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Thread: Unconstrained Optimization

  1. #1
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    Red face Unconstrained Optimization

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

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

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

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

    $\displaystyle = 10e^{16}$

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

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

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

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

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

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

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

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

    $\displaystyle = -18e^{16}$

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

    I was taught that $\displaystyle f_{xx}$ and $\displaystyle [f_{yy}$ 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 $\displaystyle f_{xx}$ and $\displaystyle f_{yy}$ having opposite signs (-ve & +ve) for $\displaystyle f(x,y) = e^{5x^2-20x-9y^2-36y}$, I get point = (2, -2), and $\displaystyle 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 $\displaystyle \exp(f(x,y))$ is also a local maximum/minimum of $\displaystyle f(x,y)$,
    and a local maximum/minimum of $\displaystyle f(x,y)$ is also a local maximum/minimum
    of $\displaystyle \exp(f(x,y))$

    RonL
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  3. #3
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    Quote Originally Posted by ashura View Post
    I'm getting $\displaystyle f_{xx}$ and $\displaystyle f_{yy}$ having opposite signs (-ve & +ve) for $\displaystyle f(x,y) = e^{5x^2-20x-9y^2-36y}$, I get point = (2, -2), and $\displaystyle 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 $\displaystyle g(x,y) = {5x^2-20x-9y^2-36y}$, 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 = $\displaystyle f_{xx} \times f_{yy} = 10e^{16} \times (-18e^{16})$) is < 0. Can't find why $\displaystyle f_{xx}$ and $\displaystyle f_{yy}$ have different signs.

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

    $\displaystyle g_x = 10x -20$

    $\displaystyle let g_x = 0$

    $\displaystyle 10x -20 = 0$

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

    $\displaystyle g_y = -18y - 36$

    $\displaystyle let g_y = 0$

    $\displaystyle -18y - 36 = 0$

    $\displaystyle -18y = 36$

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

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

    $\displaystyle f_{yy} = \frac {\partial f_y}{\partial y} = -18$
    Last edited by ashura; Dec 2nd 2006 at 03:56 PM. Reason: correction
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  5. #5
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    It seems that $\displaystyle f_{xx}$ and $\displaystyle f_{yy}$ 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 $\displaystyle f_{xx}$ and $\displaystyle f_{yy}$ 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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