Er... I guess I didn't explain that as clearly as I thought. The first diagonal (| 1j1) has length (%:2).
The idea was to rotate this segment down to the real line and then make another right triangle by drawing the line up to ((%:2) j. 1). Pythagoras tells us the hypotenuse of this triangle is (%: +/ *: (%2), 1) which is (%: +/ 2 1) or (%: 3). So basically it's just generating the square roots of each natural number in turn, and then you just square those to get the area. The line at the end was actually the answer: *:@| (|j.1:)^:(<n) 0j1 [ n=. 10 1 2 3 4 5 6 7 8 9 10 Not really a puzzle. I just thought it was cool. :) On Fri, Feb 26, 2016 at 1:51 PM, Kip Murray <[email protected]> wrote: > areas =: 1 + [: *: 1 + i. > areas 10 > 2 5 10 17 26 37 50 65 82 101 > > --Kip Murray > > On Friday, February 26, 2016, Michal Wallace <[email protected]> > wrote: > > > Imagine a compass rooted at the origin of the complex plane. > > > > Starting at the point 0j1, draw the arc down to the real line, and then > > from this point, draw vertical line segment extending upwards 1 unit, to > > arrive at the point 1j1. > > > > Repeat this process of drawing the arc down to the real line and moving > up > > 1 unit N times, creating a series of points along the horizontal line (] > > j. 1:). > > > > For each point, draw a line segment from the origin to the point, and > then > > draw a square (*:) whose sides are the length (|) of this segment. > > > > What is the area of each square? > > > > *:@| (|j.1:)^:(<n) 0j1 [ n=. 10 > > ---------------------------------------------------------------------- > > For information about J forums see http://www.jsoftware.com/forums.htm > > > > -- > Sent from Gmail Mobile > ---------------------------------------------------------------------- > For information about J forums see http://www.jsoftware.com/forums.htm ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
