- #1

bugatti79

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## Homework Statement

1) Consider the 3 norms in vector space R^3, ##\| \|_i## where i=1,2 and infinity. Given x = (2, -5,3) and y = ( -3, 2,0).

Calculate ##\|x\|_1, \|x+y\|_2, \|x-2y\|_\infty##

2)Prove Rigorously that

##\displaystyle \lim_{n \to \infty}=\frac{4n^2+1}{2n^2-1}=2##

## Homework Equations

##x=(x_1,x_2,x_3), \|x\|_1= \sum^{3}_{i=1} |x_i|, (\|x\|_2= \sum^{3}_{i=1} |x_i|^2)^{1/2}, \|x\|_\infty=\underbrace{max}_{i=1,2,3} |x_i|##

I calculate

1) ##\|x\|+1= |x_1|+|x_2|+|x_3|=10##

##\|x+y\|_2=\sqrt{(2^2)+(-5)^2+(3^2)+(-3^2)+(2^2)+(0^2)}=\sqrt{51}##

##\|x-2y\|_\infty= |3-2(-3)|=9##

2) Proof:

let ##\epsilon## be given.

Find ##\displaystyle n_0 \in \mathbb{N}## s.t ##\left | \frac{4n^2+1}{2n^2-1} -2 \right | < \epsilon \forall n> n_0##

##\left | \frac{4n^2-4n^2+2}{2n^2-1} \right | =\left | \frac{3}{2n^2-1} \right |=\frac{3}{2n^2-1}< \epsilon## iff

##1+\frac{3}{\epsilon} < 2n^2## ie

## \forall n>n_0 > \sqrt{\frac{1}{2} (1+\frac{3}{\epsilon})}## we have

##\left | \frac{3}{2n^2-1} -2 \right | < \epsilon \forall n > n_0##

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