My heuristic is to find a nearby square (here, 140^2 = 19600) and divide the discrepancy (83) by double the root of the nearby square (280). This is the number to adjust by. [^0]
[^0] The Maclaurin series expansion of (A^2 + x)^(1/2) starts A + x/(2A), which is where this comes from.
83/280 is close to 84/280, which is 12/40 or 0.3. If I was trying to show off [^1], I might adjust that -- 1/28 is a quarter of a seventh, about 0.0357, so I can take 0.004 off of my adjustment to get 0.296 or 0.297.
[^1] I mean, show off more.
(The actual answer is 140.29611 -- the "0.3" adjustment is right to one part in 36,000; "0.296" is right to one part in 1.2 million.)
My heuristic is to find a nearby square (here, 140^2 = 19600) and divide the discrepancy (83) by double the root of the nearby square (280). This is the number to adjust by. [^0]
[^0] The Maclaurin series expansion of (A^2 + x)^(1/2) starts A + x/(2A), which is where this comes from.
83/280 is close to 84/280, which is 12/40 or 0.3. If I was trying to show off [^1], I might adjust that -- 1/28 is a quarter of a seventh, about 0.0357, so I can take 0.004 off of my adjustment to get 0.296 or 0.297.
[^1] I mean, show off more.
(The actual answer is 140.29611 -- the "0.3" adjustment is right to one part in 36,000; "0.296" is right to one part in 1.2 million.)