Reputation
4,388
Top tag
Next privilege 5,000 Rep.
Approve tag wiki edits
Badges
8 20
Newest
 Yearling
Impact
~65k people reached

Apr
21
comment Fundamental Theorem of Algebra for highschool
Nice proof, I hadn't seen this before. It's not so difficult to make precise right? What's the real reason we feel that a small wiggly contour near $a_0$ can't smoothly deform into a great big contour which looks like a circle centered at $0$ traversed $n$ times, without the contour passing through $0$? We're reasonably well motivated to define the winding number at this point.
Apr
21
revised Fundamental Theorem of Algebra for highschool
added 24 characters in body
Mar
8
comment Show that $ \lim\limits_{n\to\infty}\frac{1}{n}\sum\limits_{k=0}^{n-1}e^{ik^2}=0$
Sure. It's a nice question once you see it. Actually look for a power of 2 starting 7777777..., and to get this consider the sequence $\log_{10}(2^n)$ modulo $1$.
Feb
23
answered Linear independence over $\mathbb Q$ and $\mathbb R$ of subsets of $2^{\mathbb N}$
Feb
20
comment Show that $ \lim\limits_{n\to\infty}\frac{1}{n}\sum\limits_{k=0}^{n-1}e^{ik^2}=0$
For more than you ever wanted to know about question 4, follow the links at mathoverflow.net/questions/231606/…. However, I think this question is reasonable: 4'. Show that $\{z: g_n(\alpha)\to z~\text{for some}~\alpha\} = \{z: |z|\leq 1\}$.
Feb
12
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.