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Taylor Formula for two variables

  1. Oct 22, 2009 #1
    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.
     
  2. jcsd
  3. Oct 22, 2009 #2
    The 1/2 is just 1/2!.

    The dots in the expression
    [tex]( \tmmathbf{x - a}) \cdot H (
    \tmmathbf{a}) \cdot ( \tmmathbf{x - a})^t[/tex]​
    are just matrix multiplication.
     
  4. Oct 22, 2009 #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.
     
  5. Oct 23, 2009 #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.
     
  6. Oct 23, 2009 #5
    Thanks for your help! It was really useful!
     
  7. Oct 24, 2009 #6
    You're very welcome!
     
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