Reputation
4,429
Top tag
Next privilege 5,000 Rep.
Approve tag wiki edits
Badges
8 24
Impact
~31k people reached

2d
comment Bounded pythagorean triples
I can't say I'm absolutely sure, but my gut feeling is that the integral is sometimes less, and sometimes more, depending on the $m, n$ chosen. I would be more inclined to view this asymptotically where, for given $\epsilon, \epsilon' \gt 0$, there exists $M$ so large that if $m \geq M$ and $n$ is the floor of $(1 + \epsilon)m$, then the quotient obtained by dividing the integral by the number of integer points in the region comes within distance $\epsilon'$ of $1$. I'm not sure I have the time to delve into this more deeply however (it's not really my area to begin with).
Aug
24
awarded  Self-Learner
Aug
24
awarded  Nice Question
Aug
22
answered group of units in a topological ring
Aug
20
comment Do the usual examples of “dimensive” Lawvere theories form an exhaustive list of such theories?
@ZhenLin Don't free Boolean algebras on $n$ elements have cardinality $2^{2^n}$, whereas finitely generated Booleans can have more general cardinalities $2^k$?
Aug
19
comment group of units in a topological ring
Ah well, I found a very simple solution. But there's still some time left to claim your prize, before I go ahead and post my solution.
Aug
14
awarded  Promoter
Aug
14
comment Bounded pythagorean triples
No, I'll leave it to you to compute the integral, and see whether it gives a result useful to you.
Aug
14
comment Bounded pythagorean triples
The area of $R$ can be computed using multivariable calculus. $R$ is the inverse image of the square $[m, n]^2$ under $z \mapsto z^2$, or the direct image of $[m, n]^2$ under (a branch of) the inverse $f: w \mapsto w^{1/2}$. The Jacobian $Jac(f)$ is given by the derivative $f'(w) = (1/2)w^{-1/2}$ and its determinant is $|f'(w)|^2 = (1/4)|w^{-1}|$, which for $w = u + i v$ is $\frac1{4\sqrt{u^2 + v^2}}$. Thus the area of $R$ is $$\int_m^n \int_m^n \frac{d u\; d v}{4\sqrt{u^2 + v^2}}$$
Aug
14
comment Bounded pythagorean triples
Getting an exact count is probably hopeless, but it might help to observe that if we put $z = x + i y$, then $z^2 = x^2 - y^2 + i(2x y)$, and so we are trying to estimate the number of integer points $(x, y)$ in the planar region $R = \{z: z^2 \in [m, n] \times [m, n]\}$. This number is roughly the area of the region $R$, a squarish-looking region bounded by four hyperbolas (here confining attention to positive $x, y$).
Aug
13
comment Bounded pythagorean triples
What are you hoping for here? Exact count? Asymptotics? Have you put any work into this yourself?
Aug
13
comment Is $\widehat{\mathbb{Z}} \otimes_{\mathbb{Z}} \mathbb{Q} \cong \mathbb{A}_{\mathbb{Q}}^f$ as topological rings?
Just a small comment that the adeles are complete: any locally compact Hausdorff abelian group is complete with respect to its canonical uniformity. You can find a proof outlined here: math.stackexchange.com/a/493995/43208
Aug
12
revised group of units in a topological ring
added further evidence that the exercise in question is mistaken
Aug
11
asked group of units in a topological ring
Aug
9
comment Is it true that if two open sets in a topological space intersect each other and share the same boundary, they are the same?
Another way of saying this appeals to the language of frames, which are complete Heyting algebras. Taking set-theoretic complements on the equality $\partial A = \partial B$ yields $A \cup (A \Rightarrow 0) = B \cup (B \Rightarrow 0)$ in the Heyting algebra of open sets, so $B$ is contained in a union of disjoint open sets $A$, $A \Rightarrow 0$, and therefore contained in $A$ by connectedness of $B$ and $A \cap B \neq \emptyset$. Similarly, $A$ is contained in $B$.
Jul
5
comment To fast invert a real symmetric positive definite matrix that is almost similar to Toeplitz
Cross-posting (asking the same question on multiple sites) is frowned upon, because it can lead to duplication of effort and hence a waste of people's time. You should wait for an answer at M.SE before giving up and trying here.
Jun
27
revised Can you equip every vector space with a Hilbert space structure?
added more material in response to a comment
Jun
27
comment Can you equip every vector space with a Hilbert space structure?
@Sushil I'll consider the converse a little later, when I have more time and leisure. For the other question, if $\binom{B}{\alpha}$ denotes the collection of subsets of $B$ of size $\alpha$, then by taking graphs of functions we have an injection $A^\alpha \to \binom{\alpha \times A}{\alpha}$, and since $|\alpha \times A| = |A|$ when $\alpha \leq |A|$, there is a bijection $\alpha \times A \to A$, and by composition along this bijection we get an injection $A^\alpha \to \binom{A}{\alpha}$.
Jun
27
awarded  Enlightened
Jun
27
awarded  Nice Answer