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Jim Lambers Math 105A Summer Session I 2003- Lecture 7 Examples
These examples correspond to Sections 2.5 and 2.6 in the text.
Example We wish to find the unique fixed point of the function f (x) = cos x on the interval [0, 1]. If we use Fixed-point Iteration with x 0 = 0.5, then we obtain the following iterates from the formula xk+1 = g(xk) = cos(xk). All iterates are rounded to five decimal places.
x 1 = 0. 87758 x 2 = 0. 63901 x 3 = 0. 80269 x 4 = 0. 69478 x 5 = 0. 76820 x 6 = 0. 71917.
These iterates show little sign of converging, as they are oscillating around the fixed point. If, instead, we use Fixed-point Iteration with acceleration by Aitken’s ∆^2 method, we obtain a new sequence of iterates {xˆk}, where
xˆk = xk −
(∆xk)^2 ∆^2 xk
= xk − (xk+1 − xk)^2 xk+2 − 2 xk+1 + xk
for k = 0, 1 , 2 ,.. .. The first few iterates of this sequence are
xˆ 0 = 0. 73139 xˆ 1 = 0. 73609 xˆ 2 = 0. 73765 xˆ 3 = 0. 73847 xˆ 4 = 0. 73880.
Clearly, these iterates are converging much more rapidly than Fixed-point Iteration, as they are not oscillating around the fixed point, but convergence is still linear. Finally, we try Steffensen’s Method. We begin with the first three iterates of Fixed-point Iteration, x(0) 0 = x 0 = 0. 5 , x(0) 1 = x 1 = 0. 87758 , x(0) 2 = x 2 = 0. 63901.
Then, we use the formula from Aitken’s ∆^2 Method to compute
x(1) 0 = x(0) 0 − (x(0) 1 − x(0) 0 )^2 x(0) 2 − 2 x(0) 1 + x(0) 0
We use this value to restart Fixed-point Iteration and compute two iterates, which are
x(1) 1 = cos(x(1) 0 ) = 0. 74425 , x(1) 2 = cos(x(1) 1 ) = 0. 73560.
Repeating this process, we apply the formula from Aitken’s ∆^2 Method to the iterates x(1) 0 , x(1) 1
and x(1) 2 to obtain
x(2) 0 = x(1) 0 − (x(1) 1 − x(1) 0 )^2 x(1) 2 − 2 x(1) 1 + x(1) 0
Restarting Fixed-point Iteration with x(2) 0 as the initial iterate, we obtain
x(2) 1 = cos(x(2) 0 ) = 0. 739091 , x(2) 2 = cos(x(2) 1 ) = 0. 739081.
The most recent iterate x(2) 2 is correct to five decimal places. Using all three methods to compute the fixed point to ten decimal digits of accuracy, we find that Fixed-point Iteration requires 57 iterations, so x 5 8 must be computed. Aitken’s ∆^2 Method requires us to compute 25 iterates of the modified sequence {xˆk}, which in turn requires 27 iterates of the sequence {xk}, where the first iterate x 0 is given. Steffensen’s Method requires us to compute x(3) 2 , which means that only 11 iterates need to be computed. 2
Example We apply Newton’s Method to find the roots of the polynomial
f (x) = x^3 − 9 x^2 + 26x − 24.
Choosing the initial iterate x 0 = 1, we apply Horner’s Method to compute f (x 0 ) and f ′(x 0 ), so that we can perform the step of Newton’s Method to compute x 1. We have
b 3 = 1 b 2 = −9 + (1)(1) = − 8 b 1 = 26 + (1)(−8) = 18 b 0 = −24 + (1)(18) = − 6.
We conclude that f (x 0 ) = −6, and
f (x) = −6 + (x − 1)q(x),
Figure 1: M¨uller’s Method applied to the polynomial f (x) = 16x^4 − 40 x^3 + 5x^2 + 20x + 6. The solid curve is the graph of f (x), and the dotted curves denote the quadratic polynomials generated by the method to approximate f (x) near the chosen starting iterates x 0 , x 1 and x 2. The left circle on the x-axis denotes the root x = 1.2417 and the right circle denotes the root x = 1.9704.
Figure 1 illustrates the way in which the method works. Given x 0 = 0.5, x 1 = 1 and x 2 = 1.5, the quadratic polynomial that that agrees with f (x) at these points is constructed, and the root that is closer to x 2 is computed to obtain the next iterate x 3. The quadratic polynomial is shown in Figure 1 as the dotted curve on the left. The method then repeats this process with x 1 , x 2 and x 3 to compute x 4 , and so on, until the iterates converge to the root 1.2417. If, instead, the initial iterates are chosen to be x 0 = 2.5, x 1 = 2, and x 2 = 2.25, then the quadratic polynomial that agrees with f (x) at these points is the one whose graph is the dotted curve on the right in Figure 1. The root of this polynomial that is closest to x 2 is close to the root 1 .9704 of f (x). The sequence of iterates produced by the method with the given starting values converges to this root. Finally, if x 0 = 0.5, x 1 = − 0 .5, and x 2 = 0, then the same process produces the complex root of − 0 .3561 + 0. 1628 i. Since the coefficients of f (x) are real, we can conclude that the fourth root of f (x) is the complex conjugate of this root, − 0. 3561 − 0. 1628 i. 2
