Chess Harmony

Long ago in the Gupta Empire, a great-but-greedy mathematician, Grababundel, presented to the Maharaja a new game that he had devised, called Chaturanga.

Thirty-two of the Maharaja’s subjects, sixteen dressed in white and sixteen in black, were assembled on a field divided into 64 squares. There were rajas and ranis, mahouts and magi, fortiers and foot-soldiers. The armies did battle, each trying to kill or capture the opposing raja.

The Maharaja, Branier Thanilux, was so pleased that he asked Grababundel to name his reward. He expected Big G to ask for the hand of the Crown Princess. But no! Grabundel said simply: “give me one grain of rice for the first square on the field, two for the second, four for the third, and so on, doubling each time until the 64th square.”
Now, as you may know (and see the appendix if you don’t), this comes to {2^{64} - 1}, a Mersenne — not to say mercenary — number of great magnitude. Grababundel was asking for more rice than the entire empire contained.
But Branier Thanilux saw through the ruse. He said to Grababundel: “You are too modest; if you wish, I will give you a boundless fortune, riches without limit.” The prospect of infinite wealth was irresistable to the greedy Grababundel, who instantly agreed.

So, Branier issued a proclamation: “On the first day, I will award one rupee to Grababundel. On the second day, half a rupee, on the third, one third of a rupee, and so on as long as this empire endures.”

Grababundel experienced a feeling of great harmony: on day {n} he would have

\displaystyle H_n = 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n} \ \ \mbox{rupees}.

He grinned broadly as he realised that the sequence of amounts was indeed unbounded: there would be no limit to his wealth.

But Grababundel’s smile began to fade as he started to compute these amounts. By Day 4, he would have two rupees, by Day 31 he’d have 4. But to amass just a meagre 10 rupees, Big G would wait almost 34 years, and for 20 it would take some 270 million days, not far short of a million years. His prospects of acquiring 100 rupees were slim: this would involve a wait of {1.5\times10^{43}} days, longer by far than the age of the universe.

The moral of the story: Don’t mess with the Maharaja: he might be Branier Thanilux.

Appendix

The amount of rice

The Mersenne Numbers are numbers of the form {2^n-1}. The number of grains of rice on the 64 squares of the chess-board is

\displaystyle S = 1 + 2 + 2^2 + 2^3 + \dots + 2^{63}

which is a geometric series. Multiplying this, term-by-term, by 2 gives

\displaystyle 2S = 2 + 2^2 + 2^3 + 2^4 + \dots + 2^{64}

Now subtract the first series from the second to get

\displaystyle S = 2^{64} - 1 \,.

Since {2^{10}} is about {10^3}, this is about {1.6 \times 10^{19}}. More precisely, it is {1.8 \times 10^{19}}.

If a rice grain is spherical with radius 2mm, its volume is

\displaystyle v = \textstyle{\frac{4}{3}}\pi 2^3 \approx 32 \mbox{mm}^3

If the packing density is 3/4, the total volume occupied by the rice is

\displaystyle V = \textstyle{\frac{4}{3}}\times(1.8\times 10^{19})\times 32\,\mbox{mm}^3 \approx 8\times 10^{20} \mbox{mm}^3 = 800\,\mbox{km}^3

If the rice is in a conical heap, with height {h} and also radius {h}, the volume is

\displaystyle \textstyle{\frac{1}{3}}\pi(h^2)h \approx h^3 = 800\,\mbox{km}^3

Therefore

\displaystyle h = \sqrt[3]{800\,\mbox{km}^3} = 9.3\,\mbox{km}

higher than Mount Everest (8,848 m).

The amount of money

The total amount awarded to Grababundel by day {n} was

\displaystyle H_n = 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n}

which is called the {n}-th harmonic number. As shown in any elementary book on infinite series, this series diverges. That is, {H_n} grows without bound as {n\longrightarrow\infty}. So, in principle, Big G has been rewarded with boundless riches. But the divergence of the series is excruciatingly slow.

It was shown by Euler that, for large {n},

\displaystyle H_n = \log n + \gamma + \epsilon_n

where {\gamma = 0.57721...} is the Euler-Mascheroni constant and {\epsilon_n \approx 1/2n}. Dropping this small last term, we see that the amount exceeds the value {H} when

\displaystyle H < \log n + \gamma \qquad\mbox{or}\qquad n > \exp(H-\gamma)

Now {\exp(-\gamma)\approx 0.562} so we write

\displaystyle n(H) = \bigl[ 0.562\exp(H) + \textstyle{\frac{1}{2}} \bigr]

where the square brackets mean the integer part and the additional half gives the nearest integer.

Let’s see what this means:

Amount          Number of days to wait
1                             1
2                             4
3                            11
4                            31
5                            83
10                        12,378
20                   272,662,840
100                 1.5 x 10^43

So, while Grababundel might hope for 5 rupees, maybe 10, he will never amass 20, let alone 100 rupees.


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