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 1d comment Can we say that $\sqrt{2}=2/(2/(2/(2/\ldots)))$? One of the points behind the continued fraction representation is that it converges no matter where you 'start from'. As long as $a_0\gt 0$, if you take $a_{n+1}=1+\frac1{a_n}$ then $\lim_{n\to\infty}a_n=\phi$. Your formula only works if you start with $a_1=\sqrt{2}$; it doesn't converge otherwise. 2d comment How to integrate $\frac{1}{2e^x-3e^{-x}}$? Consider a substitution $u=e^x$; you should be able to get the integral into easily-managed form. 2d revised Do you need true randomness to beat the two-envelope game? Added a clarifying note 2d comment Do you need true randomness to beat the two-envelope game? I think you have the roles of $A$ and $B$ reversed. :-) That said, there are a couple of issues with this; most specifically, that the strategy for $A$ (the number-giver) doesn't (unlike $B$'s) have to be computable; we can assume that $A$ has an oracle that enumerates the total recursive functions. (I'll make this clearer in the problem statement itself.) In particular, this means that $A$ can assure he always wins at least once. 2d revised Do you need true randomness to beat the two-envelope game? tweaked language a bit 2d awarded Good Answer Apr22 comment Simplify $\frac{(\cos \frac{π}{7}-i\sin\frac{π}{7})^3}{(\cos\frac{π}{7}+i\sin\frac{π}{7})^4}$ What is $\theta$ in your answer? Apr22 revised Is $e^{e^9}$ an integer? Fixed a matho; added 25 characters in body Apr22 answered Is $e^{e^9}$ an integer? Apr22 comment Is $e^{e^9}$ an integer? To be more specific, since $e^9\approx 8100$, the integer part of $e^{e^9}$ is only about 10K bits (plus or minus) so it wouldn't be hard to compute first $e^9$ and then $e^{e^9}$ to sufficient accuracy to prove it's not an integer. (Note that we can compute $e^x$ to $n$ bits of accuracy in time approximately $M(n)$, where $M(n)$ is the time to multiply two $n$-bit numbers). Apr21 comment Is $A_4$ isomorphic to $D_3\times \mathbb Z_2$? Nick: it's not just that $H$ has an element of order 2; rather, all elements of $H$ are of order 2. (Why?) Apr21 comment Fermat's last theorem and $\mathbb{Z}[\xi]$ @ulli It's more or less given in the first link in my comment: we have factorizations in $\mathbb{Z}[\zeta_n]$ of $x^n+y^n$ and of $z^n$, and can show that they aren't related by multiplying by units (as they would need to be if unique factorization holds). Apr21 asked Do you need true randomness to beat the two-envelope game? Apr21 comment Fermat's last theorem and $\mathbb{Z}[\xi]$ See en.wikipedia.org/wiki/… and en.wikipedia.org/wiki/… - in particular, there are only finitely many $n$ for which the cyclotomic ring is a UFD. Apr21 comment The even-numbered coefficients of the Maclaurin series of $\frac{1}{\cos(x)}$ are odd integers. @robjohn Sorry for being so late in getting back to you - your answer stands just fine, I think. I was considering adding some of the detail to mine, particularly expanding out the recurrence relation and doing the quick odd/even case analysis; it's been long enough at this point that I think the answer will be all right as it is, though, particularly with yours for deeper details. Apr21 comment What is the limit $\lim_{n\to\infty}\frac{1^n+2^n+3^n+\dots+n^n}{n^{n+1}}$? Jack: does 'from which it follows that your limit is zero' actually follow? After all, $\lim_{n\to\infty}(1+\frac1n)^m=1$ for all $m$, but that doesn't imply the same result when $m$ is replaced by the limiting variable. Isn't this the exact same situation? Apr21 comment Sequence with Prime Numbers John: global behavior implies asymptotic behavior - if you know that $a_i\lt K$ for all $i$ then you know that $\sum_{i=1}^na_i\lt nK$ for all $n$. But the converse doesn't hold: it's not necessarily the case that $\sum_{i=1}^na_i\lt nK$ for all $n$ implies $a_i\lt K$ for all $i$; it doesn't even imply that $a_i\in O(1)$ (as a function of $i$). You need a matching (global) lower bound to even possibly go from asymptotic to global - and in the case at hand, the lower bounds are so small that you can't get any useful information from them. Apr21 comment Odd perfect squares whose decimal representation consist only of 1's and o's The problem is that just because $10^k+1$ isn't the square of an integer doesn't mean that it's not a square $\mod 10^{k+1}$. Indeed, OP's original post shows that $1001$, while not a square, is a square $\mod 10^4$ (it's the square of $4251$). Apr21 comment $A^{-1}$ has integer entries if and only if the ${\rm det}\ (A) =\pm 1$ @Theresa You know that $A$ and $A^{-1}$ are matrices with integer entries; what does that say about their determinants? Apr20 comment Bonus integration problem we got at class: Integrate $\frac {x \sin x}{1+\cos^2x}$ between $0$ and $\pi$ What were you trying? There seems to be a natural 'first' step, though it gets messy quickly...