normed-spaces
Which two of the vectors $u=(-2,2,1)^T$, $v=(1,4,1)^T$, and $w=(0,0,-1)^T$ are closets to each other in distance for (a) the Euclidean norm? (b) the infinity norm? (c) the 1 norm?
I believe I know how to solve this, but I was hoping someone could confirm or deny this for me. For (a) I took $||u-v||_1$, $||u-w||_1$, and $||v-w||_1$, but what do I take for the infinity and 1 norm? How do the 1-norm and the Euclidean norm differ?
The canonical basis is $$\{(1,0,0,\dots,0),(0,1,0,\dots,0),\dots,(0,0,0,\dots,0,1)\}$$ Obviously, there any many more unit vectors.
Consider a point $P$ on the red curve and $P'$ the intersection of the ray $OP$ with the blue curve. Then $\|OP'\|_4=1$ so $\|OP\|=\frac{OP}{OP'}$. Therefore, you need to find $P$ so that $\frac{OP}{OP'}$ is maximum. Alrenatively, you need to find the smallest dilation of the blue curve that contains inside the red curve. Try some dilations ( it seems you use Geogebra) and see one that may just fit.
Added: Just played a bit with Geogebra and it seems that the curve $x^4 +y^4 = 6.57^4$ just barely contains the red curve. So the norm is approximately $6.57$.
Best Answer
It usually goes like this;
The 1-norm of a vector with components $x_n$ is $\sum_n |x_n|$
The 2-norm is the euclidean norm given by $\sqrt{\sum_n x_n^2}$
The p-norm is given by $\sqrt[p]{\sum |x_n|^p}$
The infinity norm is the limit as the powers of these things go to infinity which happens to have the nice form $max(|x_n|)$.
Here I'm using $|a|$ to mean the absolute value of a