Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm given a positive number, a unit vector $u \in \mathbb{R} ^n $ and a sequence of vectors $ \{ b_k \} _{ k \geq 1} $ such that $|b_k - ku| \leq d $ for every $ k=1,2,...$.

This obviously implies $ |b_k| \to \infty $ . But why does this imply $\angle (u,b_k) \to 0 $ ? I've tried proving it using some inner-product calculations, but without any success.

In addition, why the given data implies that there must exist $i<j$ such that $|b_i| \leq \frac{\delta}{4} |b_j| $ ?

Thanks a lot !

share|improve this question

2 Answers 2

up vote 0 down vote accepted

You have $|b_k| \leq |b_k -k u| + k=d+k$, and $|b_k| \geq k-|b_k -k u|=k-d$. This gives: $$\lim_{k \to \infty} \frac{\langle u, b_k-ku\rangle}{|b_k|} = 0, \;\; \lim_{k \to \infty} \frac{\langle u, ku\rangle}{|b_k|} = 1,$$ so it follows that $$\lim_{k \to \infty} \frac{\langle u, b_k \rangle}{|b_k|} = \lim_{k \to \infty} \frac{\langle u, b_k -ku\rangle + \langle u, ku\rangle}{|b_k|} = 1.$$

share|improve this answer
    
Thanks !! have you got any idea about my second question? –  joshua May 22 '12 at 15:59
    
Fix $i$. Since $|b_j| \geq j-d$, you can choose $j$ large enough so that $|b_j| >\frac{4}{\delta} |b_i|$. –  copper.hat May 22 '12 at 16:04
    
Great !!! Thanks !!! –  joshua May 22 '12 at 16:14

From $|b_k-k\,u|\le d$ you get that for all sufficiently large $k$ $$ k-d\le |b_k|\le k+d $$ and $$ k-d\le u\cdot b_k\le k+d. $$ Then $$ \frac{k-d}{k+d}\le\cos(\angle(u,b_k))\le\frac{k+d}{k-d}. $$ It follows that $\cos(\angle(u,b_k))\to1$ and $\angle(u,b_k)\to0$

share|improve this answer
    
Thanks !!!!!!!! –  joshua May 22 '12 at 16:43

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.