Reputation
31,725
315/100 score
 Jan 27 revised Isn't the structure of $\mathrm{Gal}(\mathbb{Q}(\sqrt2,\sqrt3)/\mathbb{Q}(\sqrt3))$ just the structure of $\mathrm{Gal}(\mathbb{Q}(\sqrt2))$? TeXified Jan 27 answered Find the limit $\lim _{n\to \infty }\left(\sum _{k=1}^n\left(\frac{k}{3^k}\right)\right)$ Jan 27 comment Find the limit $\lim _{n\to \infty }\left(\sum _{k=1}^n\left(\frac{k}{3^k}\right)\right)$ @Dr.SonnhardGraubner Because it's unmotivated; the goal in solving a problem like this generally isn't to come up with the answer, but to come up with a method. Jan 27 comment Embedding tuples of natural numbers into real numbers @Philippe This edit doesn't really change the applicability of my comment at all - just map the $(k-1)$-tuple of all but the first coordinate onto $[0,1)$ using your mapping of choice and then add your first coordinate. Jan 27 comment Embedding tuples of natural numbers into real numbers In fact you can embed the lexicographic order on all finite tuples of natural numbers (which answers every case of fixed arity at once) pretty straightforwardly; see math.stackexchange.com/questions/123969/… for more details. Jan 26 comment Determine the equation of a parabola with roots $2 + \sqrt {3}$ and $2 - \sqrt {3}$, and passing through the point $(2,5)$ Also, note that you can simplify $(x-(2+\sqrt{3}))(x-(2-\sqrt{3}))$ into a rational polynomial and it's probably worth expanding that out before you go further. Jan 26 comment Trigonometric Expression for $1 + \cos \alpha + \cos 2\alpha + \cdots + \cos n \alpha$ using complex numbers Have you tested it using a few values of $n$ and $\alpha$? For instance, what if you take $n=1$? Jan 25 comment Is there a rational surjection $\Bbb N\to\Bbb Q$? I think the gist of the idea here is okay, but the details have a lot of gaps - what's the 'highest-order term' of $1/(3x-2)$, for instance? You have to prove some results on boundedness of rational functions that are relatively straightforward but still need fleshing-out. Jan 24 comment Tips for Prime Factorization of a Given Large Interger Once you've found the first factor, it's actually easy to use hint 3 to factor the chunk $584533$ that you're left with; from $p\cdot q=584533$ and $p+q=2049-503=1546$ the quadratic formula will get you both of $p$ and $q$ quickly by just taking a square root (which can be done easily by hand). Jan 22 awarded Enlightened Jan 22 awarded Nice Answer Jan 21 answered Will sums of infinitely many primes ever fail to generate almost all natural numbers? Jan 21 comment A probability question that I failed to answer in a job interview @NateEldredge Oooh, that's a great example. Very good catch; thank you! I was missing the 'diagonal' terms in my mental products - turns out that, e.g., $p_1p_2+p_1p_3+p_2p_3$ isn't quite the same as $(p_1+p_2+p_3)^2$... :P Jan 21 revised A probability question that I failed to answer in a job interview Fixed a few logical errors Jan 21 comment A probability question that I failed to answer in a job interview @DavidKleiman Yep - I realized that after my initial post (which guessed that it was uniformly $\tilde{p}_k$) and made a quick edit. Jan 21 revised A probability question that I failed to answer in a job interview added 80 characters in body Jan 21 answered A probability question that I failed to answer in a job interview Jan 21 comment Do all rational numbers repeat in Fibonacci coding? A comment from the other question, repeated over here for folks who might look at one but not the other: representations are highly nonunique ($\frac12$ has the obvious terminating representation but also has another which arises from writing $\frac12=\frac13+\frac16$ and then applying a greedy algorithm to $\frac16$; etc.) I suspect numbers have continuum many representations but my topology isn't strong, enough to prove it, sadly. Jan 21 comment Does 1/4 eventually repeat in Fibonacci coding? @Joe Thinking about it more, representations are guaranteed to be highly nonunique, so you may want to talk about some form of canonical representation. (For instance, $\frac12$ is obviously $0.1$, but it also has a representation $0.0101001010000100000000001\ldots$, obtained by 'skipping' the obvious term and writing $\frac12=\frac13+\frac16$ and then expanding $\frac16$ greedily.) In fact, every number in $0\ldots 1$ should have infinitely many and possibly (my topology isn't great) continuum many expansions. Jan 21 comment Does 1/4 eventually repeat in Fibonacci coding? @ASKASK Since $\phi\lt 2$ it should be possible to prove that a greedy representation exists, but by the same standard it's highly unlikely that representations are unique.