In one line
Write the required number as the root of a simple equation, such as \(x^2-N=0\) for \(\sqrt N\), and apply \(x_{n+1}=x_n-\dfrac{f(x_n)}{f'(x_n)}\) to get a ready iteration formula.
Explained simply
Before calculators, people found \(\sqrt{10}\) by hand. Newton–Raphson gives a beautiful recipe: guess, then average your guess with \(N\) divided by your guess. Each round roughly doubles the number of correct digits. Computers still use formulas like these inside their square-root and division routines.
Technically, \(\sqrt N\) is the positive root of \(f(x)=x^2-N\). With \(f'(x)=2x\), Newton–Raphson gives \(x_{n+1}=x_n-\dfrac{x_n^2-N}{2x_n}=\dfrac12\left(x_n+\dfrac N{x_n}\right)\). The same idea works for other roots and for reciprocals.
The reciprocal formula is especially clever: it computes \(1/N\) using only multiplication and subtraction, with no division at all. Early computers that had no hardware for division used exactly this trick. These derivations are also a good test of whether you really understand the Newton–Raphson formula rather than just memorising it.
How it works
- Choose \(f(x)\) whose root is the desired quantity: \(x^2-N\) for \(\sqrt N\), \(x^3-N\) for \(\sqrt[3]N\), \(\dfrac1x-N\) for \(\dfrac1N\).
- Find \(f'(x)\).
- Substitute into the Newton–Raphson formula and simplify.
- Use the simplified formula with a sensible \(x_0\).
Key formula
- \(\sqrt N\): \(x_{n+1}=\dfrac12\left(x_n+\dfrac N{x_n}\right)\)
- \(\sqrt[3]N\): \(x_{n+1}=\dfrac13\left(2x_n+\dfrac N{x_n^2}\right)\)
- \(\sqrt[k]N\): \(x_{n+1}=\dfrac1k\left((k-1)x_n+\dfrac N{x_n^{k-1}}\right)\)
- \(\dfrac1N\): \(x_{n+1}=x_n(2-Nx_n)\) (no division needed)
Quick example
Find \(\sqrt{10}\) with \(x_0=3\):
- \(x_1=\frac12\left(3+\frac{10}3\right)=3.166667\)
- \(x_2=\frac12\left(3.166667+\frac{10}{3.166667}\right)=3.162281\)
- \(x_3=3.162278\)
So \(\sqrt{10}\approx3.1623\). Check: \(3.1623^2\approx10.0001\).
Reciprocal formula check: \(f(x)=\frac1x-N\), \(f'(x)=-\frac1{x^2}\), so \(x_{n+1}=x_n+x_n^2\left(\frac1{x_n}-N\right)=2x_n-Nx_n^2=x_n(2-Nx_n)\).
Common mistakes
- Choosing \(f(x)=x-\sqrt N\), which still needs \(\sqrt N\) to evaluate; the point is to avoid it.
- Algebra slips while simplifying; write the unsimplified step first.
- For the reciprocal formula, starting with \(x_0\) too far from \(1/N\) (it diverges if \(x_0\) is not between 0 and \(2/N\)).
How MU asks it
Asked as 3–5-mark parts: "Obtain the Newton–Raphson iterative formula for \(\sqrt N\) and hence find \(\sqrt{12}\)" or "Derive a formula to find \(\dfrac1N\) and evaluate \(\dfrac1{31}\)". Show the chosen \(f\), \(f'\), the substitution, the simplified formula, and two or three iterations.