- #1
sponsoredwalk
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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|| = [itex] \sqrt{ \frac{n(n \ + \ 1)(2n \ + \ 1)}{6}}[/itex]
It follows that A • B = [itex] \frac{n(n \ + \ 1)}{2}[/itex].
If I work with [itex] \frac{A \ \cdot \ B}{||A|| \ ||B||} \ = \ \frac{ \frac{n(n \ + \ 1)}{2}}{ \sqrt{n} \ \sqrt{ \frac{n(n \ + \ 1)(2n \ + \ 1)}{6}}}[/itex]
I can simplify to get
[itex] \frac{A \ \cdot \ B}{||A|| \ ||B||} \ = \ \frac{ \sqrt{6} \ \sqrt{n \ + \ 1}}{2 \ \sqrt{2n \ + \ 1}}[/itex]
and so
[itex] \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}} \ )[/itex]
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|| = [itex] \sqrt{ \frac{n(n \ + \ 1)(2n \ + \ 1)}{6}}[/itex]
It follows that A • B = [itex] \frac{n(n \ + \ 1)}{2}[/itex].
If I work with [itex] \frac{A \ \cdot \ B}{||A|| \ ||B||} \ = \ \frac{ \frac{n(n \ + \ 1)}{2}}{ \sqrt{n} \ \sqrt{ \frac{n(n \ + \ 1)(2n \ + \ 1)}{6}}}[/itex]
I can simplify to get
[itex] \frac{A \ \cdot \ B}{||A|| \ ||B||} \ = \ \frac{ \sqrt{6} \ \sqrt{n \ + \ 1}}{2 \ \sqrt{2n \ + \ 1}}[/itex]
and so
[itex] \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}} \ )[/itex]
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?