In a recent post, I started a series of posts about finding roots of equations, and how you might use MATLAB to teach some of this topic. In particular, I will introduce the idea of fixed point iteration.
clear all close all
Let me restate the example function. We'll start with a simple cubic polynomial f
which I define as an anonymous function.
f = @(x) x.^3 + x - 1; fplot(f,[-2 2]) title f grid on
One way to find a solution to f(x) = 0 is to define another function, g such that g(x) = x where the function g is related to f. If we can find a value x that solves the equation with g, this fixed point is also the zero of the function f. How do we define such a suitable g? Looking at the equation for f and setting its value to zero, we can derive an equation for x.
Here's one way we can rewrite f(x) = 0 so that x is alone on one side of the equation.
Let's call this function g1.
g1 = @(x) 1 - x.^3; fplot(g1,[-2 2]); g1str = '1-x^3'; title g1 grid on
Now let's add a straight line to represent
straightLine = @(x) x; title '' hold on fplot(straightLine, [-2 2], 'g') legend(g1str,'x','Location','SouthEast') axis([-2 2 -2 2]), axis equal
The intersection of these 2 curves is a fixed point of g1(x), and therefore a zero of f(x). Let's zoom in a bit.
axis([0 1 0 1])
That intersection looks like it happens around the same location as the answers we got from using roots and fzero in the previous post.
fzsolution = fzero(f,0.5)
fzsolution = 0.68233
First, we'll select the same starting value that we used when we called fzero. The starting point is then (0.5,0.5). And then we can easily calculate the next value of g1.
x(1) = 0.5; y(1) = x(1); x(2) = x(1); y(2) = g1(x(2));
Let's plot the trajectory of the solutions, noting the first point with a red star.
Now let's get the next estimate. We'll set the current value of g1 to the new x value, sliding onto the straight line, and then calculate the next value of g1 from this new value for x. And plot it.
x(3) = y(2); y(3) = y(2); x(4) = x(3); y(4) = g1(x(4)); plot(x,y,'r')
Let's do the same set of steps again.
n = 5; x(n) = y(n-1); y(n) = y(n-1); x(n+1) = x(n); y(n+1) = g1(x(n+1)); plot(x,y,'r')
n = 7; x(n) = y(n-1); y(n) = y(n-1); x(n+1) = x(n); y(n+1) = g1(x(n+1)); plot(x,y,'r') hold off
By now, you have surely noticed that instead of spiraling into the solution we found with fzero, our estimates are spiraling out! In the next post, we'll explore another definition for g(x) = x and see what happens from there.
Yes, I know in this post I kept overplotting lines successively. But I didn't want us to get distracted by managing the plots.
In addition to this post and the previous one, there will be one more. Until it is published, the 3rd link will not be useful.
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