Giovanni De Gaetano
Reputation
1,859
Next privilege 2,000 Rep.
7 24
Impact
~22k people reached

# 488 Actions

 20h awarded Popular Question Jan 5 comment What's the formula for the 365 day penny challenge? @SpYk3HH Thanks for the answer! Actually my question was not "how did you answer your own question?" but rather "how did you post question AND answer in less than a second?". Whereas this is the right place to deposit such a link I'm not going into, but I take the freedom to ask you to format your answer according to the standard of the website (i.e. writing the formulas in mathmode). Jan 5 comment What's the formula for the 365 day penny challenge? Your question and your answer to your question report the very same time (even the same second!). On one hand I'm curious to know "how" did you do it, on the other hand I'm also curious to know "why" did you do it... Jan 4 comment Is there a function with two different pseudo-primitives? (Motivated by FTC) Thanks for the comment! I originally commented with the first line of thought that came to my mind, but, as you point out, there are more elegant ways to do so. I think you should add this as an answer, so that the OP can accept it. Jan 4 answered Is there a function with two different pseudo-primitives? (Motivated by FTC) Dec 14 comment calculate $\small\lim_{n\to\infty}\big(\frac{1}{n^{2}}+\frac{2}{n^{2}}+…+\frac{n}{n^{2}}\big)$ The hidden answer is fantastic! Dec 14 comment If I have that $X$ is a random variable satisfying $0\leq X \leq 1$, how can I show that $P\left(X \geq \frac{E(X)}{2}\right) \geq \frac{E(X)}{2}$? @user136503 That's precisely what I meant for accepting the answers, thanks. And also your observation is correct: $P(X>t)$ is decreasing for $t$ increasing because $P$ is a (probability) measure, thus $\sup_{t \in \left[ \frac{E(X)}{2}, 1\right]} P(X > t) = P\left(X \geq \frac{E(X)}{2} \right)$. Dec 14 comment If I have that $X$ is a random variable satisfying $0\leq X \leq 1$, how can I show that $P\left(X \geq \frac{E(X)}{2}\right) \geq \frac{E(X)}{2}$? I'm glad you like it. It is a standard result considering that $1-\frac{E(X)}{2}\leq 1$. Indeed one has in general (under "niceness" hypothesis) $$\int_a^b f(x) dx \leq \sup_{x \in [a,b]} f(x) \cdot |b-a|.$$ I also seize the opportunity to recall you, since looking at your profile you may be unaware of it, that you can (and should) accept one of the answers to your question by clicking the mark next to it. I invite you to do so also for your past questions. Dec 14 comment If I have that $X$ is a random variable satisfying $0\leq X \leq 1$, how can I show that $P\left(X \geq \frac{E(X)}{2}\right) \geq \frac{E(X)}{2}$? I would like the downvoter to let me know how can I improve my answer, or else at least to explain why it has been downvoted. Dec 14 answered If I have that $X$ is a random variable satisfying $0\leq X \leq 1$, how can I show that $P\left(X \geq \frac{E(X)}{2}\right) \geq \frac{E(X)}{2}$? Dec 9 comment Find the area and perimeter of shaded region. To see such an homework question with no effort be rewarded with upvotes, and none of the answer given to it be rewarded in any way leaves me disconcerted about how people vote. -1 to the question and +1 to the deserving answers. Dec 7 comment Conjecture about natural number satisfying $m(n)^k+1\space\mid\space n^{2k}+1$ Ooops, now I see it. I missed a $2$ in the conjecture, 'sorry for the annoyance. Dec 7 comment Conjecture about natural number satisfying $m(n)^k+1\space\mid\space n^{2k}+1$ Thanks, but isn't $p^2$ the greatest proper divisor of $p^3$? If the definition were as you mean it, the conjecture would work for all the $n$ of the form $n= p^j$ for $j$ natural odd and $p$ odd prime. Is it right? Dec 7 comment Conjecture about natural number satisfying $m(n)^k+1\space\mid\space n^{2k}+1$ It is not immediate to me why you conjecture that they should all be of the form $p$ or $p^3$. Clearly the conjecture holds for all the $n=p$ for an odd prime $p$, but $p^3$? I would be grateful if you could elaborate a bit more. Dec 2 comment Is there a function with two different pseudo-primitives? (Motivated by FTC) Let $F$ and $G$ be two pseudo-primitives, we want to show that $F(x) - G(x)$ is constant for any $x \in [a,b]$. Let $y$ be another element in $[a,b]$, then $F(x) - G(x) = F(y) - G(y)$ if and only if $F(x)-F(y) = G(x) - G(y)$, but the latter condition is guaranteed by their being both pseudo-primitives (and the FTC). Nov 30 comment If $f$ satisfies $|f(z)|\le |f(z^2)|$ then $f$ is constant Ah, and obviously if you show that $f$ is constant in any neighborhood of zero, then by analyticity... Nov 30 comment If $f$ satisfies $|f(z)|\le |f(z^2)|$ then $f$ is constant Did you already bump into this Wiki page: en.wikipedia.org/wiki/Maximum_modulus_principle ? Nov 30 comment Bounded analytic functon with small derivatives Sure, therefore you want to estimate $M$ in terms of the $\epsilon_i$. Nov 30 comment Bounded analytic functon with small derivatives Perhaps I still don't get your point. Doesn't the sentence: "bounded in $D$ by absolute value by a constant $M>0$" means that you already assume \$|f(z)|