I've only ever been downtown + a few neighborhoods, like the number of times I left Hyde Park is strictly less than the number of nights I spent in the barn this year
Oh, earlier, when I was dicking around with Dirichlet-theorem-y stuff,
I just realized I didn't need it.
To show that $1+x^n$ is a factor of some polynomial with prime exponents, simply find two primes $p_1$ and $p_2$ with the same residue mod $n$ (using pigeonhole; there are only $n$ residues)
So the first time I tried that problem, I was working with a friend and saying "Yo, so I'm pretty sure that given 2 numbers, I should be able to add something to each so that they both become prime, that's probably how this problem goes"
@Zee normally I'd battle it out with you because my secondary reason for existing is to debate for the sake of it, since that's just not true. But, I think we know where that's gonna go, and I think it'll be bad for the chat if we do so, so let's agree to just not
How we'd write Maxwell's equations today isn't how Maxwell wouldn't have written them, for the simple reason that he was working before vectors were a thing.
But I mean I push back against that since Kant is, but I'll say let's steer away from that direction. We already know that you have a more society-based view of the value of a subject, and while that can lead to a debate on whether this is true, I don't really see the need for something to be beneficial to society for it to be worthwhile. This difference in values almost surely won't be reconciled.
@Daminark My main thing from reading X person on Kant is that "Humanity as an End in Itself" is by far a better representation of his ethics than the mere form of the categorical imperative.
I'd agree with that, my issues with the categorical imperative in its canonical formulation mostly revolve around the fact that I think circumstance may prevent its proper application, and that morality ought be able to give us an answer to a situation that applies to me now, even if it can't apply to all rational beings as such
@Balarka Well, presumably, if the categorical imperative is true, it would suggest that civilizational variances of morality is a result of people getting it wrong. Or am I misreading your claim?
@EricSilva then what does it mean to believe in the dialectic?
Maybe that logically reasoned arguments are at least capable of ruling out what isn't true. Much weaker than what you proposed, still not self-evident to me. @Daminark
I think the Marxian interpretation of dialectics is more like that in the process of uncovering the truth, the opposing opinions, "thesis" and "antithesis" coexist, than saying "that's how finding truth works"?
I don't know if I believe that. I'm still grappling with how much I buy into Kant but I believe the statement "X maxim is immoral" is one that can be assigned a definite truth value
@EricSilva Yeah but he talks about various things. I just think that's what his interpretations are on dialectics, whereas you are speaking of his theory of history guided by "material dialectic", which is a different beast I think
@Balarka note that this doesn't suggest that every choice is morally determined, there are some situations in which available actions are morally neutral
@Daminark yeah that's fine to me. but that there's a universal moral code, given my notions of "universal", "moral" and "code" agree with Kant's, is something I do not believe in
@Daminark I always found it really bad that there are definitely situations you can cook up (situations that aren't even far-fetched) where you can't determine what's right if you believe in the categorical imperative.
@EricSilva by "you can't determine", do you mean like, a typical person wouldn't likely be able to figure out how to operate by the CI, or that the CI does not provide an answer?
@EricSilva In a given circumstance, yes, but if CI were proven true, then even if one is unable to always figure out what the correct answer is, one would presumably aim to do so whenever possible
and I have a question that I solved by exhaustion, but I would like to have a better proof: "every commutative binary operation on a set of two elements must be associative"
I mean, one acts as close to morally as one possibly can. I do not believe that any correct moral code will have an answer which is accessible to everyone, or even anyone. Doesn't negate the importance of trying to find out whether a given one is correct, since it allows one to at least attempt to do so. The killers of Batman's parents did what was ultimately the correct utilitarian action
@Daminark I don't know much computer programing though in my first year of undergraduate school, we had the compulsory computing course which taught fortran.
but I have never used it in my physics course or research.
[Introduction] It is a well known result that the harmonic series $\sum_{n=1}^{\infty}\frac{1}{n}$ is divergent. However, it is not commonly discussed whether the Dirichlet series with powers -1. i.e. $\sum_{n=1}^{\infty}\frac{a_n}{n}$ converges for a sequence $(a_n)$ that is zero for countably many not necessary consecutive terms (i.e. the number of consecutive terms that are nonzero in any subsequence is finite.)
Only if $\sum_{n=1}^{\infty}\frac{a_n}{n}$ converges for all possible sequences $(a_n) \neq (1)$ will the notion of Base-$\omega$ make sense
Recall that base-n for $n \in \Bbb{N}, x\in \Bbb{R}$ is defined as follows:
$x=\sum_{-\infty}^{\infty}\frac{a_m}{n^m}$
Therefore, base-$\omega$, where every rational of the form $\frac{1}{n}$ are the positions in this system is given by
Based on the fact that rationals are dense, a sequence of rationals can be made to converge to any real number. Therefore, we suspect similar things can happen for a Dirichlet series of power -1
Suppose you take the set $X=\{\sum_{k \in A} \frac{1}{k}: A \in \mathcal{P}(\mathbb{N} \setminus \{1\})\}$. Suppose that we agree to introduce the symbol $\infty$ to encompass the cases where the series $\sum_{k \in A} \frac{1}{k}$ diverges (so $\infty \in X$). My question is if any irrational nu...
I am trying to see whether I can express any real number as a sum of all reciprocal positive intergers, which is basically expressing a number in a base with infinite digits
@Secret Take $S = \sum_{i=1}^{n} 1/n$ until adding $S + \frac{1}{n+1} > \pi$ and $S < \pi$. Then ommit the numbers until you get to some $a = \frac{1}{k}$ that $S + a < \pi$ keep repeating this process and it should ultimately converge to $\pi$ I think.
@Zee I played with the cantor function C that you refer me to yesterday, I noticed that if I pick any point in the cantor set e.g. x=0.2002020202020020202000020200202020 ..., then basically the value C(x) is entirely controlled by some kind of generalised dynadic rational series, and that is where the increase occurs (the increase has Lesbegue measure zero because naively speaking, there is no end to infinite digits, and the increase from one digit to another basically took place at $\lim_{N_x\to \infty}\frac{1}{2^{N_x}}=0$, so roughly speaking, uncountably many zeros piling up at the "smal…
@Dair I think I can also do the same thing by starting with the decimal representation of $\pi$, subtract from that a rational that has the largest decimal representation < $\pi$. Keep repeating that for the remaining and I think I should get the series
@Secret: Also, the question doesn't refer to every set of positive integers. It refers to any subset of integers... I'm pretty sure regardless of how you sum all reciprocals it will still diverge.
I know that the harmonic series $1 + \frac12 + \frac13 + \frac14 + \cdots$ diverges. I also know that the sum of the inverse of prime numbers $\frac12 + \frac13 + \frac15 + \frac17 + \frac1{11} + \cdots$ diverges too, even if really slowly since it's $O(\log \log n)$.
But I think I read that if ...
