When Cara Leon goes missing, Sam Mujrif is hired to investigate. Cara is 8 times taller than Sam, but evidence points to players much smaller than either of them—and technology with the potential to radically alter the balance of power between the scales.

Square roots with compass and straight edge. Given line segments AB of length x and BC of length 1, construct D so AD is length x-1, then E so DE meets AD at a right angle and AE is length x+1. Then DE is length 2√x, and can be bisected to find √x.

RT @StartsWithABang: The top quark isn’t a loner after all: “toponium” is real!. The top quark is the most massive, shortest lived particle….

Link:

Wow, I just (re)discovered Worpitzky's identity, I’m only 142 years behind the times.

“Families Like Ours” is low-key, high-bureaucracy apocalypse, with Denmark preemptively emptied due to the prospect of inundation. It’s all a bit First World Problems compared to real-life refugees, but there are flawed characters, tough decisions and believable human stories.

Just when I thought my digital experience couldn’t possibly get any more exciting!

RT @colin_fraser: I do not believe it tried to do this in any sense that is more real than the sense in which Dwight Schrute tried to oust….

RT @GaryMarcus: Daddy, daddy, what was it like doing a PhD back when people wrote their own articles? . You know, before people snuck in hi….

I should add that, with hindsight, the relation:. 1/S(d,n) = d/(d-1) (1/S(d-1,n) - 1/S(d-1,n+1)). is easily proved straight from the formula for S(d,n). Since S(d-1,1)=1, this immediately shows that:. Σ_{n=1…∞} 1/S(d,n) = d/(d-1).

Science writer Amanda Gefter spent more than a decade tracking down the people who knew him, and diving into a vast store of his unpublished work and private papers. A long read, but fascinating and moving. Link:

Peter Putnam was a promising student of John Wheeler who became obsessed with a computational model of the mind, was tormented by a pathological relationship with his wealthy mother, and ended up contributing to community politics while working blue collar jobs.

This follows from the definition of H(r,i), and the formula for S(d,n) mentioned upthread. So the infinite sum of 1/S(d,n) is d/(d-1), as we found previously.

This generalises immediately to tell us that the sum of the i=d-1 series that starts at 1/d is just 1/(d-1). But each such series is just 1/d times the reciprocals of the d-simplicial numbers! That is:. 1/(d S(d,n)) = H(d-2+n, d-1).

This in turn means that if you add up the infinite number of entries with a given i, such as those coloured red in the image, you get a telescoping sum:. 1/4 + 1/20 + 1/60 + . = (1/3 - 1/12) + (1/12 - 1/30) + (1/30 - 1/60) + . = 1/3.

The general formula is easy to prove by converting the binomial coefficients into ratios of factorials. Now, it follows that each entry [except the leftmost in a row] is the difference between the one above left and directly left, e.g. 1/12 = 1/3 - 1/4.

The key property that makes this so powerful is:. H(r,i) = H(r+1,i) + H(r+1,i+1). That is, each entry is equal to the sum of the two entries below it, e.g. 1/3 = 1/4 + 1/12. for the three green entries in the image.

To form the harmonic triangle, take the usual triangle of binomial coefficients, multiply each row by its length, then take the reciprocal. So H(r,i) = 1/[(r+1) Binom(r,i)]. where r=0,. for the rows and i=0,. ,r across each row.

I just learned from @stevenstrogatz that Leibniz himself used a beautiful construction, known as “the harmonic triangle”, to find all the infinite sums of the reciprocals of the simplicial numbers.

RT @stevenstrogatz: @gregeganSF

RT @stevenstrogatz: @gregeganSF