After reading Tim Dellinger's article[^0] on performance, I wondered what distribution the firing system he alludes to would lead to. TL;DR: under some not-too-outlandish assumptions, it certainly quacks like a Pareto distribution.
I'm a mathematician who can code and write. I love solving problems and I love writing things up in creative and engaging ways -- whether it's explaining why the Black-Scholes equation is what it is, why this bridge [^0] has weird curvy bricks, or how to make your graphs look nice in matplotlib.
If a mathematically-switched on writer can solve problems for you, hit me up. I'm open to contract work, freelance projects, or a permanent post.
I'm a mathematician who can code and write. I love solving problems and I love writing things up in creative and engaging ways -- whether it's explaining why the Black-Scholes equation is what it is, why this bridge [^0] has weird curvy bricks, or how to make your graphs look nice in matplotlib.
If a mathematically-switched on writer can solve problems for you, hit me up. I'm open to contract work, freelance projects, or a permanent post.
Location: Dorset, UK
Remote: Only
Willing to relocate: No
Technologies: Go, Python, versatile
Résumé/CV: https://www.dropbox.com/scl/fi/7rz657fdb5bwtlap5kqhn/CV_Colin_Beveridge-2.pdf?rlkey=e53nx1o6oxk7mlnj2jzxlorif&st=i6j1gdku&dl=0
Email: colin@colinbeveridge.co.uk
Hi! My name is Colin and I am a mathematician. I love solving problems and communicating the solutions.
Looking for short- to medium-term contract/consulting work, but open to discussing other opportunities.
A trick I learned -- I think from Colin Wright -- involves powers of 2.
* Write the powers of 2 up to 2^10 in lex order and pop a decimal point after the first digit of each:
* 1.28
* 1.6
* 2
* 2.56
* 3.2
* 4
* 5.12
* 6.4
* 8
The logs of these are approximately 0.1, 0.2, 0.3, etc.
This is to do with 2^10 being about 10^3; you can do something very similar with powers of 5, but I find those harder to remember.
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.)
I did it similarly: log(19,683) is about 4.3, so ln(19,683) is about 4.3 * 2.3, which is about 9.9. That jumps out as close to 9ln(3).
So the natural log of the cube root is (presumably) 3 ln(3), making the cube root itself 27.
Alternatively, the digit sum of 19,683 is 27, so the cube is a multiple of 9. Given that it's a perfect cube, 19,683 must be a multiple of 3^3 = 27. Dividing by that gives 729, or 9^3, so the cube root is 3 * 9=27.
Lastly, it's possible to dredge up that e^3 is about 20, so (27.1828...)^3 is about 20,000, and it's plausible that 27 is the cube root we're after.
[^0]: https://timdellinger.substack.com/p/hey-wait-is-employee-per...
If you're interested, do mess around with the code. (I'm not sure if this should be a Show HN.)