Taylor Formula for two variables

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

[tex]T_2 ( \tmmathbf{x}) = f ( \tmmathbf{a}) + \nabla f ( \tmmathbf{a}) \cdot<br /> ( \tmmathbf{x - a}) + \frac{1}{2} ( \tmmathbf{x - a}) \cdot H (<br /> \tmmathbf{a}) \cdot ( \tmmathbf{x - a})^t[/tex]

where
[tex]x=(x_1,x_2)[/tex] and
[tex]a=(a_1,a_2)[/tex]
and H is the Hessian

It was given by my professor, I understand the the first part just fine (until [tex]\frac{1}{2}[/tex]). 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 [tex]x-a[/tex]?

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.
 
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Juan Pablo said:
It was given by my professor, I understand the the first part just fine (until [tex]\frac{1}{2}[/tex]).
The 1/2 is just 1/2!.

Juan Pablo said:
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 [tex]x-a[/tex]?

The dots in the expression
[tex]( \tmmathbf{x - a}) \cdot H (<br /> \tmmathbf{a}) \cdot ( \tmmathbf{x - a})^t[/tex]​
are just matrix multiplication.
 
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.
 
H(a) is assumed refer to the Hessian of f at a. And yes, [tex](x-a)^t[/tex] 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.
 
Thanks for your help! It was really useful!
 
Juan Pablo said:
Thanks for your help! It was really useful!

You're very welcome!