Newton Method Visualizer

root-finding with tangent steps and convergence diagnostics

Worked example — Newton Method Visualizer

Root estimate	1.52138
Iterations	3
Residual \|f(x)\|	5.894×10⁻⁷

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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0.55 and 0.58 end up at opposite roots 🖖

Newton doubles the number of correct digits at every step, and the Fast root preset measures it: from x₀ = 1 towards √2 the error runs 0.414, 0.0858, 0.00245, 2.1 × 10⁻⁶, 1.6 × 10⁻¹². Four steps, twelve digits. The same mechanism is what breaks it. Where the derivative is nearly flat the tangent barely tilts and the step is enormous. On x³ − x, starting at 0.5774 throws the first guess to −2210 — which is why 0.55 ends up at the negative root while 0.58 ends up at the positive one.

follow the tangent to the axis 🖖

To find where a curve crosses zero, Newton's method replaces the curve with its tangent line at your current guess and jumps to where that straight line hits the x-axis. Because a smooth curve looks almost straight when you zoom in, that crossing usually lands much closer to the true root. Repeat, and you close in fast. The formula: x_n+1 = x_n − f(x_n)/f'(x_n).

computers divide without dividing 🖖

Modern CPUs often compute a/b by first finding 1/b, and Newton's method does this using no division at all. Applied to f(x) = 1/x − a it gives the iteration x_n+1 = x_n(2 − a·x_n), built from only multiplication and subtraction — operations hardware does cheaply. The same trick underlies fast reciprocal-square-root routines, including the famous one in Quake III.

NEWTON'S METHOD — WHEN IT DOUBLES YOUR DIGITS, AND WHEN IT RUNS AWAY

Which Newton Case Are You In?

Newton's method follows the tangent down to the axis and repeats: xₖ₊₁ = xₖ − f(xₖ)/f′(xₖ). When it works it is spectacular, roughly doubling the number of correct digits every step. Every way it fails comes from the same place — that denominator. A near-zero f′ throws you far away, an f′ that vanishes at the root slows you to a crawl, and starting on the wrong side of a turning point hands you a different root altogether.

Quadratic convergence — the case Newton is famous for e_k+1 ∝ e_k²

Flat slope at the root — the speed drops to linear f′(r) = 0 ⇒ e_k+1 ≈ (1 − 1/m)e_k

It runs away — the step grows instead of shrinking x → −2x

It converges — to a root you did not ask for f′(x₀) = 0

Quadratic convergence — the case Newton is famous for

What you know: f is smooth, f′ is comfortably non-zero near the root, and the starting point is close enough. Each step roughly squares the error.

Iteration: e_k+1 ∝ e_k²

Worked example: f(x) = x² − 2 from x₀ = 1 → 1.5, 1.416667, 1.41421569, 1.41421356237: 1, then 3, 6 and 12 correct digits

Open this case: fast root

$Quadratic convergence — the case Newton is famous for. Each tangent lands far nearer the root than the last, and the error squares. f is smooth, f′ is comfortably non-zero near the root, and the starting point is close enough. Each step roughly squares the error.$

Each tangent lands far nearer the root than the last, and the error squares.

Flat slope at the root — the speed drops to linear

What you know: The root is a repeated one, so f and f′ vanish together there. The tangent is nearly horizontal exactly where you are heading.

Iteration: f′(r) = 0 ⇒ e_k+1 ≈ (1 − 1/m)e_k

Worked example: f(x) = x³ from x₀ = 0.1 → 0.0667, 0.0444, 0.0296, 0.0198: each step multiplies the error by 2/3 instead of squaring it

Open this case: flat slope

Flat slope at the root — the speed drops to linear. A triple root flattens the tangent, so each step covers only a third of what is left. The root is a repeated one, so f and f′ vanish together there. The tangent is nearly horizontal exactly where you are heading. — A triple root flattens the tangent, so each step covers only a third of what is left.

It runs away — the step grows instead of shrinking

What you know: f′ collapses faster than f as you approach the root, so the correction f/f′ overshoots by more on every iteration.

Iteration: x → −2x

Worked example: f(x) = ∛x from x₀ = 0.01 → −0.02, 0.04, −0.08, 0.16: the update works out to exactly x → −2x, so the distance doubles and the sign flips forever

Open this case: diverge

It runs away — the step grows instead of shrinking. Every tangent crosses the axis twice as far out as the last, on the opposite side. f′ collapses faster than f as you approach the root, so the correction f/f′ overshoots by more on every iteration. — Every tangent crosses the axis twice as far out as the last, on the opposite side.

It converges — to a root you did not ask for

What you know: The function has several roots and the start sits near a turning point, where f′ is small and the first step is enormous.

Iteration: f′(x₀) = 0

Worked example: f(x) = x³ − x from x₀ = 0.7 settles on +1, but from x₀ = 0.5 the very first step lands on −1 — a root on the far side of two others

Open this case: wrong basin

It converges — to a root you did not ask for. Two nearby starts, two different roots: the boundary sits at the turning point, not between the roots. The function has several roots and the start sits near a turning point, where f′ is small and the first step is enormous. — Two nearby starts, two different roots: the boundary sits at the turning point, not between the roots.

Problem solved in full

The root of x³ − x − 2 by Newton's method 5 steps
Newton's method finds the root of x³ − x − 2 in three iterations. Bisection needs about twenty steps just to reach 10⁻⁶, which Newton passes on its second. Work out where the speed comes from — and what it costs.
1. The method is one idea: replace the curve by its tangent and solve that instead, because a straight line can be solved exactly. The next guess is where the tangent crosses zero.
  
  $x_{n + 1} = x_{n} - \frac{f ( x _{n} )}{f ^{'} ( x _{n} )}, f (x) = x^{3} - x - 2, f^{'} (x) = 3 x^{2} - 1$
2. From 1.5 the first step is almost the whole journey. The function is only −0.125 there, the slope is 5.75, and their ratio moves the guess by about 0.022.
  
  $x_{0} = 1.5 : x_{1} = 1.5 - \frac{- 0.125}{5.75} = 1.52173913$
3. Two more steps and the digits stop changing. The tool reports three iterations because the fourth would move the answer by less than the tolerance.
  
  $x_{2} = 1.52137981, x_{3} = 1.52137971$
4. The interesting quantity is not the estimate but the error. Track it and the pattern is unmistakable: each error is roughly the square of the one before, so the number of correct digits doubles every step — 2, then 4, then 7, then 14.
  
  $e_{n} = ∣ x_{n} - r ∣ : 2.14 \times 1 0^{- 2}, 3.59 \times 1 0^{- 4}, 9.92 \times 1 0^{- 8}, 7.55 \times 1 0^{- 15}$
5. And the constant in front is not arbitrary. Expanding f about the root shows the ratio of consecutive squared errors tends to |f″/2f′| evaluated there, which for this cubic is 0.768. The observed ratios are 0.786, 0.768, 0.768.
  
  $\frac{e _{n + 1}}{e _{n}^{2}} ⟶ \frac{f ^{''} ( r )}{2 f ^{'} ( r )} = \frac{3 r}{2 ( 3 r ^{2} - 1 )} = 0.768$
Answer

The tool prints a root of 1.52138 after 3 iterations. Quadratic convergence is why: bisection halves the interval each step and needs about twenty to reach 10⁻⁶, while Newton squares the error and is past it on the second step. The cost is that it is not guaranteed. Bisection cannot fail once it has bracketed a root; Newton has no bracket, and a starting point where the slope is near zero throws the next guess far away. That is the trade the whole of numerical root-finding is built around, and it is why production solvers bracket first and only then switch to Newton.

$e_{n + 1} \approx 0.768 e_{n}^{2} : correct digits DOUBLE each step$

Learning path

Three numerical methods, and where each one gives out

Leads to Integrating numerically

References (1)

Insight block 2 draws the tangent. Neither the tangent nor the derivative is Newton’s, which this traces: T. J. Ypma, "Historical Development of the Newton–Raphson Method." SIAM Review 37(4), 531–551, 1995 — Newton’s own procedure used no derivative and no geometric tangent; the form taught today is closer to Simpson’s.

By the 42Math team · Updated 2026-07-27

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

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Example problems

fast root - From x0 = 1 the error falls 0.414, 0.0858, 0.00245, 2.1 × 10⁻⁶, 1.6 × 10⁻¹² - each roughly the square of the last, so four steps give twelve correct digits of √2.
flat slope - x³ has a triple root at 0, where Newton crawls instead of squaring: every step just multiplies x by 2/3. After six it halts at 0.0088 and reports convergence. The root is 0.
diverge - Newton on the cube root reduces to x → -2x exactly, so the guesses run 0.01, -0.02, 0.04, -0.08, doubling and flipping sign forever. It diverges from every start but the root itself.
wrong basin - x0 = 0.55 sits on the positive side and converges to -1. The first tangent throws it to -3.60, because the derivative nearly vanishes at 1/√3 = 0.5774 - and 0.58 goes to +1 instead.
negative root - The same x² - 2 as Fast root, started at -1.2 instead of 1, converging to -1.414 in three steps. Newton finds the root you point it at, not the one you wanted.
strict tolerance - x³ - x - 2 from 1.6 reaches 1.5213797 in three iterations even at a tolerance of 10⁻⁸. Bisection would need about twenty-seven halvings to match it.