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Software engineer and long-time dabbler in mathematics; 11th in the Putnams forever ago but I've long since atrophied.


21h
comment Can Number Theory be visualized?
I should probably chew on this excellent answer more to answer this question myself, but: is there a 'traditional' reflection group at work here? Is it a Coxeter group, and if so what can be said about it?
21h
comment Evaluate $\pi$ using $\arctan(\frac{\sqrt{3}}{3})$
This is a terrible approximation - better to say that $\frac1{2n+3}\lt 1$ so we just need $6(\frac{\sqrt{3}}{3})^{2n+3}\lt 10^{-10}$, which can be easily solved to give $n\approx 20$, give or take.
1d
comment $Θ(n) + O(n) = ?$ (recurrence equation)
FWIW, I'd encourage using the notation $f(n)+g(n)\in\Theta(n)$ as it tends to lead to less confusion; sometimes (e.g., in the phrase $T(n/2)+\Theta(n)$) there's not much to be done about it, but where the set notation can be used I've found it leads to far fewer misconceptions.
1d
comment Determine all units in $\mathbb{Z}[\omega] := \{a+b\omega\mid a,b\in\mathbf{Z}\}$ where $\omega = \frac{-1 + i \sqrt{3}}{2}$
FWIW, I think this answer could use a bit of cleanup, but it's a fair bit clearer than the other one, and closer to what OP is likely to want. (Algebraic extensions are almost certainly relevant to the question at hand, given that it's a question about an algebraic extension!)
1d
comment About the properties of $x * y = \frac {xy}{x + y + 1}.$
In fact, the operation (with 'sensible' treatment where the denominator is zero) does have an identity and an inverse over a suitably changed/expanded domain. One way of making the associativity a little easier (and a guidance towards inverse/identity) is to recognize that $\frac1{x\star y}=\frac1x+\frac1y+\frac1{xy}=(\frac1x+1)(\frac1y+1)$.
1d
comment Density of primes in a polynomial
The problem is that, frankly, no one knows. If $p(x)$ is linear, then the density of $A$ is zero (and in fact, one can give $O\left(\frac{n}{\log n}\right)$ estimates for the number of elements of $A$ less than $n$); this is a consequence of Dirichlet's theorem. But as of right now, nobody knows a single polynomial of degree $\gt 1$ that is known to have even infinitely many prime values, so all the heuristics about density are only that.
1d
comment Proof $(\log(n))^{\log(\log(n))} = O(n)$
@Anthony Possibly the most fundamental tool in the asymptotics toolbox is that $\log(n)$ is in $o(n^a)$ as $n\to\infty$ for any $a\gt 0$ (and so as a result, $\log(n)^a$ is in $o(n^b)$ as $n\to\infty$ for any $a,b\gt 0$.) Informally, any power of $\log n$ grows slower than any power of $n$.
1d
answered Fascinating induction problem with numerous interpretations
1d
comment Is this a valid way to show that the recursive sequence $x_n = x_{n-1} + \frac{1}{x_{n-1}^2}$ is unbounded?
@Michael That's understandable - but you should be able to put together everything in the last few comments here...
1d
comment Is this a valid way to show that the recursive sequence $x_n = x_{n-1} + \frac{1}{x_{n-1}^2}$ is unbounded?
@Michael I think you are. Suppose that $\lim_{n\to\infty}x_n=L$. Then what is $\lim_{n\to\infty}\frac1{x_n^2}$?
1d
comment Is this a valid way to show that the recursive sequence $x_n = x_{n-1} + \frac{1}{x_{n-1}^2}$ is unbounded?
A different way of looking at vadim's hint: if the sequence has a limit, then of necessity $\lim_{n\to\infty}(x_n-x_{n-1})=0$ (why?) -- can you see why this is causes a contradiction?
1d
comment Is this a valid way to show that the recursive sequence $x_n = x_{n-1} + \frac{1}{x_{n-1}^2}$ is unbounded?
One strong indication that you're barking up the wrong tree: none of your algebraic manipulations up to where you got have made any reference to the sequence definition you were given. Now, there are sequences that are increasing but bounded (for instance, $x_n=x_{n-1}+\frac1{n^2}$) so unless you use some specific property of this sequence, then you know your argument can't go anywhere.
2d
comment One integral, two solutions?
You're right that your two final values are inequal - but it is in fact the case that the two differ by a constant. You've got terms of the form $\ln y$ and $\ln(-y)$, and the latter is $\ln((-1)\cdot y)= \ln(-1)+\ln(y) = i\pi+\ln y$...
2d
comment Why is Klein bottle non-orientable?
For the second half of your question, just try parallel transporting a tangent frame along the loop $x=1/2$.
Jan
28
comment Two example statements meant to demonstrate the importance of quantifier order don't appear to do so
Given that it appears to me to already be a duplicate (see my note above), deleting it might be in order if you're amenable to doing such; it's certainly your choice.
Jan
28
comment Two example statements meant to demonstrate the importance of quantifier order don't appear to do so
You can say that there's some $d$ such that $y=x-d$ - but you're 'assigning' $d$ (that is, effectively assigning a value to $y$) before assigning $x$. In fact, for the truth of the second statement you (of necessity) have a different $y$ (and a different $d$) for each $x$.
Jan
27
comment $S = \{n: n \text{ is an integer and } n=n^n\}$
It's definitely not necessarily that convoluted. If $n\lt-1$ then $n^n$ isn't an integer; if $n\geq 2$ then $n^n\geq n^2=n\times n\geq 2\times n\gt n$.
Jan
27
comment $S = \{n: n \text{ is an integer and } n=n^n\}$
Your argument that $n$ is an invertible element of $\mathbb{Z}$ only works if $n^{n-2}$ is in $\mathbb{Z}$ as well, and there's no a priori reason to believe that's so (it's certainly not for infinitely many $n\in\mathbb{Z}$!)
Jan
27
comment An easy example of a non-constructive proof without an obvious “fix”?
@MarcvanLeeuwen I suppose it depends on what one thinks of as 'Cantor's argument'. The traditional presentation certainly uses decimals, but that presentation is already awkward for any number of reasons, and the fact that decimals are such a miserable representation for real numbers (particularly for constructivist purposes) just makes it that much worse. But I don't think of Cantor's argument as being the decimals; I think of it as being the diagonalization itself, and that can be done on e.g. continued fraction expansions or the like.
Jan
27
comment What is the connection between slant/oblique asymptote to the polynomial part of the function and polynomial division?
@kuhaku Yes - the solution $y$ of the equation $x=\tan y$ with $-\frac\pi2\lt y\lt\frac\pi2$.