theory-poker-nash-equilibrium-choose-between-bet-or-raise-and-range-of-h
I can't understand how a division is made between the call range and the raise range in Poker texas hold em with a GTO strategy
I use some solver software to implement a GTO strategy and i want understand how to do it.
 
 
1 hour later…
6:09 PM
My new robot can help cause search and rescue missions
 
6:26 PM
?
 
 
1 hour later…
7:32 PM
plastic can be converted into petrol
 
@ÍgjøgnumMeg I'm not sure you understand the Weierstrass $M$-test. Or I'm very confused.
@user123 Use the mean value theorem and you get the bound $\int_x^{x+h} M_1 |x-\xi|\,dx$. You get the largest value when $\xi$ is at an endpoint, and then you get $\int_0^h |x|\,dx = \frac12M_1h^2$.
 
7:48 PM
@TedShifrin help me plz :'(
I'm frustrated, I can't understand
 
Alessandro! are u italian :P
 
:)
me too
riesci ad aiutarmi col mio quesito ? Sto impazzendo un pò ..
 
Purtroppo non so quasi nulla né di probabilità, né di teoria dei giochi né di poker
 
8:01 PM
capito .. è teoria dei giochi /equilibrio di nash // non riesco a capire come calcolano il range di mani da rilanciare ...
cmq ok , speriamo mi risponde qualcuno..
buona serata
 
grazie, anche a te!
 
8:18 PM
@BalarkaSen hi
 
Hi @Balarka
What kind of interesting math have you been doing recently?
 
Not super interesting stuff, but a little bit of this and that
This semester's a bore
I have been learning some optimization as part of coursework and that has been fun though
 
I have been doing hyperelliptic kolgomorov entropy calculations
 
I see
I spent the afternoon working out why a topology fact holds. Turns out that it is trivial but I wasn't thinking
 
Hah, happens
 
8:29 PM
When do the isotopic subgroups of two non-isomorphic abelian varieties commute?
if they have cars, and if the distance is not too long
 
@BalarkaSen It was like I have a partition of unity $\{f_i\}$ of $X$ and I extend them to functions $\{F_i\}$ on $rX$ for some compactification $rX$. Is it still a partition of unity?
 
Anyone wants to hear me rant about optimization? I have an exam this Monday, so just feel like talking about it :P
 
anyone going to laugh at my joke?
 
Answer is yes, as long as $\{f_i\}$ is finite, but not necessarily so in general
 
What are the functions at the points of infinity
 
8:31 PM
Aight
Yesterday while I was doing all that Gödel stuff I actually had a p-set due the following day
 
Dunno, depends on $f_i$ and the compactification
 
so at the end of the p-set I included a proof of Gödel
 
I mean, since $X$ is dense if there is an extension it's unique, right? So we don't really care what it looks like
 
Hi, how does the Weyl group of a simple Lie algebra act on the fundamental weights $\omega_p$? Is it by $(w\cdot \omega_p)(\alpha) = \omega_p(w^{-1}\alpha)$, or some other action?
 
@Alessandro I mean you need some open sets and functions at the points in $rX\setminus X$ to cover $rX$, right?
 
8:33 PM
@Ultradark ha HAAAR
 
No wait, there's no open sets involved, I just want $\sum f_i=1$, the partition of unity is not subordinate to any particular cover
 
Oh, you mean that
Well, what about gridifying $\Bbb R^2$ by squares and putting bump functions supported on each of those squares (enumerate them)
They are all $0$ at infinity
 
That's why it doesn't work with infinitely many $f_i$
 
Ya
Gotcha
 
Well I was thinking about the same bump functions example but on $\Bbb R$ and $S^1$
 
8:35 PM
That's simpler, heh. I just gave whatever I had in mind
I was confused about your setup because of this picture, didn't realize that was the question
 
For finitely many $f_i$ you can just take $x\in rX\setminus X$, $(x_\alpha)$ a net in $X$ going to $x$ and now you can swap the limit with the sum and then bring it inside the $f_i$
 
Yup, makes sense
 
(also this has nothing to do with compactifications, the only thing I'm using is that $X$ is a dense subspace and that the functions can be extended)
Now I'm thinking what happens if I try to throw a cover in the mix
Like if $f$ is supported on an open $U\subseteq X$, and I extend $f$ to some compactification $rX$, is it supported on $\tilde{U}=rX\setminus\overline{(X\setminus V)}$, that being the biggest open set in $rX$ whose intersection with $X$ is $U$?
(the closure is taken in $rX$)
 
8:51 PM
(I have started talking about optimization in garbology, hop in if you care)
 
I'm just thinking about general topology at the moment
Which is a very nice thing to think about once in a while
 
What is a group? is it like a set of elements with different symmetries?
 
9:20 PM
$xsin^2(x) $ , x goes to infinity why is it 0
 

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