Newton-Raphson Method

A LevelAQAEdexcelOCR

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.

A LevelAQAEdexcelOCR

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 $x_{n+1}$.

A LevelAQAEdexcelOCR

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 $0$.
A LevelAQAEdexcelOCR
A LevelAQAEdexcelOCR

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.

A LevelAQAEdexcelOCR

Newton-Raphson Method Example Questions

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.

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

$3x\ln{x}-7=0$

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

\begin{aligned} f'(x) &=3\ln{x}+3x\times \dfrac{1}{x} \\ &=3\ln{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 $3x\ln{x}=7$ is $2.522$ to $4$ significant figures.

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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