# Longer Calculus Proof

• Oct 24th 2013, 07:26 PM
mathnerd15
Longer Calculus Proof
this is an Apostol problem in chapter 12 and I guess it's a hypothetical definition of a norm of a vector
Assuming this definition of the norm prove these statements-

$Def. ||A||=\sum_{k=1}^{n}a_{k}, prove ||A||>0, if ||A||\neq0,||A||=0 if A=0, ||cA||=c||A||,triangle equality ||A+B||\leq ||A||+||B||,$

do I just write out the sum components in the triangle equality proof? kind of odd since normally one defines norm as the square root of the dot product of a vector
Use this definition in V2 and prove on a figure the set of all points (x,y) of norm 1- this is just the line x+y=1?

which of the above theorems/statements would hold if we defined the norm as an absolute value?

$Def. ||A||=|\sum_{k=1}^{n}a_{k}|$
anyways I feel like I don't have a lot of confidence from this forum!
• Oct 25th 2013, 04:07 AM
chiro
Re: Longer Calculus Proof
Hey mathnerd15.

I was just going to ask about the absolute value property for the norm (you should have it unless you have restrictions on the space).

With regards to the properties most will rely on the properties of absolute values and also the fact that absolute value obeys triangle inequality (since it is a norm itself).
• Oct 25th 2013, 09:25 PM
mathnerd15
Re: Longer Calculus Proof
oh do you mean this definition of norm has some truth? since norm is really defined as (AdotA)^(1/2) then you do get an absolute value in the squares of the components? but norm isn't actually the sum of the absolute value of the components of a vector, you have to use triangulation/Pythagorean to calculate the length?

anyway if you have by Apostol's other definition in the problem with |ak| for argument's sake, ||A||=1 in V2 then |x|+|y|=1 and you have square with y intercepts at 1,-1, x intercepts -1,1. equations y=-x+1, y=x+1, y=-x-1, y=x-1 where segments are constrained by absolute value to length 1.

I have health problems, is it good to do a lot of problems from Apostol or better to move on to Hubbard and Rudin? I'm also reading Strang's Linear Algebra, Diff Eq Braun, Griffiths electrodynamics
• Oct 25th 2013, 10:46 PM
mathnerd15
Re: Longer Calculus Proof
with health problems it's kind of tough to cover a lot of material, but I think my health is improving