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12:23 AM
@nbro: How do you define a function?
 
$f : A \rightarrow B$
So I guess that's an evaluation
 
You define a function by giving its values at points of its domain, right?
 
Yeah
 
12:55 AM
can a closed form used to approximate another number include a transcendental number
would this be useful though because you'd have to compute the transcendental number to approximate the other number
 
1:26 AM
Let $f(w,x,y,z) = wz − xy$. The set of points $(a,b,c,d)$ where the total derivative of $f$ is zero is finite.
How to see this?
 
in The h Bar, 34 secs ago, by Secret
This paper is interesting to me not only because it derive an inverse correspondence to go from quantum operators to classical phase space distributions, but also the various change of variables used that made the integral to be separable. May study more on these rules to see what substitution can lead to separation given an integrand of n variables
 
@Secret saw this earlier and thought you might find it interesting: arxiv.org/abs/1801.04781
 
Should have a look after I finish reading two other quantum papers
 
@Secret Figure 1.3 on page 31 seems neat
 
1:41 AM
Huh, so you can probe reaction pathways and transitions states with Bohm, hmm...
 
yeah
i mean, setting aside any interpretational questions, I find pilot wave stuff neat insofar as it suggests the possibility of visualizations which employ trajectories and nevertheless are intrinsically quantum
and that seems valuable to me, even if one ultimately concludes that those trajectories shouldn't be given any further significance than that
 
Visualising the trajectory profiles may be easier to follow dynamics compared to just having a blob of electron density map which doe not contain enough information on where the electrons are flowing
Also, for those who do Bohm by focusing on the quantum potential, those trajectories highlight which area the quantum potential is largest, and then stuff may be done about those regions to fine tune reaction dynamics and hence controlling the chemical reaction
 
The explicit non locality is what attract me the most along with the pretty trajectory diagrams in Bohm, because it give us the idea on directly do engineering on the wavefunction
 
my own view of the 'quantum potential' is that it has rather the status of the 'effective potential' you get when talking about rotating systems. it's not as fundamental as the guidance equation, but it's nevertheless effective in understanding what influence the wavefunction has
 
1:54 AM
In contrast in the orthodox framework, the wavefunction can only be implied by the distribution of observables obtained, thus it provide less insight on the direct control of its shape
Currently, the most imaginative idea that Bohmian mechanics inspires me from the context of the slow pointer experiments, is the possibility to get the environment to work with you on the quantum stuff
Since Bohmian mechanics is explicitly nonlocal, for a noisy enough system, the wavefunction will have a lot of dependence on nonlocal factors from the environment. Thus in theory if there is a way to repeatedly perform experiments with different configuration of devices and etc. to modulate the wav function, some background wavefunction contributed only by the environment may be extracted. Then the trajectory diagram of that could be useful to identify where to place devices in order to shape
the wavefunction at strategic locations taking account of the environment contributions to get a much greater control on the probability distributions, thus in effect "controls the predestination itself"
Of course, one likely possibly that such scheme will fail is the environment may decohered the system so much that it loses superposition, thus its wavefunction if any, will look like classical phase space and not interfering
But in theory, changing the experiment setup should leave whatever nonlocal factors from the environment unchanged, because you cannot signal with nonlocality even in Bohm, thus no way a local change in the experiment setup will be able to influence the environmental factors far away unless they are already entangled, thus it should be possible to probe the background wavefunction contributed entirely by the nonlocal factors
 
i like the idea that the Bohm picture, though it can't enable any behavior which QM didn't allow you before, could nevertheless be useful in designing better experimental systems
 
Hi
 
I don't know if that's realistic, but it's a nice little dream.
 
yeah I am not entirely sure either, as said, if the wavefunction becomes so decohered that it consists only of pointer states, then such background wavefunction will not be able to do anything useful
Another thing I am not sure is what will the orthodox description of the notion of "controlling predestination" be like, since the wavefunction is not necessary to be real in the orthodox's interpretations
What I knew so far is that the implementation of the "slow pointers" scenario requires weak measurement and post selection, as that surreal trajectory paper showed, and weak measurement itself is a complicated setup both in the theory and its experimental implementations
thus it could be possible that the setup is so complicated that the background cannot be resolved easily, if exist
And I do personally think that Bohm could contribute a lot in computational chemistry, though we still need to find a protocol such that we only need to compute some of the trajectories instead of all of them to describe the phenomena as otherwise it is more expensive than the orthodox
 
2:16 AM
yeah, that kind of thing makes me wonder about the 'synthetic' approach to Bohmian trajectories
 
2:26 AM
How do you solve $\zeta(s)=x$
 
7
Q: Inverse of the Riemann zeta function

Victor LiuI'm wondering if there is any information on the inverse of the Riemann zeta function (not it's reciprocal, but its functional inverse). This would obviously be a multi-valued function.

Partial results only
 
I put it into wolfram alpha and it gave me s=-2n and s=nontrivial zeros
that has to be wrong though
surprised that got 7 upvotes on mathoverflow
 
2:44 AM
@Ultradark I think that is pretty common. Just look at how many infinite series representations of some numbers contains $\pi$
The following is a list of significant formulae involving the mathematical constant π. The list contains only formulae whose significance is established either in the article on the formula itself, the article Pi, or the article Approximations of π. == Euclidean geometry == π = C d {\displaystyle \pi ={\frac {C}{d}}} where C is the circumference of a circle, d is the diameter. A = π r 2 ...
1
Q: Infinite series with e and pi

Danil KrotkovPlease, help me to prove that $$ \sum_{n=1}^\infty \frac{\cos(\pi n-\sqrt{\pi^2n^2-9})}{n^2}=-\frac{\pi^2}{12 e^3} $$ I found this fact here https://en.wikipedia.org/wiki/List_of_representations_of_e but there's no reference.

 
Note that, if it was easy to invert the Riemann zeta function, then the Riemann hypothesis wouldn’t be a big deal: just invert $\zeta(s)=0$
 
lol yes
 
So, I wouldn’t hold my breathe
 
In other news, I don't know if there a rules that governs how to deal with bare transcendentals in an infinite sum e.g.:
$$\sum_{k=1}^{\infty} \frac{\pi}{a_k} + s a_k$$
I think such sums will be pretty hard to compute since one does not know in general whether the sum is a transcendental
where s is some known transcendental number. Sums where the transcendental is not wrapped by a transcendental function are very rare though
Then of course, one can always turn it into a multiple sum by expanding the transcendental into its infinite series representation
but that is more brute force and less illuminating on the rules that governs transcendentals
Tbh, $\pi$ is such a ubiquitous number that it is unthinkable that there are no nice formulas of the form:
$$f(\pi,s) = g(\pi)$$
where $f,g$ are algebraic functions, which can be used to simplify some summands involving $\pi$ and other transcendental numbers $s$
$0 = df = f_x dx + f_y dy + f_z dz + f_w dw = (wz-y) dx + (wz-x) dy + (w-xy) dz + (z-xy) dw$
$= wzdx -ydx + wzdy - xdy + wdz -xydz + zdw - xydw$
$= w(zdx+zdy) + (zdw + wdz) - (ydx + xdy) -y(xdz - xdw)$
$= wz(dx+dy) + d(wz) - d(xy) -xy(dz - dw)$
 
 
1 hour later…
4:27 AM
A while back one of my professors mentioned a theorem in topology and described it as follows:

If you take two identical sheets of paper and lay one on top of the other, each point on the top sheet is directly above a corresponding point on the bottom sheet.

If you crumple the top sheet and place it anywhere above the bottom sheet, there is at least one pair of corresponding points still lined up vertically.

Does anyone know the name of this result?
 
Brouwer's fixed-point theorem is a fixed-point theorem in topology, named after L. E. J. (Bertus) Brouwer. It states that for any continuous function f {\displaystyle f} mapping a compact convex set to itself there is a point x 0 {\displaystyle x_{0}} such that f ( x 0 ) = x 0 ...
 
@Simplex brouwer's fixed point theorem
 
5:27 AM
Hi, for a graph G, "if complement of G is not connected, then G is connected", right?
 
No.
Actually, nevermind. I misread the question.
 
5:52 AM
Suppose $\overline{G}$ is not connected. We want to prove $G$ is connected.
To do this, we must prove that, for any two distinct vertices $v_1,v_2$, there is a path between them.
Consider two distinct vertices $v_1$ and $v_2$. If $\{v_1,v_2\} \not\in E(\overline{G})$, then $\{v_1,v_2\} \in E(G)$, so there's a path between them and we're done. Can you think of what to do in the case where $\{v_1,v_2\} \in E(\overline{G})$?
 
6:05 AM
https://math.stackexchange.com/a/356744/76284

People like this (in the comments) drive me absolutely up the wall.
 
7:03 AM
Let $f$ be a holomorphic function with domain being open disk of radius $1$ cetered at origin. Is it true that $f$ is uniquely determined by values it takes on a set like $\{1/n\}\cup \{0\}$?
 
@Silent That sounds correct. As far as I recall, any set with an accumulation point will suffice.
 
Thank you. I knew that this result for line segment or connected open subset of domain. now i know that it holds for way weaker situations! complex analysis is amazing
 
@Silent Do look it up though. I might be misremembering
 
ok. i tried looking it for on math. se and two books i have, but could not find exactly what i am looking for. but if i go the same way as those 'line segment/ open+connected' proof, i get desired result.
 
4
A: Is an analytic function determined by its values on a lattice?

Nate EldredgeNo to your first question. The Weierstrass factorization theorem asserts that for any sequence $\{a_n\}$ of nonzero complex numbers with $|a_n| \to \infty$, there is a nontrivial analytic function $f$ whose zero set is precisely $\{a_n\}$. Clearly we can enumerate all the nonzero points $a_n$ o...

In complex analysis, a branch of mathematics, the identity theorem for holomorphic functions states: given functions f and g holomorphic on a domain D (open and connected subset), if f = g on some S ⊆ D {\displaystyle S\subseteq D} , S {\displaystyle S} having an accumulation point, then f = g on D. Thus a holomorphic function is completely determined by its values on a single open neighborhood in D, or even a countable subset of D (provided this contains a converging sequence). This is not true...
 
7:15 AM
@user76284 then since $\overline{G}$ is disconnected graph we can have at least two disjoint sets with containing vertices having edges between them, and $v_1$ and $v_2$ will be on a same set. For G there will be edges connecting all vertices between these two sets and so there will be a path between $v_1$ and $v_2$ ?
 
@user76284, thank you!
 
@ThomasKlimpel "Hilbert's vision that "actual infinite" could be justified by finitistic methods was disproved by Gödel in a certain sense..." Can you explain what you mean here? — user76284 yesterday
 
@user76284 thank you very much
 
7:31 AM
@user76284 Do I get you right? You will ping people about a discussion back in April 2013? And if they respond to your ping and discuss with you today, then you will declare publicly that "People like this (in the comments) drive me absolutely up the wall."? Why?
 
$\nabla \times \mathbf F = \dfrac1{r\sin\theta} \left[ \dfrac{\partial}{\partial\theta} (\sin\theta F_\varphi) - \dfrac{\partial F_\theta}{\partial\varphi}\right] \widehat{\mathbf r} + \dfrac1r \left[\dfrac1{\sin\theta}\dfrac{\partial F_r}{\partial \varphi} - \dfrac{\partial}{\partial r} (rF_\varphi)\right] \widehat{\boldsymbol \theta} + \dfrac1r \left[\dfrac{\partial}{\partial r} (rF_\theta) - \dfrac{\partial F_r}{\partial\theta}\right] \widehat{\boldsymbol\varphi}$
 
@ThomasKlimpel If you read the comments you'd know the one I responded to was made in 2017, not 2013. Not that it should be relevant.
I don't see how the date is relevant.
And yes, it does drive me up the wall when someone says that anyone who works with infinite sets "has left the field of rational observation and entered the field of dogma".
Why does it drive me up the wall? Because

1. It's wrong (as in the statement made was objectively, factually incorrect according to the definition of the terms that were used).

2. Most people who make such a comment know or should know it is wrong in the above sense.

3. It's pejorative toward mathematicians who work in this field.
 
But he is mainly trying to explain to you why there is a significant difference between assuming the existence of the empty set, and assuming the existence of an infinite set.
 
@ThomasKlimpel Any other questions?
"...he is mainly trying to explain to you why there is a significant difference..."

No. If you read the conversation you'd know that he is mainly trying to explain that infinite sets are not "well-defined".
 
How familiar are you with mathematical logic? How familiar are you with "Wildberger's views"? You request that I explain "Hilbert's vision that "actual infinite" could be justified by finitistic methods was disproved by Gödel in a certain sense...". What can I assume as basis for my explanation? And should I expect that you will later publicly complain about me, if I try to explain it?
 
7:41 AM
The point about the empty set was merely to show that its existence is assumed just like the existence of an inductive set is assumed.
"You request that I explain "Hilbert's vision that "actual infinite" could be justified by finitistic methods was disproved by Gödel in a certain sense...""

Yes exactly. I was genuinely curious about what you meant by that.
Though that was from a different comment thread, if I recall correctly.
"should I expect that you will later publicly complain about me, if I try to explain it?"

If it's a bad explanation, I might.
 
Hilbert had written "The foundations of geometry", and proved the consistency of geometry by constructing a model of the axioms. If you could construct a model of ZFC by similar means, then you could prove the consistency of ZFC. But of course, then construction itself must also be justified, and only if it could be justified by finite means would it be sufficient to justify the "actual infinite".
 
Apr 23 at 4:31, by Rithaniel
Here's a question for you: We know that no set of axioms will ever decide all statements, from Gödel's Incompleteness Theorems. However, do there exist statements that cannot be decided by any set of axioms except ones which contain one or more axioms dealing directly with that particular statement?
Apr 23 at 4:33, by Rithaniel
"Infinity exists" comes to mind as a potential candidate statement.
Apr 23 at 4:38, by Rithaniel
If you were to show that you can attain infinity from finite things, you'd have a bombshell on your hands. It's widely accepted that you can't. If fact, I believe there are some proofs floating around that you can't attain infinity from the finite.
As far as asking any maths person knew, the proposition "whether there exists a foundation F such that there exists a finite string that encodes the derivation of the axiom of infinity as a theorem" is still open
Or in more layman terms, nobody knows whether it is possible to define an actual infinity without axiomises its existence in the first place
 
@ThomasKlimpel What do you mean by "justify by finite means"?
 
There are examples where the construction of a model works. For example, Gödel constructed a model of ZFC using only a seemingly weaker theory, his constructible universe. And certain dependent type theories have models in PA, in a similar sense as Gödel's constructible universe is a model in ZF. And reverse mathematics also managed to justify slightly stronger theories in slighly weaker theories, albeit using different approaches and justifications.
 
In what sense do you deem these to be "finitistic"?
Or you're saying they're "not finitistic"?
 
7:55 AM
As far I am currently aware, the only known foundations where axiom of infinity is a theorem all involve large cardinal axioms
I have yet to came across any foundation where an actual infinity is derived to exist by a system of axioms
 
Well, strictly speaking an axiom is also a theorem...
But I get the gist of what you mean.
 
no they are different. Given some foundation F, a system of axioms S are propositions that are assumed to hold true in F. If you switch to another F', then of course some of the S may become theorems (meaning they can be derived from the S' and nothing else)
that is, one derive theorems from a prescribed system of axioms in a foundation, assuming I understood my logic stuff correctly
 
What I meant was that every axiom of a system F is also a theorem of F.
Trivially.
 
ah ok
(I have a habit of excluding one step trivial cases, often ends up causing me to miss some scenarios because of that)
 
The meaning of "justify by finite means" is more or less understood, but people will still discuss about the exact limit. Assuming more consistency strength than PA is no longer finitistic. Assuming that a certain ordinal number is actual an ordinal number (i.e. well ordered) may be OK or not, it is up to debate. Using an axiom system with provable total functions (like PRA) must be accepted as finitistic. Whether or not PA should be considered as finitistic is up to debate, but it is not.
 
8:01 AM
@Secret This seems relevant: math.stackexchange.com/questions/2198922/…
@ThomasKlimpel You're saying a theory can be considered finitistic iff its proof theoretic ordinal is below, say, $\omega \uparrow\uparrow \omega$?
 
Strictly speaking, the OP's sequence of sets doesn't have a stratifiable description, though the Frege naturals do. The stratifiable version of the OP's sequence is actually a funny creature, and can terminate after a finite number of steps in NF. — Malice Vidrine Mar 23 '17 at 1:23
wow, I should study that more
 
@user76284 No, I tried to avoid saying that. PRA is finitistic, because it only assumes the existence of provably total function, not because its proof theoretic ordinal is below a certain limit. PA is non-finitistic, because it assumes the existence of non-total functions, but its consistency can be justified by finitistic means, because it follows from PRA and the assumption that $\epsilon_0$ is well ordered.
 
"PRA is finitistic, because it only assumes the existence of provably total function" Provably total in what?
 
@user76284 Good question. I have to leave now.
 
I'm thinking EFA.
Good night.
 
8:28 AM
Anyway, I think the axiom of infinity is no more suspect than the other axioms.
 
One of my biggest desires is hoping that some time when I am still alive, I will be able to made first contact with an actual infinite object
It is really really hard to e.g decide whether to find Ultimate L or to go for Martin's maximum when one does not even knew how a physical infinity will behave
 
$\exists x (\exists y (y \in x \land \neg\exists z(z \in y)) \land \forall y (y \in x \rightarrow \exists z (z \in x \land y \in z \land \forall w (w \in y \rightarrow w \in z))))$
It's just predicate logic and $\in$ (hope I didn't mess up parentheses).
By "suspect" I meant suspect in terms of its logical form.
 
It is no more suspect than the other axioms because they are all defined using $\in$ and 1st order logic. But the concept of actual infinity get very troublesome when one tries to do predicative mathematics
Currently it seems there is a consensus that one cannot derive an actual infinity without somehow diagonalising the collection of natural numbers with an uncomputable function
 
I don't think the axiom of infinity is considered impredicative.
"one cannot derive an actual infinity without somehow diagonalising the collection of natural numbers with an uncomputable function"
What did you mean here?
 
give me a sec when I find the original discussion block
 
8:51 AM
in Mathworks (Not the main chat!), Oct 13 '17 at 8:06, by user21820
Like what I told you earlier, the standard set theorist answer is to use Replacement, but as Noah pointed out there, it's not usually considered predicative. He didn't give any reason, but my reason is the same as for the specification schema. Both schemas give the existence of an object that reifies some definable predicate/function. The problem is that any reification of such definable things essentially has 'access' to the entire universe, but it is a member of it.
@user76284 Sorry that it took so long, the whole discussion is scattered and repeated in 3 different chat rooms many years ago
But the gist is, something is impredicative if the object can only be defined by some definable function that is an indicator function of the whole universe, which said object is a member of
 
9:16 AM
> The connections with type theory make constructive versions of NF an obvious target for investigation, but nothing publishable has emerged so far. The only known proof (Specker’s) of the axiom of infinity in NF has too little constructive content to allow a demonstration that INF (NF with a constructive instead of classical ambient logic) admits an implementation of Heyting arithmetic, but opinion is divided on whether or not this tells us that INF is significantly weaker than NF.
2
Q: Proving infinity vs Axiom of infinity

Zelos MalumI am not much of a set theorist, I deal primarely with algebra in my interest and what I study so this is toward set theorists. I am curious as to why cannot infinity be properly proven to exist? I know this is the issue which makes it be an axiom in ZFC set theory but I'd like to know in a more ...

> And then you have the "Jack-in-the-box" axioms - that just give you a set ex nihilo, that you otherwise couldn't build using the toolbox. In ZFC, there's only one - Infinity. But we often consider other similar axioms - we call them "strong axioms of infinity," or large cardinals.
> The reason Infinity has to be a jack-in-the-box is that, if we start with the emptyset, then all our toolbox axioms just keep spitting out finite sets. So we have a "gap" between the finite and the infinite.
@ZhenLin The formula "$x=x$" is stratified, so you get a universal set. — Noah Schweber Oct 9 '15 at 19:26
Thus the question is still open
as even in NF (even ignoring the fact that Speckler's 1953 proof is highly nonconstructive) the reason why one can do the proof is because we have something "larger", the universal set in NF
2
A: Is ZF${}-{}$(Axiom of Infinity) consistent?

Asaf KaragilaYour question is unclear. It is true that $\sf ZF$ cannot prove its own consistency. But $\sf ZF$ can prove the consistency of $\sf ZF-Infinity$, simply by verifying that the set of hereditarily finite sets satisfies all the axioms of $\sf ZF$ except the axiom of infinity. This set, often denot...

and then there are clear roadblocks in ZF-Inf
Thus the question "is there exists a predicative infinity" is basically about whether actual infinity can be defined without a function that quantifies the whole universe of any conceivable systems X
If the object X can be produced only by such a function f, then it is impredicative since X is in the universe, thus f must necessarily involve X, which is "circular" because X is not proved to exist yet, thus putting quantifiers on it makes no sense
-> That also basically what Noah said about Jack in the box axioms
So as far I am aware, the only way this question can be settled is either:
1. Demostration of a proof regardless of which system X one uses and showed it is impossible e.g. via incompleteness theorems, via contradictions, or some other notions of falseness
2. That proving or disproving this question will involve a generalised Godel incompleteness theorems extends to all metasystems of arbitrarily high level, thus ruling out the ability for it to be ever provable
 
9:40 AM
Though to be honest, I think NF is the closest we have get to a positive answer, since the universal set basically came for free from the stratification axiom
we just have to discard the fact we can talk meaningfully about its size and all that nonconstructive baggage
 
 
4 hours later…
1:32 PM
Is there a concept of an element in an algebraic structure being "isolated?" As in "$\gamma$ is isolated if $ab=\gamma\implies a=\gamma$ or $b=\gamma$." Similar to zero in an integral domain, just not strictly in relation to zero.
 
1:54 PM
In algebra, a prime ideal is a subset of a ring that shares many important properties of a prime number in the ring of integers. The prime ideals for the integers are the sets that contain all the multiples of a given prime number, together with the zero ideal. Primitive ideals are prime, and prime ideals are both primary and semiprime. == Prime ideals for commutative rings == An ideal P of a commutative ring R is prime if it has the following two properties: If a and b are two elements of R such that their product ab is an element of P, then a is in P or b is in P, P is not the whole ring R.This...
You can do something a lot more than that, by using prime ideals
Thus your example is a prime ideal that is a singleton
 
Ah, that's a good point.
 
The zero product property results if the prime ideal is also the additive identity or absorber
 
So, yeah, if we're talking about the multiplicative structure on a ring, then the only "singleton prime ideal" would be the zero ideal.
 
I'm working on a proof involving the Bolzano-Weierstrass Theorem, but I feel quite stuck (see the linked question and answer). When we do a proof by contradiction by assuming there is an $x_1,x_2 \in [a,b]$ such that $f(x_1)>f(x_2)$, what is it we're trying to reach in this particular case? I'm also confused at the hint given in the comment to the answer - how would a decreasing sequence conflict with $f'>0$? Any help is deeply appreciated.
1
Q: Proving $f'(x)>0$ implies growing function using Bolzano-Weierstrass Theorem

schnUsing the Bolzano-Weierstrass theorem (for every bounded sequence, there exists a convergent subsequence), how would one go about proving that $f'(x)>0$ on an interval $[a,b]$ implies that $f(x)$ is increasing on $[a,b]$, i.e. $$f'(x)>0 \implies \forall x_1,x_2\in[a,b], \ f(x_1)<f(x_2) \ \text{f...

I can see how one can divide the interval $[x_1,x_2]$ and any subsequent intervals as one wishes to create a sequence which would feed the aim of the proof. Yet am both confused at the aim and how to apply the BWT to a possible sequence.
 
 
1 hour later…
3:33 PM
Artin in his book algebra first shows that $(x^3-2)$ is kernel of substitution homomorphism $\Phi:\Bbb Q[x]\to\Bbb C$ given by $x\to\sqrt[3]2$ is $(x^3-2)$. Then claims that restriction $\Phi'$ of $\Phi$ to $\Bbb Z[x]$ has kernel that is also generated by $(x^3-2)$, and as a proof of that quotes this theorem: Let $f$ be a monic integer polynomial. If $f$ divides $g$ in $\Bbb Q[x]$, then $f$ divides $g$ in $\Bbb Z[x]$.
I can't see why restriction $\Phi':\Bbb Z[x]\to\Bbb C$ has ideal $(x^3-2)$ from above argument.
 
4:01 PM
@Mathein @Alessandro @Daminark @Ted I got the scholarship to study in Heidelberg!
@Silent $g \in (f)$ iff $f \mid g$
 
First of all, congrats @ÍgjøgnumMeg! Now, how do we know that kernel of restricted homomorphism has to be principal ideal?
 
4:18 PM
Congrats! @ÍgjøgnumMeg
 
4:37 PM
Ausgezeichnet, @ÍgjøgnumMeg.
 
@Silent it just says that $(X^3 - 2)\Bbb Z[X] \subseteq (X^3 - 2)\Bbb Q[X]$ I guess
@Ted @Alessandro danke euch!
 
 
2 hours later…
6:47 PM
Hey, I have trouble in understanding the proof of definite integrals. Can someone explain the last step in this picture?
The text is Danish. In English the first two lines says:
1. Target: A = ....
2. We know: A(x) is an integral of f(x)
 
What are the consequences of replacing $x \cup \{x\}$ with $\{x\}$ in the axiom of infinity?
 
@user76284 None, but the former is more convenient
 
So it's equivalent to, not weaker than, ZFC?
 
If, instead, you had an axiom saying "there exists a set $A$ such that for each element $x\in A$ then $\{x\}\in A$"
then you could prove that there exists a set $A$ such that for each element $x\in A$ then $x\cup\{x\}\in A$
and vice versa
 
It has to be nonempty too, right?
 
7:01 PM
Oh, right
 
Otherwise the empty set would satisfy that.
 
Yeah
The usual formulation connects to something called the "Von Neumann ordinals"
Say that we call the empty set $0$
 
Yeah, and this one to Zermelo ordinals.
But von Neumann ordinals, of course, generalize to the transfinite.
 
Ah OK you know what those are
 
The reason I was asking is because this alternative version of the axiom seems simpler.
 
7:02 PM
Right. You also have an easy criterion for when $n<m$
($n\in m$, for von Neumann)
You also know that the set $n$ has $n$ elements
so $A$ has $n$ elements iff there's a bijection $f:A\leftrightarrow n$
 
Consider the axiom: $\forall x \exists y (x \in y \land \forall z (z \in y \rightarrow \exists w (z \in w \in y)))$
Pretty compact.
I modified a bit so you must separately assert the existence of some set.
Alternatively you can have $\exists x (\exists y (y \in x \land \neg \exists z (z \in y)) \land \text{[the inductive part]})$
 
I see
Yeah I mean I guess if your goal was fewest symbols then yeah
 
Actually, I guess just $\exists x (\exists y (y \in x) \land \text{[the inductive part]})$ is sufficient since it doesn't have to start with the empty set.
Yeah, I was basically looking for the simplest axiom that entails the existence of an infinite set.
 
Try to prove that, if there's a set with all Zermelo numbers, then there's a set with all von Neumann numbers, though
(finite numbers)
Good exercise
It involves defining a function inductively and then using replacement
 
I have another question: Tarski's axiom (every set is an element of some universe) is stronger than ZFC. Is there a weaker version of Tarski's axiom that's also stronger than ZFC?
Or I guess I should say "yields a theory stronger than ZFC in conjunction with the other axioms".
I wonder if a similarly compact statement goes beyond the axiom of infinity in its strength.
What about
$$\exists x (\exists y (y \in x) \land \forall y (y \in x \rightarrow \exists z (\forall w (w \in y \rightarrow w \in z) \land z \in x)))$$
 
7:30 PM
so there are primes and semiprimes
are there "quasiprimes" (numbers with exactly 3 prime factors)?
 
in the sense that numbers like 2*3*7=42 exist, sure.
in the sense of that terminology existing already?
 
yeah I've only heard of primes and semiprimes
 
there's a definition of quasi-prime here, but i don't really understand it: encyclopediaofmath.org/index.php/Quasi-prime_number
 
I found a weird function that generated 10/10 primes
but then it decreased from there lol
after the first 60 terms of the sequence, it's at an 80 percent success rate, but I think it will continue to drop very slowly
 
7:51 PM
(not to self: Leave the reply to his question next time user 76284 is on)
 
Is Tarski's axiom implied by some large cardinal axiom? @user76284
 
@user76284 And you want $z\ne y$, right?
So, you want a nonempty set such that, for every element of the set, there's another element that's a superset of the first one
 
@AkivaWeinberger Ah, right.
 
$\exists x\ne\emptyset,\forall y\in x,\exists z\in x:y\subsetneq z$
 
It seems reasonable to me that "there exist a proper class of some large cardinal" could be enough to get universes but I'm not sure
 
7:53 PM
is how I'd write it if I could use more symbols for shorthand
 
@Semiclassical Think huge composite numbers formed in modern encryption schemes by multiplying huge prime numbers, those are likely to be quasiprimes
 
Basically I was wondering whether adding a similarly simple but different "closure condition" takes us beyond ZF.
 
You basically just wrote a variation of the axiom of infinity
 
That's the simplest version of axiom of infinity known (written entirely in 1st order logic)
 
@Secret Nice
My proposed axiom seems simpler, though?
 
7:55 PM
They're defining "simplest" in a weird way
 
@Secret what would you say is a good success rate for a function to generate primes or semiprimes
 
Quantifier alternations, I think
as opposed to, like, pedagogic simplicity
(what's easiest to teach)
 
Structurally, it looks similar, I still see either y in x, followed by a huge block of implication building the inductive part of the set
 
(to humans)
 
I see
 
7:57 PM
@Ultradark I have no idea, I am terrible with prime numbers. If I am good at them, I might have a lot of insights about transcendental number theory already
@user76284 btw, did user21820's definition of predicativity in the quote make sense to you? Sorry that it took so long to dig that up
 
Ah so a proper class of inacessibles is equivalent to Tarski's axiom because $H_\kappa$(=$V_\kappa$ for inaccessible cardinals) is a universe
 
Remind me what an inaccessible cardinal is?
 
So I guess that "there exist an inaccessible cardinal" can be considered an axiom between ZF(C) and Tarski's axiom
 
Is it just, you can't get there with power sets (or something)?
 
@AkivaWeinberger generally something that is unreachable from taking powers iteratively
 
7:58 PM
@AkivaWeinberger $\lambda<\kappa\implies 2^\lambda<\kappa$
 
@AlessandroCodenotti Ah I see
 
@Secret sounds legit
 
@AlessandroCodenotti "it is not a sum of fewer than κ cardinals that are less than κ" is that also necessary?
Or is it a redundant condition.
 
So "$\exists x,\forall y\subsetneq x,\exists f:\mathcal P(y)\to x,f\text{ injective}$"
 
@user76284 yeah sorry, it needs to be regular as well
What I wrote earlier is just called strong limit
So an inaccessible cardinal is a regular strong limit cardinal
(existence of singular strong limit cardinals is easy to prove in ZFC)
 
8:00 PM
Ah 'cause otherwise $\aleph_\omega$ would count
as would $\aleph_0$, technically
($\aleph_\omega=\aleph_0+\aleph_1+\aleph_2+\dotsb$)
 
No, it's consistent that even $2^{\aleph_0}$ is bigger than $\aleph_\omega$
 
Oh
Oh wow
Never mind then
$\beth_\omega$ then
 
But $\beth_\omega$ is a strong limit cardinal of countable cofinality
 
8:02 PM
($\beth_{\alpha+1}=\mathcal P(\beth_\alpha)$
\aleph and \alpha are so confusing
 
Does this look ok?
 
$א$
$דוגמה$
 
Weird, so small
 
trying to prepare an outline for my talk next week
 
8:03 PM
Oops forgot to include $\in$.
 
So basically counting the number of quantifiers and logical operators
 
Yeah
 
$x\implies y$ has a weight of two? ($\lnot x\lor y$)
 
Yeah I wasn't sure about whether to count it as 1 since $\{\neg,\land,\lor\}$ is already a complete basis.
Probably won't matter much.
 
This seems like the sort of thing that only an exhaustive computer search could do
and then I'm not sure a computer would be able to tell when it works
 
8:06 PM
@user76284 $\neg$ and $\lor$ is already complete (so is $\neg$ and $\land$)
 
As is $\lnand$
Aw I was hoping that'd work
$\uparrow$ (NAND)
I think I've seen it written as $\bar\land$
 
Hmm you're right.
@AkivaWeinberger That's true too, though something tells me the resulting formulas won't look very elegant with only NANDs :P
 
$\barwedge$
Huh, \barwedge works
(\wedge is \land, \vee is \lor)
$\barvee$
What!
$\veebar$
Oh weird
 
The problem is I don't know if it makes sense to add too many binary operators, e.g. $\land, \lor, \rightarrow, \veebar, \wedgebar, \oplus, \leftrightarrow, \leftarrow$, etc.
 
Oh $\veebar$ is XOR apparently (not NOR)
You could also weight them randomly, like Scrabble pieces
 
8:12 PM
I tried to fix my LaTeX within the time limit. Oh well.
 
I mean in the end I dunno how much it'll matter
 
What's a minimal basis for intuitionistic logic?
 
nands form a complete basis already?
or it might be nands and nots?
 
@Secret In classical logic it does.
 
right
 
8:17 PM
"intuitionistic connectives are not definable in terms of each other in the same way as in classical logic, hence their choice matters"
Ok, I think I can go with the above. Does this question look fine?
 
"In particular, {∨, ↔, ⊥} and {∨, ↔, ¬} are complete bases of intuitionistic connectives."
 
@Secret Where's that from?
 
wikipedia on intuitionist logic
Otherwise I am only aware it is complete like classical logic does
 
Ah. Damn, now I feel inadequate again.
 
@user76284 looks fine and good to go
 
8:24 PM
Yeah, I'll just post it as-is for now.
0
Q: Simplest axiom that entails the existence of an infinite set

user76284Let $\phi$ be a formula in first-order logic without equality and with the binary relation $\in$. Let the size $s(\phi)$ of a formula $\phi$ be given by the following inductive definition on the grammar of well-formed formulas: \begin{align} s(\forall x \phi) &= 1 + s(\phi) \\ s(\exists x \phi) &...

@Secret I should also add a link to the paper you shared.
 
sure, that will be good
that link only shows the simplest formulation of axiom of infinity in terms of the level of logic used, but yours is different because you are asking the shortest sentence that can formulate the axiom of infinity
so two different definition of shortness to compare and contrast on
 
Ah, someone's already caught up to my pet peeve with equality :P
 
9:09 PM
@Semiclassical is there a Green function $G(x,y;\chi,\xi)$ on the upper half plane $\{ (x,y) \mid y \ge 0 \}$ for the Laplacian operator $\nabla^2 G = \delta(x-\chi) \delta(y-\xi)$ that vanishes on the $x$-axis $G(x,0;\chi,\xi) = 0$?
 
Should be something like $-\ln (x-\chi)-\ln(y+\xi)$, I think?
In physics terms, a delta-function solution to the 2D Laplacian is something like an infinite line charge
And if you want to ensure zero potential along a plane, you introduce an image charge distribution across that plane of symmetry
Simplest idea is to use another copy of the infinite line charge, but at $(x,y)=(\chi,-\xi)$
And iirc the potential of an infinite line charge is like log(distance from line)
Oh. I guess then it should be $\log[(x-\chi)^2+(y+\xi)^2]$
There’s an overall constant but w/e
Well. Should be the combination of that image potential and the potential of the source
 
9:24 PM
I leave for the airport in roughly an hour
Not completely packed yet :/
 
So $$\log[(x-\chi)^2+(y-\eta)^2]-\log[(x-\chi)^2+(y+\eta)^2]$$ @LeakyNun
 
Also they're not serving food (on a 12-ish hour flight) so I have to bring my own
 
That’s my final answer I think (again up to an overall constant)
 
I'm doing a stopover in Stockholm
 
@AkivaWeinberger oof
 
9:25 PM
so I'll technically be in Norway for a bit
(Flying back to Israel from New York)
(and then I stay there 'til mid-June)
 
@Semiclassical oh and its $y$ derivative should look like $\dfrac{y}{\pi(x^2+y^2)}$
 
I want to try to do some more traveling over the next few years
 
Where should it look like that? I don’t see how it could look like that everywhere, since that expression doesn’t reference the source point $(\chi,\xi)$
Oh, like in the sense of an external field. Hrm
 
probably, idk, substitute $x$ for $x-\chi$
 
9:32 PM
Physically, I think I see what’s going on, not sure how to get the Green’s function tho
 
I don't see why your function has a zero laplacian though, because of the $y+\eta$ term
well what's going on physically?
 
Well, when you specify the value of the (normal) derivative on the boundary of Laplace’s equation
In physics terms, that amounts to specifying a charge distribution on that boundary
 
oh and what would that Green function solve?
 
Neumann boundary conditions, basically
That one is a bit of a hassle iirc tho
 
strange
because here it says that it solves the boundary condition $u(x,0) = p(x)$
$\nabla^2 u = 0$
 
9:36 PM
Moreover, I’m not confident you can specify both Dirichlet and Neumann boundary conditions simultaneously in general
So your description perplexes me
Do you have the original problem statement?
 
Suppose we wish to find a bounded solution of the 2D Laplacian equation $\nabla^2 \phi = 0$ subject to $u(x,0) = p(x)$. Show that $\phi(x,y) = p(x) \star_x q(x,y)$ for some $q$ where $q$ is a fixed function to be identified. Explain how $q$ is related to the Green's function for this problem.
 
Gotcha. So this is Dirichlet boundary conditions
Where are you getting the derivative condition from?
 
oh sorry for the confusion
the answer is that $q$ is the derivative of the Green's function
and $q$ is what I told you
 
Hmm. Interesting
Also, it looks to me like log(x^2+y^2) does have zero Laplacian (away from the origin)
And therefore that sum of logs I wrote above should as well
 
so the shifted version has zero Laplacian away from the translation point
i.e. your function doesn't
because there are two translation points
right
oh it's alright the second translation point doesn't exist
oh did I mention that it's the upper half plane
 
9:49 PM
Yeah, it’s not in the domain
You did
 
great
is there... a solution that is not guesswork?
 
Hah, probably. Talk to a physicist, get a physicist answer
 
hey @Semiclassical do you know...
 
if you want a “derivation” rather than just a constructed solution, maybe a Fourier transform approach?
You should be able to get log(x^2+y^2) that way at least
 
I tried... unfortunately log is not bounded
 
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