Newton-Raphson method

Bisection is safe but slow: it halves a bracket step by step, gaining roughly one bit of accuracy per iteration no matter how the function curves. The Newton-Raphson method throws away the safety net and instead follows the tangent line at the current guess all the way down to the axis, betting that a smooth function looks enough like a straight line nearby that this shortcut lands close to the true root. The bet usually pays off spectacularly: near a simple root the number of correct digits roughly doubles every step, so a handful of iterations reach machine-level accuracy where bisection would still be crawling. This article derives the iteration formula from the tangent-line idea, applies it to the same cubic used earlier for bisection so the speed contrast is visible, and catalogues the specific ways the method can misbehave when its one assumption — a well-behaved derivative and a decent starting guess — is violated.

The core insight

Replace $f$ near the current guess $x_n$ by its tangent line and slide to where that line crosses the axis, $x_{n+1}=x_n-f(x_n)/f'(x_n)$ , giving quadratic convergence near a simple root.

The core idea

Tangent-line approximation. Near the current guess $x_n$ , the curve $y=f(x)$ is replaced by its tangent line, which shares the same value and slope at $x_n$ .
Solve the linear model for its zero. Setting the tangent line to zero, instead of the curved $f$ , gives an equation a student can solve in one line — that solution becomes the next guess.
Iterate. Feed the new guess back in and repeat; each pass typically lands closer to the root than the last.
No bracket, no safety net. Unlike bisection or regula falsi, the method never checks that the root is trapped between two points, which is exactly why it can be fast — and exactly why it can also fail.

The tangent-line approximation near $x_n$ is

f(x) \approx f(x_n) + f'(x_n)(x-x_n)

where: $f(x_n)$ is the function value at the current guess and $f'(x_n)$ is the slope there. Setting this linear model to zero and solving for $x$ gives the iteration formula:

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

where: $x_n$ is the current approximation, $x_{n+1}$ is the next one, and $f'(x_n)\ne 0$ is required at every step. Geometrically, $x_{n+1}$ is where the tangent line to $y=f(x)$ at the point $(x_n, f(x_n))$ crosses the $x$ -axis — the figure below shows one such step.

Figure

A curve with a tangent line drawn at the current guess crossing the axis at the next guess, closer to the root

Convergence, qualitatively. Near a simple root — one where $f'(\text{root})\ne 0$ — Newton-Raphson exhibits quadratic convergence: the count of correct decimal digits roughly doubles each step, far outpacing bisection's steady linear halving of the bracket. The price is reliability: if $f'(x_n)$ is close to zero, if $x_0$ is a poor guess, or if $f$ has awkward curvature nearby, the iteration can oscillate or shoot off toward a different root entirely, with no bracket to rescue it.

Comparing the three root-finders on the same footing. Bisection needs only that $f$ change sign across a bracket, no derivative and no smoothness required, and shrinks that bracket by exactly half every step, so it is slow but essentially unbreakable. Regula falsi also needs just a bracket but usually converges faster than bisection by using a secant line instead of the plain midpoint. Newton-Raphson trades that safety for speed: it needs $f$ to be differentiable and needs only a single starting guess, not a bracket, and in exchange it can double its accuracy every step near a simple root. The syllabus expects a student to recognise this trade-off qualitatively — faster, but with no safety net — rather than just recite the iteration formula.

Derivation ANewton-Raphson iteration from the tangent line

The goal is to derive $x_{n+1}=x_n-f(x_n)/f'(x_n)$ from nothing more than the definition of a tangent line and elementary algebra, so that the formula is remembered as a picture rather than memorised as a rule, and so the sign in front of the fraction is never in doubt.

Step 1. Write the tangent line to $y=f(x)$ at the current guess $x_n$ . Since the tangent shares the value $f(x_n)$ and the slope $f'(x_n)$ with the curve at that point, the line is $L(x) = f(x_n) + f'(x_n)(x-x_n)$ — this is just the point-slope form of a straight line, with the curve's own slope standing in for the line's slope.

Step 2. Replace the hard problem $f(x)=0$ with the easy problem $L(x)=0$ . Instead of solving the (possibly curved, possibly transcendental) equation $f(x)=0$ directly, solve the linear stand-in $f(x_n) + f'(x_n)(x-x_n) = 0$ , which any student can rearrange.

Step 3. Solve the linear equation for $x$ . Move $f(x_n)$ across and divide by $f'(x_n)$ (valid only when $f'(x_n)\ne 0$ ): $f'(x_n)(x-x_n) = -f(x_n)$ , so $x - x_n = -f(x_n)/f'(x_n)$ , giving $x = x_n - f(x_n)/f'(x_n)$ .

Step 4. Rename this solution the next iterate. Call the $x$ found in Step 3 by the name $x_{n+1}$ , since it becomes the starting point for the next pass through the same three steps: $x_{n+1} = x_n - \dfrac{f(x_n)}{f'(x_n)}$ . Repeating this loop is the entire method — no new idea is needed at any later step, only the same tangent-line replacement centred on the newest guess.

Worked example

Find a root of $f(x)=x^3-x-1=0$ near $x_0=1.5$ , carrying out three iterations, and compare the speed to bisection on the same cubic.

Step 1. Differentiate: $f'(x)=3x^2-1$ .

Step 2 — iteration 1. $f(1.5)=3.375-1.5-1=0.875$ and $f'(1.5)=3(2.25)-1=5.75$ , so

x_1 = 1.5 - \frac{0.875}{5.75} = 1.5 - 0.15217 = 1.34783

Step 3 — iteration 2. $f(1.34783)\approx 2.44855-1.34783-1=0.10072$ and $f'(1.34783)=3(1.81665)-1=4.44995$ , so

x_2 = 1.34783 - \frac{0.10072}{4.44995} = 1.34783 - 0.02264 = 1.32519

Step 4 — iteration 3. $f(1.32519)\approx 2.32693-1.32519-1=0.00174$ and $f'(1.32519)=3(1.75613)-1=4.26839$ , so

x_3 = 1.32519 - \frac{0.00174}{4.26839} = 1.32519 - 0.00041 = 1.32478

Step 5 — check. The true root is $x\approx 1.32472$ , so three iterations already match it to four decimal places. Bisection on this same cubic (starting from a bracket around $x_0=1.5$ ) needed four iterations just to narrow the bracket to width $0.0625$ — Newton-Raphson's three steps land far closer, far faster. After three iterations, $x_3\approx 1.32478$ , matching the true root $1.32472$ to four decimal places.

Competitive stretch

Applying Newton-Raphson to $f(x)=x^2-5=0$ gives the iteration $x_{n+1}=\tfrac12\left(x_n+5/x_n\right)$ — the classical Babylonian square-root algorithm as a special case. Starting from $x_0=2$ , three iterations give $x_1=2.25$ , $x_2=2.23611$ , $x_3=2.23607$ , matching $\sqrt5\approx2.23607$ to five decimal places, again showing the error shrinking roughly quadratically at each step.

Common traps

Dividing by $f'(x_n)$ without checking it isn't zero or nearly zero — a near-zero derivative flings the next iterate far away instead of refining it.
Choosing a starting guess $x_0$ far from the root, or near a point where $f'\approx 0$ — with no bracket as a backstop, a poor guess can converge to the wrong root or fail to converge at all.
Getting the sign wrong in the formula — it is $x_{n+1}=x_n-f(x_n)/f'(x_n)$ , subtraction, not addition; a slip reverses every step's direction.
Assuming quadratic convergence is guaranteed everywhere — it holds near a simple root ( $f'(\text{root})\ne0$ ); at a repeated root convergence degrades to linear.
Forgetting to re-differentiate for a new function — the iteration formula is generic, but $f'(x)$ must be recomputed from scratch for each new $f$ before any iteration can start; reusing an old derivative from a previous problem is a common careless slip under exam time pressure.

Ask the tutor: If you want to see step by step what happens when a starting guess is chosen too close to a local extremum, ask the AI tutor to walk through a diverging example with you.

Why does Newton-Raphson need f prime, while bisection does not?

Bisection only ever asks whether $f$ is positive or negative at the midpoint of a bracket, so it needs no calculus. Newton-Raphson instead builds a tangent line at the current guess, and a tangent line is defined by its slope — so $f'(x_n)$ is unavoidable. That dependence is also the source of its speed: knowing the slope tells the method not just which direction to move but roughly how far, which a sign-only test like bisection's can never do.

What happens if I pick a starting guess right at a local maximum or minimum of f?

At such a point $f'(x_0)=0$ , so the formula divides by zero and the method is undefined at that exact guess. Even a guess merely close to such a point is dangerous: $f'(x_n)$ is small, so $f(x_n)/f'(x_n)$ is large, and the next iterate can be flung far from where you started, sometimes toward a completely different root or off toward infinity. Always sanity-check that $f'(x_0)$ is comfortably away from zero before starting.

Is the Babylonian square-root method really the same as Newton-Raphson?

Yes, exactly. Apply the general formula $x_{n+1}=x_n-f(x_n)/f'(x_n)$ to $f(x)=x^2-a$ , whose root is $\sqrt a$ , and $f'(x)=2x$ gives $x_{n+1}=x_n-\dfrac{x_n^2-a}{2x_n}=\dfrac12\left(x_n+\dfrac{a}{x_n}\right)$ , which is precisely the ancient averaging rule for extracting square roots by hand. It is a specific instance of Newton-Raphson, not a coincidence.

Newton-Raphson method

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads