The Newton–Raphson method

A fast iteration that uses the tangent — x_(n+1) = x_n − f(x_n)/f'(x_n) — its geometric meaning, why it converges so quickly, and the cases where it fails (a horizontal tangent or a poor starting value).

Pure · Numerical methods

The Newton–Raphson method

Why it works

The iteration

x = g(x)

works, but slowly, and only for the right rearrangement. The Newton–Raphson method is a cleverer choice of step that homes in on a root astonishingly fast, using the tangent to the curve.

Start at a guess

x_n

. Draw the tangent to

y = f(x)

at the point

(x_n, f(x_n))

. Unless the curve is doing something awkward, that tangent crosses the

x

-axis much closer to the root than

x_n

was — so take the crossing as your next estimate

x_{n+1}

, and repeat.

To find where the tangent meets the axis: it has gradient

f'(x_n)

and passes through

(x_n, f(x_n))

, so its equation is

y - f(x_n) = f'(x_n)(x - x_n)

. Setting

y = 0

and solving for

x

gives the crossing, and that is the formula:

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

It converges very fast. Near the root each step roughly doubles the number of correct digits, so two or three steps from a sensible start usually nail several decimal places — far quicker than a plain

x = g(x)

iteration. You still need

f'(x)

, so the method costs a differentiation, but the speed is well worth it.

When it fails. The geometry also shows the traps:

$f'(x_n) = 0$ — a horizontal tangent never meets the axis, and the formula

divides by zero. The method breaks down completely.

Near a stationary point — if $f'(x_n)$ is small, the tangent is almost flat and

crosses the axis a long way off, throwing the next estimate wildly out. A starting value too close to a turning point is dangerous.

A poor starting value can converge to a different root than intended, or

diverge. Newton–Raphson is fast but not foolproof — start near the root you want.