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Joshua Z proslijedio/la je Tweet
STEPHEN KING WRITING ABOUT LIVING IN NEW ENGLAND: The old man who ran the town dump communed with darkness. He kept a Hand of Glory in a 1982 Boston Bruins mug. Crows and bats were his to command. ME AFTER MOVING TO NEW ENGLAND: Jesus, I used to think Stephen King made shit up.
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Joshua Z proslijedio/la je Tweet
Jane Austen’s DRACULA is the fan fiction we didn’t know we neededhttps://twitter.com/LEBassett/status/1222153780273336320 …
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I haven't had a chance to look at Moreira's argument in detail yet, but it might be interesting if it can be adopted to give lower bounds on how many colors the greedy coloring produces if it is restricted to coloring just 1 through n. 7/n n =7.
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It seemed like this greedy coloring would be easier to analyze than the general problem above, since it only involves a specific coloring. But I don't know any proof other than Moreira's theorem that it requires infinitely many colors. 6/n
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Does the above procedure give only finitely many colors to color all integers? Even without Moreira's theorem, one would expect the answer to be no, and of course Moreira's result implies this. 5/n
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This connects with a related problem that I had discussed previously with some people. Let's color the positive integers greedily. For each n, we assign the lowest possible color number to n such that 1,2,3... n satisfy the coloring property listed above. 4/n
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This was a long-standing open problem and apparently in 2017 Joel Moreira published papers in the Annals showing that no finite set of colors would suffice. (The Annals of Mathematics is arguably the top math journal in the world, and by any reasonable metric in the top 5).3/n
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The problem is as follows: Is there some way to color the positive integers with only finitely many colors so that for any positive integers a and b, both a+b and ab have different colors, except for the case when a=b=2 (since then ab=a+b)? 2/n
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I just found out that a certain unsolved problem which I was fond of was solved 3 years ago and I didn't even notice it. 1/n
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The situation starts looking more complicated with polynomial maps from R^n to R^m (that is things like (x,y) -> (x^2, y+x). I don't know any simple characterization of what the ranges of such functions looks like. There probably is one, but I don't know it.
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For any x which is positive and small, we can take y=1/x, and then f(x,y) = (1-1)^2 + x^2 which will be small. So f(x,y) can get as close to 0 as we want. But f(x,y)=0 cannot happen. If f(x,y)=0, then we need x =0, but f(0,y) = (1-0y)^2 +0^2 = 1. 3/4
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Consider the polynomial in 2 variables f(x,y) = 1 - 2xy + x^2y^2 + x^2. f(x,y) can take on any value which is >0, but there's no real solution to f(x,y)=0. 1/2. To see this, rewrite f(x,y) = (1-xy)^2 + x^2. 2/4
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It is well known that the range of a single variable polynomial is either all real numbers, all real numbers less than or equal to some value, or or all real numbers which are at least some value. But multivariable polynomials can be more complicated. 1/4
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To be clear, by new we mean the last decade or so. But this should be getting more attention than it has been.
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There is no best way to map a spherical world to a flat surface, but some ways are better than others. Some are better or worse than others depending on the purpose. That said, a new contender has entered the ring.https://twitter.com/DiscoverMag/status/1217922015757250582 …
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E.g., Horatio Hornblower. In that context, what may have happened with the original Star Trek was substituting nauticalbabble with technobabble."Furl the jib right after we jibe" sounds similar to "raise shields right before we come out of warp" if you don't have context.
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Why is Star Trek more than many other scifi shows full of technobabble? A ossible influence that is often not discussed is the similarity of the show to 18th-19th century naval adventures. A common trope of that sort of novel is the amount of naval jargon thrown about. 1/2
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Joshua Z proslijedio/la je Tweet
Tell us a story about why cars suck that sounds like a lie but is absolutely true.
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Joshua Z proslijedio/la je Tweet
[1/5] How fiendish can a quadrilateral be? There’s a series of infamous puzzles where you’re given four of the eight angles between the sides and diagonals of a quadrilateral. The puzzle is to find one of the unknown angles—using classical elementary geometry, not trigonometry.pic.twitter.com/bBuVhL8700
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This problem was as far as I'm aware, first proposed by Richard Guy. My intuition on this problem keeps jumping back and forth on whether it is true or not. Right now, I'm leaning towards ti being false. 7/7
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