Interval Bisection (Edexcel IAL Further Maths)

Spec point — 3.1

Interval Bisection

What is interval bisection?

• Interval Bisection means splitting an interval containing a root into two halves and using a sign change test to find out which half contains the root

• For example, if $\mathrm{f} ( 1 ) < 0$ and $\mathrm{f} ( 2 ) > 0$ and $\mathrm{f} ( x )$ is continuous

  • Then a root must lie in $[ 1 , 2 ]$

  • Find the midpoint of the interval

    • $x=1.5$

  • Check the sign at the midpoint

    • For example, $\mathrm{f} ( 1.5 ) < 0$

  • Write down the half-interval that has the sign change

    • $\mathrm{f} ( 1 ) < 0$, $\mathrm{f} ( 1.5 ) < 0$ and $\mathrm{f} ( 2 ) > 0$
    • So $[ 1.5 , 2 ]$ has the sign change

• The process can be repeated

  • For example, if $\mathrm{f} ( 1.75 ) > 0$ then $[ 1.5 , 1.75 ]$ contains the sign change

How do I use interval bisection to estimate a root?

• If a root lies in the interval $[ 1 , 2 ]$, then the first approximation of the root, $x_{1}$, is simply the midpoint

  • $x_{1}=1.5$

• The second approximation, $x_{2}$, requires using Interval Bisection once

  • For example, giving $[ 1.5 , 2 ]$

  • Then find the midpoint

    • $x_{2}=1.75$

  • And so on

Exam Hint Read the question carefully to see if an approximation to the root is required, or the interval containing the root is required.

Worked Example The equation $x^{3}+x-16=0$ has a root in the interval $[ 2.1 , 2.9 ]$.

Use interval bisection to find an interval of width 0.2 that contains the root.

Find the width of the interval given in the question

$2.9-2.1=0.8$

Interval bisection halves the width To get a width of 0.2, interval bisection is needed twice Find the signs of the interval end-points

$\begin{gathered} \mathrm{f} ( 2.1 ) & = - 4.639 < 0 \\ \mathrm{f} ( 2.9 ) & = 11.289 > 0 \end{gathered}$

Find the midpoint of the interval

$x=\frac{2.1+2.9}{2}=2.5$

Find the sign of $\mathrm{f} ( x )$ at $x=2.5$

$\mathrm{f} ( 2.5 ) = 2.125 > 0$

Find the new halved interval containing the sign change Check the signs: $\mathrm{f} ( 2.1 ) < 0$, $\mathrm{f} ( 2.5 ) > 0$ and $\mathrm{f} ( 2.9 ) > 0$

$[ 2.1 , 2.5 ]$

Repeat the process Find the midpoint of the interval

$x=\frac{2.1+2.5}{2}=2.3$

Find the sign of $\mathrm{f} ( x )$ at $x=2.3$

$\mathrm{f} ( 2.3 ) = - 1.533 < 0$

Find the new halved interval containing the sign change Check the signs: $\mathrm{f} ( 2.1 ) < 0$, $\mathrm{f} ( 2.3 ) < 0$ and $\mathrm{f} ( 2.5 ) > 0$

$[ 2.3 , 2.5 ]$