Newton-Raphson Method

A LevelAQAEdexcelOCR

Newton-Raphson Method Revision

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:


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:


Substituting this into the Newton-Raphson formula:


Starting with x_0=6:


Using the formula again to find the following iterations:







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




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










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:



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:










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

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




Substituting in x=0.5:




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

Additional Resources


Exam Tips Cheat Sheet

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Formula Booklet

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Newton-Raphson Method Worksheet and Example Questions

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Newton-Raphson Method and Other Recurrence Relations

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