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 $$f(x)=0.25 x^3-0.5 x^2-x-3$$.

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.

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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

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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