How do I differentiate a vector times vector?

Quote:

Originally Posted by Twig

Hi

This came up as I was reading about Non-linear least squares fitting.

We have data points $(t_{i},y_{i}) \, i=1 \cdots \, m$ .
We wish to find the vector $\vec{x}$ of parameters that gives the best fit in the least squares sense.

So we have our model function $f(t,\vec{x}) \mbox{ , }f:\mathbb{R}^{n+1}\longrightarrow \mathbb{R}$ .

Now we define the residual function $\vec{r}:\mathbb{R}^{n}\longrightarrow \mathbb{R}^{m}$ by $r_{i}(\vec{x})=y_{i}-f(t_{i},\vec{x}) \mbox{ , }i=1 \cdots \, m$ .

Then we wish to minimize the function $\phi(\vec{x})=\frac{1}{2}\vec{r}(\vec{x})^{T}\vec{ r}(\vec{x})$ .

Here is my question, how would I get the gradient of this function?

The book says $\nabla \phi(\vec{x}) = J^{T}(\vec{x})\vec{r}(\vec{x})$ , where J denotes the Jacobian.

I don´t really follow..

thanks!

$\phi(x)=\frac{1}{2}\sum_{i=1}^n (r_i(x))^2.$ thus for any $1 \leq j \leq n$ we have $\frac{\partial \phi}{\partial x_j}=\sum_{i=1}^n r_i(x) \frac{\partial r_i}{\partial x_j},$ which is exactly $\nabla \phi(x) = J^T(x) r(x).$