Mathematics - 2021-03-08 (page 4 of 4)

Semiclassical

20:00

fair

leslie townes

my chemistry teacher had radioactive rocks. hearing the geiger counter was a kick. i think it was uranium. it would have been more fun with plutonium.

Ted Shifrin

@Leslie and @PM2Ring seem to have a very off-color sense of adventure.

leslie townes

as they said in repo man, let's go get sushi and not pay.

that's my sense of adventure.

i paid for sushi the other day and lost track of how much i had ordered and ended up feeding some of it which had been left out a little too long to my cat. i won't apologize.

Ted Shifrin

The point, @Semiclassic, is that the volume of melted water is different from the volume of ice. Not intuitively obvious at the outset.

@leslie That might not be so good for the cat. Worms and such.

Semiclassical

the volumes are, yes

leslie townes

20:04

it was a calculated risk. our cat has an eye specialist. if she needs worm treatment, we can go to a worm specialist.

Semiclassical

but to put it another way: the mass of the ice, both above and below the water, is equivalent to how much water we'd need to fill the submerged portion

Ted Shifrin

You don't need a specialist. You just need drugs @leslie. My cats were hunters outside and so I had to treat them often.

Semiclassical

which means that if the ice melts, all we're doing is replacing the displaced water with actual water

Ted Shifrin

Right. So the level doesn't change.

Semiclassical

right

PM 2Ring

20:05

There's been a few questions about plutonium and radiothermal generators on the Space Exploration SE in recent weeks. I spent a couple of hours trying to find a more accurate figure for the amount of heat generated by plutonium 238. Wikipedia only gives 2 sig figs: 0.57 watts/gram

Ted Shifrin

Did I give you the physics puzzle that a math friend gave me at a conference? I've had students figure it out in milliseconds, but I sure didn't.

leslie townes

i had a cat who would kill absolutely everything. birds, lizards, opossums, raccoons, squirrels. she'd leave these really elaborate murder scenes on our porch. like some kind of serial killer with a signature who was taunting the police.

olivia eats canned food and the occasional sushi.

Semiclassical

the way i tend to make it intuitive is to imagine two scenarios. one is like what PM posed: used some mechanical object to keep the ice submerged

Ted Shifrin

@leslie She was leaving you bountiful gifts.

Semiclassical

in that case, the ice shrinks as it melts and therefore the water level falls

user2103480

20:06

@MikeMiller we actually first look for any kind of solution that satisfies certain measurability and then prove that there's a continuous thing in the L^p equivalence class, but this is not important and you see it right

Semiclassical

alternatively, suspend the ice above the water and allow it to melt. then of course the water rises

Mike Miller

Got it, the C^0 norm messes up the contraction mapping argument

Ted Shifrin

@Semiclassic: Given two identical metal bars and absolutely no props (no hair, for example), you're told that one is magnetic and the other not. Decide which is which. You will get this in less than milliseconds 'cuz of all your teaching.

Mike Miller

So you get your L^p solution then prove regularity theorems

That's common enough though I only know how to do that in Sobolev land

user2103480

I mean, it is important and that's the part of the proof where one uses all the big guns, but you get the essence

@MikeMiller yup yup

Semiclassical

20:08

hmm.

Mike Miller

Do you get like Holder regularity theorems?

Semiclassical

move one bar around the other?

should get some lorentz force shenanigans

Ted Shifrin

How do you remove symmetry from this? No, no, no relativity. Just middle school.

Semiclassical

pfft, electromagnetism is inherently rlativistic

Ted Shifrin

Blah.

leslie townes

20:09

would licking the bars help? because i did that.

Ted Shifrin

You're making me feel better for how stupid I was.

smacks leslie

No, that's using a prop.

leslie townes

so you fiddle with the two of them. that's all you can do. fine.

i'm not sorry for licking the bars

Ted Shifrin

Especially if one was chocolate-coated.

leslie townes

just fiddle around with the pair of them until you see that one of them is a magnet

Semiclassical

yeah, breaking the symmetry between them is the hard part

leslie townes

20:11

my daughter, 2.4 years old, is obsessed with magnets. i'll ask her and report back.

Semiclassical

are we assuming the other metal is a conductor?

leslie townes

this morning she was doing the junco dance. it is inspired by the dark-eyed juncos she sees in our yard. she flaps her 'wings' and shouts 'junco dance' while jumping up and down. i think she's ready for the magnet puzzle.

PM 2Ring

@TedShifrin I got it in less than a millisecond. But I don't know whether that's because of the countless hours I've spent playing with magnets, or because I've heard it before. :)

Semiclassical

if it's something that can be magnetized by the presence of the magnet i'm dubious this works

Ted Shifrin

Yes, she should know immediately. Reminds me of an old Groucho Marx line.

copper.hat

20:12

your daughter needs some neodymium magnets (just kidding!)

leslie townes

we have them, i've kept them well away from her.

Ted Shifrin

@PM2Ring: If you've heard it before, of course. It's interesting how our minds work. Of course, I knew the answer cold, but didn't summon it. But freshman calculus students do.

Semiclassical

smart

leslie townes

my favorite groucho line is on meeting a widow, 'will you marry me? did he leave you a lot of money? answer the second question first.'

Ted Shifrin

Leslie's daughter may turn into copper's daughter when she grows up a bit.

copper.hat

20:13

she has a bit of a quandary at the moment

PM 2Ring

@leslietownes I have some Nanodot magnets. I bought them a few months before they were banned here.

Ted Shifrin

LOL, I adore the Marx Brothers. ... My line was more about his retort to Chico, who said "This is so simple a five-year old child would understand it." (Or some number of years.) "Well, then, fetch me a five-year old child and have him explain it to me." In typical Groucho lilt.

Oh oh, what have you done, @copper?

copper.hat

her passport expires in april but the uk embassy is not taking in person applications (which she needs as it is her first as an adult).

leslie townes

i went through several years where i was obsessed with the marx brothers. i read everything groucho wrote, including his tax advice pamphlet, 'many happy returns.'

Ted Shifrin

Oh, the US embassy in the UK?

copper.hat

20:15

london

Ted Shifrin

Hmm, so maybe you need political intervention?

leslie townes

the magnets i own probably should have been banned. my daughter will ingest almost anything.

Ted Shifrin

Yes, no ingestion allowed.

No marbles, either.

copper.hat

@TedShifrin i think so. i am trying to decide when to pull the emergency cord.

i was hoping to do it without invoking some of my favors.

Ted Shifrin

There must be other foreign students in the UK in a similar position?

leslie townes

20:16

is now the right time to complain about EU 'strangulations'? i am a brexiteer. there. i said it. it definitely seems like a good idea for the UK to do that to itself.

user2103480

@MikeMiller Hmm I think I can't quite answer this, but you may be right. What we do is prove that a "modified" version of the convolution with the semigroup ( $F: y \mapsto \int_{0}^{t}(t-r)^{\alpha-1} T_{t-r} y(r) d r$ with $1/2 > \alpha > 1/(2p)$ ) maps L^2p into continuous functions, by saying that it satisfies this for continuous functions and by proving that for an approximation of an L^2p function by continuous functions, the images converge uniformly

copper.hat

my consular capital is in short supply

user2103480

This all pretty much sounds like some regularity arguments but I don't grasp it enough yet

Mike Miller

Interesting, this is out of my wheelhouse completely

copper.hat

@TedShifrin probably not many.

Ted Shifrin

20:17

@leslie Next you're going to announce proudly that you're in love with Trompolini.

Semiclassical

the route i'm thinking right now is to put one vertical and pass the other horizontally above it

PM 2Ring

@Semiclassical You're getting warm

Ted Shifrin

I can tell that Semiclassic will pass this puzzle on to his students next semester.

leslie townes

i'm not in love with him, but i envy his undeniable manhood.

Ted Shifrin

wretches violently

leslie townes

20:18

the lying about his height and weight and finances. that's a man.

Semiclassical

whichever one has a force on it, is not the magnet

user2103480

@MikeMiller maybe you know the term for it, "factorization method"

Ted Shifrin

what kind of contorted misogynistic str8 view of manhood to have.

Mike Miller

Nope new to me

leslie townes

i think he could have found a better tailor.

Ted Shifrin

20:18

If you are epsilon serious, your wife needs to take you out behind the wood shed and clobber you.

Semiclassical

@TedShifrin that never give up, never think things through attitude

user2103480

Anyways, this has been helpful, thanks for the inquiry. I didn't before see the similarity to the usual PDE style of "prove there's some kind of solution then show it's got good properties"

schn

Semiclassical

i do think it needs to be some kind of relative motion thing

leslie townes

i'm not remotely serious, i think anybody who sees themselves in tr*mp needs years of therapy which they won't get because of the policies they have convinced their representatives to support.

schn

20:20

Is the Fourier transform of the unit Heaviside step function correct in the attached screenshot?

Ted Shifrin

@Semiclassic Not really. Of course, it's all about "relativity" in that sense.

Semiclassical

i will say that the classic example this brings to mind is pushing a magnet through a conducting coil

schn

$h(t)$ is the unit Heaviside step function, $H(\omega)$ is its Fourier transform.

Ted Shifrin

Go back to middle school, @Semiclassic :P

Semiclassical

where the problem is that, depending on which reference you pick, the attraction between them is either due to induced electric field or due to the motion of charges in a magnetic field

geocalc33

20:23

I'm kinda confused...if you have a series in R^2 then it depends on how you measure distances so it would depend on the metric you're using right?

Nike Dattani

This was interesting. I had no idea about some of these types of logic until doing the research to answer the question:

3

Q: How does the NOT gate generalize beyond binary?

If we are working with qudits instead of qubits, how do the NOT and CNOT gates work? If the control state for a qubit system is $|1\rangle$, what is it for a $d$-ary qudit system, and why? For instance, in a qutrit system with three bases $|0\rangle,|1\rangle,|2\rangle$: if one had a CNOT gate, w...

schn

@schn From youtube.com/watch?v=8yGs0eUqrg8 .

Semiclassical

the other reason i feel like lorentz force shenanigans should be applicable here is because it breaks Newton's third law (at least naively)

leslie townes

so how does the magnet thing work. i licked them. they tasted the same

Semiclassical

i still stand by "hold one level and drop the other in front of it"

PM 2Ring

20:25

@leslietownes Hint: what happens in the middle of a bar magnet?

Ted Shifrin

Shhhhh.

Semiclassic is still thinking.

PM 2Ring

Ok

Semiclassical

eh. i'm thinking that you could use Lorentz force to distinguish it, just as you use Lorentz force to distinguish a whole bunch of other things

Ted Shifrin

@geocalc You need a norm at the very least.

Semiclassical

i'm pretty convinced that you could use their relative forces when in motion to distinguish them. it may be that there's an answer that doesn't require motion, but i find it pretty silly to dismiss that. (the example i gave above is what Einstein uses to lead off his first paper on special relativity, for example)

leslie townes

20:30

that is a cool and admittedly hard problem. anyone who can solve it immediately should be on a watch list.

they might be x-men.

Semiclassical

if we're assuming that the other metal bar can be magnetized by the magnet, that's different

Ted Shifrin

Semiclassic isn't taking me seriously when I say middle school.

Semiclassical

hence my question way back at the beginning about whether the other bar is a conductor or something that can be magnetized

Ted Shifrin

LOL ... my original attempted solution was trying to use the magnet to magnetize the non-magnet. But then which is which?

Semiclassical

also, if you say "break it in half and see if they stick"

i count that as using an outside prop

Ted Shifrin

20:32

Yes, and how do you break it without fracturing the magnetic field?

Not allowed.

leslie townes

one time i was interviewing for a financial-adjacent firm and was posed a problem. i solved it insantly, not because i'm a genius but because it resembled soemthing i had done before. then the guy asked me to talk about my phd thesis 'to explain it to a ten year old.' i hesitated. no more interviews after that. maybe there are things that can't be explained to a ten-year-old. i solved your dumb riddle.

whatever, i own a house now.

Semiclassical

uh. putting a hairline fracture in it wouldn't fracture the magnetic field

Ted Shifrin

What axe do you break it with?

Semiclassical

sure, i'm fine with that being disallowed on mechanical grounds. i just don't agree with it calling that 'fracturing the magnetic field'

if i put a hairline crack in the metal bar, it falls apart spontaneously. if i put a hairline crack in the magnet, it'll stick together. no fracturing of the magnetic field required---just fracturing of the metallic bonds

that's still disallowed

Ted Shifrin

One of my students told me the following question he was asked at an interview on Wall Street. You have 30 minutes to make a 4-mile bicycle ride. No other requirements. Must there be a 15 minute interval in which you went exactly 2 miles?

Buraian

20:37

damn

that's an amazing question

Semiclassical

that seems like a mean value theorem thing (or intermediate value, i'm forgetting which)

leslie townes

i had something similar about arrival times of buses. it was stupid and humiliating. can you just talk to me like maybe i have something to offer you? and not just goofy crap? is your office staffed with goofballs?

Ted Shifrin

He got all their other puzzles, but not that one. Since I had taught him a fair amount, I shamed him for not getting it immediately.

leslie townes

he was right not to go there, for not being a goofball.

Thorgott

yeah, it's IVT

Ted Shifrin

20:39

This is standard at financial places and consulting places. They want to see how you problem-solve (and react to not knowing things cold) on your feet.

amWhy

Hey, @robjohn! St. Patrick's day is around the corner! I'm looking forward to seeing the "mean green Shamrock machine"!

Buraian

4/30 = 1/7.5 , now 15/7.5 =2 . So, if you go faster then you hit your destination faster then you will still cover more than 2 miles in that time

That sound right?

Ted Shifrin

Yes. You need to arrive exactly 30 minutes after you depart.

leslie townes

i was also asked about the number of gallons of gas sold in the united states. i was right, as to the order of magnitude. i think they mostly pushed me out because i become hostile to non-technical questions.

Semiclassical

my somewhat silly mental picture is to start with the position vs. time graph

Ted Shifrin

20:41

Wow. I'd have no clue on that one, Leslie.

Semiclassical

ugh, Fermi problems

Thorgott

what kinda question is that

amWhy

Hello, Ted! You going to paint your, uhm, whatever (looks like shells) green for St. Pat's day? But I do not want to interrupt, which it seems I'm doing....

leslie townes

while answering it i realized, i do not want to work here.

Semiclassical

that's a classic Fermi problem. you see them in physics from time to time

Ted Shifrin

20:42

Erase, robjohn.

I know you can do it.

All these people ruining puzzles.

Semiclassical

i don't know how one would set it up, but the go-to example is Fermi's piano tuner problem: grc.nasa.gov/www/k-12/Numbers/Math/Mathematical_Thinking/…

Ted Shifrin

But @Buraian nailed me on a technicality omitted from the statement. So he gets a point :)

robjohn

@TedShifrin Sorry, it's my bad for not reading back far enough

geocalc33

@TedShifrin okay let's say I have a metric and a norm in R^2 and also in R^2_+. And I have a map between R^2 and R^2_+. Can I "transport the series" from R^2 to R^2_+ in some way? I know that I can transport the metrics...

leslie townes

i did it from inferences from the number of gas stations in iowa city and the population of iowa city, and estimates about the population of america.

Ted Shifrin

20:43

We have two puzzles on the floor, @robjohn.

Semiclassical

1) From the almanac, we know that Chicago has a population of about 3 million people.

2) Now, assume that an average family contains four members so that the number of families in Chicago must be about 750,000.

3) If one in five families owns a piano, there will be 150,000 pianos in Chicago.

4) If the average piano tuner serviced four pianos every day of the week for five days rested on weekends, and had a two week vacation during the summer, then in one year (52 weeks) he would service 1,000 pianos.

leslie townes

i think that's a phenomenally dumb way to hire people.

Buraian

hehe :-)

Ted Shifrin

That's precisely the thinking they are looking for, @leslie. I still couldn't begin to do it.

I think seeing how people approach problem-solving is a valid thing to do. How many hairs are on your head? That's another one.

Semiclassical

I can respect it but I just sorta hate it

Ted Shifrin

20:44

Ability to think with order-of-magnitude reasoning, not formulas.

leslie townes

red headed people have fewer hairs than others. that question is hairist.

Semiclassical

@TedShifrin hair on your head has a lot fewer assumptions along the way, though

geocalc33

I think there are 20,000 hairs on my head

Semiclassical

estimate hair per square inch, estimate the size of a head, estimate the area on your head

Ted Shifrin

Or, in my case, maybe a couple thousand hairs left. :P

geocalc33

20:45

okay I was wrong there are approx. 100,000

Semiclassical

now dimensional analysis problems, them i like

PM 2Ring

Quick Fermi problem: What's the square root of the ocean? That is, what volume is $v$, such that $n$ is the number of molecules of water in $v$ and $n^2$ is the number of water molecules in the ocean, IOW, the volume of the ocean is $nv$.

Ted Shifrin

Lunchtime for me. Bye, all.

leslie townes

i stopped being good at numbers and just talked shit through interviews so i could get jobs in the legal industry. who knows. i'll say 20 million. or billion. or trillion.

amWhy

@TedShifrin Not until you tell me if you're going to don green on March 17th!

geocalc33

20:48

@PM2Ring I'm gonna say 239 billion

Thorgott

i think the number of hairs on my head is less of a "problem" and more of a "fact no one should ever care about"

Semiclassical

lol

leslie townes

i have relatives who were bodyguards for english judges during a very bad time in ireland, and other relatives who bore arms against people for stuff like that. i have mixed feelings about donning green (and perhaps more importantly orange) on any holiday.

Semiclassical

the one thing i'll say for it is that it does get you to think about estimating the whole from the small

leslie townes

just tell me how great i look. i don't need follicles to pull this look off.

Semiclassical

20:50

if you want to figure out how many hairs are on your head, start by figuring out how many are in a small portion of your scalp

then extrapolate

PM 2Ring

@geocalc33 If that's an estimate of $n$, you're way too small.

I want $v$, preferably in millilitres, but I'll settle for other units. :)

leslie townes

my scalp kind of sucks right now. it was fine 20 years ago

PM 2Ring

And it's a Fermi problem, so you can be off in either direction by an order of magnitude or two.

geocalc33

does a series change, when you change the metric and norm?

Edward Evans

@Thorgott I wish I could get that question

Matthew Christopher Bartsh

20:56

First let the bar magnet and the iron bar stick to each other. Whichever one you can rotate 90 degrees without changing the attraction in the iron bar.

Edward Evans

cuz my head looks like a cue ball

Matthew Christopher Bartsh

*is the iron bar

Semiclassical

that's changing the problem, though. from metal bar to iron bar

geocalc33

my head is the solution to a moduli space problem

Matthew Christopher Bartsh

The metal bar is attracted to the magnet, right?

Semiclassical

21:00

a dime is metallic. but it doesn't react to a magnet. (I've got a neodynium one on hand)

this is why i was asking about the kind of metal.

if it's ferrous, then that changes things

Matthew Christopher Bartsh

Okay, I see it says two identical metal bars. My mistake. But if they are identical, how can one be magnetic and the other not? That would be a difference.

Semiclassical

in other places where i'm seeing it asked, it's uniformly a bar magnet and an iron bar

aka a strong magnet and a magnetizable metal

PM 2Ring

@MatthewChristopherBartsh The bars have identical composition. But one is magnetised and the other isn't. So the unmagnetised one is attracted to the magnetised one.

Semiclassical

yeah, uh, good luck using a neodynium magnet to pick up stainless steel silverware

Matthew Christopher Bartsh

So is the answer that it is a lie when it is said that one is magnetic and the other is not?

Semiclassical

21:03

no

Matthew Christopher Bartsh

Why not?

PM 2Ring

Also, the magnetic one is a normal bar magnet, with a single north pole at one end and a single south pole at the opposite end.

Semiclassical

one has a fixed magnetic field which won't change

(ignoring demagnetization and all that)

the other is an iron bar, which becomes magnetized in the presence of an external magnetic field

mostly i'm just annoyed, because not all metals contain iron

Matthew Christopher Bartsh

So my first answer is right?

Semiclassical

so i don't think the problem was stated right

if the problem is "distinguish between a bar magnet and an iron bar" then fine

but "distinguish between a bar magnet and an arbitrary metal bar"? i call foul on that

Matthew Christopher Bartsh

21:05

I think I got the problem as it was stated correct with my second answer.

Semiclassical

(and if we're invoking a certain magnetic field as part of the answer, well---it may be a familiar part of everyday life, but within the context of physics it's an external field like any other.)

Matthew Christopher Bartsh

@PM2Ring Why did you find yourself multiplying hex numbers in the course of your programming?

@PM2Ring I thought hex numbers in programming were data.

PM 2Ring

@Semiclassical If you could use props, then yes, you could use the Earth's magnetic field to figure out which bar is the magnet. But no props are allowed, just the two magnets.

Semiclassical

is your finger a prop, hah

Matthew Christopher Bartsh

My first answer uses no props.

Semiclassical

21:11

but, again, i'm annoyed at the switch from "a metal bar" to "an iron bar"

PM 2Ring

@MatthewChristopherBartsh You need to do additions & subtractions to figure out memory addresses. You rarely need multiplication, eg when you have an array of structures, but you can cheat, since you know the structure size and number of array elements in decimal.

@MatthewChristopherBartsh Your first answer isn't completely clear to me. How are the two bars aligned?

Here's the original question again:

1 hour ago, by Ted Shifrin

@Semiclassic: Given two identical metal bars and absolutely no props (no hair, for example), you're told that one is magnetic and the other not. Decide which is which. You will get this in less than milliseconds 'cuz of all your teaching.

Semiclassical

i call BS on that problem statement

Matthew Christopher Bartsh

@PM2Ring So why were you able to multiply in hex? Did you just learn to do that for fun?

Semiclassical

if the other bar is just metal, that is not enough to say anything about their magnetism

copper.hat

are you allowed to use the earth's magnetic field?

Semiclassical

21:15

it may seem like i'm being defensive about this (and maybe i am). but it's the equivalent of someone posing the IVT problem from earlier and saying "i've got a function f such that f(0)=0 and f(30)=4. must i have f(t+15)=f(t)+2 for some t?"

And the answer in that case is no, because I've said nothing about $f$ being continuous

once i set it into a motion context, then the inference to continuity is fine

but just saying "a function" is not going to cut it, in the same way that being metal is not going to cut it

PM 2Ring

@MatthewChristopherBartsh I don't think I said that I multiplied in hex. But I certainly did lots of addition & subtraction. I did know a guy (one of my first programming teachers) who could multiply in hex. He was an expert in both hardware and software. And he was also studying ancient Greek.

Matthew Christopher Bartsh

@PM2Ring So why did he know how do that?

PM 2Ring

@Semiclassical THe other bar is magnetisable, and is attracted to the magnet. It has to be, since it's otherwise identical to the magnet. Apart from not being magnetised.

Semiclassical

people have hobbies

yeah, uh, pure iron bars? they're not permanent magnets

PM 2Ring

@MatthewChristopherBartsh He was good at memorising stuff. He was also very fast at reading punch cards.

Semiclassical

21:21

if a material is magnetizable in the way you want it to be, then it's not going to be a permanent magnet

Matthew Christopher Bartsh

@PM2Ring Did he learn to multiply hex numbers in order to be better at programming?

PM 2Ring

@Semiclassical Let's say the two bars are made of the same alloy which has appropriate magnetic properties. Imagine you're at a magnet factory. You've been given two identical bars. One bar has been through the machine that makes magnets, the other one hasn't.

@MatthewChristopherBartsh No, he did it for the pure geek cred. :)

Matthew Christopher Bartsh

@PM2Ring Could you put me in touch with him?

PM 2Ring

@MatthewChristopherBartsh I haven't seen him in over 40 years.

Semiclassical

that's not how magnets are made. you have hard iron and soft iron. hard iron retains magnetic field, soft iron doesn't. if you have two hard iron bars, one magnetized and the other unmagnetized, then putting one in the presence of the other will render it as having the same magnetic field

the problem only makes sense if one of the items retains magnetic field and the other doesn't, and those require different compositions

Matthew Christopher Bartsh

21:28

@PM2Ring Is my unified system for pronunciation of all bases that are a power of two an original idea?

PM 2Ring

@MatthewChristopherBartsh I think so.

:57278532 Yes. I know what you mean. I've played with magnets and paper clips. That doesn't matter (very much) for the solution to this problem.

Matthew Christopher Bartsh

@PM2Ring Is it the best system for pronunciation of bases that are powers of two ever proposed?

PM 2Ring

@MatthewChristopherBartsh I don't know.

Matthew Christopher Bartsh

@PM2Ring I mean when you are counting, measuring, and doing calculations in those bases.

@PM2Ring Even in base 10 we pronounce telephone numbers as a string of numerals, but not when pronouncing a number that is a quantity.

Edward Evans

Why is that a problem

Matthew Christopher Bartsh

21:41

@PM2Ring Have you heard of a system that is better for the purposes I described?

@EdwardEvans Why is what a problem?

Edward Evans

Having to pronounce telephone numbers as a string of numerals

Matthew Christopher Bartsh

It's not problem.

Who said it was?

Edward Evans

idk what your purposes are since I just read your last couple of messages but I thought you're trying to solve some kind of problem lol

PM 2Ring

@MatthewChristopherBartsh I'm not familiar with your system.

Matthew Christopher Bartsh

I realized that you can use the subsets of the names used for binary as the names for the powers of two used for base 4, base 8, base 16 and every other base that is a power of two.

@PM2Ring Thus binary 10000 (sixteen) and hex 10 can both be called 'ri'.

@PM2Ring So you can combine and mix bases without any conflicts, and without any new names being needed. Plus all the binary names in the first place can be deduced. You only need to memorize the rules for generating the infinite set of names.

@PM2Ring Thus people who are interested in using one or more bases that are powers of two can do so in a way that is analogous to how we use base ten when counting and calculating, and not only in a way that is analogous to reciting telephone numbers. By calling binary 10 "one oh" , you are treating it like a phone number, which is not appropriate if you want to compare it to base ten in a way that compares like with like.

@PM2Ring My system could be useful in education whenever base 8 is taught, and is compared to base ten, for example. As it is, like is not compared with like. Base eight is read out like a phone number while base ten is read out like a quantity making a false comparison.

@EdwardEvans Edward, you can learn about my system from my answer, rated -2, here: math.stackexchange.com/questions/65760/…

wklm

22:03

Hello Everyone :) Why does a cartesian product can't form an algebra?
From what i've researched it's not closed under complementation
but I can't figure out why
could someone provide a counterexample or intuition?

Thorgott

I don't understand your question. What exactly doesn't form an algebra? And what meaning of the word "algebra" are you using?

Matthew Christopher Bartsh

Hex counting: fifteen, ri, ri one, ri two, ri three.

PM 2Ring

@MatthewChristopherBartsh Ok. I remember seeing it the other day. I didn't downvote it, but I can't justify upvoting it either, since it doesn't seem to be really answering the OP's question, which is about current standard methods of pronouncing non-decimal numerals. I've only briefly skimmed parts of your answer, but I will look at it more closely over the next few days.

Matthew Christopher Bartsh

@PM2Ring Binary counting: eight four two one, ri, ri one, ri two, ri two one

wklm

sorry, it got to fulfill those 3 conditions en.wikipedia.org/wiki/Field_of_sets

apparently a cartesian product isn't closed under complementation

so it's not an algebra. But I can't see why is that

Matthew Christopher Bartsh

22:10

@PM2Ring Great. I look forward to hearing your thoughts.

Thorgott

"A cartesian product" isn't the type of data to which this definition applies

you have to specify a set and a family of subsets

wklm

I want to keep it as general as possible

Thorgott

but you're not asking a question yet

wklm

If I made it a cartesian of two measurable spaces[1] would it make more sense?

[1] https://en.wikipedia.org/wiki/Measurable_space

Thorgott

what's the family of subsets on the product you are considering, be precise

wklm

22:31

$\langle X, \mathcal{F}\rangle$ and $\langle Y, \mathcal{G}\rangle$ to be $\langle X\times Y, \mathcal{H}\rangle$, where $\mathcal{H}$ is the intersection of all algebras on $X\times Y$ that contain the collection ${A\times B\subseteq X\times Y: A\in\mathcal{F}, B\in\mathcal{G}$

uh, sorry, I thought matjax would compile it

Thorgott

you forgot a closing bracket, I think

in any case, $\mathcal{H}$ very much is an algebra on $X\times Y$, because you defined it as an intersection of algebras and an intersection of algebras is an algebra

Ted Shifrin

22:56

@Semiclassical I thought I said two identical bars, one magnetized, one not.

Semiclassical

yeah, and my point is that that's not physically sensible. if they're the same composition, then they're both either hard or soft magnets

if they're both hard magnets, then exposing the one with magnetization to the other will just result in a second hard magnet with identical magnetization

if they're both soft magnets, well, that's not a permanent magnet

the problem is fine if one insists that one is a hard magnet and the other is soft, but identical in mass/volume/shape

copper.hat

sounds like a silly interview question to me.

why are manhole covers round...

Semiclassical

so you can roll them along the ground?

copper.hat

many are rectangular, but the fermi popular answer is that it cannot fall down into the hole.

but i doubt very much if that was an engineering decision.

i mean a reason for,

a question a colleague got when he defected to the quants was "estimate the number of gas stations in the usa". that is a much more reasonable question :-).

BigSocks

23:16

@copper.hat what was his answer and rationale if you remember

Thorgott

we just had that question earlier as well

why are interviewers so obsessed with gas stations smh

copper.hat

@BigSocks that was a long time ago, i've forgotten.

lack of imagiation

geocalc33

23:40

what does "behave well" mean

copper.hat

didn't your mother explain that to you?

geocalc33

I mean "behave well" when used in mathematics lol

copper.hat

I think you need to supply a little more context :-)

geocalc33

0

Q: Why is Euler product destroyed?

Consider the lower branch of the hyperbola $f(x)=\frac{1}{x}.$ We can add up the images on the curve corresponding to pre-images equal to $-{n}$ for $n=1,2,3,\cdot\cdot\cdot.$ Upon doing this we get, $h(x)=-\sum_{n\ge1} n^x.$ It can be shown with some algebra that $h(x)$ can be expressed as a pro...

number-theory prime-numbers analytic-number-theory euler-product

the commenter used "behave well"

I asked him about it but he/she hasn't responded yet

copper.hat

I have a hard time parsing your first sentence, so I will pass on this.

geocalc33

23:49

it's just "adding the y-values for natural number x-values" on the hyperbola

for ex. on the upper branch you could do 1+1/2+1/3+...

love_sodam

Is there some example that zariski topology is discrete?

Thorgott

zariski topology on k^0

love_sodam

what is k^0?

Thorgott

a zero dimensional vector space over k

aka a point

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