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Math Help - Unconstrained Optimization

  1. #1
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    Red face Unconstrained Optimization

    I'm getting f_{xx} and f_{yy} having opposite signs (-ve & +ve) for f(x,y) = e^{5x^2-20x-9y^2-36y}, I get point = (2, -2), and f_x  \left (i.e.  \frac{\partial f}{ \partial x} \right ) = (10x - 20) \cdot e^{5x^2 -20x -9y^2 -36y}

    f_{xx} = \frac{\partial (f_x)}{ \partial x}

     = 10 \cdot e^{5x^2-20x-9y^2-36y} \{10 (x - 2)^2 + 1 \}

    at point 2, -2 , f_{xx} = 10 \cdot e^{5 \cdot 4-40-36+72} \{10(2-2)^2 + 1 \}

    = 10e^{16}

    f_y (i.e. \frac{\partial f}{ \partial y})

     = (-18y -36) \cdot e^{5x^2-20x-9y^2-36y}

    f_{yy} = \frac{\partial (f_y)}{ \partial y}

    f_{yy} = -18 \{[(y + 2) \cdot e^{5x^2-20x-9y^2-36y} \times  (-18y -36)] + e^{5x^2-20x-9y^2-36y} \times 1 \}

    f_{yy} = -18 \cdot e^{5x^2-20x-9y^2-36y} \{-18(y + 2)^2 + 1 \}

    f_{yy} = -18 \cdot e^{5x^2-20x-9y^2-36y} (-18y^2 - 72y - 72 + 1 )

    f_{yy} = -18 \cdot e^{5x^2-20x-9y^2-36y} (-18y^2 - 72y - 71 )

    at point 2,-2 f_{yy} = -18 \cdot e^{5x^2-20x-9y^2-36y} (-18 \cdot 4 + 144 - 71 )

    = -18e^{16}

    so f_{xx} = 10 \cdot e^{16} and f_{yy} = -18 \cdot e^{16} This gives a -ve sign for delta.

    I was taught that f_{xx} and [f_{yy} should have the same sign but i'm not getting that here.
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  2. #2
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    Quote Originally Posted by ashura View Post
    I'm getting f_{xx} and f_{yy} having opposite signs (-ve & +ve) for f(x,y) = e^{5x^2-20x-9y^2-36y}, I get point = (2, -2), and f_x \left (i.e. \frac{\partial f}{ \partial x} \right ) = (10x - 20) \cdot e^{5x^2 -20x -9y^2 -36y}
    You might make things easier for yourself with the observation that a local
    maximum/minimum of \exp(f(x,y)) is also a local maximum/minimum of f(x,y),
    and a local maximum/minimum of f(x,y) is also a local maximum/minimum
    of \exp(f(x,y))

    RonL
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  3. #3
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    Quote Originally Posted by ashura View Post
    I'm getting f_{xx} and f_{yy} having opposite signs (-ve & +ve) for f(x,y) = e^{5x^2-20x-9y^2-36y}, I get point = (2, -2), and f_x \left (i.e. \frac{\partial f}{ \partial x} \right ) = (10x - 20) \cdot e^{5x^2 -20x -9y^2 -36y}
    It looks like (2,-2) is a saddle point to me. The attachment shows a plot of g(x,y) = {5x^2-20x-9y^2-36y}, from which you can see the saddle.

    (In the plot red corresponds to the highest and blue the lowest values)

    RonL
    Attached Thumbnails Attached Thumbnails Unconstrained Optimization-gash.jpg  
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  4. #4
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    Smile

    I also concluded that (2,-2) is a saddle point because delta(determinant = f_{xx} \times f_{yy} = 10e^{16} \times (-18e^{16})) is < 0. Can't find why f_{xx} and f_{yy} have different signs.

    g(x,y) = 5x^2 - 20x -9y^2 -36y

    g_x = 10x -20

    let g_x = 0

    10x -20 = 0

    x = \frac{20}{10} = 2

    g_y = -18y - 36

    let g_y = 0

    -18y - 36 = 0

    -18y = 36

    y = -\frac{36}{18} = -2

    f_{xx} = \frac {\partial f_x}{\partial x}  = 10

    f_{yy} = \frac {\partial f_y}{\partial y}  = -18
    Last edited by ashura; December 2nd 2006 at 03:56 PM. Reason: correction
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  5. #5
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    It seems that f_{xx} and f_{yy} don't have to be the same sign according to this site:

    Where is the max, min or saddle point in f(x,y)=x<sup>2</sup>-y<sup>2</sup>?

    So my solution could be correct.
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  6. #6
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    Quote Originally Posted by ashura View Post
    It seems that f_{xx} and f_{yy} don't have to be the same sign according to this site:

    Where is the max, min or saddle point in f(x,y)=x<sup>2</sup>-y<sup>2</sup>?

    So my solution could be correct.
    If you look at the plot of the function you will see that a section through
    the plot along the line y=-2, that the function has a minimum so the first
    derivative wrt x, is zero at x=2, and the second derivative is positive.

    If you look at the plot of the function you will see that a section through
    the plot along the line x=2, that the function has a minimum so the first
    derivative wrt y, is zero at y=-2, and the second derivative is negative.

    That the first partials at a point are zero and the second have oposite signs
    is guarantees that the point is a saddle point.

    RonL
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  7. #7
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    Wink

    Thanks for the explanation, where can I get the grafting software?
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  8. #8
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    Quote Originally Posted by ashura View Post
    Thanks for the explanation, where can I get the grafting software?
    You can get 2-D graphing software for free.
    Free Software Downloads and Software Reviews - Download.com
    (Spyware Free)

    But I do not know if there are 3-D graphing software for free. They usually are expensive.
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  9. #9
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    Quote Originally Posted by ashura View Post
    Thanks for the explanation, where can I get the grafting software?
    What I use is Euler, which is a data analysis language, which is probably too
    complex for your needs.

    RonL
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