Limiting Value of Theta as n --> ∞

[adm_strat]

[SOLVED] Vector Problem

Given

Suppose a and b are vectors in Vn and theta is the angle between them. If a=<1,1,...,1> and b=<1,2,...,n>, find the limiting value of theta as n-->infinity


Relevent equations:
a*b = |a|*|b| * cos(theta)
or
theta = arccos ((a*b)/(|a|*|b|)



I know:

1) as n --> infinity, a*b = 1*1 + 1*2 + 1*3 + ... + 1*infinity
Therefore a*b equals infinity as n approaches infinity

2) as n --> infinity, |a|*|b| = sqrt(1 + 1 +...+ 1) * sqrt(1 + 4 + 9 + 16 +...+ (infinity)^2)


I can make all kinds of assumptions from here but I don't know where to go with this problem. Just looking at what I have so far it looks as if |a|*|b| goes to infinity faster then a*b, but I don't know how to show that and I am stuck. Thanks for the help

 

[Mr.Brown]

nono you got the scalar product all wrong :) |a|*|b| -> max(a_i*b_i) = n2 because n*n is the biggest summand in the scalar product.

so the numerator goes to n^2.

Now you got to take a better approch to calculating a*b=1+2+3+...=n*(n+1)/2

so you have arccos((n^2+n/2)/n^2)) goes to arccos(1) = 0, Pi , 2Pi,... so on :)

i guess it goes that way doesn´t it ?

 

[D H]

I know:

1) as n --> infinity, a*b = 1*1 + 1*2 + 1*3 + ... + 1*infinity
Therefore a*b equals infinity as n approaches infinity

2) as n --> infinity, |a|*|b| = sqrt(1 + 1 +...+ 1) * sqrt(1 + 4 + 9 + 16 +...+ (infinity)^2)

Find expressions for $\vec a \cdot \vec b$, $|\vec a|$, and $|\vec b|$ for finite n, form $\cos\theta$, and finally take the limit as $n\to\infty$.

Now you got to take a better approch to calculating a*b=1+2+3+...=n*(n+1)/2

Good so far. How about the rest?

so you have arccos((n^2+n/2)/n^2)) goes to arccos(1) = 0, Pi , 2Pi,... so on :)
i guess it goes that way doesn´t it ?

Nope. Not that way.

 

[adm_strat]

I agree with the $\vec a \cdot \vec b$ = $$\frac{n*(n+1)}{2}$$


I don't agree with the part where you say that $|\vec a|$ $|\vec b|$
= n$$^{2}$$

My reasoning is $|\vec a|$ = $$\sqrt{1 + 1 +...+ 1}$$ and as $n\to\infty$, the limit goes to $$\sqrt{n}$$

-- Also shouldn't the sum of the series: $$\sqrt{1 + 4 + 9 + 16 +...+ \infty^{2}$$ be soemthing like $$\sqrt{ \frac{(2n+1)*(n+1)}{2}}$$ as $n\to\infty$ ?

I am not sure if the $$\sqrt{ \frac{(2n+1)*(n+1)}{2}}$$ is correct or not and it is the last thing holding me up. I looked through my calculus book and I can't find it in the series section. Can anyone confirm this or tell me what it auctually is? Thanks in advance.

 

[Dick]

Sum of the first n squares is n*(n+1)*(2n+1)/6. How did you arrive at your guess??

 

[adm_strat]

I was just trying to remember from previous classes that I took. Simply a guess.

 

[adm_strat]

I think that means that as $n\to\infty$, then $$\frac{\vec a \cdot \vec b }{|\vec a| |\vec b|}\to\frac{\sqrt{3}}{2}$$

That means that $\arccos\frac{\sqrt{3}}{2} \to \frac{\pi}{6}$, which is the angle between the two vectors

Thanks for the help