Section 9.1 Newton's method
The idea of Newton's method is geometrically natural. We want to solve f(x)=0 but since f may be complicated, we replace it by its linear approximation (tangent line at a certain point). The resulting linear equation is easy to solve. Of course, its solution is probably not a solution of the original equation, but perhaps it is close to it. The process repeats, using the previously found solution as a new point at which to draw a tangent line. A static image of this process is below. Interactive visualization is also available.
g(x)=xβf(x)fβ²(x)=xfβ²(x)βf(x)fβ²(x)
Note that any point where f(x)=0 and fβ²(x)β 0 is a fixed point of g. In fact, it is a super-attracting point of g because
gβ²(x)=1βfβ²(x)fβ²(x)βf(x)fβ³(x)(fβ²(x))2=f(x)fβ³(x)(fβ²(x))2
which is zero when f(x)=0.
Example 9.1.2. Find and simplify the function g of Newton's method.
Given \(f(x) = x^3 - x\text{,}\) find and simplify the function \(g(x) = x - f(x)/f'(x)\text{.}\)
Solution
Bringing \(g\) to common denominator often helps:
\begin{equation*}
g(x) = \frac{xf'(x) -f(x)}{f'(x)} = \frac{x(3x^2-1) - (x^3-x)}{3x^2-1} = \frac{2x^3}{3x^2-1}
\end{equation*}
g = @(x) 2*x^3/(3*x^2-1); x0 = 0.6;we get βFound x = 1 after 11 stepsβ. If the starting point is different, the process may converge to a different root of the equation x3βx=0. For example with
x0 = 0.57
we get βFound x = -1 after 14 stepsβ. A small change of initial condition produced a rather different solution; we went from finding x = 1 to finding x = -1. The pattern of what starting points produce which solution can be very complicated: see the illustrations on Wikipedia page Newton fractal.