Numerical solution of equations — common mistakes

To show a root is correct to $n$ decimal places, you must show a change of sign in an interval of width $10^{-n}$ that contains your answer. For example, to show $\alpha = 2.153$ (3 d.p.), you must test the interval $[2.1525, 2.1535]$. Simply showing that your iterations are constant to $n$ d.p. is not sufficient proof for the final marks.

The Newton-Raphson method can converge very quickly, but it's sensitive to the starting value. If $f'(x_n)$ is close to zero (i.e., you are near a turning point), the fraction becomes very large and can send your next approximation far away. Always be mindful of the graph's shape.

First, use the change-of-sign rule to find a small interval $[a, b]$ that contains the root. A good choice for $x_0$ is the midpoint of this interval, $(a+b)/2$. Alternatively, sketch the graph to see which end of the interval looks closer to the root.

The iterative method $x_{n+1} = g(x_n)$ works by finding a fixed point, where the input equals the output ($x = g(x)$). This is graphically where the curve $y=g(x)$ intersects the line $y=x$. This intersection point is also a root of the original equation $f(x)=0$. Simply iterating with $f(x)$ (e.g., $x_{n+1} = f(x_n)$) doesn't have this property and won't converge to a root of $f(x)=0$.

Both show how an iteration converges. A staircase diagram happens when $0 < g'(x) < 1$, and the approximations approach the root from one side. A cobweb diagram happens when $-1 < g'(x) < 0$, and the approximations oscillate around the root, spiralling inwards.

Not always. Newton-Raphson usually converges much faster (it has 'quadratic convergence'), but it is more sensitive to the starting value and can fail spectacularly if you start near a turning point. The standard $x=g(x)$ iteration is often more robust, even if it's slower.