# Newton Method Direction

• Apr 28th 2010, 12:59 PM
lpd
Newton Method Direction
I'm a bit stuck on this problem:

Consider the function http://www.sosmath.com/CBB/latexrend...035a5858ec.gif.

a) Verify that the Newton direction at (x,y) = (0,0) exists and is a descent direction.

This part...do i just find the negative of the inverse of the Hessen with points (0,0) substitued in and multiplied by the gradient fuction and I got a result. [0.5 -1.5]^T

Which exist because there's a value for it?

b) Evauate the eigenvalues of Hessen*f(0,0) and show that this matrix is not positive definite. Comment on the relationship between this result and your answer in a). Does this contradict to the Lemma (conditions for the Newton Direction) if the grad of f(x^k) and hessen of f(x ^k) is postive definite, then hessen of f(x^k) is invertible and the newton direction is a descent direction for f at x^k.

well, i went and showed that the eigenvalues of Hessen of f(0,0) = positive or negative sqrt of 8. So since the eigen values are not both postive, so therefore the matrix is not positive definite. Rather its neither a negative or postive definite matrix.

But I don't know how to explain the conditions for the Newton Direction. Like, it was possible to find a descent direction. But maybe it only exist for certain values?

Any help will be appriciated!!

Thanks.
• Apr 28th 2010, 01:53 PM
Quote:

Originally Posted by lpd
I'm a bit stuck on this problem:

Consider the function http://www.sosmath.com/CBB/latexrend...035a5858ec.gif.

a) Verify that the Newton direction at (x,y) = (0,0) exists and is a descent direction.

This part...do i just find the negative of the inverse of the Hessen with points (0,0) substitued in and multiplied by the gradient fuction and I got a result. [0.5 -1.5]^T

Which exist because there's a value for it?

b) Evauate the eigenvalues of Hessen*f(0,0) and show that this matrix is not positive definite. Comment on the relationship between this result and your answer in a). Does this contradict to the Lemma (conditions for the Newton Direction) if the grad of f(x^k) and hessen of f(x ^k) is postive definite, then hessen of f(x^k) is invertible and the newton direction is a descent direction for f at x^k.

well, i went and showed that the eigenvalues of Hessen of f(0,0) = positive or negative sqrt of 8. So since the eigen values are not both postive, so therefore the matrix is not positive definite. Rather its neither a negative or postive definite matrix.

But I don't know how to explain the conditions for the Newton Direction. Like, it was possible to find a descent direction. But maybe it only exist for certain values?

Any help will be appriciated!!

Thanks.

1) To show it's a descent direction you have to show that $f'(x^k) s^k$ < 0. (Note it should be an upside down $\Delta$ in front of the $f(x^k)$ but I don't know the latex for it so just used a ')

2) I'm not 100% but I believe the direction will always be descent if the Hessian is +ve definite but MAY be descent if the Hessian is not.
As in, Hessian +ve definite => Descent but...

=> Descent does not always imply positive definite
• Apr 28th 2010, 02:36 PM
(Note it should be an upside down $\Delta$ in front of the $f(x^k)$ but I don't know the latex for it so just used a ')
\nabla $\nabla$