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Thread: Closed Mapping - Non-Linear Programming

  1. #1 Closed Mapping - Non-Linear Programming 
    Forum Freshman holysword's Avatar
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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:

    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.


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  3. #2  
    Forum Professor serpicojr's Avatar
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    So this is saying, given a point x in X, take any sequence of points x<sub>k</sub> of X converging to x. Now take any sequence of points y<sub>k</sub>, each y<sub>k</sub> being in the set A(x<sub>k</sub>), 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(x<sub>k</sub>) "converge" to the set A(x). (Nothing here guarantees that there exists a sequence y<sub>k</sub> that actually converges.)


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  4. #3  
    Forum Freshman holysword's Avatar
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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 ><"
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    Here is an example. Consider the sequence x<sub>k</sub> = 1 ⁄ k, where k is a positive integer. This is the sequence 1, <sup>1</sup>⁄<sub>2</sub>, <sup>1</sup>⁄<sub>3</sub>, <sup>1</sup>⁄<sub>4</sub>, …

    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, <sup>1</sup>⁄<sub>2</sub>, <sup>1</sup>⁄<sub>3</sub>, <sup>1</sup>⁄<sub>4</sub>, … converges to 0.

    The formal definition is this: We say that a sequence of numbers x<sub>k</sub> converges to x iff given any ε > 0, there is a positive integer K such that |x<sub>k</sub>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 x<sub>k</sub> be within that specified error margin of the limit x.
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  6. #5  
    Forum Professor serpicojr's Avatar
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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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    Forum Freshman holysword's Avatar
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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...
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  8. #7  
    Forum Professor serpicojr's Avatar
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    So if you understand convergence, then what don't you understand about the phrase "sequence of points x<sub>k</sub> of X converging to x"?
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  9. #8  
    Forum Freshman holysword's Avatar
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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?
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  10. #9  
    Forum Professor serpicojr's Avatar
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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 x<sub>k</sub> -> x in X, this just means x<sub>k</sub> is a sequence of real numbers converging to x in the sense that you're already familiar. Now we're suppose to let y<sub>k</sub> be an element of A(x<sub>k</sub>) = [-|x<sub>k</sub>|,|x<sub>k</sub>|], and we're assuming that the sequence y<sub>k</sub> 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 x<sub>k</sub> converging to x, and given a sequence y<sub>k</sub> such that |y<sub>k</sub>| ≤ |x<sub>k</sub>| for all k and such that y<sub>k</sub> converges to some point y, then |y| ≤ |x|. This is easy:

    |y| = lim |y<sub>k</sub>| ≤ lim |x<sub>k</sub>| = |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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