# Newton-Raphson Method

## 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.

**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:

Substituting this into the **Newton-Raphson formula**:

Starting with x_0=6:

x_1=6-\dfrac{6^2-8(6)+11}{2(6)-8}=6.25Using the formula again to find the following iterations:

x_2=6.25-\dfrac{6.25^2-8(6.25)+11}{2(6.25)-8}=6.236111111x_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.

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

**[5 marks]**

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.

**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