Lecture 5
Fixed-Point Iteration¶
Key results¶
- \(c\) is a fixed point of \(g\) if \(g(c) = c\): where \(y = g(x)\) meets \(y = x\).
- \(f(c) = 0 \iff c\) is a fixed point of \(g(x) = x - f(x)\); iterate \(x_{n+1} = g(x_n)\).
- If \(g\) is continuous and \(g([a,b]) \subseteq [a,b]\), then \(g\) has a fixed point in \([a,b]\).
Fixed points and roots¶
A number \(c\) is a fixed point of \(g\) if \(g(c) = c\). Graphically it is where the curve \(y = g(x)\) crosses the diagonal \(y = x\).
If \(f(c) = 0\), then \(c\) is a fixed point of \(g(x) = x - f(x)\); conversely \(g(c) = c\) makes \(c\) a root of \(f(x) = x - g(x)\). Every root-finding problem can be recast as a fixed-point problem, and Newton's method is one: \(x_{n+1} = g(x_n)\) with \(g(x) = x - f(x)/f'(x)\).
The iteration and cobweb diagrams¶
Pick \(x_0\) and feed each output back in:
If \(x_n \to c\) and \(g\) is continuous, then \(c = \lim x_{n+1} = g(\lim x_n) = g(c)\): the limit is a fixed point.
A cobweb diagram draws the iteration: from \(x_n\) go vertically to the curve (height \(g(x_n) = x_{n+1}\)), then horizontally to the diagonal, and repeat. For \(g(x) = \cos x\) from \(x_0 = 1\) (radians) the values are \(0.5403,\ 0.8576,\ 0.6543,\ 0.7935, \ldots\) and they settle on \(c = 0.7390851\ldots\), the number whose cosine is itself.
Example: four rearrangements¶
\(x^3 - 2x - 5 = 0\) has a root \(r \approx 2.0945515\) in \([2,3]\). It can be written as \(x = g(x)\) in many ways, all with the same fixed point:
Starting from \(x_0 = 2\):
| \(n\) | \((2x+5)^{1/3}\) | \(\sqrt{(2x+5)/x}\) | \((2x+5)/x^2\) | \(x^3 - x - 5\) |
|---|---|---|---|---|
| \(0\) | \(2.000000\) | \(2.000000\) | \(2.0000\) | \(2\) |
| \(1\) | \(2.080084\) | \(2.121320\) | \(2.2500\) | \(1\) |
| \(2\) | \(2.092351\) | \(2.087348\) | \(1.8765\) | \(-5\) |
| \(3\) | \(2.094217\) | \(2.096517\) | \(2.4857\) | \(-125\) |
| \(4\) | \(2.094501\) | \(2.094017\) | \(1.6139\) | \(-1953005\) |
| \(5\) | \(2.094544\) | \(2.094697\) | \(3.1590\) | \(\cdots\) |
| \(6\) | \(2.094550\) | \(2.094512\) | \(1.1342\) | \(\cdots\) |
The cube-root form climbs a staircase to \(r\) (each error about \(0.15\) times the previous one); the square-root form spirals in; \((2x+5)/x^2\) spirals out; and \(x^3 - x - 5\) explodes. Near the fixed point, a curve gentler than the diagonal closes in and a steeper one flies apart. Lecture 6 makes this precise with \(g'\).
When does a fixed point exist?¶
If \(g\) is continuous on \([a,b]\) and \(g([a,b]) \subseteq [a,b]\), then \(g\) has at least one fixed point in \([a,b]\).
Proof. Let \(h(x) = g(x) - x\). Since \(g\) takes values in \([a,b]\), \(h(a) = g(a) - a \ge 0\) and \(h(b) = g(b) - b \le 0\). By the Intermediate Value Theorem \(h(c) = 0\) for some \(c \in [a,b]\), i.e. \(g(c) = c\).
For \(g(x) = (2x+5)^{1/3}\): \(g\) is increasing, \(g(2) = 9^{1/3} \approx 2.0801\) and \(g(3) = 11^{1/3} \approx 2.2240\), so \(g([2,3]) \subseteq [2,3]\) and a fixed point is guaranteed.
Algorithm¶
input g, x0, tol, maxit
for n = 0 to maxit − 1
x(n+1) ← g(xn)
if |x(n+1) − xn| < tol : return x(n+1)
end
error: no convergence within maxit