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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.