Newton-Raphson Method (Edexcel IAL Further Maths)

Spec point — 3.1

Newton-Raphson Method

How do I apply the Newton-Raphson method?

• The Newton-Raphson method is a process for finding roots of equations

  • Equations must be rearranged into the form $\mathrm{f} ( x ) = 0$

• If $x=α$ is a root of the equation $\mathrm{f} ( x ) = 0$, then you need to choose $x_{0}$, the initial (first) approximation

  • This is usual given in the question (and is a value near to the root)

• You can then find successive approximations $x_{1}$, $x_{2}$, $x_{3}$, ... using the Newton-Raphson formula:

  • $x_{n + 1} = x_{n} - \frac{ \mathrm{f} ( x_{n} )}{ \mathrm{f} ' ( x_{n} )}$

  • This is an iterative formula

  • $\mathrm{f} ' ( x )$ is the derivative of $\mathrm{f} ( x )$

    • Work this out beforehand

• Increasing the number of approximations improves the accuracy

  • The Newton-Raphson method usually converges quickly to the root compared to other methods

Exam Hint The formula for the Newton-Raphson method is given in the Formulae Booklet.

How do I apply the Newton-Raphson method on my calculator?

• If your calculator has a recursion function

  • Input the formula, start value and number of steps
  • The output is a table showing all the approximations

• Alternatively:

  • Type in the value of $x_{0}$ and press =

    • This stores it in the "Ans" (answer) button

  • Type $\text{Ans} - \frac{ \mathrm{f} ( \text{Ans} )}{ \mathrm{f} ' ( \text{Ans} )}$ into your calculator and press =

    • This finds $x_{1}$

  • Without pressing any other button, press = again

    • This finds $x_{2}$

  • Repeatedly pressing = gives further approximations

    • The more you press it, the closer to the root it becomes

How does the Newton-Raphson method work geometrically?

• The method works by drawing a tangent to the curve $y = \mathrm{f} ( x )$ at $x=x_{0}$

  • then finding where the tangent cuts the x-axis
  • This x-intercept is the new approximation, $x_{1}$

• This process is then repeated

  • Draw the tangent to $y = \mathrm{f} ( x )$ at $x=x_{1}$
  • Find its x-intercept and call it $x_{2}$

• The approximations get closer and closer to the root

How do I know if the Newton-Raphson method will fail?

• The Newton-Raphson method fails if the initial value, $x_{0}$, satisfies $\mathrm{f} ' ( x_{0} ) = 0$

  • Algebraically, this is because the formula $x_{n + 1} = x_{n} - \frac{ \mathrm{f} ( x_{n} )}{ \mathrm{f} ' ( x_{n} )}$ cannot divide by zero

  • Geometrically, this is because $x_{0}$ is at a stationary point on the curve $y = \mathrm{f} ( x )$

    • A tangent drawn at a stationary point is horizontal
    • This will never intersect the x-axis

• The Newton-Raphson method also fails if the sequence $x_{1}$, $x_{2}$, $x_{3}$, ... diverges

  • This can happen if $x_{0}$ is chosen:

    • too far away from the root
    • or at a point where the gradient is very small

• The Newton-Raphson method is sometimes avoided when $\mathrm{f} ( x )$ is too tricky to differentiate

Worked Example The equation $2x^{4}-4x^{3}=-1$ has a solution in the interval $[ 1 , 2 ]$.

(a) Using $x_{0}=2$ as the first approximation to the solution, apply the Newton-Raphson method to find a second approximation.

Rearrange the formula into the form $\mathrm{f} ( x ) = 0$

$2x^{4}-4x^{3}+1=0$ so $\mathrm{f} ( x ) = 2 x^{4} - 4 x^{3} + 1$

(You could also use $-1+4x^{3}-2x^{4}$) Find $\mathrm{f} ' ( x )$ using differentiation

$\mathrm{f} ' ( x ) = 8 x^{3} - 12 x^{2}$

Substitute $\mathrm{f} ( x )$ and $\mathrm{f} ' ( x )$ into the Newton-Raphson formula $x_{n + 1} = x_{n} - \frac{ \mathrm{f} ( x_{n} )}{ \mathrm{f} ' ( x_{n} )}$

$x_{n + 1} = x_{n} - \frac{( 2 x_{n}^{4} - 4 x_{n}^{3} + 1 )}{( 8 x_{n}^{3} - 12 x_{n}^{2} )}$

Substitute $x_{0}=2$ into the formula to get $x_{1}$

$x_{1} = 2 - \frac{( 2 \times 2^{4} - 4 \times 2^{3} + 1 )}{( 8 \times 2^{3} - 12 \times 2^{2} )} = 2 - \frac{1}{16} = 1.9375$

Check this solution lies in the interval $[ 1 , 2 ]$

$x_{1}=1.9375$

$\frac{31}{16}$is also accepted

(b) Explain why the midpoint of the interval $( 1 , 2 )$ cannot be used as the first approximation when applying the Newton-Raphson method.

The Newton-Raphson method fails if $x_{0}$ satisfies $\mathrm{f} ' ( x_{0} ) = 0$ Find the midpoint of the interval

$x_{0}=\frac{1+2}{2}=1.5$

Check if $\mathrm{f} ' ( 1.5 ) = 0$

$\mathrm{f} ' ( 1.5 ) = 8 \times 1.5^{3} - 12 \times 1.5^{2} = 0$

Write a conclusion either about dividing by zero, or about having a horizontal tangent

Final answer: $\mathrm{f} ' ( 1.5 ) = 0$, but you cannot divide by zero in the formula$x_{n + 1} = x_{n} - \frac{ \mathrm{f} ( x_{n} )}{ \mathrm{f} ' ( x_{n} )}$ So 1.5 cannot be used as the first approximation

You could also say that$x=1.5$is at a stationary point where the tangent is horizontal, meaning it cannot intersect the x-axis to make$x_{1}$