Taylor Formula for two variables

Juan Pablo
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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<br /> ( \tmmathbf{x - a}) + \frac{1}{2} ( \tmmathbf{x - a}) \cdot H (<br /> \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.
 
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Juan Pablo said:
It was given by my professor, I understand the the first part just fine (until \frac{1}{2}).
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 x-a?

The dots in the expression
( \tmmathbf{x - a}) \cdot H (<br /> \tmmathbf{a}) \cdot ( \tmmathbf{x - a})^t​
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, (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.
 
Thanks for your help! It was really useful!
 
Juan Pablo said:
Thanks for your help! It was really useful!

You're very welcome!
 
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