Understand · explore · practise
Understand Newton–Raphson failure: zero derivatives, domain escape, two-cycles, divergence and unsuitable starts. Explore contrasting cases, explain tangent behaviour and recognise when an exact root has already been found.
Before you startNewton–Raphson method; function domains
01 / Check the reason before repeating the formula
Newton’s method is a rule for choosing estimates, not a guarantee of convergence from every start. An equation may have a perfectly good real root while a chosen sequence never reaches it.
At each step, check whether f is defined, whether a root has already been found, and whether the derivative permits another tangent intercept.
x0 = 0; f(x0) = 2; f′(x0) = −2.
Next estimate: 0 − 2/(−2) = 1.
Both tangents are defined, but the sequence cycles between 0 and 1.
Values: 0.
Each button press performs one substitution. This explorer displays up to six substitutions; reaching that display limit is not itself a failure. A large step alone does not prove eventual failure: compare the quadratic example, whose estimates later recover. Keep function-domain failure separate from a zero slope or an already-solved equation.
02 / A horizontal tangent at a nonroot gives no update
f(0) = −1, so the current input is not a root
The equation remains unsolved.
f′(x) = 3x², so f′(0) = 0
The tangent is horizontal.
Its equation is y = −1
It never meets the horizontal axis.
Newton’s quotient would divide −1 by zero
Choose a different suitable start instead of calculating an infinite estimate.
For f(x) = x² − 4 and x₀ = 0, explain why Newton’s update is unavailable.
Calculate f(0) and f′(0).
f(0) = −4 and f′(0) = 0. The horizontal tangent y = −4 has no x-intercept; the formula divides a nonzero number by zero.
03 / Stop when an exact root has already been found
f(0) = 0
Zero is an exact root.
f′(0) = 0 as well
The usual update would contain 0/0.
There is no need for another update
Report the root and stop.
This is different from a horizontal tangent at a nonroot
Check f first, rather than labelling every zero derivative a failed solve.
For f(x) = (x − 3)⁴, what should you do if the current estimate is exactly 3?
Evaluate the function before forming a quotient.
Stop: 3 is an exact root, even though f′(3) = 0. Do not evaluate 0/0.
04 / A small slope can produce a large jump
f(0.1) = −0.99 and f′(0.1) = 0.2
The current point is far below the axis with a shallow slope.
x₁ = 0.1 − (−0.99)/0.2 = 5.05
The tangent intercept is far to the right.
This first step moves farther from the positive root 1
It is not an immediate improvement.
Later positive iterates approach 1
A large first step can be a warning without proving eventual divergence.
For the same function, calculate x₁ from x₀ = 0.01.
Use the equivalent formula (x + 1/x)/2.
x₁ = (0.01 + 100)/2 = 50.005.
For f(x) = 10⁻¹²(x − 2), the derivative is tiny everywhere. Does Newton necessarily make a huge jump?
The numerator carries the same scale factor.
No. f/f′ = x − 2, so any input updates directly to 2. The ratio, not the derivative magnitude alone, determines the step.
05 / A sequence may revisit the same two estimates
f′(x) = 3x² − 2
Both slopes below are nonzero.
At 0: f = 2 and f′ = −2, so x₁ = 1
The first tangent meets the axis at 1.
At 1: f = 1 and f′ = 1, so x₂ = 0
The second tangent returns to the original start.
The sequence repeats 0, 1, 0, 1, …
Neither value is a root.
There is a real root between −2 and −1
f(−2) = −2 and f(−1) = 3; the failed sequence does not imply no root.
Pause, replay or seek freely. The notes explain the same idea and stay in view.
For this cubic, write the tangent equations at x = 0 and x = 1.
Use y − f(a) = f′(a)(x − a).
At 0: y = 2 − 2x, whose intercept is x = 1. At 1: y = x, whose intercept is x = 0.
Starting from x₀ = −2 for the same cubic, calculate x₁.
f(−2) = −2 and f′(−2) = 10.
x₁ = −2 − (−2)/10 = −1.8. This starts near the sign-bracketed negative root; convergence still needs checking.
06 / A valid starting input can lead outside the domain
The real domain is x > 0; f′(x) = 1/x
The starting value is allowed.
xₙ₊₁ = xₙ − xₙ ln xₙ
This formula follows from Newton’s method on the positive domain.
x₁ = 3 − 3 ln3 ≈ −0.295836866004
The tangent has a negative intercept.
ln x₁ is not real
The next Newton evaluation is unavailable. The real root 1 still exists.
For f(x) = ln x and a positive current input x, when is the next estimate x(1 − ln x) positive?
The factor x is already positive.
Exactly when 1 − ln x > 0, or 0 < x < e. At x = e the next estimate is zero; at x > e it is negative. Both leave the logarithm’s real domain.
Newton’s formula for 1/x − 2 simplifies to 2x − 2x². Is x = 0 now an allowed input for the original Newton calculation?
The original function is still evaluated in the method.
No. The algebraic simplification does not remove the original restriction x ≠ 0.
07 / Growing estimates can move away indefinitely
f′(x) = 1/[3(∛x)²]
This derivative is positive for every nonzero real x.
f(x)/f′(x) = 3x
Use the real cube root for negative inputs too.
Newton gives xₙ₊₁ = −2xₙ
The magnitude doubles and the sign reverses.
From 1: 1, −2, 4, −8, …
The estimates diverge from the root 0.
At exactly zero the equation is solved
The derivative there is not finite, but no update is required.
For this cube-root example, start at x₀ = −0.25. Find the next three estimates.
Multiply by −2 at every step.
0.5, −1 and 2. The distance from zero doubles at each step.
08 / A repeated root can slow convergence
f′(x) = 2(x − 2)
The quotient is defined if x ≠ 2.
xₙ₊₁ = xₙ − (xₙ − 2)/2 = (xₙ + 2)/2
The error relative to 2 halves each time.
This converges, but only linearly in its error
It does not show the usual rapid local improvement associated with a simple root.
Stop if an exact root is reached
Do not continue into a 0/0 quotient.
For f(x) = (x − 1)³ and x ≠ 1, express the next error xₙ₊₁ − 1 in terms of xₙ − 1.
Cancel two powers in f/f′.
xₙ₊₁ = xₙ − (xₙ − 1)/3, so the next error is (2/3)(xₙ − 1). This still converges, but the error shrinks by a fixed factor.
09 / Use the required root and physical interval
An equation may have several real roots. Choose a start informed by a graph, a sign bracket and the interval requested in the problem. After iteration, check that the result belongs to that interval and satisfies the original equation.
If a model represents positive time, a negative root might be mathematically valid but irrelevant to the situation. Do not silently accept it merely because the numerical calculation converged.
For x² − 4 = 0, compare one Newton step from x₀ = 3 and x₀ = −3.
Use xₙ₊₁ = (xₙ + 4/xₙ)/2.
The next values are 13/6 and −13/6 respectively. The starts move towards different roots, +2 and −2.
10 / Change the method for a stated reason
When a Newton step is undefined or unsuitable, explain the cause. A different start may help. If a continuous function has a valid sign bracket, bisection can refine that interval while retaining the root guarantee.
A practical safeguard is to accept a proposed Newton step only when it is defined and suitable for the bracket; otherwise take a bisection step. Keep the sign bracket updated. This is a method choice, not a claim that Newton itself always converges.
For x³ − 1, Newton fails at x₀ = 0. Give a valid sign bracket that could be bisected.
Choose endpoints on opposite sides of the root 1.
[0, 2] works: f(0) = −1 and f(2) = 7, with a continuous polynomial. Its first midpoint is 1, an exact root.
For f(x) = 1/x, may you switch to sign-change bisection on [−1, 1]?
The fallback still needs continuity.
No. The function is discontinuous at zero and has no real root. Opposite signs across the pole do not justify bisection.
11 / Report the cause and the next sensible action
Classify: (a) f(a) ≠ 0 but f′(a) = 0; (b) f(a) = 0; (c) two distinct nonroot estimates repeat forever.
Separate an undefined update, success and a cycle.
(a) No finite Newton update at this input. (b) An exact root has been found; stop. (c) A nonconvergent cycle, even though each individual step may be defined.
Section 1 of 11 · Check the reason before repeating the formula