Mathematics

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)

Akiva Weinberger

1:11 AM

Hm: Arithmetic can be defined by strings

user76284

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.

Akiva Weinberger

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)

Ted Shifrin

DogAteMy, you mean $x^p-x$.

In a field of degree $q=p^n$, the polynomial is $x^q-x$.

Akiva Weinberger

Yes, thanks

Stan Shunpike

hello

@TedShifrin Hi Ted!

Ted Shifrin

1:22 AM

Heya Stan.

Do I owe you an email?

Stan Shunpike

No, I am stuck and have been pondering how to phrase what I'm stuck on.

Ted Shifrin

I was waiting for you for you to fix the double-perp, etc., problems.

Stan Shunpike

yeah that's the one

Ted Shifrin

Oh, stuck whereupon?

Stan Shunpike

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

Ted Shifrin

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.

Stan Shunpike

Do those axioms include infinite dimensional ones

Ted Shifrin

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.

Stan Shunpike

ok

Ted Shifrin

I was just warning you that your proof couldn't rely on finite-dimensionality.

ÍgjøgnumMeg

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?

Ted Shifrin

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.

ÍgjøgnumMeg

@Ted I'll have a think, thanks for the tip :)

I tried just hitting it with Morera's theorem

Stan Shunpike

@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?

Ted Shifrin

Nah. I don't think that helps.

Stan Shunpike

@TedShifrin I will then proceed on problem 17 and try again.

ÍgjøgnumMeg

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

Ted Shifrin

1:28 AM

Yes, @Stan. Section 3.6.

@ÍgjøgnumMeg: Upper half plane is crucial.

Ultradark

What's an example of a completely multiplicative function s.t. $f(ab)=f(a)f(b)$ for $a,b \in \Bbb R$

ÍgjøgnumMeg

@Ted okay :D Thank you for the tips, I'll go and bang my head against the page

Ted Shifrin

LOL ... here's a hint: Think about $|f_n|$. @ÍgjøgnumMeg.

ÍgjøgnumMeg

Those guys are all bounded right?

I mean.. $f_n$ are all bounded because of the upper half plane condition I guess

Ted Shifrin

Yeah. Can you control the bounds?

ÍgjøgnumMeg

1:30 AM

err

Stan Shunpike

@TedShifrin excellent. ok I will email you soon. I just saw Schumann's 3rd symphony performed. Highly recommend it! youtu.be/OfR8d3aJKEs

ÍgjøgnumMeg

I'm not sure if I know what you mean by "controlling" the bounds

Ted Shifrin

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?

ÍgjøgnumMeg

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

Stan Shunpike

@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!

Ted Shifrin

1:38 AM

Uniformly on COMPACT subsets.

Bye all.

ÍgjøgnumMeg

:) Bye @Ted thanks for the help, COMPACT

ÍgjøgnumMeg

2:03 AM

Hate to admit it, but I'm really enjoying doing lots of complex analysis atm

Ultradark

@ÍgjøgnumMeg number theory question: Does $\sum_{n=1}^\infty \frac{1}{\phi(n)^s}$ have a euler product for any multiplicative function $\phi(n)?$

ÍgjøgnumMeg

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

Ultradark

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.

Rrjrjtlokrthjji

Hi all

Ultradark

(this is for the euler phi function)

Akiva Weinberger

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 "…

ÍgjøgnumMeg

@Ultradark I would expect this to be a super difficult question to answer tbh

Akiva Weinberger

(Er - the quote there got cut off at the end)

Rrjrjtlokrthjji

@TedShifrin glad to find you still active here Prof, really benefited from your lectures and messages here, ty

Ultradark

@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

ÍgjøgnumMeg

Establishing the existence and convergence etc. of an Euler product is hard at the best of times lol

Leaky Nun

2:51 AM

claymath.org/sites/default/files/riemann1859.pdf

how on earth do you read this

@ÍgjøgnumMeg kannst du es lesen?

ÍgjøgnumMeg

@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

Leaky Nun

lol

Akiva Weinberger

@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.)

Secret

3:21 AM

(will get back on you on the above shortly)

Adaptive stepsize

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

Akiva Weinberger

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?

Ultradark

math.stackexchange.com/questions/3436183/… already answered!

Akiva Weinberger

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)

ÍgjøgnumMeg

@Ultradark cool :)

Akiva Weinberger

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

Adam

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

Secret

Is conf(T) the same as con(T), that there is a contradiction in T

Akiva Weinberger

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

Secret

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

Akiva Weinberger

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)"

TanMath

@anakhro ah i see that... It seems like a pretty good geometrical picture of one-forms though

Akiva Weinberger

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

Adam

4:32 AM

this is ok right? $n=a^2+b^2 \land a^2 \equiv 1 \pmod 4 \land b^2 \equiv 1 \pmod 4 \Rightarrow n \equiv 2 \pmod 4$

ÍgjøgnumMeg

sure

Adam

ok ill do the other bit tomorrow then if I have to ask I should be in bed

The East Wind

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

Akiva Weinberger

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')"

ÍgjøgnumMeg

6:10 AM

Morning @Alessandro

Akiva Weinberger

6:28 AM

There are only three types of islands: shipping hazards, resources, and potential markets

Ted Shifrin

@Rrjrjtlokrthjji Um, thanks. Who were you before? :P

Akiva Weinberger

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

We’re Rewarding the Question Askers

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. …

ÍgjøgnumMeg

@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

Adam

6:44 AM

need help here plz math.stackexchange.com/questions/3402693/…

Adam

6:57 AM

Talk about it with my advisors and professors? um strangers on the internet is the same thing

pff

Secret

@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

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$Mathematics$ Mathematics