revised
Comparing two power-series
New proof
Loading…
comment
Comparing two power-series
You're right, the substitution step is false. Apologies for posting nonsense. Abobe is a new try to deal with the convergence issue, hoppefully correct to the end this time.
revised
Comparing two power-series
explanation expanded
Loading…
comment
Comparing two power-series
Yes, of course. I've added more details. Sorry for being too succinct. It's not easy to give the exactly right amount of hint.
revised
Comparing two power-series
superfluous bracket removed
Loading…
revised
Comparing two power-series
typos corrected
Loading…
answered
Loading…
comment
Average distance of the mean of $n$ random complex numbers in a unit disc
@sajjad veeri: for $n=3$ see my answer to the follow-up post by Moritz Firsching.
comment
Expected absolute value of the average of two points from the disc
@Moritz Firsching: I have now found the exact value of $\operatorname{exp\_abs}(3)$ via an amazing result of Borwein and co-workers, please see above.
revised
Loading…
revised
Expected absolute value of the average of two points from the disc
Added part on exact value
Loading…
comment
Expected absolute value of the average of two points from the disc
The best is a closed form expression. It looks like $I_3=\frac{\pi^2}{6}\, W_3(1,1)$, where $W_3(1,1)=\frac{476}{525}A+\frac{52}{7\,\pi^2}\frac{1}{A}$ with $A=\frac{3}{16}\frac{2^{1/3}}{\pi^4}\Gamma(\frac{1}{3})^6$ is the expected radial distance $\mathbb{E}|X+Y+Z|$ for $X,Y,Z$ uniform on $S^3$, as given in scholarship.claremont.edu/jhm/vol6/iss1/7 (on page 100). (Timothy Budd pointed to this paper in the related MO post.)
revised
Average distance of the mean of $n$ random complex numbers in a unit disc
added 35 characters in body
Loading…
Loading…
comment
Bounding the max-loaded bin using${m \choose k} \|A\|_k^k$
(1) Quick question: yes, thanks, corrected (2) I will write you an email (tomorrow, I hope)
revised
Bounding the max-loaded bin using${m \choose k} \|A\|_k^k$
typo corrected
Loading…
revised
Bounding the max-loaded bin using${m \choose k} \|A\|_k^k$
added 603 characters in body
Loading…
comment
Bounding the max-loaded bin using${m \choose k} \|A\|_k^k$
Please let me know if anything is unclear or too succinct. I tried to be verbose, but of course it always depends.
revised
Bounding the max-loaded bin using${m \choose k} \|A\|_k^k$
essential typo corrected
Loading…