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1d
answered Proving that an infinite dimensional space is closed.
1d
answered Proving two domains are not conformally equivalent
1d
answered Class of subsets which is not a $\sigma$-ring
1d
comment Each number in $S\subseteq \{1,\ldots,2n\}$ does not divide another one, with $|S|= n$. In how many ways?
Needed $3 (j/2)$ not to be in $S$.
1d
answered Sum of three cross products is zero.
1d
revised Each number in $S\subseteq \{1,\ldots,2n\}$ does not divide another one, with $|S|= n$. In how many ways?
added 131 characters in body
1d
comment Evaluating definite integral of $e^{i t^2}$
The integral converges as an improper Riemann integral, i.e. $\lim_{M \to -\infty, N \to +\infty} \int_{M}^N e^{it^2}\; dt$.
1d
comment Why should quaternions exist?
The existence of a mathematical object (the quaternions) is not contingent on any aspect of physical reality. Rather, the algebra generated (over the reals) by the $2 \times 2$ matrices $i\sigma_x$, $i \sigma_y$, $i \sigma_z$ is a representation of the Quaternions. The fact that these Pauli matrices can be used in describing certain physical particles is interesting, but not relevant.
1d
comment Cramer, $P(S_n\geqslant na)\sim e^{-n I(a)}$
I don't know if it's possible in general to say in which cases the $o(n)$ is $O(1)$, but the fact that it doesn't happen for the normal distribution makes me guess that this will be extremely rare, if indeed it ever occurs.
1d
comment Cramer, $P(S_n\geqslant na)\sim e^{-n I(a)}$
$|b_n| < n \epsilon$ when $\left| \dfrac{\ln(a_n)}{n} - c \right| < \epsilon$
2d
comment Can this congruence be simplified?
... and those seem to be all the solutions; at least, there are no others with $p+q \le 10^5$
2d
comment Can this congruence be simplified?
Also for consecutive terms in A032908, with $5$ instead of $6$.
2d
comment Can this congruence be simplified?
Here's something interesting: it looks like each pair $(a(n),a(n+1))$ of consecutive terms in OEIS sequence A101265 (oeis.org/A101265) satisfy the condition, with $(p+q)(p+q+1)/(pq) = 6$
2d
comment Can this congruence be simplified?
I suspect there are only finitely many solutions for any fixed $p$, and for most $p$ there will be none. The only solutions for $1 \le p \le q \le 1000$ are $$[p,q] = [1, 1], [1, 2], [2, 2], [2, 3], [2, 6], [3, 6], [6, 14], [6, 21], [14, 35], [21, 77], [35, 90], [90, 234], [77, 286], [234, 611]$$
2d
answered What is the distribution of a binomial variable where the number of trials is itself random?
2d
answered Can the product of an $4\times 3$ matrix and a $3\times 4$ matrix be invertible?
2d
comment Can this congruence be simplified?
For example, try $p = 6$, $q = 3$, $pq = 18$: $(p+q)(p+q+1) = 90 \equiv 0 \mod 18$, but $18$ doesn't divide either $p+q = 9$ or $p+q+1 = 10$.
2d
comment Can this congruence be simplified?
If you factor $pq = P_1^{d_1} \ldots P_n^{d_n}$ where $P_1, \ldots, P_n$ are primes, then each $P_j^{d_j}$ must divide exactly one of $p+q$ and $p+q+1$. In general some will divide $p+q$ and others will divide $p+q+1$, and $pq$ itself won't divide either.
2d
comment Can this congruence be simplified?
Not if $\gcd(p+1, pq) > 1$.
2d
comment Can this congruence be simplified?
I'm not sure what you mean by that.