Newton-Raphson Method

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Newton-Raphson Method

The Newton-Raphson Method is a different method to find approximate roots. The method requires you to differentiate the equation you’re trying to find a root of, so before revising this topic you may want to look back at differentiation to refresh your mind.

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Using the Newton-Raphson Method

Finding roots of an equation in the form $f(x)=0$ , requires you to find $f'(x)$ and then use the following formula:

$Large{x_{n+1}=x_n-dfrac{f(x_n)}{f'(x_n)}}$

This method iteratively finds the $x$ -intercept of the tangent to the graph of $f(x)$ at $x_n$ and then uses this value as $x_{n+1}$ .

Why the Newton-Raphson Method Can Fail

Similar to other iteration formulas, if your starting point of $x_0$ is too far away from the actual root, the Newton-Raphson method may diverge away from the root.
The Newton-Raphson method can also fail if the gradient of the tangent at $x_n$ is close or equal to $textcolor{red}{0}$ . This is shown in the diagram below, where the tangent has a gradient very close to $0$ , so the point where it meets the $x$ -axis will be very far away from the root, so the sequence of iterations may diverge.

Furthermore, if the tangent at a point on $f(x)$ is horizontal, i.e. $x_n$ is a stationary point, then the Newton-Raphson method will fail. This is because the tangent will never meet the $x$ -axis, so there will be no further iterations. In addition to this, the tangent is horizontal when $f'(x)=0$ , so the formula would not work as you cannot divide by $0$ .

Example: The Newton Raphson Method

Find a root of the equation $x^2-8x+11=0$ to $5$ decimal places using $x_0=6$

First we need to differentiate $f(x)=x^2-8x+11$ :

$f'(x)=2x-8$

Substituting this into the Newton-Raphson formula:

$x_{n+1}=x_n-dfrac{x^2-8x+11}{2x-8}$

Starting with $x_0=6$ :

$x_1=6-dfrac{6^2-8(6)+11}{2(6)-8}=6.25$

Using the formula again to find the following iterations:

$x_2=6.25-dfrac{6.25^2-8(6.25)+11}{2(6.25)-8}=6.236111111$

$x_3=6.236111111-dfrac{6.236111111^2-8(6.236111111)+11}{2(6.236111111)-8}=6.236067978$

$x_4=6.236067978-dfrac{6.236067978^2-8(6.236067978)+11}{2(6.236067978)-8}=6.236067977$

Thus a root of $x^2-8x+11=0$ is $6.23607$ to $5$ decimal places.

Newton-Raphson Method Example Questions

Question 1: Use the Newton-Raphson method with $x_0=1$ , to find a root of the equation $x^3-2x^2-5x+8=0$ to $5$ decimal places.

[4 marks]

Firstly we need to differentiate $f(x)=x^3-2x^2-5x+8$

$f'(x)=3x^2-4x-5$

Now we need to apply the Newton-Raphson formula, starting with $x_0=1$ :

$x_1=1-dfrac{1^3-2(1)^2-5(1)+8}{3(1)^2-4(1)-5}=dfrac{4}{3}$

$x_2=dfrac{4}{3}-dfrac{(dfrac{4}{3})^3-2(dfrac{4}{3})^2-5(dfrac{4}{3})+8}{3(dfrac{4}{3})^2-4(dfrac{4}{3})-5}=1.362962963$

$x_3=1.362962963-dfrac{(1.362962963)^3-2(1.362962963)^2-5(1.362962963)+8}{3(1.362962963)^2-4(1.362962963)-5}=1.36332811$

$x_4=1.36332811-dfrac{(1.36332811)^3-2(1.36332811)^2-5(1.36332811)+8}{3(1.36332811)^2-4(1.36332811)-5}=1.363328238$

So a root of $x^3-2x^2-5x+8=0$ is $1.36333$ to $5$ decimal places.

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Question 2: Use the Newton-Raphson method with $x_0=2$ , to find a root of the equation $3xln{x}=7$ to $4$ significant figures.

[5 marks]

Firstly, we need to rearrange the equation so it is in the form $f(x)=0$ :

$3xln{x}-7=0$

Then we need to differentiate $f(x)=3xln{x}-7$ , to do this we will need to use the product rule:

$begin{aligned} f'(x) &=3ln{x}+3xtimes dfrac{1}{x} &=3ln{x}+3 &=3(ln{x}+1) end{aligned}$

Now we need to apply the Newton-Raphson formula starting with $x_0=2$ :

$x_1=2-dfrac{3(2)ln{2}-7}{3(ln{2}+1)}=2.559336473$

$x_2=2.559336473-dfrac{3(2.559336473)ln{2.559336473}-7}{3(ln{2.559336473}+1)}=2.522322342$

$x_3=2.522322342-dfrac{3(2.522322342)ln{2.522322342}-7}{3(ln{2.522322342}+1)}=2.522182638$

$x_4=2.522182638-dfrac{3(2.522182638)ln{2.522182638}-7}{3(ln{2.522182638}+1)}=2.522182636$

So the root of $3xln{x}=7$ is $2.522$ to $4$ significant figures.

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Question 3: Explain why starting with $x_0=0.5$ for the equation $-x^2+x+12=0$ will fail when using the Newton-Raphson method.

[2 marks]

Differentiating $f(x)=-x^2+x+12$ :

$f'(x)=-2x+1$

Substituting in $x=0.5$ :

$f'(0)=-2(0.5)+1=0$

Thus, the Newton-Raphson method will fail because you cannot divide by $0$

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Additional Resources

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Specification Points Covered

I2 – Solve equations using the Newton-Raphson method and other recurrence relations of the form $x_{n+1}=g(x_{n})$ ; understand how such methods can fail
I4 – Use numerical methods to solve problems in context