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Thread: 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

    ($\displaystyle -cdF/dw_1,...,-cdF/dw_M$)

    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

    ($\displaystyle -cdF/dw_1,...,-cdF/dw_M$)

    where c is a positive constant of proportionality.

    thanks
    Because that is the gradient of a diffrenciable function.
    ---
    I can prove it for $\displaystyle f(x,y)$.
    Let $\displaystyle \nabla f_0\not = 0$ at a point.
    Let, $\displaystyle \bold{u}$ be any unit vector.
    Then, directional derivative of,
    $\displaystyle D_u f=\nabla f_0 \cdot \bold{u} = ||\nabla f_0||\cdot ||\bold{u}||\cos \theta$
    The largest possible value of,
    $\displaystyle \cos \theta =1$
    Thus, maximum is,
    $\displaystyle ||\nabla f_0||(1)$
    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,
    $\displaystyle f:\mathbb{R}^n\to \mathbb{R}$.
    At point, $\displaystyle P(x_1,x_2,...,x_n)$
    You have, $\displaystyle \nabla f\not = \bold{0}$.
    Consider, a unit vector,
    $\displaystyle \bold{u}=<u_1,...,u_n>$ and, $\displaystyle \sum_{k=1}^n u_k^2 = 1$

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