An Elephant, a Dog and an Ass,
Tried to see who would pee the fastest,
On the referee’s go,
All three unleashed their flow,
And learnt: Pee time is constant over mass*.

*assuming mass > 3 kg The first sentence of this paper reads: “We filmed the urination of 16 animals and obtained 28 videos of urination from YouTube, listed in SI Appendix. Movies S1–S4 show that urination style is strongly dependent on animal size.” http://www.pnas.org/content/111/33/11932.full

An interesting story hit the news yesterday about three mysterious individuals at an obscure quant fund who secretly gave away billions of dollars to fund research into diseases such as Huntington’s disease. What they are doing is admirable, not just in terms of leaving a tremendous impact on science but also the manner in which they have gone about their philanthropy, quietly and without fanfare. A legend has grown around the campus of their quant fund, which apparently hires Ph.D.s and computer programmers. One sample interview question was given as: “For any prime number larger than 3, prove that

Today, some of us were playing a game. Everyone contributed to the creation of a bank of questions which included “What kind of music do you like to listen to?” and “What is the strangest thing that you have eaten?”. A script was created to draw from a total of 72 unique questions at random with replacement and no option would be given to pass on questions. The idea was that we would take turns on our birthdays to answer these questions in an up close and personal session. Of course, there was the issue of how many questions one should

compoundchem:

It’s often said that cooking is like chemistry, or vice versa. So here’s a look at some of the organic compounds you can find in various herbs and spices. Whilst they are all a complex mixture of various organic compounds, this highlights those that contribute significantly to their taste, flavour or aroma.

You can read more about each of the compounds here: http://wp.me/p4aPLT-8c

Many years ago, I received a sticker from a good friend on Pi Day. It was a commemorative design by the the Maryland Science Center with Pi digits. Upon closer scrutiny, I found two missing digits in the 8th line from the top between 6 and 8. I tried unsuccessfully to mark the spot with an asterisk. A second smudged asterisk at the bottom indicates the missing two digits, ‘79’. It is easy to imaging what had happened. The missing digits are the 99th and 100th digit of Pi and these digits were likely lost during a

We celebrate Pi day with a poem. Why, you ask? Because it puts the  PI in PoetIc, and that is your clue. Scroll to the end of the poem if you’d like the answer.

Poe, E. Near a Raven Midnights so dreary, tired and weary.      Silently pondering volumes extolling all by-now obsolete lore.  During my rather long nap – the weirdest tap!      An ominous vibrating sound disturbing my chamber’s antedoor.          “This”, I whispered quietly, “I ignore”.   Perfectly, the intellect remembers: the ghostly fires, a glittering ember.      Inflamed by lightning’s outbursts, windows

Today, we take a quick look at two methods for computing pi using your humble PC. These two methods are infinite sums, meaning that the value of the sum approaches pi as you increase the number of terms added. Here they are. 

1. Ramanujan’s Formula

2. Plouffe’s spigot formula To save you the calculator work, you can use the following python code to execute both calculations. I have listed the results for 10 iterations of each formula below. from __future__ import division import numpy as np import random import math np.set_printoptions(precision=15) def pi_plouffe(n): return np.cumsum([x for x in[(4./(8.*k+1.) -2./(8.*k+4.)

When I was young, I thought I could derive the value of pi as follows. Imagine a badly drawn hexagon as below. Every hexagon is the combination of 12 right angled triangles like the one shown in red. 

The following is known from trigonometry. If we extrapolate from a hexagon to a polygon of N sides, we get the following. Note that 2Nx/2r is the ratio of the hexagon’s perimeter to its ‘radius’. As N approaches infinity, the polygon approaches a circle and the value of 2Nx/2r should approach pi.

Using the same principle, you can also easily derive