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Math Help - Nonlinear programming

  1. #1
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    Nonlinear programming

    I am trying to do a non-linear programming work. We have two functions: f(x)=x^2/2+e^{-x} and g(x) = x^4-32x^2. I just want to know how exactly one can get the minimum of x, x*. For example using the newton method:

    Code:
    k    ak          F'(x)    F''(X)       k     ak            g'(x)     g''(x)
    1    2,0000    1,8647    1,1353         1    0,2000    -12,7680    -63,5200
    2    0,3576    -0,3417    1,6993        2    -0,0010    0,0645    -64,0000
    3    0,5587    -0,0132    1,5719        3    0,0000    0,0000    -64,0000
    4    0,5671    0,0000    1,5672                    
    5    0,5671    0,0000    1,5671
    So I caluclate and get this but then how can I determine the minimum of x, x*?
    Last edited by Keep; December 12th 2010 at 12:46 PM.
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  2. #2
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    Well the minimum for the function f(x) will be where f'(x)= 0 and f''(x*)>0.

    Same goes for g(x)
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    OK I understand f'(x) but what did you mean by f"(x*) or you meant f"(x)? Because I am looking for the minimum, x*, so how can I find f"(x*) when I don't even have x*? If you mean the minimum, x*, is where f'(x) = 0 and f"(x)>0, then yeah I understand. So that would be x* = 0.5671 for f(x) but then I don't know how I can get x* for g(x) because although we have a place where f'(x) = 0, we have no where where f"(x)>0?

    Thanks in advance.
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    OK I read a some stuff on the internet and I found out that the minimum is where f'(x) = 0 and f''(x)>0. If f''(x)<0, then that is a maximum point but in the question we were asked to find out the minimum, not the maximum so I am confused. When f'(x) = 0, f"(x)<0 so I am wondering how I can find the min. for this equation.
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  5. #5
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    Quote Originally Posted by Keep View Post
    OK I understand f'(x) but what did you mean by f"(x*) or you meant f"(x)? Because I am looking for the minimum, x*, so how can I find f"(x*) when I don't even have x*? If you mean the minimum, x*, is where f'(x) = 0 and f"(x)>0, then yeah I understand. So that would be x* = 0.5671 for f(x) but then I don't know how I can get x* for g(x) because although we have a place where f'(x) = 0, we have no where where f"(x)>0?

    Thanks in advance.
    x* is the solution to f'(x)=0.
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  6. #6
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    Quote Originally Posted by Keep View Post
    OK I read a some stuff on the internet and I found out that the minimum is where f'(x) = 0 and f''(x)>0. If f''(x)<0, then that is a maximum point but in the question we were asked to find out the minimum, not the maximum so I am confused. When f'(x) = 0, f"(x)<0 so I am wondering how I can find the min. for this equation.
    g(x) = x^4-32x^2

    g'(x)= 4x^3-64x

    making g'(x)=0 to solve

    0 = 4x^3-64x

    0 = 4x(x^2-16) = 4x(x-4)(x+4)

    x^* = 0,-4,4

    Now find g(x^*)
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  7. #7
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    OK thanks. Substituting we get: for x* = 0, g(x*) = 0, for x* = +/-4, g(x)= -256. So out of those three values for x* which one is the minimum?
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