[math-fun] Puzzle: Find limiting angle as dimension -> oo
Let v_n be the vector in R^(n-1) defined as (1/n, 2/n,...,(n-1)/n). Let w_n be the vector in R^(n-1) with the same coordinates in reverse order. Find the limit as n -> oo of the angle between v_n and w_n. (Try this without using numerical approximation or summation-of-powers formulae.) Is there a simple geometric reason for the answer? I don't know. --Dan
I'll guess the limit is 90 degrees on the grounds that random vectors in high dimensional spaces tend to be orthogonal.
From: asimovd@aol.com
Let v_n be the vector in R^(n-1) defined as (1/n, 2/n,...,(n-1)/n).
Let w_n be the vector in R^(n-1) with the same coordinates in reverse order.
Find the limit as n -> oo of the angle between v_n and w_n.
(Try this without using numerical approximation or summation-of-powers formulae.)
Is there a simple geometric reason for the answer? I don't know.
--Dan
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<HTML><FONT FACE=arial,helvetica><FONT SIZE=2 FAMILY="SANSSERIF" FACE="Arial" LANG="0">Let v_n be the vector in R^(n-1) defined as (1/n, 2/n,...,(n-1)/n). <BR> <BR> Let w_n be the vector in R^(n-1) with the same coordinates in reverse order.<BR> <BR> Find the limit as n -> oo of the angle between v_n and w_n.<BR> <BR> (Try this without using numerical approximation or summation-of-powers formulae.)<BR> <BR> Is there a simple geometric reason for the answer? I don't know.<BR> <BR> --Dan</FONT></HTML>
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Let v_n be the vector in R^(n-1) defined as (1/n, 2/n,...,(n-1)/n).
Let w_n be the vector in R^(n-1) with the same coordinates in reverse order.
Find the limit as n -> oo of the angle between v_n and w_n.
(Try this without using numerical approximation or summation-of-powers formulae.)
Is there a simple geometric reason for the answer? I don't know.
--Dan
Ok, I cheated (numerical approximation) and found limiting angle pi/4 (45 degrees). I don't know a simple geometric reason for this, but perhaps someone can use the following to find one. Permute the coordinates of v_n (or w_n). All such points lie in an (n-2)-plane, and their convex hull forms a "permutahedron". Permutahedra tile Euclidean space. I consider them generalized hexagons (the 3-dimensional member is the truncated octahedron). v_n and w_n are diametrically opposite points of the permutahedron, so v_n - w_n is twice its circumradius. A normal to the permutahedron hyperplane through the origin intersects the permutahedron center c_n = (m/n, m/n, ..., m/n) where m = n/2. Does some relationship between permutahedra circumradii and the distance between the origin and the hyperplace containing the permutahedron give any insight? - Scott
1. Handwaving solution: first vector tends to first coordinate axis, second to the last; answer pi/2. 2. I don't see why one shouldn't use very elementary summations (but see 3. below) The scalar product of (the numerators of) the two vectors is 1*(n-1) + 2*(n-2) + ... + (n-1)*1 which is the (n-1)th tetrahedral number (see Fig 2.34 on p.46 of the Book of Numbers, where one sums a diagonal of the multiplication table): (n+1)*n*(n-1)/6 so the scalar product is (n+1)*(n-1)/6n The lengths are just the sum of the first (n-1) squares, divided by n^2: n*(n-1)*(2n-1)/6n^2 So cosine of angle twixt the two is (n+1)*(n-1)/6n / (n-1)*(n-1)*(2n-1)*(2n-1)/36n^2 = 6n*(n+1)/(n-1)*(2n-1)*(2n-1) which tends to 0. 3. Now, with hindsight, we don't need formulas: The product of the numerators, or the sum of their squares, are each of order n^3 so the order of the actual quantities is n and n / n*n ---> 0. R. On Tue, 17 Dec 2002, Scott Huddleston wrote:
Let v_n be the vector in R^(n-1) defined as (1/n, 2/n,...,(n-1)/n).
Let w_n be the vector in R^(n-1) with the same coordinates in reverse order.
Find the limit as n -> oo of the angle between v_n and w_n.
(Try this without using numerical approximation or summation-of-powers formulae.)
Is there a simple geometric reason for the answer? I don't know.
--Dan
Ok, I cheated (numerical approximation) and found limiting angle pi/4 (45 degrees).
I don't know a simple geometric reason for this, but perhaps someone can use the following to find one.
Permute the coordinates of v_n (or w_n). All such points lie in an (n-2)-plane, and their convex hull forms a "permutahedron". Permutahedra tile Euclidean space. I consider them generalized hexagons (the 3-dimensional member is the truncated octahedron).
v_n and w_n are diametrically opposite points of the permutahedron, so v_n - w_n is twice its circumradius. A normal to the permutahedron hyperplane through the origin intersects the permutahedron center c_n = (m/n, m/n, ..., m/n) where m = n/2.
Does some relationship between permutahedra circumradii and the distance between the origin and the hyperplace containing the permutahedron give any insight?
- Scott
participants (4)
-
asimovd＠aol.com -
John McCarthy -
Richard Guy -
Scott Huddleston