1,012 reputation
523
bio website about.me/LucasCAPS
location Rio de Janeiro, Brazil
age 26
visits member for 3 years, 7 months
seen Oct 17 at 18:09

Graduated in Applied Mathematics and taking Master's in Statistics at UFRJ.


Mar
29
comment Proving $\big(n!^{\frac1n}\big)_{n\in\mathbb N^*} \to \infty$
Makes sense, but that $a^{n−a}$ shouldn't be there, should it? I edited it out, feel free to rollback if needed. Also, you assume $a\in\mathbb N$, but that's without loss too; one just has to replace $a$ with $\lfloor a\rfloor$.
May
20
comment Prime numbers stretch to infinity, but what about the distance between them?
Polignac turned out to be right: golem.ph.utexas.edu/category/2013/05/….
May
14
comment A suspicious way to conclude convergence
Yes. I thought a real divergent sequence had to go to $\pm\infty$.
May
13
comment A suspicious way to conclude convergence
Indeed, $S$ neither converges nor diverges.
May
7
comment Is there a need for another integration technique?
Next time I'll know better than to not draw the freaking region. =D
Sep
21
comment Splitting field of a slightly general polynomial
I've done exercises with $x^n-a$, $n$ and $a\neq1$ known, and indeed, the Galois group does not turn out to be abelian. Based on your answer, I have another idea: when $a$ is $1$, $\omega\mapsto\omega^k$ is invertible if and only if $k$ is invertible modulo $n$; but the set of such numbers is an abelian group, so I could argue $\mathrm{Gal}(F)\simeq\mathbb Z_n^\times$.
Jul
25
comment Finding a pair of elements to satisfy an inequation
@JackSchmidt I had an insight as soon as I read your comment. A field is a domain, and this one has at least two non-zero distinct elements. Pretty easy, I should've thought of that by myself. Thanks.
Mar
20
comment A sequence of nested fractions with a counter-intuitive limit
@Gerry Good idea. youtube.com/watch?v=GFLkou8NvJo#t=1m And there is the spoiler that wau is 1.
Dec
22
comment Proving Hahn's Extension Theorem for a finite initial measure
Ooooh, so it was an infimum argument! Like I said, Bartle goes way too fast.
Dec
16
comment Existence and uniqueness of a morphism that completes an alternate path
@DylanMoreland In order to make sense, $p$ and $q$ have to be integers; but $\mathbb Z_{-k}=\mathbb Z_k$, and $\mathbb Z_0$ is useless (empty, I think), so I restricted $p$ and $q$ to the positive integers.
Dec
14
comment Existence and uniqueness of a morphism that completes an alternate path
@DylanMoreland It turns out that's not minor; it's a pretty big part of the solution. I'll post it in a few days.
Dec
12
comment Non-unital rings: a few examples
@FredrikMeyer Even shorter proof than I thought. I'd said "provably", BTW. Thanks, now I'm 100% sure of the uniqueness. I'll edit that part.
Nov
20
comment Measure Product Theorem: may non-$\sigma$-finiteness result unique product?
@DavideGiraudo Hmmm, you're right. I had actually found that question before, but not read it. I thought it was a "counter-counter-example", but it's about a counter. I will edit my question and ask about a double counter instead.
Oct
31
comment Tricky congruence question
Argh, I already knew $p|b$, I forgot to put in the "it can be shown" part.
Sep
19
comment Removing a hypothesis when generalizing the Lebesgue measure
Um… You didn't change the type of bracket. I just did (feel free to roll back, even though I made pretty small changes).
Sep
18
comment Independence of discrete random variables
I don't think you need to transform into Bernoulli. Calculate the probability that $X=x$ given that $Y=y$; it has to be the same as without conditioning on $Y$.
Sep
18
comment The number of functions with a certain property
Since, for all $n$ values of $k$, $f(k) \leq k$, $f(k)$ can take $k$ values, which means there are $1\cdot 2\cdot\dots\cdot n=n!$ such functions. So, you are indeed not wrong on that bit.
Sep
18
comment Removing a hypothesis when generalizing the Lebesgue measure
The general idea makes a lot of sense, but I think I spot two small mistakes. 1, $f$ has to be continuous to some side or necessarily to the right? 2, isn't $\bigcap_{n\in\mathbb N}[a, b + h_n) = [a, b]$, since $b\in[a, b + h_n)$ for every $h_n$?
Sep
13
comment Defining a measure as a supremum
Much simpler than I thought. Thanks. I'll erase my incorrect approach from the question.
Sep
13
comment Defining a measure as a supremum
@kahen But no-one was seeing the hideousness, and now the space between the lines that define $\mu_\sup$ is a bit large. align is a good idea, though.