- #1
sponsoredwalk
- 531
- 5
Okay, here's a cool question I'm just not able to get:
If θ is the angle between vectors A = (1,1,...,1) & B = (1,2,...,n) then find
the limiting value of θ when n → ∞, where n is the dimension of the space.
Okay, I am using A • B = ||A|| ||B|| cos θ.
Since A = (1,1,...,1) we have ||A|| = √n.
Since B = (1,2,...,n) we have ||B|| = \sqrt{ \frac{n(n \ + \ 1)(2n \ + \ 1)}{6}}
It follows that A • B = \frac{n(n \ + \ 1)}{2}.
If I work with \frac{A \ \cdot \ B}{||A|| \ ||B||} \ = \ \frac{ \frac{n(n \ + \ 1)}{2}}{ \sqrt{n} \ \sqrt{ \frac{n(n \ + \ 1)(2n \ + \ 1)}{6}}}
I can simplify to get
\frac{A \ \cdot \ B}{||A|| \ ||B||} \ = \ \frac{ \sqrt{6} \ \sqrt{n \ + \ 1}}{2 \ \sqrt{2n \ + \ 1}}
and so
\theta \ = \ \lim_{n \to \infty} \ \cos^{-1} \ ( \frac{A \ \cdot \ B}{||A|| \ ||B||} \ ) \ = \ \ \lim_{n \to \infty} \ \cos^{-1} \ ( \frac{ \sqrt{6} \ \sqrt{n \ + \ 1}}{2 \ \sqrt{2n \ + \ 1}} \ )
But this isn't any better, I don't know where to go from here.
The answer is pi/6 but I don't know how to get it! Any idea's?
If θ is the angle between vectors A = (1,1,...,1) & B = (1,2,...,n) then find
the limiting value of θ when n → ∞, where n is the dimension of the space.
Okay, I am using A • B = ||A|| ||B|| cos θ.
Since A = (1,1,...,1) we have ||A|| = √n.
Since B = (1,2,...,n) we have ||B|| = \sqrt{ \frac{n(n \ + \ 1)(2n \ + \ 1)}{6}}
It follows that A • B = \frac{n(n \ + \ 1)}{2}.
If I work with \frac{A \ \cdot \ B}{||A|| \ ||B||} \ = \ \frac{ \frac{n(n \ + \ 1)}{2}}{ \sqrt{n} \ \sqrt{ \frac{n(n \ + \ 1)(2n \ + \ 1)}{6}}}
I can simplify to get
\frac{A \ \cdot \ B}{||A|| \ ||B||} \ = \ \frac{ \sqrt{6} \ \sqrt{n \ + \ 1}}{2 \ \sqrt{2n \ + \ 1}}
and so
\theta \ = \ \lim_{n \to \infty} \ \cos^{-1} \ ( \frac{A \ \cdot \ B}{||A|| \ ||B||} \ ) \ = \ \ \lim_{n \to \infty} \ \cos^{-1} \ ( \frac{ \sqrt{6} \ \sqrt{n \ + \ 1}}{2 \ \sqrt{2n \ + \ 1}} \ )
But this isn't any better, I don't know where to go from here.
The answer is pi/6 but I don't know how to get it! Any idea's?
