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bio website twitter.com/SimeonLePoisson
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visits member for 1 year, 8 months
seen 7 hours ago

Jul
15
awarded  Enlightened
Jul
15
awarded  Nice Answer
Jul
13
comment What is an algorithm for making text form a circle
@MPW: good remark (+1)!
Jul
13
answered What is an algorithm for making text form a circle
Jul
11
answered Series Convergence/Divergence $\frac{n^n}{(n+1)^{n+1}}$
Jul
11
comment If $\lim\limits_{x \to \infty} f(x) = 1$, can we have function $f(x)$, such that $\int_0^{\infty}f(x)dx$ converges
@Bhoot: a continuous function is bounded on compacts.
Jul
11
comment If $\lim\limits_{x \to \infty} f(x) = 1$, can we have function $f(x)$, such that $\int_0^{\infty}f(x)dx$ converges
The answer is NO. Actually you can even prove that the Cesaro mean $\frac{1}{T}\int_0^T f(t)\,dt$ converges to $1$ as $T \to +\infty$.
Jul
11
comment If $A$ is compact and connected, then is $\Bbb R^2\setminus A$ connected?
@Matteo: maybe you thought simply connected instead of connected?
Jul
9
comment The asymptotic behavior of the CDF of Binomial distribution
Yes. Actually it is much easier to prove under the assumption $q > \frac{1}{2}$, because $2^{\alpha n} < \frac{1}{2} 2^n$, which implies that $\ell \leq \frac{n}{2}$.
Jul
9
revised The asymptotic behavior of the CDF of Binomial distribution
added 37 characters in body
Jul
9
comment The asymptotic behavior of the CDF of Binomial distribution
@MBP: Given your question, I assumed that $\alpha$ was fixed and bounded away from $1$. Can you edit it to make it more accurate?
Jul
9
revised The asymptotic behavior of the CDF of Binomial distribution
added 230 characters in body
Jul
9
reviewed Approve suggested edit on Taking the sin of arccos
Jul
9
answered The asymptotic behavior of the CDF of Binomial distribution
Jul
9
awarded  Nice Question
Jul
9
revised The probability that x birthdays lie within n days of each other
added 374 characters in body
Jul
9
comment The probability that x birthdays lie within n days of each other
@Evert: drawing a picture might be more convincing, but here is the reasoning. If $E_m$ and $E_{m'}$ both happen and $m < m'$, it implies that $m'$, which is one of the birthdays, lies in $[m,m+n]$. Similarly $m \in [m',m'+n]$. Since $m < m'$, this is only possible if $m' + n - 365 \geq m$, which in turn implies $2n \geq 365$. This is absurd since we assumed $n < 365/2$. Notice that this bound cannot be improved.
Jul
7
comment The probability that x birthdays lie within n days of each other
@mjqxxxx: sorry, I did not see your comment relative to this restriction.
Jul
7
revised The probability that x birthdays lie within n days of each other
added 476 characters in body
Jul
7
comment The probability that x birthdays lie within n days of each other
@Evert: is it any better now?