The h Bar

5:01 PM

'The following words are attributed to Hilbert:
“If I were to awaken after sleeping for a thousand years, my first question would be:
Has the Riemann Hypothesis been proven?”'

Anonymous

@bolbteppa Whoa, I vaguely remember someone saying that but couldn't remember who.

Anonymous

Hilbert, nice

Anonymous

Visualizing the Riemann zeta function and analytic continuation

►

Anonymous

BTW my knowledge of RH is limited to Wiki. But I found that video really enlightening ^

bolbteppa

Think I tried to watch it and gave up, will try again

danielunderwood

5:04 PM

@Blue I think it's decent so far. I think it's meant for people with a bit of QM knowledge and little SR and tends to go on the safe side, so it can be slow at times. I got a similarly titled book on LQG and it was in the complete opposite direction. Like a couple chapters went from "this is a Lorentz transformation" to "this is how we define a Yang-Mills theory"

Anonymous

@danielunderwood Hah, so it doesn't need/presume GR?

Leaky Nun

The Riemann zeta function is heavily connected with primes

(there's this one key connection between complex analysis and number theory: Residue theorem)

it's used to prove the prime number theorem, that there are approximately $\frac x {\ln x}$ primes below $x$

Sir Cumference

I've got to do a presentation about stars next week, anyone know about protostars?

Emilio Pisanty

@JohnRennie not sure what it was, but something jogged this from my memory: theguardian.com/politics/2014/nov/27/…. Wasn't 2014 fun?

Sir Cumference

I gotta make sense of them

Leaky Nun

5:10 PM

We define the prime counting function $\pi(x) := \#\{p \le x \mid p~\text{prime}\}$, and the prime number theorem states $\pi(x) \sim \frac x {\ln x}$

to prove that, we introduce some auxiliary functions

$\psi(x) := \displaystyle \sum_{p \le x} \log p \left\lfloor \frac {\log x}{\log p} \right\rfloor$ and $\psi_1(x) := \displaystyle \int_0^x \psi(t) \ \mathrm dt$

because these functions are easier to work with than the $\pi$ function

danielunderwood

@Blue Not that I've seen. It does use tensor index notation that people with some knowledge of GR may be more comfortable with, but I haven't seen anything other than that. I don't think the book actually treats gravity much other than the weak field limit $g = \eta + h$. I got a preview of Polchinski from amazon and it looks like it requires GR and QFT, which is a bit out there for me at the moment

Leaky Nun

and it turns out that $\psi_1(x) = \displaystyle -\frac1{2\pi i} \int_{c-i\infty}^{c+i\infty} \frac{x^{s+1}}{s(s+1)} \frac {\zeta'(s)} {\zeta(s)} \ \mathrm ds$

and this is one of the examples of the connections between Riemann zeta function and prime numbers

@Blue ^

Anonymous

@LeakyNun Very interesting

Anonymous

I need to read it a few more times

Anonymous

But getting the idea

danielunderwood

5:16 PM

@EmilioPisanty "ingrained liberal bias"...come to the US and you can party like it's 2014 again

Anonymous

How exactly do they get that form for $\Psi_1(x)$ tho?

Emilio Pisanty

@danielunderwood it's just how much easier it was to take things in good fun given how much lower the stakes were

I mean, all sorts of things went wrong in 2014, but they mainly happened in other places

bolbteppa

Why can't number theory be easy and obvious, even simple things are impossible :(

Emilio Pisanty

Crimea & whatnot

it's much easier to go "look at those guys, if you gave them any power they'd break democracy" than it is to make fun of those same guys after they got that power and the went ahead and broke democracy

(* caveats apply. obviously.)

Anonymous

@danielunderwood I see. Could give it a try then :P

Emilio Pisanty

5:20 PM

but anyways. let's not talk politics. or at least I'll stop here.

Leaky Nun

@Blue A generalization of "functions like $\zeta$" is called Dirichlet functions, i.e. functions that are given by a series in the form of $\sum f(n) n^{-s}$ (for $\zeta$ we have $f(n) = 1$). And it turns out that you can differentiate a Dirichlet function term by term, and multiplying two Dirichlet functions together can be performed using Dirichlet convolution (e.g. $(f \ast g)(10) = f(1) g(10) + f(2) g(5) + f(5) g(2) + f(10) g(1)$). Using that knowledge, we convert $\frac{\zeta'}{\zeta}$ into

a Dirichlet series, and then interchange integral with summation

and we express $\psi_1$ also in terms of a series

and then use some Residue theorem to prove that they're equal

danielunderwood

@Blue yeah I've picked up a handful of books that were like "woah, how am I going to get through this?" and it certainly wasn't one of those

Anonymous

@LeakyNun Interesting. So residue theorem is the key connection here

Leaky Nun

20 mins ago, by Leaky Nun

(there's this one key connection between complex analysis and number theory: Residue theorem)

Anyway, from here:

13 mins ago, by Leaky Nun

and it turns out that $\psi_1(x) = \displaystyle -\frac1{2\pi i} \int_{c-i\infty}^{c+i\infty} \frac{x^{s+1}}{s(s+1)} \frac {\zeta'(s)} {\zeta(s)} \ \mathrm ds$

you can use some heavy estimation to bring that integral to a small loop around $s=1$

The $\dfrac{x^{s+1}}{s(s+1)}$ gives $\dfrac{x^2}{2}$ (since this is non-zero finite number, we can move it out)

Anonymous

Ah, I see

Leaky Nun

5:29 PM

the residue of $\dfrac{\zeta'(s)}{\zeta(s)}$ at $s=1$ is $-1$, by the argument principle, since $\zeta$ has a simple pole at $s=1$

so the whole integral becomes $\dfrac{x^2}2$

Anonymous

I should probably go through the individual steps though. Know of any place where this is explained in full?

Leaky Nun

so $\psi_1(x) \sim \dfrac{x^2}2$

and with less work we can establish the prime number theorem from this

Anonymous

Makes sense

Leaky Nun

@Blue there are quite a lot of proofs out there, and I've looked through many, and I prefer chapter 2 of this paper

I don't know about your taste, but you might like this "short proof"

I don't prefer short proofs because it gives me not much understanding

Anonymous

Awesome, thanks :) I don't know much number theory, but since this is more of complex analysis, it seems I can give it a try.

Anonymous

5:33 PM

(Will ping you if I get stuck)

Leaky Nun

@Blue glad to help anytime

be warned that the whole proof is estimation upon estimation upon estimations

Anonymous

Hehe

Anonymous

Will keep that in mind

Leaky Nun

6:00 PM

@Blue This table ^ from here might give you more of an understanding of the connection between prime numbers and Riemann zeta function

$\sigma$ is the real part of the input of the Riemann zeta function

(the convention is $s = \sigma + it$)

(you won't believe how long I tried to re-find this table)

Semiclassical

6:15 PM

Iirc the reflection formula for the zeta function guarantees that if $\sigma + it$ is a zero, then so is $1-\sigma+it$ is as well. (I think you need t nonzero for this to work, so you don’t get doubling of the trivial zeros on the negative real line)

coniferous_smellerULPBG-W8ZgjR

Does anyone know what happened to this guy? physics.stackexchange.com/users/12951/mark-rovetta He stopped to login to all stack exchange websites on the same october 2015 date

his profile indicates he wanted to be active forever

Semiclassical

So there being no zeros for sigma>1/2 is enough to ensure that all nontrivial zeros also have 1-sigma<1/2, and therefore sigma = 1/2

Which is the more familiar form

Avantgarde

@vzn Huh?

Emilio Pisanty

@coniferous_smellerULPBG-W8ZgjR there's a facebook profile that looks like the same person, with updates from 2017

so it doesn't look like a catastrophe

coniferous_smellerULPBG-W8ZgjR

thanks @EmilioPisanty

Emilio Pisanty

6:20 PM

ditto LinkedIn

@coniferous_smellerULPBG-W8ZgjR have you given any thought to removing the MathJax compilation error from your Location field?

coniferous_smellerULPBG-W8ZgjR

yes, but unfortunately I don't really like the idea to change my location. 1st because it is harmless, the only downside is 1 cm of lost space in the page where my name is listed, which is nowhere near a catastrophe. 2nd: if that's really troublesome, I hope this makes dev actually fix the bug (assuming it is a bug). if I simply comply, they might just not fix it and who knows what other people can do with that bug, assuming it is one

but if that gets me banned, sure I'll comply

Leaky Nun

@Semiclassical that's right

coniferous_smellerULPBG-W8ZgjR

and btw it was just a random string of chars, as I once wrote to blue in this chat @EmilioPisanty , you can search for "random" or "zsh" I suppose if you want the details

I had 0 intention to "break" anything

Emilio Pisanty

@coniferous_smellerULPBG-W8ZgjR it's not a bug. You can't disable MathJax because other users do have intentional MathJax in their location profiles.

the location field's length breaking the formatting, on the other hand, is indeed a bug, and it's on the devs to fix it.

seriously, though - if you're going to include active characters in the charset you're randomizing from, wouldn't it make more sense to escape those active characters?

The only change that's required is changing the $s to \$ and they will display as 'intended'.

Physics Meta

0

Q: May I ask a big-list question?

While writing a meta-section in my thesis I wanted to give some examples of important theory proposals preceding experimental verification but I couldn't name very many from the top of my head. For example, I knew that Wolfgang Pauli proposed the existence of the neutrino which was experimentall...

discussion

Emilio Pisanty

6:31 PM

just as a suggestion.

vzn

6:46 PM

@Avantgarde just replying to your own question, do you (really) care or not?

7 hours ago, by Avantgarde

What is the meaning of life?

bolbteppa

35

Q: Sir Michael Atiyah's conference on the Riemann Hypothesis

An announcement of a conference by Sir Michael Atiyah that is to take place next Monday between 9:45 and 10:30 is circulating in several websites. Sure it should always be a thing to listen to him, but this announcement stands out in that he is claiming to have a simple proof of the Riemann Hypot...

riemann-zeta-function riemann-hypothesis

user280247

Are all facts which allow us to postulate electrons as waves, related to many electrons?

user280247

Also in what regards to photons, i.e beam of light

user280247

Cause it seems to me that an electron is a particle, not a wave, but many electrons can behave as waves. Is this correct?

user280247

(btw, I know abt de broglie, etc etc)

bolbteppa

6:52 PM

Motl says in disq.us/p/1vxfaj1 he e-mailed Atiyah and gives the response he got

ACuriousMind

@santimirandarp No. The double slit experiment, for instance, also works when only one electron at a time passes through the slits.

Anonymous

@LeakyNun Aha, cool!

user280247

@ACuriousMind where did you read it? Some days ago I saw a proof on the contrary, I can show you it

user280247

@ACuriousMind en.wikipedia.org/wiki/…

Anonymous

@santimirandarp Show it?

user280247

6:56 PM

show the proof? @Blue

Anonymous

Yes

user280247

Yes, thats on wikipedia

user280247

XD

Anonymous

Where?

Anonymous

Link?

user280247

6:57 PM

@Blue look at the link I appended

ACuriousMind

@santimirandarp That's it. You can see that even though only one electron at a time passes, the interference pattern still builds up.

So it is indeed the wavefunction of each individual electron interfering with itself.

user280247

No, but each dot is an electron @ACuriousMind

Anonymous

@santimirandarp I can only see an image, which confirms that even a single electron can create inteference

ACuriousMind

@santimirandarp Yes, each dot is an electron. What's your point?

user280247

@Blue I interprete it as the contrary

user280247

6:58 PM

@ACuriousMind the interference pattern arises with many particles

user280247

@ACuriousMind so just one electron behaves as a particle, many as a wave...

Anonymous

@santimirandarp Read the section on "Interference of individual particles"

user280247

@Blue but am I wrong?

Anonymous

> Sending particles through a double-slit apparatus one at a time results in single particles appearing on the screen, as expected. Remarkably, however, an interference pattern emerges when these particles are allowed to build up one by one (see the adjacent image).

Anonymous

@santimirandarp Yes

Anonymous

7:00 PM

If individual electrons behaved as particles there wouldn't be any interference pattern

ACuriousMind

@santimirandarp No - each electron behaves as a quantum object, which is neither a classical particle nor a wave. What the single-electron double slit shows is that you see the interference pattern even if no electron can interact with any other electron. You may need many measurements of the single electrons to see the patterns, but these did not interact, so each of them individually must have a wavefunction that can interfere with itself.

user280247

hmmm I see...I supposed this experiment was made differently

Anonymous

Try throwing paint balls at a wall and check if you get any interference pattern

Anonymous

(or use slits if you want to)

user280247

@Blue well don't be rude...it's just a question

ACuriousMind

7:02 PM

Perhaps a more drastic demonstration is that the interference pattern vanishes when you add "which-way information" to each electron by additionally measuring directly at the slit.

Anonymous

@santimirandarp Umm...what?

Anonymous

I meant it for real

bolbteppa

quora.com/…

He apparently made 2 claims that were questioned recently, not just the sphere one...

user280247

@ACuriousMind that's surprising (for me at least)

Anonymous

Delayed-choice quantum eraser

A delayed-choice quantum eraser experiment, first performed by Yoon-Ho Kim, R. Yu, S. P. Kulik, Y. H. Shih and Marlan O. Scully, and reported in early 1999, is an elaboration on the quantum eraser experiment that incorporates concepts considered in Wheeler's delayed-choice experiment. The experiment was designed to investigate peculiar consequences of the well-known double-slit experiment in quantum mechanics, as well as the consequences of quantum entanglement. The delayed-choice quantum eraser experiment investigates a paradox. If a photon manifests itself as though it had come by a single path...

Anonymous

7:06 PM

Actually read the second one

Anonymous

It's basically an extension of the YDSE

user280247

So single electrons produce an interference pattern. If there is only one photon we will see just one dot isn't it?

Anonymous

If you allow them to build up, yes

Anonymous

The point is that they can interfere with themselves

user280247

Yes, or more plainly they behave as waves @Blue

Anonymous

7:08 PM

@santimirandarp Well, you're missing the point

Anonymous

Yes, if you send a single electron you'll notice one dot on the screen

Anonymous

BUT

user280247

with many, an interference pattern, I know

Anonymous

If you send several electrons you'll notice an interference pattern building up

Anonymous

NOW

Anonymous

7:09 PM

the point is, if a single electron behaved as a particle

Anonymous

this wouldn't be possible

user280247

Yes, and neither do photons, isn't it?

ACuriousMind

@santimirandarp The thing is that quantum objects are neither particles nor waves. But when you make a position measurement of one, like when it hits a detector screen, you force it to look like a classical particle that has a definite position.

Anonymous

@santimirandarp Yeah, neither electrons nor photons behave as particles (in this context)

user280247

Hmm I don't like that @ACuriousMind

ACuriousMind

7:10 PM

@santimirandarp Thankfully, nature has never cared for what we humans like ;)

user280247

@ACuriousMind XD

user280247

Just want to say it

Anonymous

Measurement is a bit confusing to understand though (intuitively)

user280247

All I've been taught is: look it is an interference pattern, so they behave as waves

Anonymous

That's non-sequitur really

user280247

7:12 PM

do you mean it is an excuse?

user280247

I have just one more question...

Anonymous

I mean the fact that they behave as waves doesn't follow directly from the fact that they form interference patterns on the screen :P

Anonymous

Anyhow, that's good for a first physics class

user280247

No, that's exactly what was confusing...

user280247

yes, and it was physics for chemists so it is not very deep

user280247

7:15 PM

So the first electron is happily going towards the slit

user280247

then it interferes and we measure

user280247

most probable way to find the dot would be the center of the screen isn't it?

user280247

I hope to be clear

user280247

And the chemist was fired off

coniferous_smellerULPBG-W8ZgjR

I kinda enlarged the userpage

big time this time

physics.stackexchange.com/users?tab=Reputation&filter=month

Anonymous

7:23 PM

10

A: What is the wavefunction of the Young Double Slit experiment?

Well, there are many things you could do. You could: consider two gaussian beams (the linked article is for electrodynamics) apply some paraxial approximation (which would be more appropriate to treat electrons with a high forward momentum) do a cheap/cheesy symmetric point source approximation...

user280247

@Blue :)

user280247

not related at all but this video is really cool: youtube.com/…

user280247

from edward teller, about schrodinger's cat

Anonymous

@santimirandarp That's actually a good question

Anonymous

12

A: Double Slit experiment with just one photon or electron

Has anyone done an experiment where they send just one particle? (And end the experiment) Do you observe two points on the screen in that case or just one? You don't observe two points - you get a probability. This is what the video was trying to indicate by showing dots on the screen. Tho...

user280247

7:29 PM

thanks, I can't read it right now but I've favorited them

Anonymous

YDSE is something I get a new perspective for, every time I revisit it

Anonymous

So, yeah, thanks for the questions!

Anonymous

@santimirandarp One thing to emphasize I guess is: There isn't anything special about two slits. You could use 200 slits and still you'd get a single dot on the screen when you send a single electron

coniferous_smellerULPBG-W8ZgjR

7:46 PM

@EmilioPisanty are you here?

I haev figured out a way to bug the userpage, it's like 2 km long now. BUT

I have noticed that there is an CSS limit nevertheless. it's just set way too high

I mean, look at the userpage. completely demolished physics.stackexchange.com/…

Anonymous

@coniferous_smellerULPBG-W8ZgjR Nice discovery XD

coniferous_smellerULPBG-W8ZgjR

I haven't discovered it, it's @EmilioPisanty who did

Avantgarde

@vzn So the meaning of life is given in that novel?

coniferous_smellerULPBG-W8ZgjR

hopefully this will make the dev lower that CSS limit

Avantgarde

I know the novel. It's pretty popular, though I haven't read it myself.

Anonymous

7:50 PM

@Avantgarde Dude it's even given in my pocket dictionary

Anonymous

(except that it is too fat to actually fit in my pocket)

Lozansky

Man it's pretty cool how matlab has transfer functions and feedback connections built-in

Emilio Pisanty

@coniferous_smellerULPBG-W8ZgjR that.... is one heck of a bug.

@coniferous_smellerULPBG-W8ZgjR Keep in mind that it's a weekend. This is unlikely to get dev-team attention before Monday.

it might be a good idea to try and catch them on the MSE chat then, though.

coniferous_smellerULPBG-W8ZgjR

8:12 PM

just want to say thank you @EmilioPisanty . I'll leave it as it is... for now.

Emilio Pisanty

no worries.

coniferous_smellerULPBG-W8ZgjR

MSE is math stack exchange? or main?

Emilio Pisanty

in Tavern on the Meta on Meta Stack Exchange Chat, 28 secs ago, by E.P.

@coniferous_smellerULPBG-W8ZgjR the main meta

coniferous_smellerULPBG-W8ZgjR

perfect

Avantgarde

8:28 PM

Oh, Atiyah proved the Riemann hypothesis?

I'm late to the show.

Lozansky

Why is it that when drawing a root locus, the value on $K$ for which the systems starts oscillating is always in the interval on the real axis (pertaining to the root locus) closest to origin?

Emilio Pisanty

@Avantgarde claims to

Avantgarde

@EmilioPisanty Yeah I read he'll present it on Monday.

Emilio Pisanty

he's scheduled to present the proof at the Heidelberg Laureate forum, the talk's on Monday morning

whether he has a proof or an almost-a-proof-but-not-quite, remains to be seen

I hope he has a proof

Lozansky

Like here

Emilio Pisanty

8:31 PM

@Lozansky roots for what?

Lozansky

Is it obvious (a priori) $K$ is in the interval $[-1,0]$ and not in $[-3,-2]$?

@EmilioPisanty Poles to the closed system's transfer function

Emilio Pisanty

@Lozansky no idea what you're talking about.

Avantgarde

Yeah sounds interesting, though I don't know anything about the hypothesis

Emilio Pisanty

what closed system, and what transfer function?

Lozansky

@EmilioPisanty The transfer function is $G_c(s) = \dfrac{K(s+2)}{s(s+1)(s+3)+K(s+2)}$

Emilio Pisanty

8:33 PM

@Lozansky if you're hoping for replies from folks that don't already have their noses soaked in exactly the same formalism as you are, that's nowhere near enough information to start helping you.

Anonymous

@Lozansky Well, the image shows the oscillation of what? What does the transfer function represent/correspond to physically?

Lozansky

@EmilioPisanty I'm bad at explaining, but most of the formalism can be found here: en.wikipedia.org/wiki/Root_locus (look at definition)

So basically we find poles to the transfer function as a function of $K$

coniferous_smellerULPBG-W8ZgjR

"1. Let's assume that GR time dilation is caused by the laws of thermodynamics. " "2.Now let's try the opposite, let's say that GR time dilation is caused by the laws of thermodynamics. " LOL

the opposite? Looks the same to me....

Lozansky

@Blue Oscillations to a system that looks like this:

And $G_o(s) = \dfrac{K(s+2)}{s(s+1)(s+3)}$

The root locus is a tool to figure out what values of $K$ make this system stable. What I don't understand is how to find $K$ such that the system becomes oscillating

Emilio Pisanty

oh f*s

please stop

Lozansky

8:40 PM

@EmilioPisanty I like the pun

Emilio Pisanty

@Lozansky what pun?

Lozansky

"In light of the advance[...]" :P

Emilio Pisanty

@Lozansky nah, that's just bait

flattery

Lozansky

But it's clever bait. A hook with a pun on it

Emilio Pisanty

I don't see the pun, sorry

Lozansky

8:44 PM

What journal are they from?

Emilio Pisanty

@Lozansky there are two roots on [-1,0] and a single root on [-3,-2]. The oscillations start when the two roots meet. A single root won't shoot off into nonzero imaginary part.

@Lozansky they're pushing "American Journal of Optics and Photonics" this time

I've had multiple emails from SciencePG

they go straight in the bin so I probably can't tell whether the previous ones were pushing the same journal

the ones from the past three months are. before that, the records are gone.

Lozansky

@EmilioPisanty Hmm makes sense I guess. Do you understand how they found $K=0.4186$?

And the pun was that a journal on optics writes "In light of the advance" :P

Emilio Pisanty

@Lozansky that's when the two roots meet.

(assuming you're plotting roots. if you're plotting poles, take the reciprocal.)

when roots meet, you get a double zero

double zeros have a zero derivative

so if you're plotting the roots of some $f_K(s)=0$, the value $K=0.4186$ corresponds to the point where $f_K(s)=0$ and $f_K'(s) =0$ have a joint solution.

Lozansky

9:02 PM

@EmilioPisanty Yup, that does it

Emilio Pisanty

@Lozansky ;-)

Emilio Pisanty

9:20 PM

hmmmmmmm

@DavidZ @dmckee Do you know if there is a record on Physics Meta of when (and why) [books] questions were banned?

i.e. the change described here

Emilio Pisanty

9:44 PM

oh, and for the benefit of our resident paradigm-shift fan

Analytical chemistry group pins the biochemistry of a 500-million-year-old fossil as containing cholesterol-like steroids, apparently settling the question of whether they're plants or animals

David Z

10:08 PM

@EmilioPisanty I actually thought they were still on topic albeit somewhat "controlled" - they just go by the name of resource-recommendations now. But perhaps dmckee could clarify what he meant there. I might be missing something.

Anonymous

5

Q: New book policy: What should we do about the old questions?

We have a new book/resource recommendation policy. However, it is of a form that it can only be applied to new questions. We have many old questions with answers that do not satisfy the policy, what should we do about them?

discussion books

Anonymous

I suppose dmckee was referring to this

Anonymous

Note: For now, if I come across old book questions (example), I'm closing them with the intent of revisiting them once this discussion gets over. — Manishearth ♦ Nov 15 '13 at 16:24

bolbteppa

10:23 PM

Promoting the integral expression for the factorial of an integer
\begin{align}
n! = \int_0^{\infty} e^{-x} x^n dx
\end{align}
to the Gamma function (with $x$ complex)
\begin{align}
\Gamma(s) = \int_0^{\infty} e^{-x} x^{s-1} dx
\end{align}
we expect to be able to find an expression for $n^s$ by sending $x \to nx$ since
\begin{align}
\Gamma(s) = \int_0^{\infty} e^{-nx} n^{s-1} x^{s-1} dn x
\end{align}
gives the expression
\begin{align}
\dfrac{\Gamma(s)}{n^s} = \int_0^{\infty} e^{-nx} x^{s-1} d x

Obviously related to the Riemann Zeta paper claymath.org/sites/default/files/ezeta.pdf

Emilio Pisanty

10:44 PM

@DavidZ no, the policy decision previous to that one. If I understood the timeline correctly, they were initially on-topic, and then there was (when?) a decision to blanket-ban all list questions, which was subsequently turned over into some-restricted-questions allowed following this and this threads.

I'm looking for (a) confirmation that that timeline is accurate (or a correction if it isn't), and (b) a meta post (or equivalent) that spells out the transition from on-topic to the blanket ban, particularly as regards (b1) the timing of that transition and (b2) the arguments laid down for why that should be done.

@Blue No. That's from after this answer. There @dmckee states

> like I did during the period when books was on-topic.

which implies that book questions were considered off-topic on July 2013 but had previously been on-topic.

David Z

@EmilioPisanty ah, OK I see what you mean. Unfortunately I don't remember the timeline in any detail. But I don't remember much discussion about book/resource recommendation questions happening anywhere other than meta, so theoretically it should be there.

Emilio Pisanty

@DavidZ yeah, I figured as much, but my searches came up empty

they're certainly not tagged [books], there's very little with that tag pre-2013

it's just figuring out a way to fish them out

David Z

All I can think of to find them would be sorting by date and browsing, which is pretty inefficient.

Emilio Pisanty

with the added inconvenience that the SE "advanced" search doesn't seem to have selection by date ¬¬

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