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


Jul
20
revised Fibonacci number that ends with 2014 zeros?
Changed parenthesis to angle brackets for ordered pairs
Jul
18
comment Numbers whose digits sum to 7
@J.J. Ahhh, I see, you're just 'stars-and-bars'ing it. I need to mentally double-check, but that makes a lot of sense.
Jul
18
comment Numbers whose digits sum to 7
It's not quite as standard as you think - it's an ordered permutation, which is a fairly unusual way of counting. (1+1+2+3 is different from 1+2+3+1, but not from 1b+1a+2+3) They're easy enough to enumerate by hand for a small case like this, but AFAIK they're not particularly well-studied.
Jul
18
comment Number of ways in a Nim game such that First Player always wins
@rayu after any move $c_i\mapsto c_i'$, the value of the new position is $p\oplus c_i\oplus c_i'$ (this is because we 'remove' $c_i$ from the position and replace it with $c_i'$). This is zero (i.e., the initial move was a winning one) iff $p\oplus c_i\oplus c_i'=0$ iff $c_i' = p\oplus c_i$ (again, using that $x\oplus x=0$ for all $x$). That $p\oplus c_i\lt c_i$ is then just the condition for the move to be legal - that is, to leave fewer coins in that pile than it started with.
Jul
18
answered Number of ways in a Nim game such that First Player always wins
Jul
18
comment Distinguishing properties of $\mathbb{Q}$ and $\mathbb{R} \setminus \mathbb{Q}$ that lead to differing cardinalities?
@blue That's an information-theoretic distinction (or, arguably, a set-theoretic one); I think the germ of the OP's question here is looking for a topological distinction between $\mathbb{Q}$ and $\mathbb{R}\setminus\mathbb{Q}$, and the 'infinite data' explanations of the irrationals don't really provide that.
Jul
18
comment Distinguishing properties of $\mathbb{Q}$ and $\mathbb{R} \setminus \mathbb{Q}$ that lead to differing cardinalities?
You can talk about arbitrarily small lengths, but you can't talk about infinitesimally small lengths - there are no such things as infinitesimals in $\mathbb{R}$ (or $\mathbb{Q}$, for that matter). There are nonstandard theories that do have such things as infinitesimal numbers, but they're not generally regarded as 'the' theory of the reals, for instance.
Jul
18
comment Distinguishing properties of $\mathbb{Q}$ and $\mathbb{R} \setminus \mathbb{Q}$ that lead to differing cardinalities?
There's no such thing as 'infinitesimally small length' (in $\mathbb{R}$). You hit the nose on the head; every rational is in fact contained in uncountably many of these intervals (as might be obvious since for instance, you can just take the intervals $(q-r, q+r): r\in\mathbb{R}^+$ ). In fact, for any $\epsilon\gt0$, every rational is contained in uncountably many intervals of length $\epsilon$.
Jul
18
answered Dice Rolling 4d10 with a twist
Jul
17
comment Prove that the cyclic group of order $3$ is isomorphic to $\Bbb Z_3$ under addition. Show answer using a proof and justification for each step.
You can't really say 'the' isomorphism, of course, because $x\mapsto[2]\bmod 3$ is an equally good isomorphism (More abstractly, you can always compose with the nontrivial automorphism of $\mathbb{Z}/3\mathbb{Z}$).
Jul
16
comment Is there a nice/clever way to visualize $\mathcal{S}\times \mathbb{R}^2$?
Given that you're talking about a 4-dimensional entity, you shouldn't expect an embedding into anything less than five-dimensional space (and to some extent, it's 'lucky' that you even get that), so unless your visualization skills are exceptionally good this is going to be a tall order.
Jul
16
comment Examples of sub-exponential functions that aren't exponential functions when chained by a polynomial
I didn't downvote, but your comment about 'still a function in $O(2^n)$ probably isn't what you mean to say - I think you mean that 'there is some polynomial such that $f(p(n))\in\Omega(2^n)$. ($O(2^n)$ bounds above; $2^{n^{1/3}}$ is still in $O(2^n)$.)
Jul
15
comment $\aleph_\omega$ and large ordinals?
Asaf and @goblin: I don't mean to suggest that fixed point -> regular, but the opposite - i.e., 'not fixed point' -> singular, since it gives us a concrete sequence witnessing that cof($\kappa$) != $\kappa$.
Jul
15
comment $\aleph_\omega$ and large ordinals?
(And another tiny comment: the reason why they're fixed points is because they're regular cardinals - that is, they're not the limit of a sequence $\{\kappa_\alpha\} : \alpha\lt\beta$ with 'size' ($\beta$)less than the cardinal itself; the technical name for this size, if you want to learn more about it, is the cofinality of the cardinal.)
Jul
15
comment $\aleph_\omega$ and large ordinals?
The comment-sized answer: virtually all large cardinals are fixed points of the aleph function; that is, they're 'numbers' $\kappa$ for which $\kappa = \aleph_\kappa$.
Jul
15
comment Prove $\int_0^{\infty} \left(\sqrt{1+x^{4}}-x^{2}\right)\ dx=\frac{\Gamma^{2}\left(\frac{1}{4}\right)}{6\sqrt{\pi}}$
Have you looked at the substitution $u\mapsto x^2$? That seems like it would transform this into a relatively classical and manageable elliptic integral...
Jul
15
comment Closure of $\mathbf{\Sigma_{n}^{0}} $ under finite cartesian products
I wouldn't call these 'cross' products - that has connotations, especially with respect to points.
Jul
14
comment $\lim_{n \rightarrow \infty} n ((n^5 +5n^4)^{1/5} - (n^2 +2n)^{1/2})$
And yes, the need to use more than one term is because you have effectively a factor of $n^2$ in front, so you need to expand to at least $o(n^{-2})$ to be sure of the limit.
Jul
14
comment $\lim_{n \rightarrow \infty} n ((n^5 +5n^4)^{1/5} - (n^2 +2n)^{1/2})$
@Timof Not quite - to find the second term in $(1+x)^{(1/k)}$ you need $\frac12\frac1k\cdot\left(\frac1k-1\right)$, not $\frac12\frac1k\cdot\left(\frac1{k-1}\right)$. In particular, you should find that the second term is negative, not positive; $ 1+\frac xk$ OVERestimates $(1+x)^{(1/k)}$.
Jul
14
answered $\lim_{n \rightarrow \infty} n ((n^5 +5n^4)^{1/5} - (n^2 +2n)^{1/2})$