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Math Help - Gradient Vector max/min

  1. #1
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    Gradient Vector max/min

    I need to show that a differentiable function f(x_1,x_2...x_n) decreases most rapidly at X in the direction opposite to the gradient vector, that is, in the direction of -\bigtriangledown f(X)

    I begin by writing the derivative as a dot product:

    D_u f(X)=u\cdot\bigtriangledown f(X)

    =\mid u\bigtriangledown f(X)\mid \cos\theta_{u,f}


    \cos\theta_{u,f} is smallest when \theta_{u,f}=\pi+2n\pi i.e when \cos\theta_{u,f}=-1. The derivative D_u f(X) becomes -\mid\bigtriangledown f(X)\mid=-\bigtriangledown f(X).

    Have I shown that the minimum of a directional derivative at a point X is the negative of the gradient? Is this what they're asking for?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by adkinsjr View Post
    I need to show that a differentiable function f(x_1,x_2...x_n) decreases most rapidly at X in the direction opposite to the gradient vector, that is, in the direction of -\bigtriangledown f(X)

    I begin by writing the derivative as a dot product:

    D_u f(X)=u\cdot\bigtriangledown f(X)

    =\mid u\bigtriangledown f(X)\mid \cos\theta_{u,f}


    \cos\theta_{u,f} is smallest when \theta_{u,f}=\pi+2n\pi i.e when \cos\theta_{u,f}=-1. The derivative D_u f(X) becomes -\mid\bigtriangledown f(X)\mid=-\bigtriangledown f(X).

    Have I shown that the minimum of a directional derivative at a point X is the negative of the gradient? Is this what they're asking for?
    u is a unit vector,

    D_u f(x)=u. \nabla f (X)=|\nabla f(X)| \cos(\theta_{u,f})

    which has a minimum when \cos(\theta_{u,f})=-1 which is when u and \nabla f(X) point in oposite directions.

    CB
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