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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.
Aug
22
comment The case of Captain America's shield: a variation of Alhazen's Billard problem
As we've seen now the case of two bounces is pretty straightforward. The case of three bounces is also pretty straightforward, and also has a solution of the form $\text{quadratic in}~\sin\theta = c$. In general for an even number of bounces you want the particle to be moving vertically after half the bounces, and for an odd number of bounces you want the middle bounce to be at one of the two points on the $x$-axis (if $C$ is on the $x$-axis), and this plus some geometry will give you an algebraic relationship between $c$ and $\sin\theta$.
Jul
24
comment existence of a subgroup of a solvable group G of order d for any divisor d of |G|, with d<|G|^(1/2)
@OfirSchnabel I wasn't sure whether $6 < \sqrt{|A_4|5}$, but I was quite sure that $6 < \sqrt{|A_4|101}$.
Jul
24
answered existence of a subgroup of a solvable group G of order d for any divisor d of |G|, with d<|G|^(1/2)
Jul
10
awarded  Civic Duty
Jul
3
comment If $x_{n+1}\leq x_n + 1/n^2$ then $x_n$ converges
You should think of $x_n$ as approximately decreasing.
Jun
2
answered Extension of $|\cdot|_\infty$ on $\mathbb R$ to $\mathbb C$
May
12
comment Steinhaus theorem for topological groups
I just mean if you take any compact neighbourhood $K$ of the identity and let $H$ be the subgroup of $G$ generated by $K$ then $H$ is an open $\sigma$-compact subgroup of $G$. The Haar measure of $G$ is then just the sum of its restrictions to the cosets of $H$, so by translating $A$ and $B$ we may assume they intersect $H$ in sets of positive measure. In effect we may assume that $G$ is $\sigma$-compact so certainly $\sigma$-finite.
May
12
comment Steinhaus theorem for topological groups
The integrand is zero for $(x,y)$ outside the compact set $AB\times A$, so such problems do not arise. Alternatively you can always just restrict to a subgroup $H$ of $G$ generated by a compact neighbourhood of the identity: this is an open $\sigma$-compact subgroup.