# Taylor Formula for two variables

1. Oct 22, 2009

### Juan Pablo

I'm attempting to understand this notation (involving the Hessian) for the quadratic Taylor series for two variable.

$$T_2 ( \tmmathbf{x}) = f ( \tmmathbf{a}) + \nabla f ( \tmmathbf{a}) \cdot ( \tmmathbf{x - a}) + \frac{1}{2} ( \tmmathbf{x - a}) \cdot H ( \tmmathbf{a}) \cdot ( \tmmathbf{x - a})^t$$

where
$$x=(x_1,x_2)$$ and
$$a=(a_1,a_2)$$
and H is the Hessian

It was given by my professor, I understand the the first part just fine (until $$\frac{1}{2}$$). I'm not sure what to do with the Hessian there. Do I take the determinant? What does the t means? Should I transpose the vector matrix of $$x-a$$?

I would like to put it in a more simple way that doesn't involve vectors so I can take the partial derivatives.

Any sort of guidance would be greatly appreciated.

2. Oct 22, 2009

### foxjwill

The 1/2 is just 1/2!.

The dots in the expression
$$( \tmmathbf{x - a}) \cdot H ( \tmmathbf{a}) \cdot ( \tmmathbf{x - a})^t$$​
are just matrix multiplication.

3. Oct 22, 2009

### Juan Pablo

Thanks for your input. I mentioned 1/2 as a delimiter of what I understand, of course I understand 1/2.

I'm not terrobly familiar with matrices. I do know multiplication, transpose and such but not much more. Doesn't the Hessian take a function as its argument? Does the superscript "t" mean I should transpose the matrix? How am I supposed to get an scalar function out of a function containing a matrix?

Sorry for all the questions, I'm really confused here.

4. Oct 23, 2009

### foxjwill

H(a) is assumed refer to the Hessian of f at a. And yes, $$(x-a)^t$$ is the transpose of (x-a).

As to getting a scalar out of this, notice that (x-a) can be thought of as a 1x2 matrix. Then we have a 1x2 matrix times a 2x2 matrix times a 2x1 matrix, which leaves a 1x1 matrix, i.e. a scalar.

5. Oct 23, 2009

### Juan Pablo

Thanks for your help! It was really useful!

6. Oct 24, 2009

### foxjwill

You're very welcome!