Pi turns up uninvited — page 2
The ball that is largest in five dimensions
A disc of radius one has area π, a ball of radius one volume 4π/3, and in each further dimension the unit ball grows — until five, where its volume is 5.264, after which it shrinks towards nothing. The recursion that shows it is the ring dissection of the disc, done one dimension at a time, and the shrinking is not the ball getting small: it is almost all of a high-dimensional cube lying outside the ball, and almost all of the ball lying in a thin rind at its surface.
A series that waits on π
Add 1/(n³ sin² n) for n = 1, 2, 3, … and the terms are mostly tiny, except where n is almost a multiple of π and sin n is almost nought. Ten million terms add to 30.3145, four-fifths of it from the single term at n = 355. Whether the sum is finite depends on how closely fractions can approach π — on a number called its irrationality measure — and the best proof available says only that the measure is below 7.1, where the series needs it below 2.5.
The sine of a figure eight
Walk round a circle and read the height against the distance walked, and the reading is the sine. Do the same on Bernoulli's figure eight and the reading is a new function, with an arc integral that has t⁴ where the circle's has t², a length Gauss found inside an average, two periods instead of one — and exactly the same list of equal divisions that compass and straightedge can draw.
The sine, rebuilt from its zeros
A polynomial is a product of factors, one for each root. Euler treated the sine the same way — one factor for each place the wave crosses the axis — and the product he wrote down is correct, though the zeros alone do not justify it. Multiplied out, it hands over the sum of the reciprocal squares, π²/6, and every even power after it.
The cosines of the powers of two
Start at cos 1 and apply x ↦ 2x² − 1 over and over. The map is chaotic and exactly solvable: the k-th point is cos(2ᵏ), the cosine of two to the k radians. Iterate it in floating point and the computed orbit leaves the true one after 51 steps. Read the binary digits of 1/(2π) instead and the true orbit can be followed for as long as digits can be computed — here 262,080 steps — and at that length it is as fair as any orbit can look. Whether it stays fair is a question about those digits, and nobody can answer it.
A third of random games have no pure equilibrium
Fill a game's payoff table with random numbers and ask whether some cell is each player's best reply to the other. Exactly one cell is expected to be, at every size, and the chance that none is climbs from one in eight to 1/e — the same constant that counts the shuffles in which nobody gets their own hat back, and for the same reason.
The length round an ellipse
An ellipse has an area anyone can write down, πab, and a perimeter nobody can: it is an integral with no expression in the functions of school mathematics. Kepler guessed it, Euler guessed better, and in 1914 Ramanujan wrote down a formula whose error falls like the fifth power of how far the ellipse is from a circle — off by four hundredths of a per cent even for a segment. Gauss had already found the way to compute it exactly, doubling the correct digits at every step.
Turning a slow series geometric
Leibniz's series 1 − 1/3 + 1/5 − 1/7 + … adds up to π/4, and it takes five billion terms to give ten decimal places. Euler found a rule that rewrites it, using exactly the same terms, as a series whose terms halve at every step — and forty of those give thirteen places. Aitken's and Shanks's later rules do better still. None of them works on every slow series, and the reason is that each one is a guess about the shape of the error.
The error that predicts itself
Archimedes' polygons close on π slowly — six doublings of the sides for two decimals. But the error does not fall at random. Each doubling adds almost exactly a quarter of the area the doubling before it added, so the rest of the slivers can be summed before they are drawn. Liu Hui saw the quarter in 263; Snell and Huygens turned it into bounds; and the 96-gon that gave Archimedes two decimals gives six once its error is known.
What a random chord is
Draw a chord of a circle at random. How often is it longer than the side of the inscribed equilateral triangle? Pick two random points on the rim and the answer is 1/3. Pick a radius and a random point on it, and the answer is 1/2. Pick a random midpoint in the disc, and the answer is 1/4. All three calculations are correct, and Bertrand published them in 1889 to warn that the phrase had no meaning. Eighty-four years later Jaynes argued that one answer is right after all.
A table that never reaches π
Fill a square table with 1/1, 1/2, 1/3 along the rows, each row starting one place further on. Every row is a piece of the harmonic series, and yet the table stretches no vector by as much as π. At a thousand rows the stretch is 2.42, at a trillion it is near 2.94, and one integral explains both the number and the failure to arrive.
The curve that bends as far as it has gone
Say how sharply a curve bends at each distance along it, and the curve is fixed. Bending in proportion to distance gives Euler's spiral, whose arms are infinitely long and end a finite distance away. Bending that repeats decides at once whether the curve ever closes, and taken in steps the spiral is a Gauss sum.
Two blocks and a wall count π
Slide a heavy block towards a light one resting by a wall, and count every collision until they part for good. A heavy block 100 times the mass gives 31; ten thousand times, 314; a million, 3,141. The digits are π because the two blocks are secretly one billiard ball in a wedge, and a ball in a wedge bounces as many times as the wedge's angle fits into half a turn.