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Defining a Gradient

by Black Orpheus
Tags: defining, gradient
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Black Orpheus
#1
Feb21-06, 08:13 PM
P: 23
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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0rthodontist
#2
Feb21-06, 08:18 PM
Sci Advisor
P: 1,253
Start by writing out the definition for ||x - a||
Black Orpheus
#3
Feb21-06, 08:21 PM
P: 23
forgot to add that it's for all x not equal to a

Black Orpheus
#4
Feb21-06, 08:31 PM
P: 23
Defining a Gradient

||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])?
benorin
#5
Feb21-06, 08:45 PM
HW Helper
P: 1,025
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.


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