Answer:
Explanation:
Newton's method is an iterative method for finding the roots of a differentiable function. Let's consider finding a root of a function f(x) using Newton's method:
x_{n+1} = x_n - f(x_n)/f'(x_n)
We can write this iteration as a fixed point iteration:
x_{n+1} = g(x_n)
where g(x_n) = x_n - f(x_n)/f'(x_n)
To show that Newton's method converges when written as a fixed point iteration, we need to show that g(x) is a contraction mapping in some neighborhood of the root.
First, let's find the derivative of g(x):
g'(x) = (f(x)f''(x))/(f'(x))^2
Since we are interested in finding the roots of f(x), let's assume that f(x) has a root at r. We can then write:
f(x) = f(r) + f'(r)(x-r) + O((x-r)^2)
Substituting this into the formula for g(x), we get:
g(x) = x - [f(r) + f'(r)(x-r) + O((x-r)^2)]/[f'(r) + f''(r)(x-r) + O((x-r)^2)]
We can simplify this expression by dropping the higher order terms and assuming that x is close to r:
g(x) = x - [f(r) + f'(r)(x-r)]/[f'(r) + f''(r)(x-r)]
= (xf'(r) - f(r))/[f'(r) + f''(r)(x-r)]
Now, let's consider the behavior of g'(x) in a neighborhood of r:
g'(x) = (f(x)f''(x))/(f'(x))^2
= [f(r) + f'(r)(x-r) + O((x-r)^2)][f''(r) + O(x-r)]/([f'(r) + f''(r)(x-r) + O((x-r)^2)])^2
= [f''(r)f(r) + O(x-r)]/[f'(r) + f''(r)(x-r) + O((x-r)^2)]^2
Since f(r) = 0, we can simplify this expression further:
g'(x) = f''(r)/(f'(r))^2 + O(1/(x-r))
As x approaches r, the term O(1/(x-r)) approaches zero, and we are left with:
g'(r) = f''(r)/(f'(r))^2
If g'(r) is less than one in absolute value, then g(x) is a contraction mapping in a neighborhood of r, and Newton's method will converge to r.
Let's consider two cases:
g'(r) = 0
If g'(r) = 0, then g(x) is not a contraction mapping, and Newton's method may not converge to r.
g'(r) ≠ 0
If g'(r) ≠ 0, then we can rewrite g(x) as:
g(x) = r + [g(x) - r]
= r + [(x-r) - f(x)/f'(x)]/[1 + g'(r)(x-r)/f'(r)]
Now, if we choose an initial guess x_0 sufficiently close to r, we can ensure that |g'(r)(x_0-r)/f'(r)| is less than one, and therefore g(x) will be a contraction mapping in a neighborhood of r. Therefore, Newton's method will converge to r in this case.
In summary, we have shown that if g(x) = x - f(x)/f'(x), then Newton's method can be written as a fixed point iteration x_{n+1} = g(x_n), and under certain conditions, g(x) is a contraction mapping in a neighborhood of the root, which implies convergence of the iteration to the root. Specifically, if g'(r) is less than one in absolute value, where r is the root of f(x), then Newton's method will converge to r.