[math-fun] Limit puzzle
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
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
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.
--rwg _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
In the light of day, I see that I left out a negative sign. The correct argument is below. It's amusing that my original erroneous argument gave the right answer. Let theta = arctan n/x, where x is the least solution to the problem for n. Then exp(i*n*theta) + (-1)^n*exp(-i*n*theta(n)) = 0, since ni + x = sqrt(n^2 + x^2)*exp(i*n*theta). This gives 2*i*n*theta = (n - 1 - 2*k)*pi*i, where k is an integer. Or theta = (n-1 - 2*k) pi/(2*n). Take the tangent of both sides and solve for x: x = n * cot((n-1-2*k) pi/(2*n)). Now cot is decreasing so the right value of k is the one that makes n-1-2*k <= n-1, and is largest. So k=0, and x = n*cot((n-1)*pi/(2*n)) = n*tan(pi/(2*n)). Since tan x ~ x as x approaches 0, we get a limit of pi/2. Thus n*theta(n) = pi/2, or x = n*tan(pi/(2*n)). Now it's clear that the limit is pi/2. On Thu, Jan 16, 2014 at 12:35 AM, Victor Miller <victorsmiller@gmail.com>wrote:
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.
--rwg _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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