Taylor Formula for two variables

In summary, the Hessian is a matrix that takes a function as an argument. The superscript "t" means transpose, and the dots in the expression (\tmmathbf{x - a}) \cdot H (\tmmathbf{a}) \cdot (\tmmathbf{x - a})^t represent matrix multiplication.
  • #1
Juan Pablo
40
0
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
( \tmmathbf{x - a}) + \frac{1}{2} ( \tmmathbf{x - a}) \cdot H (
\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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  • #2
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 (
\tmmathbf{a}) \cdot ( \tmmathbf{x - a})^t[/tex]​
are just matrix multiplication.
 
  • #3
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
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.
 
  • #5
Thanks for your help! It was really useful!
 
  • #6
Juan Pablo said:
Thanks for your help! It was really useful!

You're very welcome!
 

1. What is the Taylor Formula for two variables?

The Taylor Formula for two variables is a mathematical formula used to approximate a function at a specific point by using the function's derivatives at that point. It is an extension of the Taylor Formula for one variable, which is used to approximate a function at a single point.

2. How is the Taylor Formula for two variables calculated?

The Taylor Formula for two variables is calculated by taking the function's partial derivatives with respect to each variable at the specific point, as well as the mixed partial derivatives. These derivatives are then evaluated at the point and used in the formula to approximate the function.

3. What is the purpose of the Taylor Formula for two variables?

The purpose of the Taylor Formula for two variables is to provide a more accurate approximation of a function at a specific point than just using the function's value at that point. It takes into account the behavior of the function in multiple directions, rather than just one variable.

4. Can the Taylor Formula for two variables be used for any function?

Yes, the Taylor Formula for two variables can be used for any function that is differentiable at the specific point. However, the accuracy of the approximation may vary depending on the properties of the function and the number of derivatives used in the calculation.

5. What are the limitations of the Taylor Formula for two variables?

The Taylor Formula for two variables can only provide an approximation of a function at a specific point. It does not take into account the behavior of the function outside of that point, and the accuracy of the approximation decreases as the distance from the point increases. Additionally, the calculation can become more complex and time-consuming for functions with multiple variables and higher order derivatives.

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