# Derivative of a function involving square root of sum of squares

1. Oct 11, 2012

### onako

Provided is a function $$f(x)=\sum_{j=1}^n ||x-x_j||$$, for x being a two dimensional vector, where ||.|| denotes the Euclidean distance in 2D space. How could one obtain a derivative of such a function?

2. Oct 11, 2012

### Einj

If you are just looking for a mechanical derivation I think you can do in this way. If $||\cdot||$ is the Euclidean norm then:

$$||x-x_j||=\sqrt{\sum_{k=1}^N{{(x-x_j)}_k^2}}$$

where N is the dimension of the Euclidean space. So:

$$\frac{d}{dx}||x-x_j||=\frac{1}{2\sqrt{\sum_{k=1}^N{{(x-x_j)}_k^2}}}\sum_{k=1}^N{2{(x-x_j)}_k}=\frac{\sum_{k=1}^N{{(x-x_j)}_k}}{||x-x_j||}$$

and so:

$$\frac{df}{dx}=\sum_{j=1}^n{\frac{\sum_{k=1}^N{{(x-x_j)}_k}}{||x-x_j||}}$$

3. Oct 12, 2012

### onako

Thanks. Now, faced with the problem of minimizing f(x) for provided 2D parameters x1, x2, x3, ..., x_k, one sets the derivative to zero, and computes for x. However, in case of more than one dimension this problem is non-trivial, I think. What would be the minimizer of f(x), provided 2D parameters x1, x2, x3, ..., x_k?

4. Oct 12, 2012

### haruspex

That it will be messy can be seen by considering just 3 points in 2 dimensions. If any pair subtends an angle > 120 degrees at the third then the answer will be that third point. Otherwise, it is the point at which each pair subtends that angle.

5. Oct 12, 2012

### hotvette

Won't the minimizer be the same if you don't take the square root?

6. Oct 12, 2012

### onako

It then means you're squaring each term, and not the function itself. If a function is squared, then these would be equivalent.

Given a set of points in 2D, a point that minimizes the sum of squared distances to such points is the barycenter; I'm not sure about the sum of distances (so, not squared).

7. Oct 12, 2012

### hotvette

All I'm saying is that I believe

$$f(x)=\sqrt{g(x)^2}$$

has the same minimizer as

$$f(x)^2 = g(x)^2$$

I remember from basic calculus that minimizing the distance from a point to a curve is the same as minimizing the distance squared, which is a lot easier to deal with. I think that's also why least squares problems are specifically formulated the way they are. Minimizing the sum of squares is a whole lot easier than minimizing the square root of the sum of squares, and yields the same answer.