Geniuses 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 !
Feynman on Fermat's Last Theorem (2016)
21–30 of 32 posts
Re: Feynman on Fermat's Last Theorem (2016)
#22>Richard Feynman was probably one of the most talented physicists of the 20th century. Starts off bold.
Do you disagree? Would it help for me to show you the probability is strictly non-zero?
Re: Feynman on Fermat's Last Theorem (2016)
#23>Richard Feynman was probably one of the most talented physicists of the 20th century. Starts off bold.
One 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?)
Re: Feynman on Fermat's Last Theorem (2016)
#24Sigh, 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. Th…
> 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.
You're mixing up "N is an nth power" with "there exists an N that is an nth power". The purpose over summing over all the integers is to obtain the latter from the former, and it's not necessarily very small even if the former is small.
Re: Feynman on Fermat's Last Theorem (2016)
#25Sigh, 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. Th…
I have to agree that this argument is utterly unconvincing, even as a reason to believe FLT may be true intuitively. It simply dismisses with a wink and a nod the idea that there might be one single counterexample out there for one single exponent (or, a finite number of counterexamples for a finite number of exponents). In other words, it does nothing to argue against the set of counterexamples being of measure 0. I…
> It simply dismisses with a wink and a nod the idea that there might be one single counterexample out there for one single exponent (or, a finite number of counterexamples for a finite number of exponents). In other words, it does nothing to argue against the set of counterexamples being of measure 0.
Those "other words" are substantially weaker than the statement before them. If x^n + y^n equaled z^n for every integer x, y, z, and n, the set of counterexamples to Fermat's last theorem would still be of measure 0. Nothing can argue against the set of counterexamples being of measure 0, because the entire problem space itself has measure 0, and the set of counterexamples is necessarily a subset of the problem space.
Re: Feynman on Fermat's Last Theorem (2016)
#26Earlier quoted context omitted.
I have to agree that this argument is utterly unconvincing, even as a reason to believe FLT may be true intuitively. It simply dismisses with a wink and a nod the idea that there might be one single counterexample out there for one single exponent (or, a finite number of counterexamples for a finite number of exponents). In other words, it does nothing to argue against the set of counterexamples being of measure 0. I…
I haven't read it closely, but it looks to me as if the calculation estimates the expected number of counterexamples rather than the "measure" of them (however you've chosen to define that).
The set of possible counterexamples has Lebesgue measure zero.
Re: Feynman on Fermat's Last Theorem (2016)
#27Here'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
Re: Feynman on Fermat's Last Theorem (2016)
#28Previous HN discussion around the time of publication just over a year ago: https://news.ycombinator.com/item?id=12018221
Re: Feynman on Fermat's Last Theorem (2016)
#29Earlier quoted context omitted.
Yeah, 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.
That'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 probabilit…
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Re: Feynman on Fermat's Last Theorem (2016)
#30[deleted]