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.
--rwg _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun