# Homework Help: Defining a Gradient

1. Feb 21, 2006

### Black Orpheus

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. Feb 21, 2006

### 0rthodontist

Start by writing out the definition for ||x - a||

3. Feb 21, 2006

### Black Orpheus

forgot to add that it's for all x not equal to a

4. Feb 21, 2006

### Black Orpheus

||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])?

5. Feb 21, 2006

### benorin

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

we have

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

which is a unit vector in the direction of x-a.

Last edited: Feb 21, 2006