A slight mistake: we must have theta(n) = arctan n/x. So we have x = n*cot((2*k+1) pi/(2*n)) for some integer k. Since x is positive and the smallest possible solution, k must be the greatest integer such that 2*k+1<= n (if it's larger, cot becomes negative). So 2*k+1 = 2*floor(n/2) - 1. Thus x = n*arctan((pi/(2*n)*( (n mod 2) + 1)). However this doesn't have a limit. If n ranges over even integers the limit is pi/2. But if n ranges over odd integers the limit is pi. On Thu, Jan 16, 2014 at 12:06 AM, Victor Miller <victorsmiller@gmail.com>wrote:
Let theta(n) = arctan x/n, where x is the least solution to the problem for n. Then
exp(i*n*theta(n)) + exp(-i*n*theta(n)) = 0, since ni + x = sqrt(n^2 + x^2)*exp(i*n*theta(n)).
This gives 2*i*n*theta(n) = (2*k + 1)*pi*i, where k is an integer.
Or n*theta(n) = (2*k + 1) pi/2. However arctan is increasing so that we must have k = 0.
Thus n*theta(n) = pi/2, or x = n*tan(pi/(2*n)). Now it's clear that the limit is pi/2.
Victor
On Wed, Jan 15, 2014 at 8:50 PM, Bill Gosper <billgosper@gmail.com> wrote:
DWilson>
For integer n >= 2, let x(n) be the smallest positive solution x of
(ni + x)^n + (ni - x)^n = 0
where i is the imaginary unit.
What is lim n->inf x(n)? --------
NeilB just got π/2, but refuses to post.
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