Evelyn Lamb さんはインスタンス mathstodon.xyz のユーザーです。アカウントさえ持っていればフォローしたり会話したりできます。 もしお持ちでないなら こちら からサインアップできます。
26daf5f9e4bd4ac5

Evelyn Lamb @evelynjlamb

@j2kun Huh, just played around with common ancestors and was really surprised that you and I are much more closely related than my husband and I are. It's my impression that his area is much closer to mine than yours is (e.g. he has written a paper with my advisor).

Evelyn Lamb ブーストされました

How to fold a bunny: youtube.com/watch?v=GAnW-KU2yn

From Erik Demaine and Tomohiro Tachi's work showing that any 3d triangulated surface can be folded from origami; MIT press release at news.mit.edu/2017/algorithm-or and SoCG 2017 paper at erikdemaine.org/papers/Origami

Evelyn Lamb ブーストされました

The theoretical computer science (TCS) community just launched a new conference: the Symposium on Simplicity in Algorithms (SOSA)

While simplicity is always appreciated in mathematics and TCS, it's never been so explicitly encouraged. I'll be looking forward to see what simplifications come out of this.

windowsontheory.org/2017/06/17

Evelyn Lamb ブーストされました
Evelyn Lamb ブーストされました

In fact, one of the primary tools in the process being debated in this case is the use of a mathematical measure called the Efficiency Gap. More here: arxiv.org/pdf/1705.10812.pdf

Evelyn Lamb ブーストされました

This is huge! The supreme court is going to hear a case on partisan gerrymandering, in which they may rule about what process (including what mathematical techniques) can be used in a case that a partisan gerrymander is illegal.

cnn.com/2017/06/19/politics/su

This is going to happen in October, just after I attend the Gerrymandering Workshop at Tufts. Exciting times!

A tweep suggests it's a cuboctahedron with diagonals drawn in on some of the squares. en.wikipedia.org/wiki/Cuboctah

Saw this cool solid as a playground climbing structure a couple weeks ago. Does anyone know if it has a name?
I spent longer than I care to admit searching for it in the list of Johnson solids before realizing it can't be a Johnson solid because it has 6 triangles around some vertices, which would lie flat if they were equilateral.
instagram.com/p/BVhlNfCHiCG/ mathstodon.xyz/media/GoLT-gr5A

Evelyn Lamb ブーストされました

Buckets of fish! jdh.hamkins.org/buckets-of-fis

A cute combinatorial game that always eventually terminates, despite the players' ability to make games arbitrarily long. And despite the infinite game tree, there's a simple trick that makes its strategy easy.

Evelyn Lamb ブーストされました

Help wanted : We're looking for examples of the terms and conditions that the different Mastodon instances have used. Are they collected somewhere, or do we need to visit each instance separately and collect/collate them "by hand"?

We're looking for something close to what we want and which can then be twoke for our needs.

Suggestions?

TIA.

Just posted a tiling on Instagram for on Saturday. instagram.com/p/BVRO1XZgDdq/
I get a cube illusion when I look at this tiling. But I just realized maybe I shouldn't because this is not a look that's possible w/cubes. No face of a cube looks like a square unless you're looking at it straight on & then you don't see other sides. I guess that trip to the Picasso Museum must have worked!
mathstodon.xyz/media/SCNgom-_B

@jennytrustad Are you on Instagram? I post a lot of foundmath there, and I'm always excited to follow other foundmath finders!

Evelyn Lamb ブーストされました

I just reread this transcript of Francis Su’s retiring MAA president address from January, “Mathematics for human flourishing.” It spoke to me even more this time than the first time I read it. He keeps coming back to this Simone Weil quote: “Every being cries out silently to be read differently.” I’m going to be thinking about that for a while. mathyawp.wordpress.com/2017/01

@gammafunction I think instead of dollar signs you use \ ( and \ ) (without the spaces), but I'm sure @christianp and @ColinTheMathmo can say for sure how it works.

@j2kun @byorgey @ColinTheMathmo OK, I'm not finding 7 with ellipses, though I don't see anything saying it's impossible.

Evelyn Lamb ブーストされました

The probability that a power of 2 starts with the digit d is $log(d+1) - log(d)$.

Observe that $2^n$ has first digit d if there is some non-negative integer k such that $d10^k \leq 2^n < (d+1)10^k$.

Applying the base 10 logarithm to this inequality, we get $log(d) + k \leq nlog(2) < log(d+1) + k$.

Taking the fractional part of this inequality, we get $log(d) \leq \{nlog(2)\} < log(d+1)$.

But by Weyl's Criterion, $\{nlog(2)\}$ is equidistributed in [0,1). The result follows.