Using the Newton–Raphson method, obtain an iterative formula for \(\sqrt N\) and hence find \(\sqrt{30}\) correct to four decimal places.

Answer: \(x_{n+1}=\frac12\left(x_n+\frac{N}{x_n}\right)\); \(\sqrt{30}\approx5.4772\).

Let \(x=\sqrt N\Rightarrow f(x)=x^2-N=0\), \(f'(x)=2x\).

\[x_{n+1}=x_n-\frac{x_n^2-N}{2x_n}=\frac{x_n^2+N}{2x_n}=\frac12\left(x_n+\frac N{x_n}\right).\]

For \(N=30\): \(5^2=25<30<36=6^2\), take \(x_0=5.5\).

\(x_1=\frac12\left(5.5+\frac{30}{5.5}\right)=\frac12(5.5+5.454545)=5.477273\)

\(x_2=\frac12\left(5.477273+\frac{30}{5.477273}\right)=5.477226\)

\(x_3=5.477226\) (no change in the first six decimals).

\(\therefore\ \sqrt{30}\approx\mathbf{5.4772}\).