Defining a Gradient

  1. Feb 21, 2006 #1
    One last question for tonight... If you let f: R^n ---> R (Euclidean n-space to real numbers) and f(x) = ||x-a|| for some fixed a, how would you define the gradient in terms of symbols and numbers (not words)?
     
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  3. Feb 21, 2006 #2

    0rthodontist

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    Start by writing out the definition for ||x - a||
     
  4. Feb 21, 2006 #3
    forgot to add that it's for all x not equal to a
     
  5. Feb 21, 2006 #4
    ||x-a|| = sqrt[(xsub1 - a)^2 +...+ (xsubn - a)^2]

    so gradient = (partial derivative of sqrt[(xsub1 - a)^2] , ... , partial derivative of sqrt[(xsubn - a)^2])?
     
  6. Feb 21, 2006 #5

    benorin

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    For [tex]f(x)=\| x-a\| = \sqrt{(x_1-a_1)^2+\cdots +(x_n-a_n)^2}[/tex]

    we have

    [tex]\nabla f(x) = \left< \frac{\partial }{\partial x_1},\ldots , \frac{\partial }{\partial x_n}\right>\cdot \sqrt{(x_1-a_1)^2+\cdots +(x_n-a_n)^2} [/tex]
    [tex]\left< \frac{\partial }{\partial x_1}\sqrt{(x_1-a_1)^2+\cdots +(x_n-a_n)^2},\ldots , \frac{\partial }{\partial x_n}\sqrt{(x_1-a_1)^2+\cdots +(x_n-a_n)^2}\right> [/tex]
    [tex]= \left< \frac{x_1-a_1 }{\sqrt{(x_1-a_1)^2+\cdots +(x_n-a_n)^2}},\ldots , \frac{x_n-a_n }{\sqrt{(x_1-a_1)^2+\cdots +(x_n-a_n)^2}}\right> [/tex]
    [tex]= \frac{1}{\sqrt{(x_1-a_1)^2+\cdots +(x_n-a_n)^2}}\left< x_1-a_1 ,\ldots , x_n-a_n \right> [/tex]
    [tex]= \frac{x-a}{\| x-a\| }[/tex]

    which is a unit vector in the direction of x-a.
     
    Last edited: Feb 21, 2006
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