Rearranging equations into iteration form

By the end you can rearrange an equation $f(x) = 0$ into the form $x = F(x)$ , link a given iteration back to the equation it solves, use it to find a root to a stated accuracy, and recognise when an iteration fails.

Take the equation $x^3 - 2x - 5 = 0$ . Rearrange it so that $x$ is on its own on the left:

$x^3 = 2x + 5 \quad\Longrightarrow\quad x = \sqrt[3]{2x + 5}.$

This is still the same equation, just written as $x = F(x)$ , with $F(x) = \sqrt[3]{2x + 5}$ .

Now turn it into an iteration by putting labels on the two $x$ 's: $x_{n+1} = \sqrt[3]{2x_n + 5}$ . You put a guess in on the right and get a new guess out on the left.

Why does the limit solve the equation? If the guesses settle down to some value $\alpha$ , then eventually $x_n$ and $x_{n+1}$ are both (almost exactly) $\alpha$ . So $\alpha = \sqrt[3]{2\alpha + 5}$ . Cube both sides: $\alpha^3 = 2\alpha + 5$ , i.e. $\alpha^3 - 2\alpha - 5 = 0$ . The limit is a root of the original equation.

Rearrange $x^3 - 2x - 5 = 0$ into the form $x = F(x)$ in three different ways.

Rearrange $x^2 - 4x + 1 = 0$ into the form $x = F(x)$ by moving the $4x$ term to one side on its own and then dividing by 4. Type it like x = (x^2 + 1)/4.

Rearrange $x^3 + 3x - 2 = 0$ into the form $x =$ by taking a cube root.

Where iteration formulas come from