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.
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
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
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.
@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
@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.
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.
@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. :)
@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.
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.
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.'
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.
@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
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"
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.
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
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...
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)
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.
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
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?
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?
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.
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.
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....
@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...
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.
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 $nv$.
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.
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.
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.
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.
@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.
(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.)
@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.
@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?
@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.
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
@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.
@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 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. :)
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
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.
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?
@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.
$\langle X, \mathcal{F}\rangle$ 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 ${A\times B\subseteq X\times Y: A\in\mathcal{F}, B\in\mathcal{G}$
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
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
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 :-).
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...
