Develop an M-file to locate a minimum with the golden-section search. Rather than using th...

Develop an M-file to locate a minimum with the golden-section search. Rather than using the maximum iterations and Eq. (7.9) as the stopping criteria, determine the number of iterations needed to attain a desired tolerance. Test your function by solving Example 7.2 using E_a,d= 0.0001.

Equation (7.9):

Example 7.2:

Golden-Section Search

Problem Statement. Use the golden-section search to find the minimum of

within the interval from x_l = 0 to x_u = 4.

Solution. First, the golden ratio is used to create the two interior points:

The function can be evaluated at the interior points:

Because f (x₂) < f (x₁), our best estimate of the minimum at this point is that it is located at x = 1.5279 with a value of f (x) = –1.7647. In addition, we also know that the minimum is in the interval defined by x_l, x₂, and x₁. Thus, for the next iteration, the lower bound remains x_l = 0, and x₁ becomes the upper bound, that is, x_u = 2.4721. In addition, the former x₂ value becomes the new x₁, that is, x₁ = 1.5279. In addition, we do not have to recalculate f (x₁), it was determined on the previous iteration as f (1.5279) = –1.7647.

All that remains is to use Eqs. (7.8) and (7.7) to compute the new value of d and x₂:

The function evaluation at x₂ is f (0.9943) = −1.5310. Since this value is less than the function value at x₁, the minimum is f (1.5279) = −1.7647, and it is in the interval prescribed by x₂, x₁, and x_u. The process can be repeated, with the results tabulated here:

Note that the current minimum is highlighted for every iteration. After the eighth iteration, the minimum occurs at x = 1.4427 with a function value of −1.7755. Thus, the result is converging on the true value of −1.7757 at x = 1.4276.

Equation (7.8):

Equation (7.7):