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[edit] Newton-Raphson Method and Hessian Matrix

NOTE TO SELF: THERE WAS A TYPO IN THE HESSIAN MATRIX IN THE BOOK. THE INVERSE OF THE HESSIAN IS CORRECT, BUT THE HESSIAN ITSELF WAS NOT.

THE HESSIAN NOW IS CORRECT BELOW.


I posted a question recently about the Hessian Matrix on the Math Talk page and received some good replies. I posted a follow-up question later in the Math and Science Question section, but then found the error in the book myself. Below deals with using the Hessian Matrix and Newton-Raphson Method to locate the minima or maxima of a function. The example in REA's Problem Solvers book "Operations Research", p. 739-740, contains an error in the Hessian Matrix.

The function in the example contains two unknowns, x1 and x2, and is:

4x12 + 2x1x2 + 2x22 + x1 + x2.


The problem is to find the values of x1 and x2 that provide the minima.

The matrix of partial derivatives is: f(x) = \begin{bmatrix}8x1 + 2x2 + 1 \\2x1 + 4x2 + 1\end{bmatrix} = 0.

The Hessian Matrix is:

H(x) = \begin{bmatrix}8&2\\2&4\end{bmatrix};

Not:

H(x) = \begin{bmatrix}8&2\\8&4\end{bmatrix} as in the book.

The third matrix is indeed the inverse of the Hessian Matrix, and is given correctly in the book as:

H-1(x) = (1/14) * \begin{bmatrix} 2&-1\\-1&4\end{bmatrix}.


The interative formula used to find the values of x1 and x2 corresponding to the minima of the original function is:

x(n+1) = x(n) - H-1(x) * f(x);


and gives the values x(n+1)1 = -1/14 and x(n+1)2 = -3/14.


I understand, at least mechanically, the substitutions and algebra used to arrive at those answers given the formula for x(n+1).

Note that:

H(x) * H-1(x) = \begin{bmatrix} 1&0\\0&1\end{bmatrix}.


Note that the eigenvalues of the Hessian Matrix are 8.8284271 and 3.1715729, which I think corresponds to a minima, which is what the REA "Operations Research" book said was the case with this particular function.


-- Mark W. Miller 06:49, 27 November 2005 (UTC)

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