Sunday, March 24, 2013

Solution to the Plouffe Puzzle

Yesterday I posted the following puzzle from Simon Plouffe: why are there so many zeroes in the base-10 expansion of sqrt(51)/14 = 0.51010203061020 ... ?

Here's the answer. Let f(x) = (sqrt(1-4x2) + 2x - 1)/(2x - 4x2). Then it is not hard to see that f can be expanded as a power series
f(x) = 1 + x + 2x2 + 3x3 + 6x4 + 10x5 + 20x6 + ...
where the coefficients are binom(n, floor(n/2)), the central binomial coefficients.
It now follows that
f(1/t)/t = 1/t + 1/t2 + 2/t3 + 3/t4 + ...
= sqrt(t2/4 - 1)/(t-2) - 1/2.
When we substitute t = 100, we get
sqrt(51)/14 - 1/2 = 1/100 + 1/1002 + 2/1003 + 3/1004 + ...
which explains the expansion we see.

We can get even more of the central binomial coefficients by taking t to be larger powers of 10. For example, for t = 1000, we get
sqrt(249999)/998 = 0.501001002003006010020035070126252462 ... .

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