Revise Numerical solution of equations for Cambridge 9709 (2.6)

The first step in solving an equation numerically is often to find a small interval where a root is known to exist. We can do this by looking for a change of sign. If a function $f(x)$ is continuous (it has no breaks or jumps) and you find that $f(a)$ is positive and $f(b)$ is negative (or vice versa), the graph of $y=f(x)$ must cross the x-axis somewhere between $a$ and $b$. This means there is at least one root in the interval $(a, b)$.

Once we have located a root, we can use an iterative method to find a more accurate approximation. This involves rearranging the equation $f(x)=0$ into the form $x=g(x)$. This rearrangement is then used to create an iterative formula. By substituting an initial guess ($x_0$) into the right-hand side, we generate a new, hopefully better, approximation ($x_1$). We repeat this process, feeding each new approximation back into the formula, until the values converge.

x_{n+1} = g(x_n)

A common exam question asks you to find a root to a certain number of decimal places. Iterating until the value stabilises is not sufficient proof. To prove that a root $\alpha$ is equal to a value, say $1.23$, correct to 2 decimal places, you must show that the root lies in the interval $[1.225, 1.235]$. This is done using the sign change method on the bounds of this interval.