Sorry -- forgot to take the square root! I was looking at the scaled up (by a factor of n) version, and mistakenly though that that might be less easy for people to see. You're quite right. Limit is 1/2. R.

And I incorrectly threw in a square root when doing dot products. Doing them correctly I also get 1/2 for the cosine and angle pi/3. - Scott

On Wed, 18 Dec 2002 asimovd@aol.com wrote:

...

Hi, Richard. You wrote:

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.

Richard, since you've already multiplied the (equal) lengths of the integer vectors to get the sum of the first n-1 squares n*(n-1)*(2n-1)/6, there's no need to square this again in the denominator. And since (1,2,...,n-1) and (n-1,...,2,1) are just scaled versions of the vectors I asked about, they will have the same angle and we may as well just consider these. According to your dot product and length calculation, the cosine at the nth stage is

(n+1)*n*(n-1)/6 divided by n*(n-1)*(2n-1)/6, and so the limit as n --> oo is 1/2.

Regards,

Dan Asimov