# Math Help - norm definition..

1. ## norm definition..

i cant understand this definition
$
\left \| x \right \|_p=(\sum_{i=1}^{n}\left | x_i \right |^p)^{\frac{1}{p}}
$

norm describes the length of a vector
so if x is our vector

what do we sum in the formula?
why there is absolute value in a vector?
what is x_i?

2. Hello,

x is a vector of $\mathbb{R}^n$
And in this case, we have $x=(x_1,x_2,\dots,x_n)$

the xi's are the coordinates of the vector.

3. so what are we doing whit those coordinates
$
x=(x_1,x_2,\dots,x_n)

$

?

4. Originally Posted by transgalactic
so what are we doing whit those coordinates
$
x=(x_1,x_2,\dots,x_n)

$

?
You do exactly what the formula said:
1) Take the absolute value of each of them.
2) Take the $p^{th}$ power of each.
3) Sum them all.
4) Take the $p^{th}$ root of that sum.

For example, if p= 1, that is just the sum of the absolute values.

If p= 2, it is just the usual "Euclidean" norm on $R^n$.

If p= 3 , n= 4, $x_1= 3$, $x_2= 2$, $x_3= 1$, and $x_4= 2$, the norm is $\left(3^3+ (-2)^3+ 1^3+ 2^3\right)^{1/3}= \left(27+ 8+ 1+ 8\right)^{1/3}= \sqrt[3]{44}$.

If p= 2 or any even power, you don't need the absolute value. But with odd p, odd powers could cancel- and we don't want that. (1, -1) is not the 0 vector so it shouldn't have 0 norm. But if we used p= 3 without the absolute value, we would have $\sqrt[3]{1^3+ (-1)^3}= \sqrt[3]{0}= 0$. With the absolute value that becomes $\sqrt[3]{|1|^3+ |-1|^3}= \sqrt[3]{1+ 1}= \sqrt[3]{2}$

5. Originally Posted by transgalactic
i cant understand this definition
$
\left \| x \right \|_p=(\sum_{i=1}^{n}\left | x_i \right |^p)^{\frac{1}{p}}
$

norm describes the length of a vector
so if x is our vector

what do we sum in the formula?
why there is absolute value in a vector?
what is x_i?
here is another definition
http://i35.tinypic.com/qxqln9.jpg

why smaller and equal?
the inner priduct gives the same resolt

for me its the same thing

wher is my mistake?