Numerical Methods  

Iterative methodsThe equation $F(x)=0$ can be transformed into the form, \[ \begin{equation}\label{gl2} x=f(x). \end{equation} \]This can be used to seek the zeros of $F(x)$ by iteration. \[ \begin{eqnarray}\label{gladd} x_{0} & & \mbox{ Starting value} \nonumber\\ x_{1} & = & f(x_{0}) \nonumber\\ x_{2} & = & f(x_{1}) \nonumber\\ & . & \nonumber\\ & . & \nonumber\\ x_{t+1} & = & f(x_{t}) \qquad (t=0,1,2,\ldots) \end{eqnarray} \]If the method converges, this iteration can lead arbitrarily close to the exact solution of the problem (within roundoff error), \[ \lim_{t \to \infty} x_{t} \to x_{exact} \]There is no guarantee that this method will converge. The convergence depends strongly from the way in which $F(x)=0$ is reformulated into $x=f(x)$. One formulation is $f(x)=F(x)+x$ however other formulations can sometimes be found that converge better. Consider the function $F(x)=x^{3}x5$ in the vicinity of $x=2$. Three reformulations of this equation into a form $x=f(x)$ are, \[ a)\quad x=x^{3}5 \qquad b) \quad x=\frac{5}{x^{2}1} \qquad c) \quad x=\sqrt[3]{x+5} \]Iterating the first two forms results in a diverging series while $c)$ converges quickly to a root of $x^{3}x5=0$. A condition for convergence is $\left\frac{df}{dx}\right < 1$ at the root. Code that iterates these equations is given below.
