I2PureStretch

The Newton-Raphson Method

Q: What happens if f'(x) is zero or very small at my starting guess?

The formula divides by f'(x_n), so a near-zero derivative produces an extremely large (or undefined) jump — the method can fail completely or send you far from the true root. Avoid starting near a stationary point.

Standard iteration can take many steps to converge, and depends on finding a good rearrangement. The Newton-Raphson method takes a completely different, geometric approach — follow the TANGENT at your current guess down to the x-axis, and that intersection is usually a dramatically better guess.

32 min Video by TLMaths Numerical Methods

The big picture

This is the fastest root-finding method in the course, and it needs no algebraic rearrangement at all — just the original function and its derivative. The idea is purely geometric: at your current estimate, draw the tangent line (which you can find because you can differentiate), and follow it down to where it crosses the $x$ -axis. Because the tangent hugs the curve closely near the point of contact, this new crossing point is usually a MUCH better approximation to the true root than the standard iteration formula would give in the same number of steps.

A-Level Maths I2-06: Locating Roots: Introducing the Newton-Raphson MethodWatch the walkthrough, then work through the notes below.

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What you'll be able to do

Derive and use the Newton-Raphson iterative formula
Apply the method to find a root to a required degree of accuracy
Recognise situations where the Newton-Raphson method fails or converges poorly
Compare the efficiency of Newton-Raphson with basic iteration

The Newton-Raphson formula

At an estimate $x_{n}$ , the tangent to $y = f (x)$ has gradient $f^{'} (x_{n})$ . Following this tangent down to the $x$ -axis (where $y = 0$ ) gives a new, usually better estimate $x_{n + 1}$ .

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

The Newton-Raphson iterative formula.

f (x) = x^{3} - x - 1

, so

f^{'} (x) = 3 x^{2} - 1

f (1.5) = 3.375 - 1.5 - 1 = 0.875

;

f^{'} (1.5) = 3 (2.25) - 1 = 5.75

3Apply the formula:

x_{1} = 1.5 - \frac{0.875}{5.75}

Answer

x_{1} \approx 1.3478

(compare to the true root

\approx 1.3247

— already very close after just one step)

Tip — The formula is really just "follow the tangent to the x-axis": rearranging $y - f (x_{n}) = f^{'} (x_{n}) (x - x_{n})$ with $y = 0$ gives exactly the Newton-Raphson formula.

Repeating for greater accuracy

Applying the formula repeatedly (using each new $x_{n + 1}$ as the next $x_{n}$ ) typically converges MUCH faster than basic iteration — often doubling the number of correct decimal places with every single step.

1Using

x_{1} \approx 1.3478

f (1.3478) \approx 1.347 8^{3} - 1.3478 - 1 \approx 0.0448

f^{'} (1.3478) = 3 (1.3478)^{2} - 1 \approx 4.4501

x_{2} = 1.3478 - \frac{0.0448}{4.4501} \approx 1.3378

Answer

x_{2} \approx 1.3378

(continuing to converge toward

\approx 1.3247

)

Why this works

Each Newton-Raphson step roughly SQUARES the error compared to the previous step (once close to the root) — this quadratic convergence is why the method needs so few iterations compared to basic fixed-point iteration.

When Newton-Raphson fails

The method can fail or converge very slowly if the starting value is chosen near a stationary point of $f$ (where $f^{'} (x_{n}) \approx 0$ , making the tangent nearly horizontal and sending $x_{n + 1}$ far away), or if the starting value is too far from the true root, sometimes jumping to a completely different root.

1Near a stationary point,

f^{'} (x_{n}) \approx 0

2The formula divides by

f^{'} (x_{n})

, so a very small denominator produces a huge (or undefined) jump in

x_{n + 1}

AnswerThe tangent is nearly flat, so it crosses the x-axis very far away (or not usefully at all) — the method can fail completely or converge extremely slowly.

Tip — A quick sketch of the curve near your chosen starting value helps confirm the tangent is well-behaved (a reasonably steep gradient, not near a turning point) before committing to several iterations.

Think like an examiner

You must differentiate the function correctly to find $f^{'} (x)$ before applying the formula — this is often where marks are lost.
Work to sufficient decimal places at each step (usually one more than the required final accuracy) to avoid compounding rounding errors.
If asked to comment on a poor starting value, mention the risk of $f^{'} (x_{n})$ being close to zero (near a stationary point).

Common misconceptions

Forgetting to recompute both

f (x_{n})

AND

f^{'} (x_{n})

at each new value of

x_{n}

Both must be evaluated freshly at the CURRENT estimate every single iteration — using an old derivative value is a common error.

Using too few decimal places during intermediate calculations.

Carry extra decimal places through each step; only round the final answer to the accuracy the question requests.

Assuming Newton-Raphson always converges regardless of the starting value.

A poorly chosen starting value (especially near a stationary point of f) can cause the method to fail or converge to the wrong root entirely.

Newton-Raphson method

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

The core iterative formula.

Stretch yourself

Use the Newton-Raphson method with $x_{0} = 2$ to find $x_{1}$ and $x_{2}$ for a root of $f (x) = x^{2} - 7$ , giving answers to 4 decimal places.

Hint — Differentiate f(x) first, then apply the formula twice in succession, using each result as the input to the next iteration.

Questions students ask

Why is Newton-Raphson usually faster than basic iteration?

It uses the tangent (gradient information) at each step, which typically approximately squares the error at each iteration once close to the root — this quadratic convergence needs far fewer steps than fixed-point iteration for the same accuracy.

What happens if f'(x) is zero or very small at my starting guess?

The formula divides by

f^{'} (x_{n})

, so a near-zero derivative produces an extremely large (or undefined) jump — the method can fail completely or send you far from the true root. Avoid starting near a stationary point.

Do I need to rearrange the equation first, like with basic iteration?

No — Newton-Raphson works directly with the original function

f (x)

and its derivative

f^{'} (x)

, with no algebraic rearrangement required.

Key takeaways

Newton-Raphson formula: $x_{n + 1} = x_{n} - \frac{f ( x _{n} )}{f ^{'} ( x _{n} )}$ .
Uses the tangent at each estimate to find a (usually much) better next estimate.
Converges faster than basic fixed-point iteration for well-chosen starting values.
Fails or converges poorly if the starting value is near a stationary point of f (where f'(x) ≈ 0).

How this fits the course

Build on

The Trapezium Rule

Leads to

The Trapezium Rule

Test yourself

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The Newton-Raphson Method

32 min Video by TLMaths Numerical Methods