# Vector Space R^3 and rigorous proof

1. Mar 5, 2012

### bugatti79

1. The problem statement, all variables and given/known data

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

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

2. Relevant 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

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

Last edited: Mar 6, 2012
2. Mar 5, 2012

### lanedance

are x and y given in the question? the way its written is pretty confusing

3. Mar 6, 2012

### Staff: Mentor

Where did the 10 come from? Are you given a specific vector x? If so, you didn't include this information in the problem statement.
The 2nd problem looks fine.

Last edited: Mar 6, 2012
4. Mar 6, 2012

### bugatti79

I have updated original post. Thanks

5. Mar 6, 2012

### Fredrik

Staff Emeritus
You need to get into the habit of reading the stuff you write an extra time before you post it. It took me a while to figure out that when you wrote $\|x\|+1$, you meant $\|x\|_1$. Assuming that I'm right about that, you did that one correctly.

Your calculations of $\|x+y\|_2$ and $\|x-2y\|_\infty$ are however horribly wrong. You showed in the other thread that you don't know what x+y means. Here you're making it clear that you don't know what 2y means. You will not be able to solve any problem that involves expressions like x+y or 2y until you have made sure that you understand what they mean. So please, look up the definition of the addition and scalar multiplication operations on ℝ2 and ℝ3. Forget everything else, and just use your books to try to answer this:

If x = (2,-5,3) and y = (-3, 2,0), then what is
a) x+y
b) 2y
c) x-2y

Edit: By "scalar multiplication operation", I mean the rule for how to multiply a vector by a number. (In particular, I don't mean the rule for how to multiply two vectors to get a number. That operation is often called a "scalar product". I prefer the term "inner product" for that, so that it sounds less similar to "scalar multiplication", which is just multiplication by a scalar (i.e. a number)).

Last edited: Mar 6, 2012
6. Mar 6, 2012

### bugatti79

x+y = (-1,-3,3)
2y = (-6,4,0)
x-2y = (8, -9,3)

above ok? Will have a look at other thread.

7. Mar 6, 2012

### Staff: Mentor

#1 is incorrect and #3 is incorrect. #2 looks fine.

For #1, it's not ||x|| + 1 (which you have already been told - please read the responses you get more carefully); it's ||x||1. IOW, it's the "1" norm (or taxicab norm.
For #3, evaluate x - 2y (you already did), and take the infinity norm of that vector. I don't see how you came up with |-1 - (-5)|, which by the way, happens to be 4, not 5. In any case, neither 4 nor 5 is the answer.

IMO, you spend too much time crafting your stuff in LaTeX, and not enough time dealing with the actual mathematics. It is preferable to have something crude-looking that is correct, than something very nicely formatted that is completely wrong.

8. Mar 6, 2012

### bugatti79

Disastrous typos.

It should read |2-(-6)|=8 for #3. Ie, I have taken the maximum calculated value from (8,-9,3)

Ok, but I thought having crude looking stuff people wont read it, they'll just skim over it and exit.

9. Mar 6, 2012

### Fredrik

Staff Emeritus
It looks like you're still not thinking about how the various notations you're working with are defined. $\max_i |x_i|$ is the largest member of the set $\big\{|x_1|,|x_2|,|x_3|\big\}$.

I think you're confusing yourself by trying to do several things at once. To evaluate $\|x-2y\|_\infty$, you must first use the definitions of x, y, scalar multiplication and addition to find x-2y. Now you can rewrite $\|x-2y\|_\infty$ in the form $\|(a,b,c)\|_\infty$. Then you use the definition of $\|\ \|_\infty$.

Last edited: Mar 6, 2012
10. Mar 6, 2012

### Staff: Mentor

I don't think that they are merely typos.
Where does 2 - (-6) come from?

To evaluate this expression: ||x - 2y||
focus on one thing at a time.

1. Evaluate x - 2y. This is a vector in R3. It is NOT the difference of two numbers.
2. Calculate the maximum norm of this vector.
I guarantee that people will be more impressed by something that makes sense, over something that makes no sense, but isn't quite as pretty. Certainly it's nice to have both, but if you have to choose, lean toward getting the mathematics right.

Besides, and I've said this before to you, posts with lots and lots of LaTeX take an inordinate amount of time to load in some browsers, and that ticks me off when it takes forever for a page to load. For that reason I tend to use LaTeX only where I need to use it.

11. Mar 6, 2012

### Fredrik

Staff Emeritus
Is that really still a problem? Some old version of IE (that very few people use) had problems before, but I thought it was fixed by the recent upgrade of MathJax. I'm using Firefox, and I've never had any problems.

12. Mar 6, 2012

### Staff: Mentor

Yes, it's still a problem in IE9, which is not an old version.

13. Mar 6, 2012

### bugatti79

1) $x-2y=(x_1-2y_1, x_2-2y_2, x_3-2y_3)$

2)$\|x-2y\|_\infty=max (x_1-2y_1, x_2-2y_2, x_3-2y_3)$ for i=1,2,3

$=(8, -9, 3)$

$=8$?

14. Mar 6, 2012

### Staff: Mentor

= ? You are given specific vectors.
You haven't taken the max norm yet, so why did that go away?
No, but at least you're not committing grievous errors.

The max norm is the maximum |xi|, for i = 1, 2, 3.

15. Mar 6, 2012

### Fredrik

Staff Emeritus
This is a good start. Edit: But as Mark said, you were given specific vectors x and y, so you should use the numbers you've been given.

This notation is weird. The words "for i=1,2,3" add no information. Either write $\max_{i\in\{1,2,3\}}\{|x_i-2y_i|\}$ or $\max\{|x_1-2y_1|,|x_2-2y_2|,|x_3-2y_3|\}$.

A real number is never equal to a triple of real numbers.

A triple of real numbers is never equal to a real number. Also, you seem to have forgotten about the absolute value symbols in the definition of $\|\ \|_\infty$.

16. Mar 6, 2012

### bugatti79

It is equal to (8,-9,3)

So the maximum norm is 9?

Not sure I follow what you are trying to say.

17. Mar 6, 2012

### Staff: Mentor

It's not clear to me what "it" refers to.

||<8, -9, 3>|| = 9
What Fredrik was saying is that you are saying that incomparable things are equal. A vector is not a number; the norm of a vector is a number. You can't compare (i.e., with =) a vector with a number.

Boiled down a bit, what you said was
||<x1 - 2y1, x2 - 2y2, x3 - 2y3>|| = <8, -9, 3> = 8

The first thing above is a number. The second thing is a vector. The third thing is a number. Again, a number can never be equal to a vector in R3 and vice-versa.

Also notice that I used no LaTeX in the above.

18. Mar 6, 2012

### bugatti79

Ok, one of my problems is that I'm very sloppy with definitions. I need to buckle up.

Thanks, at least this thread is finish.

19. Mar 6, 2012

### Fredrik

Staff Emeritus
OK, since we have arrived at the correct final result (but still no acceptable way of arriving at that result), I will show you how you should have done this.

Since x = (2,-5,3) and y = (-3, 2,0), we have x-2y=(2,-5,3)-2(-3,2,0)=(8,-9,3). So
$$\|x-2y\|_\infty=\|(8,-9,3)\|_\infty=\max\big\{|8|,\,|-9|,\,|3|\big\}=9.$$ As you can see, this is a trivial problem if you just use the definitions and do one thing at a time.

20. Mar 6, 2012

### Staff: Mentor

And it wouldn't hurt to spend some time reviewing vector algebra. It doesn't seem that you have a good handle on that area, which is preventing you from making progress in the area you're currently studying.