Pure 9.3 - Newton-Raphson and recurrence relations | Edexcel A-Level Mathematics

Use recurrence relations to approach roots, and derive Newton-Raphson by following a tangent to the axis. You will calculate successive approximations, simplify a requested recurrence and recognise when a fast method becomes unreliable.

Use a recurrence to chase a fixed point

Can a straight line give a better root estimate than your current guess? Before using tangents, try a simpler repeated instruction: start at $2$ , then take $\sqrt{2x+3}$ . The outputs $2.645751\ldots$ and $2.879497\ldots$ head towards $3$ , a root of $x^2-2x-3=0$ .

You already know

A fixed point satisfies $\alpha=g(\alpha)$ . An iteration $x_{n+1}=g(x_n)$ gives such a point if it converges and $g$ is continuous there. A valid fixed-point equation does not itself guarantee convergence.

An iterate is one member of this sequence of approximations. Follow the question's starting index carefully.

Worked example 1: a square-root recurrence

Question. Use $x_{n+1}=\sqrt{2x_n+3}$ , with $x_1=2$ , to find $x_2$ , $x_3$ and identify the possible positive limit.

Plan. Calculate with stored values, then solve the fixed-point equation.

\begin{aligned} x_2&=\sqrt7=2.645751311\ldots && \text{substitute }x_1 \\ x_3&=\sqrt{2(2.645751311\ldots)+3} \\ &=2.879496939\ldots && \text{repeat} \\ \alpha^2-2\alpha-3&=0 && \text{square a positive limit} \end{aligned}

Why this step? The square-root output is non-negative, so only the positive solution of the quadratic can be a limit.

Answer. $x_2\approx2.6458$ , $x_3\approx2.8795$ ; the possible limit is $3$ .

Check. Later values $2.959559744\ldots$ , $2.986489492\ldots$ move towards $3$ , and $\sqrt{2(3)+3}=3$ .

Change only the constant $3$ to $4$ . The same reasoning now identifies a different fixed point.

Worked example 2: preserving the branch condition

Question. Use $x_{n+1}=\sqrt{2x_n+4}$ , with $x_1=2$ , to find $x_2$ , $x_3$ and the possible positive limit.

Plan. Repeat the substitutions, then solve the resulting quadratic.

\begin{aligned} x_2&=\sqrt8=2.828427125\ldots \\ x_3&=3.107547948\ldots && \text{use stored }x_2 \\ \alpha^2-2\alpha-4&=0 && \text{fixed-point equation} \\ \alpha&=1+\sqrt5 && \alpha\ge0 \end{aligned}

Why this step? Squaring creates the candidate $1-\sqrt5<0$ , which cannot equal the non-negative square root.

Answer. $x_2\approx2.8284$ , $x_3\approx3.1075$ ; the possible limit is $1+\sqrt5\approx3.2361$ .

Check. $(1+\sqrt5)^2=2(1+\sqrt5)+4$ , and both sides of the unsquared equation are positive.

Checking a limit against the unsquared recurrence prevents an algebraic extra solution slipping through.

Watch out

Wrong: every root of the squared equation is a possible limit. Right: enforce the original recurrence's domain and sign restrictions as well.

For comparison, $x_{n+1}=\sqrt{5-x_n}$ from $x_1=2$ gives $1.732050808\ldots$ , $1.807746993\ldots$ , $1.786687719\ldots$ , alternating about its positive fixed point. Your turn changes only the coefficient of $x_n$ in the last worked recurrence from $2$ to $-1$ .

Derive Newton-Raphson from a tangent

At a current estimate $x_n$ , replace the curve by its tangent at $(x_n,f(x_n))$ . The tangent's intercept with the $x$ -axis becomes $x_{n+1}$ . When the tangent follows the curve well, that intercept can be much closer to a root.

[DIAGRAM: asset_slug: p1_9_3_solve_equations_using_the_newton_raphson_method_and_other_recurrence_relations__diagram_01; asset_name: Tangents improve an estimate or cause a large jump; caption: Newton-Raphson follows a tangent to its axis intercept; a small gradient can send it far away; description: Left panel shows tangents to x squared minus 2 at 2 and 1.5 with successive axis intercepts. Right panel shows the nearly horizontal tangent to (x minus 1) squared minus 0.04 at x=1.03 meeting the axis near 1.68167, beyond the nearer root 1.2.]
Newton-Raphson follows a tangent to its axis intercept

The tangent equation gives the formula rather than merely a picture to memorise:

\begin{aligned} y-f(x_n)&=f'(x_n)(x-x_n) && \text{point-gradient form} \\ -f(x_n)&=f'(x_n)(x_{n+1}-x_n) && \text{set }y=0 \end{aligned}

Divide by the gradient, provided it is non-zero.

Newton-Raphson iteration

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

Use where $f$ is differentiable and $f'(x_n)\ne0$ . Convergence to the intended root is not guaranteed.

In the formula booklet.

A numerical tangent calculation makes the geometry concrete.

Worked example 3: deriving one step geometrically

Question. For $f(x)=x^2-2$ , use the tangent at $x_0=2$ to find $x_1$ .

Plan. Find the point and gradient, then the tangent's intercept.

\begin{aligned} f(2)&=2,\qquad f'(2)=4 && \text{point and slope} \\ y-2&=4(x-2) && \text{tangent equation} \\ -2&=4(x_1-2) && y=0 \\ x_1&=1.5 \end{aligned}

Why this step? The intercept is where the tangent predicts the curve reaches zero.

Answer. The next estimate is $1.5$ .

Check. $\sqrt2\approx1.4142$ ; $1.5$ is closer than the start $2$ .

Change only the constant in the curve to $3$ . The slope at $2$ remains $4$ , but the point's height changes.

Worked example 4: a changed tangent height

Question. For $f(x)=x^2-3$ , use the tangent at $x_0=2$ to find $x_1$ .

Plan. Use the new function value with the same derivative.

\begin{aligned} f(2)&=1,\qquad f'(2)=4 \\ y-1&=4(x-2) && \text{new tangent} \\ x_1&=2-\frac14=1.75 && \text{axis intercept} \end{aligned}

Why this step? The correction is height divided by gradient, so changing the height changes the step.

Answer. $x_1=1.75$ .

Check. This is closer to $\sqrt3\approx1.7321$ than $2$ was.

The derivative is the slope at the current estimate, not at the root you are trying to find.

Key idea

The correction $-f(x_n)/f'(x_n)$ is the horizontal movement from the current point to the tangent's axis intercept.

Apply that geometry with new numbers.

Apply the Newton-Raphson formula

First write $f(x)=0$ , differentiate, then insert the current estimate into both numerator and denominator. Keep the subtraction and brackets visible: a negative residual can make the estimate increase.

Worked example 5: two Newton-Raphson steps

Question. Use $x_1=1.5$ to find $x_2$ and $x_3$ for $x^3-x-2=0$ .

Plan. Differentiate the cubic and repeat the supplied formula using stored values.

\begin{aligned} f'(x)&=3x^2-1 && \text{differentiate} \\ x_2&=1.5-\frac{1.5^3-1.5-2}{3(1.5)^2-1} \\ &=1.521739130\ldots && f(1.5)=-0.125 \\ x_3&=x_2-\frac{x_2^3-x_2-2}{3x_2^2-1} \\ &=1.521379806\ldots \end{aligned}

Why this step? Reusing the unrounded $x_2$ preserves the accuracy of the second tangent calculation.

Answer. $x_2=1.521739130\ldots$ , $x_3=1.521379806\ldots$ .

Check. The residual at $x_3$ is about $0.000000589$ , far smaller in magnitude than $-0.125$ .

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Last Accessed

Pure Mathematics (Papers 1 and 2)

Statistics (Paper 3, Section A)

Mechanics (Paper 3, Section B)

Pure 9.3 - Newton-Raphson and recurrence relations

Use a recurrence to chase a fixed point

Derive Newton-Raphson from a tangent

Apply the Newton-Raphson formula

Continue this lesson.

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