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 Yearling
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comment Show that $ \lim\limits_{n\to\infty}\frac{1}{n}\sum\limits_{k=0}^{n-1}e^{ik^2}=0$
Actually not quite tangent.
Feb
11
comment Show that $ \lim\limits_{n\to\infty}\frac{1}{n}\sum\limits_{k=0}^{n-1}e^{ik^2}=0$
I wrote down question 4 thinking that the answer was going to be obviously the whole unit disk or obviously just $\{0,1\}$, or something like that, but actually I think the question is rather subtle! Let $L$ be the set. Then I can prove that $L$ is a closed, convex subset of the unit disk whose boundary touches the unit circle only at $1$ and is tangent to it there.
Feb
8
answered Show that $ \lim\limits_{n\to\infty}\frac{1}{n}\sum\limits_{k=0}^{n-1}e^{ik^2}=0$
Jan
27
awarded  Yearling
Jan
26
revised girth of Cayley graphs on abelian groups
added 14 characters in body
Jan
26
comment girth of Cayley graphs on abelian groups
Having put it that way, it's clear that the proof you link to has at least one typo. For example take $G = \mathbf{Z}/n\mathbf{Z}$ and $S = \{-1,0,1\}$. Then the Cayley graph $X(G,S)$ is an $n$-cycle, possibly with some double edges or self-loops depending on definitions, so there are no $4$-cycles. Maybe the identity is not allowed to be in a "Cayley set", but I don't think this is standard terminology.
Jan
26
answered girth of Cayley graphs on abelian groups
Oct
7
reviewed Approve Finding a probability on an infinite set of numbers
Aug
25
awarded  Nice Answer
Aug
24
comment How did author do this algebraic manipulation?
Write $D=(D+1)-1$ and expand.
Aug
22
comment The case of Captain America's shield: a variation of Alhazen's Billard problem
Sorry, of course that's 3 bounces.
Aug
22
comment The case of Captain America's shield: a variation of Alhazen's Billard problem
This is not the only solution for $2$ bounces, at least if $c$ is sufficiently large. Here is another one: If $c>1/2$ then rotate an inscribed equilateral triangle until it passes through $C$. Now fire along the triangle. This gives two further solutions which are not symmetric in the $x$-axis.
Aug
22
comment Trajectories on a circular billiards table
Here's another idea that your comment suggests: Consider rotating a regular pentagon inscribed in the unit circle. For every position for which this regular pentagon passes through $C$ we get a valid trajectory which returns in $5$ bounces. For $c<\cos(2\pi/10)$ we get $0$ trajectories in this way, but for $\cos(2\pi/10)<c<1$ we get $2$ essentially different trajectories. Similarly for a regular pentagram, with two other transition values of $c$ ($c=\cos(2\pi/5)$ and whatever the distance from $0$ to a vertex of the inner pentagon is). This doesn't seem to be the whole story though.
Aug
22
revised Trajectories on a circular billiards table
added 35 characters in body
Aug
22
comment Trajectories on a circular billiards table
@Tad Good point. I'm struggling to decide whether the cases $c=1$ and $c\approx 1$ are really the same, but already what you say shows that my paragraph starting "For small $n$..." is bogus, at least the "only if" parts.
Aug
22
revised Trajectories on a circular billiards table
fixed typo in def of f(n,1)
Aug
22
comment The case of Captain America's shield: a variation of Alhazen's Billard problem
I've spun this question off here: math.stackexchange.com/questions/1405746/…
Aug
22
asked Trajectories on a circular billiards table
Aug
22
comment The case of Captain America's shield: a variation of Alhazen's Billard problem
(continued) Another question: Does the answer depend on $c$? It's not obvious to me that there aren't some critical points $c$ at which the answer changes.
Aug
22
comment The case of Captain America's shield: a variation of Alhazen's Billard problem
(continued) Are you just interested in seeing this algebraic mess in all its glory, or do you want an algorithm to numerically compute $\theta$ in terms of $c$? Maybe it's more interesting to ask a related qualitative question like this one: How many different directions can Cap shoot in such a way that the shield returns to him after exactly $n$ bounces? Ignoring the two points on the $x$-axis (which always work), and working only up to reflection in the $x$-axis, the answer is $1$ for $n=2$, $1$ for $n=3$, but I think $3$ for $n=4$. This question is much more likely to gain traction here.