Sean Eberhard
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 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)