which works well enough...if you can avoid issues of perspective

with basic projective geometry and a few reference points (e.g. known heights or widths) you can usually figure out where the camera was relative to its subject. this doesn't answer the bigger question of what day it was, where the sun was, where north was. but it's a start.

a good article on this topic is by byers and henle, appropriately titled "where the camera was." citeseerx.ist.psu.edu/viewdoc/…

I find it remarkable how many photos of galaxies and on give no clue as to the orientation of the photo or galaxy or whatever. I think the photo should have celestial north straight up.

for instance, suppose you drop a ball and track its vertical motion. that's not too bad: the ball remains the same distance from the camera as it falls, so getting z(t) accurately is easy enough

by contrast, tossing a ball into the air doesn't work so well

it's hard to avoid the ball moving toward/away from the camera along the way

we had some machinery in my physics class for attempting to measure velocities of rolling objects. it was not great. it was also noisy.

and that small discrepancy can make it look like the object is moving faster/slower at the start/end, and thus changing acceleration

it could sort of mostly get the velocity but the acceleration values were useless

well, accelerations are always hard

because numerical differentiation is hard

differentiation sucks.

lol

even non numerical differentiation sucks.

i do like to point out to my students

that as much as they may think that integral calculus is harder than differential calculus

i tell everyone until i'm blue in the face, turn it into something about integration. reason with the derivative and integrate it. do not differentiate anything.

when it comes to numerical analysis, integration is way easier

almost every function space of interest is closed under antidifferentiation, and breaks down at some point under differentiation. just don't do it.

for instance, if your position uncertainty is something like $\sigma_x$, and you sample the positions at a rate of $1/\Delta t$

then your uncertainty in velocity will scale like $\sigma_x/(\Delta t)$...in which case making $\Delta t$ smaller makes the situation worse, not better :P

and even more so for acceleration. the horror of $\sigma/(\Delta t)^2$.

Is it true that the rank of $O_K$ as a $\Bbb Z$-module is the same as the degree $|K\colon \Bbb Q|$? ($K$ a number field)

there is some neat math underlying this, though

the program i use for getting the position data has some built-in algorithms for computing velocity and acceleration from position data

for velocity it's just $v_n:=\frac{1}{\Delta t}(x_{n+1}-x_{n-1})$ for the velocity at time $t=n\Delta t$

which implies that $v_n$ and $v_{n-1}$ are negatively correlated

it gets even weirder when you look at the acceleration algorithm

nearest-neighbor samples are uncorrelated. but next-to-nearest-neighbor samples are negatively correlated :3

i'd love to see what algorithm the thing was using in my physics class. it must have just been complete chaos in there.

(or, more properly, the deviations of the samples from their 'true' value are like that)

you'd roll a ball down a plank and the acceleration graph would look like a saw wave.

that doesn't sound so far off from what i'm describing, actually

i wonder what the salespeople did to demo those products.

What are the dotted lines I see crossing the page here once in a while?

let me generate a quick numerical example

that's a good question, maybe they indicate page breaks in the official printed record of the proceedings of Mathematics chat

JEE main results declared

@AlessandroCodenotti yes. Let {b1, ..., bn} be a basis of K. Then, for each bi there is non-zero integer di such that bi di in OK, so {bi di} is a linearly independent subset of OK of cardinality n, so n <= rank OK.

OK_(0) = K

OK_O=K

O_K_O=K

O_K ill stop

your brain on algebra

Algebra brain

@leslietownes Since I am not a crank of the first order, and we are both in the siblinghood of cranks of secon order could you advise me to whom I might show my Medium article about my new low memorization unified pronunciation system for all bases that are a power of two.

*the second order

*low-memorization

@BalarkaSen beggars can't be choosers, Balarka, this ain't chipotle

good question. if there is some kind of society of base 2 devotees, i'm assuming founded and largely populated by englishmen (and i do mean men and not women), they might be into it.

is JEE main essentially determinative of where people go to school for some subjects? or just a part of it. i don't know the indian system very well.

Reject $f(x)$. Return to $x(f)$

How does men .v women come into it?

@leslietownes JEE main is essential for all engineering students who wish to pursue a good career in Technological programmes....other colleges which support Bsc or Bstat or Bmath take entrance tests like KVPY/IISER

And why Englishman?

i'm being stereotypical, but the prototypical enthusiast of something unorthodox is an english man.

there were some in my family. maybe that's where i got it.

@Snapdragon-X so this is a big day for a very large number of people. interesting.

@leslietownes no...only 10000 people out of 660000

that still seems like a lot of people to me.

The rest have to settle for mediocre colleges and are not treated well at home (although some supportive parents do treat well)

@leslietownes Note: I wasn't being serious about being a second-order crank. I was just playing along with your joke. I don't think I am a crank and I don't think you are either.

@leslietownes Not compared to the second figure

my undergrad had two paths to entrance because it was a public school. if you scored high enough on exams, you just got in, at least if you lived in the state. there was another more holistic route involving essays and high school versions of resumes. grad school was very holistic because the mass-administered exam (the math GRE) was something where, at the time, about 20% of takers got a perfect score. it was meaningless as a differentiator.

I don't consider my ideas to be unorthodox, only poorly understood. As you pointed out correctly, I am not denying any established truths.

i think they have tried to rehabilitate the math GRE but it is really hard to mass-administer math.

No one is arguing that base ten is ideal.

No one is arguing that we should pronounce binary a certain way, or at least there is no orthodox position on that.

matt i think you would really like augustus demorgan's "budget of paradoxes." it is in the public domain, you can also find hardcover versions published about 100 years ago by dover. demorgan is very sympathetic to what he calls 'paradoxers' who simply do not buy the orthodoxy. he is less sympathetic to people who, for some reason, deny established math and then abuse people in public. it's a fun read.

I see

@leslietownes did it look like that by any chance

you're definitely focused on something that has merit. one of the main knocks on binary is the large number of symbols needed to represent even fairly small numbers of interest. which for a computer, who cares, but humans need to remember things. a system of pronouncing them might aid memory and make it a more feasible base to use.

Is Budget about mathematicians?

yes it looked exactly like that. did you visit my physics class in 1998?

lol

it's paradoxers of all types. many are mathematical, but there are also people with astronomical theories, flat earthers, etc. his father in law was a paradoxer about negative numbers. i think it made him sympathetic to unorthodox thought.

@LeakyNun Nice, thanks. I have one more doubt if you have time

sure

what i did there was create Gaussian random noise for times t=0 to T for a particular sampling rate, and took those as position samples for a stationary object

Does paradoxer mean unorthoxer?

it was very fun to work with that equipment. i thought i understood, you know, parabolas and stuff. then i saw that and wondered if everything i had been taught was a lie. but as you indicate, it's just numerical noise.

and then applied the aforementioned algorithm to it. the way that works is to compute local five-point parabolic best-fits to that

which is about as smart of a way to get individual accelerations as one can with just the time series

Do you know how I could make contact with the StackExchange editor called @PM@Ring?

Without the @

per de morgan, "I use the word in the old sense: a paradox is something which is apart from general opinion, either in subject-matter, method, or conclusion."

Now I have $K=\Bbb Q(\sqrt{7},\sqrt{10})$, and I showed that if $\alpha_i=(1\pm\sqrt{7})(1\pm\sqrt{10})$ for $i=1,\ldots,4$, then $3$ does not divide any power of any $\alpha_i$, but $3$ divides $\alpha_i\alpha_j$ for all $i\neq j$. Now I'm asked to deduce that $O_K$ is not generated by a single element over $\Bbb Z$, do you see an obvious way to do that? Because I found the same exercise in a book and it is a bunch of nontrivial intermediate steps before being able to deduce that

anyways, one neat feature of those algorithms is that you tend to get that spikiness

I have reason to believe he or she would be sympathetic.

mostly because next-to-nearest-neighbor samples are negatively correlated

PM@Ring I mean.

*PM2Ring

i picked $\sigma_X=\Delta t=0.1$ in that case, so the noise comes out as roughly $\sigma_X/(\Delta t)^2=10$

which is pretty bad if you're trying to show that the acceleration is zero :P

How does one StackExchange editor send a personal message to another?

@AlessandroCodenotti chissa Dedekind--Kummer aiutera?

@leslietownes anyways, this is why I always tell my students to determine best-fit curves for position and never for acceleration

the former doesn't get destroyed by position uncertainty

@AlessandroCodenotti or maybe they want you to look at the discriminant because there's a formula for monogenic rings or something like that

@LeakyNun Would make sense, it was the previous part of the exercise (I didn't know it was called Dedekind-Kummer however)

How should we pronounce Kummer?

i've mostly heard it in the US as rhyming with boomer.

the english r and german r sound very different to me, but that's what we do in the states.

it's always slightly upsetting to learn the native pronunciation of something that you hear in your mind in a bastardized form.

Is filling tax in your country easy?

@AlessandroCodenotti IIRC you should look at $\alpha_i = f_i(\alpha)$ and show that the minimal polynomial $f$ of $\alpha$ divides $f_if_j$ in $\Bbb F_3$

i live in the US. i think it's fairly easy if you don't have a lot of forms of income other than one job, and nothing that would qualify for deductions from income or tax owed. it gets more complicated if you accumulate property and children and have multiple obligations.

i work with a lot of people in the entertainment industry, where every project is its own job. if you work in editing TV shows, for example, you might have 30 different forms you need to collect and file. and then deduct the costs of business-related personal expenses. it can be a nightmare even if the total amount of income is very small.

I "invited" @PM 2Ring or PM 2Ring to "Mathematics" just now. I hope that was the right thing to do. I think he or she would be sympathetic because I saw a post by him or her a few years ago where he indicated interest in a new pronunciation system for binary.

@leslietownes I see, I frequently observe my dad filling up taxes and I find it super tedious. I even theorized it being the reason for tax evasion LOL

@Astyx right, that's what the book is suggesting too. I guess I have no choice

i've been very lucky, most of my family's income is in the form of wages from two employers. there is a tiny amount of investment income and some small deductions for interest on mortgage payments (a hotly contested issue). i could work it out by hand if i had to.

I thought Euler rhymed with "ruler" until a year or two ago, when I found out it rhymes with boiler.

@leslietownes I am not talking about explicit calculations. I meant is the process of filing the IT form easy?

And I thought Neumann was pronounced like Newman.

Perhaps there's some other way. But it seems intuitive to reduce the problem mod 3 given what you've proven

it really depends on the identity of the taxpayer. for some people there is a 1-page form, called the 1040-EZ. for others, there's more of it. software can automate the filling-in, although usually this is at a cost imposed by private companies. it is fairly moronic that the process is not simpler in the US, because most payors have withholding and reporting obligations which mean that the tax authority already knows the numbers i'm reporting to it.

it takes me probably 2 hours, once all of my family's forms are in. the government already knows what i'm telling it, so it's dumb that it takes two hours. and i pay $50 to some private company to fill the forms in for me.

oops, i used the wrong algorithm for computing accelerations

@leslietownes Exactly! so it's the same everywhere lol...

i understand that in some countries the government just sends you a statement akin to a bill, and you can evaluate that on your own time, but often it's correct. we could do that in the US, but that would put the tax preparers out of business. so we can't do that in the US.

And I had just started conforming to say Seuss to rhyme with Zeuss, when I found out yesterday that it rhymes with voice, which was how I pronounced it as a child.

i liked filing taxes when i was a student. it was the one page. 'i have nothing, i owe nothing. maybe you owe me $100 in excess withholding.' now it is more complicated.

that's more like it.

more representative (as if it looks any different)

haha LOL...I have recently entered that phase though ("I have nothing, owe nothing")

@ Semiclassical, What are you making?

the software i use, when i reported that i was filing jointly for the xth year in a row, congratulated me on my status as a joint filer. it was like "congratulations on being married for several years," but in tax language.

it was very romantic.

Nearest neighbours...Kmeans algorithm?

@Snapdragon-X showing what random noise does to a certain numerical-differentation scheme

@leslietownes xD

aka why taking second-derivatives of data is hard

you basically get a readout on what the source of noise is. i love it.

@Semiclassical i get it ! I didn't get what you said earlier xD

@Semiclassical ... make it nondifferentiable?

@leslietownes yeah. of course, if the expected acceleration varies with time

then it's not as bad

because you can hope to distinguish the large expected change in acceleration from the small statistical fluctuations

but intro physics is so much "constant acceleration ftw" that you don't really get that scenario

i sometimes play an electric guitar and have surprised myself with unexpected noise sources. it can sound really great. it might be less instructive to teach people physics. if noise is dominating the picture that is supposed to empirically illustrate what is happening.

by contrast, it's not so bad for velocity vs. time graphs. smaller statistical noise on top of linear time-series data

acceleration vs. time graphs are usually pretty hosed tho

if i were selling a product i wouldn't even offer the option.

one case where i have had decent results is with circular motion

but there i really have to make students lower the sample rate

and, well, that's precisely a case where the acceleration is not constant. (the magnitude is, but not the components)

so it's about the best-case scenario for an acceleration graph

in my lab in 1998 we also had some equipment to use to measure the mass of an electron. it emitted an enormous amount of ozone and we all got headaches, but nobody was saying anything. eventually i said, 'does anybody else's head hurt?' and we all had headaches.

got a pretty good value for the mass of the electron, though.

@MatthewChristopherBartsh Hi. I guess it's an interesting idea, for people into that kind of thing. :) Personally, I prefer to use hex rather than binary, because it's less error-prone. I'm a little bit dyslexic, and it's very easy for me to mix up bits unless the numbers are small. But if a bitstring is broken up into blocks of 4 bits I can instantly convert it to hex.

Hi.

It's an honor to talk to you.

@MatthewChristopherBartsh FWIW, Morse code operators used to pronounce Morse code using "dit" or "di" for "dot" and "dah" for "dash". Apparently, during WWII, it was common to hear them chatting in Morse in the pub. :)

My system includes a system for pronouncing hex.

And every other base that is a power of two.

I mean, base 2, base 4, base 8, hex, base thirty-two, ... ad infinitum.

@MatthewChristopherBartsh in general, there is no method to communicate personally on SE. Moderators can send messages to people regarding out-of-band situations, but normal communication is via posts (main and meta), chat, and comments.

All using one small set of rules to minimize memorization.

Mathew you can generate a RSA communication and hope that mods dont kick you LOL

What is 'out-of-band'?

RSA is an encryption system, right?

I used to know a guy who could (mentally) do addition in hex quite fast. I used to be reasonably fast at it, when I was doing low-level assembler & machine code, but that was a few decades ago, and I'm a bit rusty these days. Also, age has taken the edgo off my general mental arithmetic skill. :) But I can still multiply 2 digit numbers in my head at a reasonable speed, and do reasonable approximations of square roots.

I've never had those mental arithmetic skills, although mine are better than the average human's, presumably.

@MatthewChristopherBartsh Yep. I got that from your earlier messages. It sounds interesting, but it's not something I'd learn myself, I'm a bit too set in my ways for that. But maybe I would have, if I heard about it 50 years ago...

Am I allowed here to link to a blog article I wrote on Medium.com?

People have been calling me a crank.

I'm not quite sure why this is mathematics, but we sure have plenty of off-target discussions in here anyway.

@PM2Ring I tried standardizing hex usage while doing problems( about 7 years ago, when I was in grade 6)...but it was of no good use for me...so i eventually forgot.

@TedShifrin I guess I shouldn't ask about the weather there, then ;-p

@MatthewChristopherBartsh yeah, the public and private key one

@TedShifrin Well, it's notation. Kind of. Alternative ways of representing numbers are certainly worth exploring, IMHO. OTOH, getting everyone to adopt a new improved scheme for writing or pronouncing numbers is likely face a lot of resistance ad inertia. :)

I've posted a long StackExchange answer here:math.stackexchange.com/questions/65760/…

@PM2Ring We have enough issues in English when people say "one hundred and fifty seven" instead of "one hundred fifty seven." :D

It contains more or less all my ideas on the topic.

@robjohn Canonical So. Cal.

@TedShifrin quatre vingt dix neuf

@Snapdragon-X Oh I see. Ha.

hi

Mais quoi d'autre? @Astyx

It amazes me how much number theory the Alexandrians like Euclid, Archimedes, etc, managed to do, considering the numeral system they were encumbered with. The old Greek system was even more cumbersome than Roman numerals, IMHO.

I think I once read about how the French numeration system came about. Or do you prefer nonante et neuf, @Astyx?

nonante neuf is not french, is quebecois

@TedShifrin I'm Australian, and we tend to use the British conventions with number names, so I'd never say "one hundred fifty seven."

septante and nonante don't exist in France

Well, and I believe that in the Arabic tradition fractions were understood as sums of distinct reciprocals :)

And Belgian.

one hundred and fifty seven

@Astyx: I know that!

Ok, just making sure :p

Walloons in Belgium say septante, huitante, nonante.

Deutsch number names

Now I see what respect you have for my French, Astyx. I'm truly hurt.

einhundertvierunddreizig

I think we should merge with the Linguistics chat.

and latex

german is very coherent with its numbering, apart from the weird switch between units and tens

@TedShifrin Certainly the Egyptians used unit fractions. I'm not sure when Arabic & Persian mathematics escaped from that.

@Snapdragon-X It wouldn't surprise me that some americans would learn septante and nonante and prefer those (since they're "closer" to the english language)

I encountered that fact (and did read up a little on it decades ago) when I worked the cute exercise in Spivak's Calculus book that every positive rational number could be written as a sum of distinct reciprocals. It blew my mind.

@Astyx wait...I am neither german, nor english...I havent heard septante and nonante in conversations or written text

oops sorry

Wrong @

It was directed at Ted

Most ire in this room is directed at Ted.

Ted Sheeran

My favorite Dies Irae is from the Verdi Requiem, @Astyx.

@Snapdragon-X Ack!

It's good :)

@TedShifrin Egyptian fractions are fun. Even after thousands of years, we still don't have a simple algorithm that can always find the best unit fraction representation that keeps the number of terms and the denominators small. Even Erdős did some work on them, IIRC.

@PM2Ring I have been unable to locate a certain forum where there was a thread with a title something like "How to pronounce binary", and in that thread someone suggested pronouncing binary, "one, two, two one, four, four one, ..." and you replied saying that a naming scheme of new names would be needed for the bigger powers of two like 128 and 1024 and so on. Can you help me find it? Has the site disappeared? Or what? I've tried both Google and Duck Duck Go.

@TedShifrin what seems crazy to me about that isn't so much that it's true, but that it's provable

oops xD

@Semiclassic I don't understand that remark. Amplify?

@PM2Ring There's an obvious algorithm, which I've even programmed in BASIC. It seems to give the optimal solution in the terms you describe.

i'm exaggerating a bit, i suppose. but given that the harmonic series grows arbitrarily large (but very very slowly) it feels plausible that all positive rational numbers could be generated

Yes, so why is it crazy that it's provable, @Semiclassic?

but somehow the idea that it's a question that can be easily proven feels weird to me

There's actually a very cute numerical observation that is the key, Semiclassic.

hmmmmm

When you take a rational number $p/q$ less than $1$ and subtract from it the largest possible reciprocal, the numerator of the resulting fraction must be $<p$.

right.

So you get a terminating process.

what about greater than 1? i'm guessing it's simple

Just add consecutive terms of the harmonic series to get close to the number.

ah, yeah

hmm

@MatthewChristopherBartsh It might have been on the old XKCD forum. Sadly, they got hacked a couple of years ago, and were taken offline. But they're archived in the Wayback Machine: web.archive.org/web/20190620104825/http://forums.xkcd.com/…

all hail the wayback machine

i do wonder at what point the wayback machine will simply outpace its space limitations

On a sidenote, who is archiving Wayback Machine

is it apparent that I lack sleep?

learned that analysts prove $$\left( \int_0^tf(s) \, \mathrm d s \right)^p \leq t^{p-1} \int_0^t f(s)^p \, \mathrm d s$$ using Hölder's.... don't hmu for a while...

@PM2Ring Anyway, I have created the naming scheme that you suggested was missing. Not because you said it (I had already come up with the idea independently), but you did beat me to it, regarding publishing the idea.

@TedShifrin Sure, it's easy to to implement the obvious "greedy" algorithm, but the denominators tend to blow up badly. David Epstein has a lot of info about Egyptian fractions, and various strategies for finding them. ics.uci.edu/~eppstein/numth/egypt

How is it greedy to take the largest possible reciprocal (hence smallest denominator) possible?

@TedShifrin until the reciprocal of the next consecutive integer is too big.

I see. So it might be more "economical" in the long run to skip the optimal one and keep looking. Thanks.

This is so far afield from my mathematical interests that this is enough to satisfy me :)

(semi-pun intended)

reminds me, i had a variation on the elementary question i posed yesterday

Thanks, @PM2Ring.

elementary question was: if you sample (with replacement!) a box of n different objects repeatedly, how many times do you expect to sample before getting all outcomes at least once

@PM2Ring I couldn't find the thread at the Wayback Machine page linked to. I have never used the Wayback Machine and I am clueless.

@user2103480 How else?

which, as @robjohn reminded me, is just $n H_n$

@TedShifrin Jensen's. So much cleaner

Oh, @Semiclassic, I taught this stuff in my probability class. I did it in terms of the prizes given out at a fast food place. Expected number of times you have to go to get every prize.

@user2103480 Sorry, why should I care?

Divide the integral by t and lebesgue gives a probability measure, x^p is convex for p>=1

Aha, @user2103480. I rarely think of that (not being a bona fide analyst). But I agree that log-convexity is powerful.

@TedShifrin No worries. It's an excellent site. I read all the articles on the Algorithms page ics.uci.edu/~eppstein/numth/egypt/intro.html a few years ago, and implemented some of them in Python. I can't remember the details, but I remember being impressed by the "reverse greedy" strategy.

I don't know that it's any cleaner, though.

@MikeMiller ...did I talk to you, like, specifically?

No, I mean, why is this inequality interesting?

I'm not trying to give you a hard time.

Sorry, I see how it read that way.

I guess it gives $L^1$ bounds in terms of $L^p$ bounds.

@MikeMiller Just one out of a bunch of inequalities we use to get universal bounds on norms

@PM2Ring By the way, the guy you were replying to appears to have been the first to publish the idea of counting in binary: "one, two, two one, four, four one, ..." and so on, correct me if I'm wrong.

I think having a variable $t$ in the limit is something that shows up in practice, but not in the basic courses. :)

In the context of the usual fixed point proofs of existence of (local) solutions of certain differential equations

however, the context i was thinking about this is in is a bit more involved. suppose that, when you sample, you have the option of taking more than one ball at a time. but when you do so, you can only keep one 'prize' at at each step

@MatthewChristopherBartsh Oh well. That thread might be there. But I'm no expert in using the Wayback Machine either. I can't think of another mathematics space where we may have run into each other, unless it was one of the Math chat rooms here. Or maybe the SO Python chatroom, where we occasionally chat about mathematics.

@Semiclassic: You have that option with no penalty?

@Ted This gets pretty ugly: $\frac{22}7=\frac11+\frac12+\frac13+\frac14+\frac15+\frac16+\frac17+\frac18+\frac19+\frac1{10}+\frac1{11}+\frac1{12}+\frac1{26}+\frac1{844}+\frac1{10862280}$

right, that's where it gets interesting

@robjohn why would you do such a thing

Hmmm, I don't remember any of that stuff!

if you're only interested in minimizing the number of times you pull from the box, --you'd of course just take all the balls at each step--

@LeakyNun Ted suggested it. Don't look at me!

I believe you that it is useful but I'll admit it's not obvious to me why unless you're specifically interested in $L^p$ solutions, and I tend not to be sure why one is interested in those unless you're screwing around with Sobolev multiplication.

@TedShifrin depends on the mode of thinking I guess. There's more need to justify the steps when using Jensen's, but the calculation itself is just $$\left(\int_0^t f(s) \, \mathrm d s \right)^p = t^p \left( \int_0^t f(s)/t \, \mathrm d s \right)^p \leq t^p \int_0^t f(s)^p/t \, \mathrm d s$$

Your original contribution was to point out that no "naming scheme" had been created and one was needed. But you didn't say anything about what form the scheme might take. I'm fairly sure you used the phrase "naming scheme". I was able to recall your name because I took it to be a pun on "A M Turing". Am I right about that?

bah, i can't remember how to strikethrough

i swear i forget it every time

strike = ---strike---

Yeah, @robjohn, I get your point. Spivak does 27/31 as an example in his exercise. And I remember getting 1/239 in a simple example.

oh derp, i forgot to turn chatjax on

@MikeMiller gimme a minute to give a coherent account of why we do that

"naming scheme" sounds like an AG nightmare

sigh

oh well

@leaky This is how the Egyptians first worked with rational numbers!!!

on the first run, there's no reason to take all the balls: you can just pick one, since it's guaranteed to be unique

on the second run, you'd pick two (to ensure that you get at least one new one0

and so on

this will of course get all n outcomes in exactly n steps

So how do you penalize taking them all to make this a meaningful question?

right. the most obvious penalty is to count the number of balls removed from the box

@PM2Ring We didn't run into each other. I merely looked at the web site. I was never a member. I would like to know who the guy you were replying to was. His name was not unforgettable, unlike yours.

@TedShifrin However, I think the Egyptians would be happy with $3+\frac17$

in which case the 1, 2, 3, ... approach is terrible because it scales linearly

@user2103480 I will be teaching shortly so I will read much later. Sorry to challenge then leave!

I would assume they only did this for numbers less than $1$, @robjohn.

the best approach in that case is simply to take one sample at a time

@MikeMiller No worries, I'll need some time anyways and it's a good way to revise my knowledge

...at least i think so

@TedShifrin Indeed, though, as you say, the divergence of the harmonic series allows it to be done for all positive rationals.

Anyhow, when I first encountered this, @robjohn, I was incredulous.

But where I'm not at all sure is what happens when you impose a limit on the number of times you can take a ball. (say, put a $1 cost on each removal of a ball, and say that you don't have infinite money)

Well, then getting $n$ at once is cheaper, so that's the wrong way to do it.

right

@PM2Ring Your original contribution was to point out that no "naming scheme" had been created and one was needed. But you didn't say anything about what form the scheme might take. I'm fairly sure you used the phrase "naming scheme". I was able to recall your name because I took it to be a pun on "A M Turing". Am I right about your name by the way? Posting this a second time because I forgot to ping you the first time.

the numerical example I have in mind (drawn from a video game, lol) has 130 possible outcomes and a budget of $1000

@MatthewChristopherBartsh I took it to be a pun on "A M Turing". That's correct.

so you definitely can't pay 1+2+3+... with that, whereas you can pay 1+1+1... and almost surely get all outcomes

but that doesn't seem like the optimal approach

i do have infinite money. so that is counterfactual

PM Turing. i like that. Turing After Dark.

Is there also the idea of "Ring me in the afternoon. Don't phone before noon" ?

(the real version of this is even a bit more annoying, because the probabilities for getting a new ball in the game aren't just "sample as many as you want, but keep 1 at a time". but i don't want to think about that)

i suppose the simplest thing would be to have some utility function

Turing Nights. did anyone else watch Baywatch Nights? it had plots involving time travel and was spun off of Baywatch. it should have been the next MAS*H.

like, assigning a small penalty to "sample 1 over and over again"