Stop Putting Commas In Your Numbers

or Why you need to read Le Système international d’unités (8e édition)

How do you write very large or very small numbers? How, for example, would you write the speed of light out in full?

If you would write c = 299,792,458 m/s then please stop, because you’re doing it wrong. You can argue all you want about tradition, and “the way things have always been done” but you are still totally, absolutely, unequivocally wrong. There is a right way, an official, standardised way, to write very large and very small numbers, and it’s not with commas in them.

“Following the 9th CGPM (1948, Resolution 7) and the 22nd CGPM (2003, Resolution 10), for numbers with many digits the digits may be divided into groups of three by a thin space, in order to facilitate reading. Neither dots nor commas are inserted in the spaces between groups of three.”

The correct way to write the speed of light is c = 299 792 458 m/s. Ideally you’d use a special Unicode character, known as “NARROW NO-BREAK SPACE (U+202F)”, which stops text from wrapping around half-way through a number, but this isn’t very well supported, so the better-supported “THIN SPACE (U+2009)” or even just a normal space will do.

The reason for this is that the decimal point isn’t always a decimal point. Only 60% of countries use a full stop, whereas other countries use other marks. For example, a number that would traditionally be written in the UK as 123,456,789.01 would be written in France, Germany, Spain and many other countries as 123.456.789,01 and in Canada as either, depending on whether you’re working in English or French. This confusion (see this for example) was deemed undesirable and as such the scientific community declared in 2003 that:

The 22nd General Conference [of the BIPM],
considering that a principal purpose of the International System of Units is to enable values of quantities to be expressed in a manner that can be readily understood throughout the world …
reaffirms that “Numbers may be divided in groups of three in order to facilitate reading; neither dots nor commas are ever inserted in the spaces between groups”, as stated in Resolution 7 of the 9th CGPM, 1948.

Remember that thousand separators are also used when dealing with very small numbers. I’ve provided some examples below if you’re struggling to get your head around them.

 Incorrect Correct Incorrect Correct 123 123 0.123 0.123 1234 1234 0.1234 0.1234 12,345 12 345 0.12345 0.123 45 123,456 123 456 0.123456 0.123 456 1,234,567 1 234 567 0.1234567 0.123 456 7 12,345,678 12 345 678 0.12345678 0.123 456 78

The Fibonacci Sequence and Converting from Miles to Kilometres

The Fibonacci Sequence is a very famous sequence of numbers, named after Italian mathematician Leonardo Fibonacci (though the sequence had already been described by Indian mathematicians). Each term in the sequence is composed of the sum of the previous two terms:

F = 1, 1, 2, 3, 5, 8, 13, 21, 34, 55 and so on …

As the sequence grows longer the ratio of each term to the previous term becomes closer and closer to the golden ratio of 1.618 (to four significant figures). This is helpful because the ratio of kilometres to miles is 1.609, which differs by only 0.556 percent.

Therefore, to quickly convert from miles to kilometres you only need to find the value in miles in the Fibonacci sequence and look at the next number in the sequence: 21 miles is 34 kilometres, 34 miles is 55 kilometres and so on. If converting in the other direction, look at the previous term to convert from kilometres to miles.

Jerk, Jounce, Snap, Crackle and Pop

Physicists are obsessed by rates: how quantities change over time. The rate of change of the number of nuclei in a radioactive sample tells us how radioactive something is; the rate at which the chemicals in a reaction change tells us how reactive something is; and so on.

If we start by looking at the displacement of an object (i.e. the distance from where it started to where it currently is) then when we look at the first derivative (by time) of displacement, (i.e. dividing the displacement of an object for how long it took to be displaced) we have calculated the object’s velocity.

$v = \frac{dx}{dt}$

If we look at the rate of change of velocity, the second derivative (by time) of the object’s displacement (i.e. the rate of change of the rate of change of its displacement), then we have calculated the object’s acceleration.

$a = \frac{dv}{dt} = \frac{d^2x}{dt^2}$

If we now look at the rate of change of acceleration, the third derivative of the object’s displacement (i.e. the rate of change of the rate of change of the rate of change of its displacement) then we have calculated the object’s jerk.

$j = \frac{da}{dt} = \frac{d^2v}{dt^2} = \frac{d^3x}{dt^3}$

The first two derivatives of displacement, velocity and acceleration, are well known and reasonably well-understood by most people. But jerk is a little bit more difficult to understand. If we apply a force to an object it will accelerate, and we usually assume that this force is applied instantaneously. But this is not correct – it takes time to apply a force. As a result, the rate of acceleration will not be constant, and thus we have the jerk.

It may be easier to understand the concept of a third derivative by looking at an example from economics: inflation. US President Richard Nixon once famously said “the rate of increase of inflation is decreasing”, using a third derivative in the process.

The rate of inflation is the rate at which prices increase over time, and this is therefore the first derivative of price. The rate of the increase in inflation is a second derivative, and if this itself is decreasing then that is a third derivative. That is, in Nixon’s case, prices were increasing (i.e. inflation was positive), and this rate of inflation was itself also increasing, but the rate at which it increasing was decreasing.

The fourth derivative of an object’s displacement (the rate of change of jerk) is known as snap (also known as jounce), the fifth derivative (the rate of change of snap) is crackle, and – you’ve guessed it – the sixth derivative of displacement is pop. As far as I can tell, none of these are commonly used.

Mathematics as the language of physics

It is often said that “mathematics is the language of physics“. But what does this mean? In this post I’m going to try to explain, by using one of my favourite proofs as an example.

The Problem

The problem is this:

A ball is placed atop a sphere and released. At what angle to the vertical does the ball lose contact with the sphere?

You would be excused for thinking that the answer is 45° or 90°, but the correct answer is more complicated than that.

Optimal stopping

Imagine a conveyor belt in front of you, on which are placed one hundred various-sized piles of money. You are allowed to stop the belt at any point and take the pile of money in front of you, but you cannot take any pile that has already passed you. Which pile should you take?

There is a mathematical solution to this problem, (sometimes called the Sultan’s Dowry Problem or the Fussy Suitor Problem) which is quite elegant.

1. Wait until 37% of the piles have gone past you. (The figure of 37% is the reciprocal of e, the base of the natural logarithms.)
2. Pick the next pile that is better than all the other piles so far.

Here are one hundred randomly-generated piles of money under £100:

£52.33, £80.83, £27.39, £84.75, £63.87, £1.66, £96.82, £76.51, £22.77, £90.94, £24.08, £60.41, £10.38, £95.59, £92.98, £46.80, £85.86, £21.96, £92.22, £29.19, £59.08, £72.22, £45.08, £63.39, £16.38, £71.49, £29.59, £78.62, £30.05, £97.98, £70.95, £3.79, £19.22, £77.52, £1.78, £48.74, £48.71, £35.95, £79.48, £11.50, £47.33, £32.83, £99.19, £3.23, £10.59, £58.22, £21.15, £61.37, £42.78, £25.27, £58.86, £32.82, £91.75, £13.04, £21.76, £72.29, £85.48, £58.81, £8.70, £91.63, £93.30, £23.00, £13.49, £11.67, £95.27, £21.37, £67.27, £90.99, £50.88, £77.22, £9.51, £10.63, £28.23, £63.94, £89.51, £90.12, £68.53, £76.98, £76.83, £92.04, £19.21, £73.82, £71.31, £99.94, £26.96, £86.92, £33.94, £8.25, £13.70, £74.44, £60.08, £11.54, £42.75, £78.67, £41.92, £92.36, £8.25, £92.89, £37.31 and £36.62.

The “best” value from the first thirty-seven piles is £97.98, so you should proceed through the remaining piles until you reach a value greater than this. This means stopping at the 43rd pile: £99.19.

Looking at the data, this strategy doesn’t actually yield the best value – waiting until the 84th pile would yield £99.94, which is slightly more. For large numbers of piles the 37% rule yields the perfect result in only 37% of cases, but this is a greater percentage than any other solution and it usually results in a very good result (i.e. one close to the perfect result).