colinbeveridge

I understand that Professor Yaffle -- the woodpecker bookend in the classic kids' TV show Bagpuss -- was loosely based on Russell.

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.

[^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.)

SEEKING WORK

Location: Dorset, England (Remote) CV: https://colinbeveridge.co.uk/cv/CV_Colin_Beveridge_mathemati... Email: colin@colinbeveridge.co.uk

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.

[^0]: https://chalkdustmagazine.com/features/a-canal-runs-through-...

Location: Dorset, England Remote: Yes, only Willing to relocate: No Technologies: Mainly go and python, but willing to learn. CV: https://colinbeveridge.co.uk/cv/CV_Colin_Beveridge_mathemati... Email: colin@colinbeveridge.co.uk

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.

[^0]: https://chalkdustmagazine.com/features/a-canal-runs-through-...

  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
  

Looking for short- to medium-term contract/consulting work, but open to discussing other opportunities.

SEEKING WORK - Dorset, UK, remote - mathematician

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.

Location: Dorset, UK Remote: Only Willing to relocate: No Technologies: Go, Python, versatile Résumé/CV: https://www.dropbox.com/scl/fi/7rz657fdb5bwtlap5kqhn/CV_Coli... Email: colin@colinbeveridge.co.uk

For 13ths, multiply your base number by 77, subtract 1, and concatenate the number that makes it up to 999.

So, for 4/13:

* multiply 4 by 77 to get 308

* subtract 1 to get 307

* concatenate 692 to get 307692

4/13 is 0.[307692] recurring. (This trick works largely because 77 × 13 = 1001).

Seventeenths are trickier, but you can get a decent approximation if you multiply the numerator by 6/100, because 1/17 is 6/102.

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.

For example:

Oi oi, fallible developer! I, idiot programming enthusiast, frequent Hacker News, mixed subject extravaganza, do you? No way!

This uses each Scrabble tile once (with blanks M and X).

If you're interested in pangrams, you might like Dave's Scrabblegrams: https://twitter.com/dc_scrabblegram

Use each tile from a standard Scrabble set once to make a meaningful sentence.

Nice. I like ln(3) and ln(5) (1.1 and 1.6) as well.

For the square root, I get 140.3.

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.

Location: UK (South-west)

Remote: only

Willing to relocate: no

Technologies: Mainly mathematics. Python, JS, some FORTRAN and SQL long ago.

CV: https://flyingcoloursmaths.co.uk/cv/CV__Colin_Beveridge.pdf

email: colin@flyingcoloursmaths.co.uk

I am an applied mathematician and writer looking for remote consulting opportunities.

My academic background is in three-dimensional models of the Sun's magnetic atmosphere, but I've spent the last decade writing about, talking about and tutoring maths, while raising a family and consulting when possible. There are few things I enjoy more than taking a problem, breaking it down into manageable chunks, and solving them — then explaining what I did so others can repeat it.

Here's a nice way to play with the parameters: https://www.desmos.com/calculator/3wrid37iir

There's a nice demonstration of a zero-knowledge protocol here: http://aperiodical.com/2012/03/using-a-zero-knowledge-protoc...