Root Finding: Bisection & Newton's Method

≈ 25 min

When an equation $f(x) = 0$ has no tidy formula, we find roots numerically.

Bisection. Start with an interval $[a, b]$ where $f(a)$ and $f(b)$ have opposite signs (so a root lies between them). Test the midpoint; keep the half that still brackets the root; repeat. Slow but sure.

Newton's method. Follow the tangent line down to where it hits the $x$ -axis:

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

Worked example (finding $\sqrt 2$ ). With $f(x) = x^2 - 2$ , $f'(x) = 2x$ , the update simplifies to $x_{n+1} = \tfrac{1}{2}\left(x_n + \tfrac{2}{x_n}\right).$ From $x_0 = 1$ : $x_1 = 1.5$ , then $x_2 \approx 1.4167$ — already close to $\sqrt 2 = 1.4142\ldots$

To start bisection on $[a, b]$ , the values $f(a)$ and $f(b)$ must…

have opposite signs

have the same sign

both be zero

both be positive

Apply one Newton step to $f(x) = x^2 - 2$ starting from $x_0 = 1$ . Find $x_1$ .

Bisection on $f(x) = x^2 - 2$ over $[1, 2]$ : the first midpoint is $1.5$ , and $f(1.5) = 0.25 > 0$ while $f(1) < 0$ , so the root lies in $[1, 1.5]$ . What is the next midpoint?

Continue Newton's method for $\sqrt 2$ from $x_1 = 1.5$ . Find $x_2$ (to 3 decimal places).