1. PF Insights is off to a great start! Fresh and interesting articles on all things science and math. Here: PF Insights

Defining a Gradient

  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)?
  2. jcsd
  3. 0rthodontist

    0rthodontist 1,253
    Science Advisor

    Start by writing out the definition for ||x - a||
  4. forgot to add that it's for all x not equal to a
  5. ||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. benorin

    benorin 1,025
    Homework Helper

    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
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook

Have something to add?