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Math Help - steepest descent

  1. #1
    AMR
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    steepest descent

    how can we proof this :
    direction of steepest descent of a differentiable function F of M variables is the vector

    ( -cdF/dw_1,...,-cdF/dw_M)

    where c is a positive constant of proportionality.

    thanks
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  2. #2
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    Quote Originally Posted by AMR
    how can we proof this :
    direction of steepest descent of a differentiable function F of M variables is the vector

    ( -cdF/dw_1,...,-cdF/dw_M)

    where c is a positive constant of proportionality.

    thanks
    Because that is the gradient of a diffrenciable function.
    ---
    I can prove it for f(x,y).
    Let \nabla f_0\not = 0 at a point.
    Let, \bold{u} be any unit vector.
    Then, directional derivative of,
    D_u f=\nabla f_0 \cdot \bold{u} = ||\nabla f_0||\cdot ||\bold{u}||\cos \theta
    The largest possible value of,
    \cos \theta =1
    Thus, maximum is,
    ||\nabla f_0||(1)
    Thus, it is in the in the direction of the gradient. Thus, it is a multiple of the vector, hence the constant of proportionality.
    ---
    I am thinking about several variables.
    I guess the proof is based on finding the critical points. Since the critical points are determined by partial derivatives they appear over here. I will think more about it.
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  3. #3
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    You have a function,
    f:\mathbb{R}^n\to \mathbb{R}.
    At point, P(x_1,x_2,...,x_n)
    You have, \nabla f\not = \bold{0}.
    Consider, a unit vector,
    \bold{u}=<u_1,...,u_n> and, \sum_{k=1}^n u_k^2 = 1

    You need to find, what the components of \bold{u} such as, D_u f (directional derivative along this vector) is extremized.
    But,
    D_u f =\nabla f \cdot \bold{u}
    Thus, you want to maximize,
    g(u_1,...,u_n)=\nabla f\cdot \bold{u}
    Thus,
    g(u_1,...,u_n)=\sum_{k=1}^n \frac{\partial f}{\partial x_k}u_k
    With constraint curve,
    c(u_1,...u_n)=\sum_{k=1}^n u_k^2=1
    Thus, Lagrange Multipliers,
    \nabla g=\kappa \nabla c
    Thus,
    \left< \frac{\partial f}{\partial x_1},...,\frac{\partial f}{\partial x_n} \right>=\left<2\kappa u_1,...,2\kappa u_n\right>
    Since,
    \nabla f\not = \bold{0} thus, \kappa \not = 0
    Thus,
    \bold{u}=\frac{1}{2\kappa}\left< \frac{\partial f}{\partial x_1},...,\frac{\partial f}{\partial x_n} \right>=\frac{1}{2\kappa} \nabla f
    Thus, we see that every extremized directional derivative is a multiple of the gradient of the function.

    ---
    Note, I used Lagrange Multipliers. The problem might be that the prove of Lagrange Multipliers for multi-variables more than 2 is based on the fact of the max/min property of the gradient. Thus, my proof was not a proof because it uses a fact that depends on itself.
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  4. #4
    AMR
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    thanks alot
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