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Summary

Newton-Raphson method

In a nutshell

The Newton-Raphson method is a way to quickly find a root. It uses differentiation and tangential lines to locate roots. The method is quick when it works, but runs into issues if the starting point $x_0$ is not chosen carefully.

Newton-Raphson diagram

The method uses tangent lines to find the root. Below is the function $f(x)=0.25 x^3-0.5 x^2-x-3$ .

Maths; Numerical methods; KS5 Year 13; Newton-Raphson method

To construct the diagram, you take an $x_0$ value and find the point $(x_0,f(x_0))$ . You then create a tangent line to $y=f(x)$ at the point $(x_0,f(x_0))$ . This line will have the slope $f'(x)$ . The point where this line intersects the $x$ -axis becomes $x_1$ . You then take a tangent to $y=f(x)$ at the point $(x_1,f(x_1))$ and repeat until you find the root. Below is what the diagram should look like.

Newton-Raphson formula

The Newton-Raphson formula is $x_{n+1}=x_n-\dfrac{f(x_n)}{f'(x_n)}$ . This formula will allow you to find subsequent $x_n$ values as they become closer to the root.

Procedure

1.	Differentiate $f(x)$ to find $f'(x)$ .
2.	Choose a value to be $x_0.$
3.	Use $x_0$ to find $x_1$ with the formula $x_1=x_0-\dfrac{f(x_0)}{f'(x_0)}$
4.	Use $x_1$ to find $x_2$ , and repeat until you find the root.

Failures of the Newton-Raphson method

You have to be careful picking $x_0$ values for the Newton-Raphson method to work. If your chosen $x_0$ value is too close to a turning point on the curve, the slope of the tangent $f'(x)$ will be close to zero. The subsequent $x_1$ value will be far away from the root.

Another issue arises if a value chosen for $x_0$ , or if any value for $x_n$ at all, is directly on the turning point. The method fails, as to complete $f'(x_n)$ would result in a division by zero. The tangent at a turning point will not intersect with the $x$ -axis and will not facilitate the finding of a root.

Example 1

Use the Newton-Raphson method on the function $f(x)=1.5x^3+2x^2-8x-4$ to find a root rounded to $3\space d.p$ . Take $x_0=5$ as your starting point.

First, find $f'(x)$ by differentiation.

$\begin{aligned}f(x)&=1.5x^3+2x^2-8x-4\\f'(x)&=(3)(1.5)x^{3-1}+(2)(2)x^{2-1}-(1)(8)x^{1-1}-(0)(4)\\f'(x)&=4.5x^2+4x-8\\\end{aligned}$

Use the Newton-Raphson method to find the root.

$\begin{aligned}x_{n+1}&=x_n-\dfrac{f(x_n)}{f'(x_n)}\\\\f(x)&=1.5x^3+2x^2-8x-4\\f'(x)&=4.5x^2+4x-8\\x_0&=5\\x_1&=5-\dfrac{f(5)}{f'(5)}\\\\x_1&=5-\dfrac{1.5(5)^3+2(5)^2-8(5)-4}{4.5(5)^2+4(5)-8}\\x_1&=3.4458\\x_2&=3.4458-\dfrac{f(3.4458)}{f'(3.4458)}=2.5414\\\\x_3&=2.5414-\dfrac{f(2.5414)}{f'(2.5414)}=2.1185\\\\x_4&=2.1185-\dfrac{f(2.1185)}{f'(2.1185)}=2.0077\\\\x_5&=2.0077-\dfrac{f(2.0077)}{f'(2.0077)}=2.0000\\\\x_6&=2-\dfrac{f(2)}{f'(2)}=2\\\\\end{aligned}$

The root of $f(x)$ is $\underline{2}$ .

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Unit 1

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Unit 2

Locating roots

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Unit 3