Newton-Raphson method for root finding: Question bank

True or false — justify

The formula finds where the curve crosses zero in one step.

False — it finds where the tangent line crosses zero, not the curve. That landing point is only closer to the true root, so we must iterate.

f^{'} (x_{n}) = 0

the method still works, just slower.

False — a zero slope means a horizontal tangent that never reaches the axis, so

f (x_{n}) / f^{'} (x_{n})

divides by zero and the step is undefined (or launches

x_{n + 1}

to infinity).

Newton-Raphson always converges whenever a root exists.

False — it is a local method. With a poor starting guess it can cycle (like

x^{3} - 2 x + 2

x_{0} = 0

) or diverge; you need a good enough guess and well-behaved

f^{'}

At a simple root the number of correct digits roughly doubles each step.

True — this is quadratic convergence: the new error is proportional to the square of the old error, so accuracy in digits doubles per iteration.

At a double root the convergence is still quadratic.

False — the quadratic proof assumed

f^{'} (root) \neq = 0

. At a double root

f^{'}

also vanishes there, and convergence drops to merely linear (errors roughly halve, not square).

The method needs the function to be differentiable.

True — the update explicitly uses

f^{'} (x_{n})

, the slope of the tangent. Without a derivative there is no tangent line to ride down.

If two guesses

x_{n}

and

x_{n + 1}

are equal, we have found a root.

True (essentially) —

x_{n + 1} = x_{n}

forces

f (x_{n}) / f^{'} (x_{n}) = 0

, so

f (x_{n}) = 0

: we have landed exactly on a root.

Newton-Raphson can only find one root of a function.

False — different starting guesses can steer you to different roots. Which root you reach depends on where you start, sometimes chaotically.

The method works on complex functions to find complex roots.

True — the same update applies over the complex numbers; the intricate boundaries between which start-point finds which root are the famous Newton fractals.

Bisection is always faster than Newton-Raphson.

False — bisection is slower (linear) but guaranteed; Newton is faster (quadratic near a simple root) but not guaranteed. See Bisection method.

Spot the error

A student writes

x_{n + 1} = x_{n} + \frac{f ( x _{n} )}{f ^{'} ( x _{n} )}

. What is wrong?

The sign is flipped — it must be minus. Adding moves you away from the root; check with

2

1 + 1/1 = 2

overshoots, while

1 - (- 1) /2 = 1.5

moves toward

1.414

A student stops the moment

f (x_{n})

is negative, thinking they overshot.

A negative

f (x_{n})

just means the guess is on the other side of the axis; it is normal and not a stopping condition. Stop when

∣ x_{n + 1} - x_{n} ∣ < ε

∣ f (x_{n}) ∣ < ε

A student concludes "

f^{'} (x_{n})

small is fine, I never hit exactly zero."

A near-zero slope is nearly as dangerous as an exact zero: the tangent is almost horizontal, so

f / f^{'}

becomes huge and hurls

x_{n + 1}

far away. Always guard against tiny derivatives.

A student derives the formula by setting the curve equal to zero in point–slope form.

The point–slope equation describes the tangent line, not the curve. We set the tangent's

y

to zero; equating the curve to zero would just restate the unsolved problem

f (x) = 0

A student uses the recurrence

x_{n + 1} = \frac{1}{2} (x_{n} + 2/ x_{n})

to find a root of

x^{2} - 3

That recurrence is specialised to

a = 2

(it computes

2

). For

x^{2} - 3

you need

x_{n + 1} = \frac{1}{2} (x_{n} + 3/ x_{n})

— the constant inside must match the number whose root you seek.

A student claims the tangent-line derivation and the Taylor derivation give different formulas.

They give the same formula. Taylor just makes the error term visible (

O ((r - x_{n})^{2})

), explaining why the method is fast; the recipe is identical. See Taylor series.

Why questions

Why do we subtract

f / f^{'}

instead of some other multiple of

f

Because

f / f^{'}

is exactly the horizontal distance the tangent travels from the current height down to the axis (rise over slope gives run). It is the geometrically correct jump, not an arbitrary scale.

Why is the tangent line the right thing to approximate the curve with?

Because near a point a smooth curve looks almost straight, and the tangent is the best straight-line fit there (matching value and slope). A straight line has one obvious axis-crossing, which algebra can solve instantly. See Tangent line and linear approximation.

Why does the error roughly square each step rather than shrink by a fixed factor?

The Taylor expansion shows the term we discard is proportional to

(r - x_{n})^{2}

. So the next error is proportional to the square of the current one — halving the error once means quartering it next, etc.

Why can Newton-Raphson be seen as a special fixed-point iteration?

Rewriting the update as

x_{n + 1} = g (x_{n})

with

g (x) = x - f (x) / f^{'} (x)

makes it a fixed-point scheme; a fixed point of

g

is exactly a root of

f

. Its derivative

g^{'} (root) = 0

at a simple root is what gives the extra-fast convergence. See Fixed-point iteration.

Why does a bad starting guess sometimes send the iteration to a different root than expected?

The method follows local slopes, and far from any root the tangent can point toward a completely different basin. The map from start-point to final root can be wildly sensitive — no continuity guarantee.

Why do we still need the derivative even for a polynomial whose roots we want?

Newton needs

f^{'}

to build each tangent. For polynomials

f^{'}

is easy to compute term-by-term, which is one reason the method is popular for Roots of polynomials. See Derivatives — definition and rules.

Edge cases

What happens if you start exactly at a point where

f^{'} (x_{0}) = 0

The very first step divides by zero and is undefined — the tangent is horizontal and never meets the axis. You must pick a different starting guess.

What happens if you start exactly at a root,

f (x_{0}) = 0

The correction

f / f^{'}

is zero, so

x_{1} = x_{0}

: the iteration stays put. It correctly recognises you are already done.

What happens between two roots where the tangent points back toward the starting region — can it loop?

Yes — a symmetric setup can make the tangents bounce back and forth, producing a cycle (e.g.

x^{3} - 2 x + 2

with

x_{0} = 0

oscillates

0 \to 1 \to 0 \to \dots

). Convergence never happens.

What if the function has no real root at all, like

f (x) = x^{2} + 1

With real starting guesses the iterates wander (or jump around) without settling, since there is no real point where

f = 0

. Over the complex numbers, though, it converges to

\pm i

What happens at an inflection point where

f^{''} = 0

near the root?

A vanishing second derivative actually helps — it makes the discarded Taylor term even smaller, so convergence can be faster than plain quadratic there. It is a benign, not a dangerous, edge case.

What if two consecutive iterates straddle a vertical asymptote of

f

The tangent slope can swing to extreme values, causing a huge jump across the asymptote and possibly divergence. Newton assumes smoothness on the region it traverses; asymptotes break that.

Recall One-line summary of every trap

Newton rides tangents, not the curve; it needs a nonzero, not-tiny slope; it converges quadratically only at simple roots, linearly at double roots; and it is local — good guesses win, bad guesses cycle or diverge.

Connections

Newton-Raphson method for root finding — the parent method these traps probe.
Tangent line and linear approximation — why linearising is the whole idea.
Taylor series — source of the quadratic-convergence claim.
Derivatives — definition and rules — supplies the $f^{'}$ every step needs.
Fixed-point iteration — Newton as a fast fixed-point map $g = x - f / f^{'}$ .
Bisection method — the slower-but-safe alternative.
Roots of polynomials — the classic application domain.