12 ms·
Feynman on Fermat's Last Theorem (2016)
- chmaynard 9y agoApparently the author also hosts this interesting website: http://www.fermatslibrary.com http://www.fermatslibrary.com
- throwanem 9y ago> the main job of theoretical physics is to prove yourself wrong as soon as possible. The main job of practical engineering, too.
- auggierose 9y agoNah, I'd say the job of practical engineering is to provide reliable, correct solutions in a timely, in-budget manner. If you are talking about theoretical engineering, I agree :D
- throwanem 9y ago> the job of practical engineering is to provide reliable, correct solutions in a timely, in-budget manner Sure. And I'm sure you'll agree this job is easier to do in direct proportion to the celerity with which we eliminate impractical approaches.
- auggierose 9y agoYou can eliminate impractical approaches all day long. At some point you have to start with the practical approach that you know will work a-priori.
- pg_is_a_butt 9y agoNERRRRRRRRRRRRRRRDDDDDDS&!^#%^&(%!#^&(%!#(^&!#%
- throwanem 9y agoWhen you have the option, sure! Not all problems are so tractable.
- auggierose 9y agoGranted. Still, I think there is a fundamental difference: In theoretical physics, a good theory of how things work (or how they definitely don't work) is enough as the result of your work, you don't need a working product in the end. In practical engineering, you do! Let me also add that I am serious about theoretical engineering, actually that's one of my favourite occupations, together with practical engineering.
- faitswulff 9y agoFeynman's approach seems very reminiscent of Jake Vanderplas's Statistics for Hackers talk[0] as opposed to the purely theoretical physicist approach that the author notes at the end. [0]: https://www.youtube.com/watch?v=Iq9DzN6mvYA https://www.youtube.com/watch?v=Iq9DzN6mvYA
- merricksb 9y agoPrevious HN discussion around the time of publication just over a year ago: https://news.ycombinator.com/item?id=12018221 https://news.ycombinator.com/item?id=12018221
- dang 9y agoAlso https://news.ycombinator.com/item?id=14355834 https://news.ycombinator.com/item?id=14355834.
- Gregaros 9y ago>Richard Feynman was probably one of the most talented physicists of the 20th century. Starts off bold.
- joombaga 9y agoDo you disagree? Would it help for me to show you the probability is strictly non-zero?
- Gregaros 9y agoSarcasm. "Probably one of the most" is quite a few weasel words for somebody universally included in any list of significant physicists.
- kgwgk 9y agoOne of how many? "One of the three most talented" would be a bold statement, "one of the 300 most talented" would be an understatement. (Edit: now it occurs to me that you might find that opening sentence an understatement?)
- Gregaros 9y agoYou missed my sarcasm. I did indeed think it was quite an understatement / poor writing to be so afraid to say "one of the best physicists of the 20th century".
- jordigh 9y agoSigh, this proof bugs me so much. Compute an approximate distance between nth powers, interpret this as the probability of an integer being an nth power, integrate this probability over the sum x^n + y^n, see that the probability of this being an nth power is also very low. I guess this is close enough for government work, but it's so utterly fallacious. For example, the distance between n^2 and (n-1)^2 is 2n - 1. That "means" that the "probability" of N being a perfect square is about 1/(sqrt(2N - 1)). This probability also goes to zero in the limit as N goes to infinity. Not very quickly, but it does. Does that mean that square numbers don't exist? We have many examples of conjectures being disproved by very large counterexamples: https://www.quora.com/What-is-an-example-of-a-conjecture-that-was-proven-wrong-for-very-large-numbers https://www.quora.com/What-is-an-example-of-a-conjecture-tha...
- ykler 9y agoYou forgot to sum over N
- jordigh 9y agoYeah, okay, I messed up. Let's argue in a different way: there are approximately sqrt(N) squares between 1 and N, so let's call sqrt(N)/N = 1/sqrt(N) the "probability" that N itself is square. That also goes to zero as N goes to infinity. If we pick higher powers of N, it goes to zero much quicker.
- nshepperd 9y agoThat's not what ykler meant. 1/sqrt(N) is the "probability" that N itself is square. We don't care about that, we want the probability that there is any solution to x^2 = N, which is sum{N=0..∞} (1/sqrt(N)). That does not go to zero. ETA: Well, actually the above is the expected number of solutions, so naturally it diverges because there are (infinite) solutions. A more proper way would be to calculate the probability that there are no solutions, which indeed goes to zero. But the probability of there being a solution, and the expected number of solutions, are the same if it is << 1.
- kumarvvr 9y agoGeniuses stand on the shoulders of giants and look further ahead, while I look at giants and assume a fetal position and cower in fear. Sigh !
- jwilk 9y agoHere's a more straight-forward (but equally hand-wavy) way to calculate the probability that N is a perfect power: P(N) ≈ (number of perfect powers near N) / (size of the neighborhood) ≈ (ⁿ√(N + r) − ⁿ√N) / r, for some smallish r ≈ d/dN (ⁿ√N) = ⁿ√N / nN
- deleted 9y ago[deleted]