# 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)