990 reputation
320
bio website about.me/LucasCAPS
location Rio de Janeiro, Brazil
age 25
visits member for 2 years, 11 months
seen 4 hours ago

UFRJ Applied Mathematics graduate.


Nov
2
suggested suggested edit on Is Lebesgue's Dominated Convergence Theorem a logical equivalence?
Nov
2
accepted Is Lebesgue's Dominated Convergence Theorem a logical equivalence?
Nov
2
suggested suggested edit on Is Lebesgue's Dominated Convergence Theorem a logical equivalence?
Nov
1
accepted Removing a hypothesis when generalizing the Lebesgue measure
Nov
1
asked Is Lebesgue's Dominated Convergence Theorem a logical equivalence?
Oct
31
revised Finding a Fermat number with a given prime factor
added a step to the reasoning
Oct
31
accepted Tricky congruence question
Oct
31
revised Tricky congruence question
added info I should have put earlier
Oct
31
comment Tricky congruence question
Argh, I already knew $p|b$, I forgot to put in the "it can be shown" part.
Oct
30
asked Tricky congruence question
Oct
30
revised Is there a combinatorial proof of this congruence identity?
improved formatting
Oct
30
suggested suggested edit on Is there a combinatorial proof of this congruence identity?
Oct
30
revised Finding a Fermat number with a given prime factor
improved formatting
Oct
3
awarded  Teacher
Sep
19
revised Removing a hypothesis when generalizing the Lebesgue measure
changed brackets and a few details in the text
Sep
19
suggested suggested edit on Removing a hypothesis when generalizing the Lebesgue measure
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 $h_n$?