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| holysword |
Posted: Fri Apr 11, 2008 4:53 pm Post subject: Closed Mapping - Non-Linear Programming |
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 Forum Freshman

Joined: 05 Apr 2008 Posts: 10
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Hi there
I'm studyin' some mathematics fundamentals for optimizations methods of non-linear programming.
I didn't understood the "Closed Mapping" thing, which is defined in my book ( Luenberg ) like this:
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A point-to-set mapping A from X to Y is said to be closed at x ∈ X if the assumptions
i) x_k -> x, x_k ∈ X
ii) y_k -> y, y_k ∈ A(x_k)
imply
iii) y ∈ A(x)
The point-to-set map A is said to be closed on X if it is closed at each point of X |
I would appreciate any help. _________________ "Nolite arbitrari quia venerim mittere pacem in terram non veni pacem mittere sed gladium"
Yeshua Ha Mashiach |
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| serpicojr |
Posted: Fri Apr 11, 2008 7:09 pm Post subject: |
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 Forum Ph.D.

Joined: 17 Jul 2007 Posts: 871 Location: JRZ
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| So this is saying, given a point x in X, take any sequence of points xk of X converging to x. Now take any sequence of points yk, each yk being in the set A(xk), and suppose this sequence converges to a point y. Then y must be in A(x). In a very loose sense, this says that the sets A(xk) "converge" to the set A(x). (Nothing here guarantees that there exists a sequence yk that actually converges.) |
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| holysword |
Posted: Sat Apr 12, 2008 5:40 am Post subject: |
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 Forum Freshman

Joined: 05 Apr 2008 Posts: 10
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Thanks for your explanation serpicojr. But I didn't get it yet. What do you mean by "sequence of points xk of X converging to x"??
There is any graphical way to represent all those things? I think that its always easiest to understand with graphical explanations.
Sorry man, I know that's all basic concepts, but I missed all those basic concepts 'cause I was sick ><" _________________ "Nolite arbitrari quia venerim mittere pacem in terram non veni pacem mittere sed gladium"
Yeshua Ha Mashiach |
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| JaneBennet |
Posted: Sat Apr 12, 2008 6:15 am Post subject: |
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 Forum Junior

Joined: 06 Apr 2008 Posts: 257 Location: UK
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Here is an example. Consider the sequence xk = 1 ⁄ k, where k is a positive integer. This is the sequence 1, 1⁄2, 1⁄3, 1⁄4, …
Notice that the numbers get closer and closer to 0? In fact, by choosing large enough integers k, we can make the numbers 1 ⁄ k get as close to 0 as we like. We say that the sequence 1, 1⁄2, 1⁄3, 1⁄4, … converges to 0.
The formal definition is this: We say that a sequence of numbers xk converges to x iff given any ε > 0, there is a positive integer K such that |xk − x| < ε for all integers k > K. The number ε can be viewed as a sort of “error” margin. In other words, given any specified error margin, we can make all but finitely many numbers of the sequence xk be within that specified error margin of the limit x. _________________ “A problem worthy of attack
Proves its worth by fighting back.” – Piet Hein
Why can’t a bull see red – literally can’t? Did You Know?
Last edited by JaneBennet on Sat Apr 12, 2008 12:35 pm; edited 1 time in total |
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| serpicojr |
Posted: Sat Apr 12, 2008 11:15 am Post subject: |
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 Forum Ph.D.

Joined: 17 Jul 2007 Posts: 871 Location: JRZ
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| Okay, so what sort of math background do you have? At my school, in order to take nonlinear programming, you would have needed to take a sequence of courses including 3 or 4 semesters of calculus, a linear algebra course, and linear programming. In calculus--specifically, a second semester calculus course--you would have learned about sequences, limits, and convergence. So, as far as my understanding goes, you should have seen this material before this class. If this is the case, and if it's been a while since you've seen sequences and limits, then now is the time to refresh your memory. If you've never seen these topics... well, let us know. |
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| holysword |
Posted: Sat Apr 12, 2008 5:24 pm Post subject: |
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 Forum Freshman

Joined: 05 Apr 2008 Posts: 10
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Thank you Janet ^^
I got this part, I know about convergence. The problem is that I can't see the relationship between this and the non-linear programing.
The first thing I can conclude from that definition, is that a closed mapping is always an unlimited set - which makes no sense in this case, since we're talking about non-linear problems with their constraints...
serpicojr, I'm in mastership degree. In graduation I studied 4 calculus, and in the forth I learned convergence. As I said above, the problem is not to understand the convergence, is realise what it means in this context...
I studied 3 linear algebra ( last one I got Jordan's Canonic Form, Jordan's blocks and etc... )
I'm studyin' linear programming. I've already studied it before, but in mastership we're going more deep in the problem... _________________ "Nolite arbitrari quia venerim mittere pacem in terram non veni pacem mittere sed gladium"
Yeshua Ha Mashiach |
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| serpicojr |
Posted: Sun Apr 13, 2008 8:53 am Post subject: |
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 Forum Ph.D.

Joined: 17 Jul 2007 Posts: 871 Location: JRZ
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| So if you understand convergence, then what don't you understand about the phrase "sequence of points xk of X converging to x"? |
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| holysword |
Posted: Wed Apr 30, 2008 5:14 am Post subject: |
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 Forum Freshman

Joined: 05 Apr 2008 Posts: 10
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Hum... I think that I get this part of convergence, its likely "in a neiborhood of x"??
I think its easiest to see it in an example, don't you know some? _________________ "Nolite arbitrari quia venerim mittere pacem in terram non veni pacem mittere sed gladium"
Yeshua Ha Mashiach |
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| serpicojr |
Posted: Wed Apr 30, 2008 10:18 am Post subject: |
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 Forum Ph.D.

Joined: 17 Jul 2007 Posts: 871 Location: JRZ
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Think of a neighborhood of x as being an open interval containing x, and when something says "for all neighborhoods of x", think of open intervals of x getting smaller and smaller.
So let's consider an example. This won't be coming from a nonlinear programming example, as I don't know the subject, but I'll try to make it applied.
For any real number x, let's define A(x) to be the domain of the function f(z) = sqrt(x^2 - z^2). It shouldn't be too hard to see that the domain of this function is the closed interval [-|x|,|x|]. (I could have just defined A(x) = [-|x|,|x|], but I figured defining it the way I did shows how one might "naturally" come up with this function.)
I claim that A is a closed mapping. So, let's check the definition. The set X is just the set of real numbers, so if we let xk -> x in X, this just means xk is a sequence of real numbers converging to x in the sense that you're already familiar. Now we're suppose to let yk be an element of A(xk) = [-|xk|,|xk|], and we're assuming that the sequence yk converges to some number y. For A to be a closed mapping, we must show that y is an element of A(x).
Well, first I'd like to note that for any real numbers a and b, a is in the set [-|b|,|b|] if and only if |a| ≤ |b|. So, really, I'm trying to prove that given a sequence xk converging to x, and given a sequence yk such that |yk| ≤ |xk| for all k and such that yk converges to some point y, then |y| ≤ |x|. This is easy:
|y| = lim |yk| ≤ lim |xk| = |x|
where these limits are as k goes to infinity. So |y| ≤ |x|, and hence y is an element of A(x) = [-|x|,|x|]. Thus A is a closed mapping.
In a rough sense, this says to me: "The function f varies by some continuous parameter, and its domain varies in a 'continuous' sense." |
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