shrughes

What is the infimum of n²|sin(n)|, over positive integers n?

Is it zero, or is it greater than zero?

Ok, it’s time to spoil the answer.

With some lack of rigor, I think the probability is 1 that the minimum value exists and is greater than zero.

Suppose two people play a number-picking game, where two players pick numbers, and then compare them against one another, infinitely many times. Suppose the first person picks a (real) number uniformly on [0,1] and the second person picks 1/2^k for his kth choice. What’s the expected number of times the first person will pick a number less than the second person does, for the rest of time? The answer is 1 — there’s a 1/2 chance of that happening for k=1, a 1/4 chance for k=2, and so on, so the expected number of times is 1/2 + 1/4 + 1/8 + …, which equals 1.

If we let r(1), r(2), r(3), … be the sequence of random numbers the first person picks, we would therefore expect there to be one value of k for which r(k)/(1/2^k) is less than 1. Maybe there are two, or three such values of k. The probability that there are infinitely many such values is zero.

Now consider |sin(n)|. For small values of ε, and random values of x the probability that |sin(x)| < ε is 2ε/π. That’s because the slope of |sin(_)| is 1 or -1 when it touches zero, and it touches zero at multiples of π. Now let’s ask ourselves: How many times is n²|sin(n)| less than 1? Well, that’s the same as playing a number guessing game where player one chooses |sin(n)| and player two chooses 1/n². And |sin(n)| is effectively a random number function. What’s the probability of player one choosing a value smaller than player two? Approximately 2(1/n²)/π. Or (2/π)/n². If we sum these probabilities, we get (2/π)(1/1² + 1/2² + 1/3² + …) which equals (2/π)(π²/6) which equals π/3. So the expected number of times n²|sin(n)| is less than 1 is about 1.04. This is not accurate — for example, we assumed small values of ε when computing 2ε/π, and early on in the sequence, we have large values of ε, but we would still expect the expected number of times n²|sin(n)| is less than 1 to be finite.

What does this mean? It means that, assuming the fractional parts of multiples of π are random-looking, n²|sin(n)| is less than 1 only for finitely many positive integers n. Which means that of this finite number, there is a minimum value. One that is of course greater than zero. So the infimum is not zero.

I’d say there’s a good chance that sin(1) is the only value in the sequence n²|sin(n)| that is less than 1. If we check the first ten million values in the sequence, we find that sin(1) is the only one so far.

Dated: Monday, February 22nd, 2010 at 11:56. Category: Cool.
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