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bio website gottwurfelt.wordpress.com
location Atlanta, GA
age 30
visits member for 4 years, 4 months
seen 7 hours ago

Data scientist and math blogger.


Dec
11
comment Does $\int_0^\infty |\cos(x^2)| \mathrm dx$ converge?
It diverges. More specifically, $|cos(x)|$ has average value $2/\pi$ (over a single period, or any whole number of periods) and it seems reasonable that $|cos(x^2)|$ should have the same property since the squaring shouldn't perturb any single half-period "too badly". So I'd guess that $\int_0^t |cos(x^2)| \: dx \approx (2/\pi)t$, which agrees with numerical results.
Dec
3
awarded  Organizer
Dec
3
revised A method for generating random math problems based on performance
edited tags
Dec
2
answered Compute $\lim\limits_{a \to 0^+} \left(a \int_1^{\infty} e^{-ax}\cos \left(\frac{2\pi}{1+x^{2}} \right)\,\mathrm dx\right)$
Nov
22
comment Do the Möbius function, totient function, sum of divisors and number of divisors uniquely specify a number?
I would vote up this answer multiple times if I could. Incidentally, my opinion (as a probabilist who is amused by number-theoretic heuristics) is that examples should get rarer as the numbers involved get larger, but I won't hazard a guess off the top of my head as to whether the number of counterexamples is finite or infinite.
Nov
22
awarded  Critic
Nov
20
awarded  Enlightened
Nov
20
answered Maximize normal density function over a subset
Nov
20
awarded  Nice Answer
Nov
19
answered $C(n,p)$: even or odd?
Nov
19
comment Probability that a vector in $\mathbb{Z}^n$ is primitive
Thanks, Chandru. I couldn't add a reference because I was teaching!
Nov
19
answered Probability that a vector in $\mathbb{Z}^n$ is primitive
Nov
19
answered $x^y = y^x$ for integers $x$ and $y$
Nov
19
answered Taking Seats on a Plane
Nov
16
revised Why is $1^{\infty}$ considered to be an indeterminate form
added comment about \infty^0
Nov
16
comment Why is $1^{\infty}$ considered to be an indeterminate form
Thanks, Chandru. (Now maybe one of these days I'll get to teach calculus again.)
Nov
16
revised Why is $1^{\infty}$ considered to be an indeterminate form
fixed notation
Nov
16
answered Why is $1^{\infty}$ considered to be an indeterminate form
Nov
15
answered A root? Or two roots?
Nov
14
comment Questions on a Test
The sum of k exponentials is gamma-distributed, as you've pointed out. But as k goes to infinity, the gamma distribution, appropriately rescaled, converges to normal.