02 Mar 2026

Roots of Functions

Numerical meaning of roots, bracketing, bisection, Newton-Raphson method, and convergence checks.

msc semester-i numerical-methods roots

A root-finding problem appears when a physical or mathematical condition is written as an equation. For example, equilibrium may require force to vanish, a turning point may require velocity to vanish, and an allowed energy may be obtained from an equation whose zero must be found.

The problem is therefore reduced to finding a value $x=\alpha$ for which

\[f(\alpha)=0.\]

The graph of $f(x)$ crosses or touches the $x$-axis at a root. Numerical methods give successive approximations to this crossing point.

Bracketing idea

If $f(a)$ and $f(b)$ have opposite signs, then a continuous function has at least one root between $a$ and $b$.

\[f(a)f(b)<0.\]

This is the basis of the bisection method.

Bisection method

The midpoint is

\[c=\frac{a+b}{2}.\]

If $f(a)f(c)<0$, the root lies in $[a,c]$; otherwise it lies in $[c,b]$. Repeating this process gives a narrower interval.

The reason the method is reliable is simple: after each step the root is still trapped inside the new interval. The length of the interval is halved at every step, so the uncertainty decreases steadily.

Newton-Raphson method

Newton’s method uses the tangent line at the present approximation:

\[x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}.\]

It is usually faster than bisection, but it needs a derivative and a good starting value.

This formula comes from replacing the curve by its tangent near $x_n$. The tangent line is followed until it cuts the $x$-axis, and that intercept is taken as the next approximation.

Stopping criteria

An iteration may be stopped when either

\[|x_{n+1}-x_n|<\epsilon\]

or

\[|f(x_n)|<\epsilon.\]

Here $\epsilon$ is the chosen tolerance.

Bisection example

Find a root of

\[f(x)=x^3-x-2\]

in the interval $[1,2]$.

Since

\[f(1)=-2,\qquad f(2)=4,\]

there is at least one root in $[1,2]$. The first midpoint is

\[c_1=\frac{1+2}{2}=1.5, \qquad f(1.5)=3.375-1.5-2=-0.125.\]

Therefore the root lies in $[1.5,2]$. The next midpoint is

\[c_2=\frac{1.5+2}{2}=1.75, \qquad f(1.75)=5.359375-1.75-2=1.609375.\]

Now the root lies in $[1.5,1.75]$. Repeating the process gives a narrower interval around the root. The known value is approximately

\[\alpha \simeq 1.521.\]

Newton step for the same equation

For the same equation,

\[f(x)=x^3-x-2,\qquad f'(x)=3x^2-1.\]

Starting with $x_0=1.5$,

\[x_1=x_0-\frac{f(x_0)}{f'(x_0)} =1.5-\frac{-0.125}{5.75} =1.521739.\]

Thus Newton’s method reaches a good approximation in one step. A final substitution check gives

\[f(1.521739)\approx 0.002.\]

Key points

• Bisection is slow but reliable.
• Newton-Raphson is fast but can fail for poor starting values.
• Always check the final value by substituting it into $f(x)$.

Practice questions

1. Define a root of a function.
2. State the condition for bracketing a root in an interval.
3. Use two bisection steps to approximate a root of $x^3-x-2=0$ in $[1,2]$.
4. Derive the Newton-Raphson formula from the tangent line.
5. Why can Newton-Raphson fail when the starting value is poorly chosen?