Rootfinding for nonlinear equations

Rootfinding for nonlinear equations

Here we want to solve an equation of the kind $$f(x)=0$$.

Suppose that there exist $$\alpha \in \mathbb{R}$$ such that $$f(\alpha)=0$$. Then we want to construct a sequence $$x_k$$, with $$k \in \mathbb{N}$$, such that
 * $$\lim_{k\to\infty}x_k=\alpha$$

The number $$\alpha$$ is called root (of the function $$f$$).

Convergence
If the sequence defined by the numerical method converges, we can ask what the rate of convergence is. With this aim, we define the order of convergence of a sequence :  Definition (Order of convergence). A sequence $$x_k$$ converges to $$\alpha$$ with order $$p\geq 1$$ if
 * $$|x_{k+1}-\alpha | = C|x_{k}-\alpha |^p, \quad \forall k>0,$$

$$p$$ is the order of convergence of the numerical method that has generated the sequence $$x_k$$. The constant $$C$$ is called rate of convergence. If $$p=1$$ and $$C$$ is between 0 and 1, the convergence is said to be linear convergence.Under the linear convergence condition,if $$C=0$$, the sequence is said to converge superlinearly and if $$C=1$$, then the sequence is said to converge sublinearly. The quantity
 * $$e_{k+1}\equiv |x_{k+1}-\alpha |$$

represents the error at step $$k$$. In general, with a numerical method, we do a finite number of iterations and for this reason we seek only an approximation $$x_{k+1}$$ of the true root $$\alpha$$. In particular, we can define a tolerance $$\epsilon$$ such that if $$|x_{k+1}-\alpha|\leq\epsilon$$,we can stop the iteration and conclude that $$x_{k+1}$$ is the approximation of the true root.

Example
Suppose that the sequence $$x_k$$ converges to  $$\alpha$$ with order 2, where the constant $$C=1$$, and suppose that the initial error  $$e_0 = |x_0-\alpha|=10^{-1}$$. Consider a tolerance $$\epsilon = 10^{-10}$$, then we can find the approximation of the true root by using the definition of convergence.Hence,


 * $$e_1 = |x_1-\alpha|=|x_0-\alpha |^2=10^{-2}$$

which is greater than the tolerance, so we should keep going until the error is less than the tolerance. We can get
 * $$e_2 = |x_2-\alpha|=|x_1-\alpha |^2=10^{-4}$$

which is also greater than the tolerance, thus we should do the iteration to calculate
 * $$e_3 = |x_3-\alpha|=|x_2-\alpha |^2=10^{-8}$$

which is still larger than the tolerance. Similarly,
 * $$e_4 = |x_4-\alpha|=|x_3-\alpha |^2=10^{-16}$$

which is less than the tolerance. Therefore, we can stop here and can conclude that $$x_4$$ is the approximation of the true root.

YangOu (talk) 02:44, 26 October 2012 (UTC)