Linear Interpolation (Edexcel IAL Further Maths)

Spec point — 3.1

Linear Interpolation

What is linear interpolation?

• Linear interpolation draws a straight line between the points A and B with coordinates $( a , \mathrm{f} ( a ) )$ and $( b , \mathrm{f} ( b ) )$, where $a<x<b$ is an interval containing a root to the equation $\mathrm{f} ( x ) = 0$

  • As $\mathrm{f} ( a )$ and $\mathrm{f} ( b )$ have different signs, A and B are on either side of the x-axis

  • The approximation to the root is where this straight line cuts the x-axis

    • This is at the point X with coordinates $( x , 0 )$
    • Points A, X and B lie on the same straight line

• It helps to use the gradient formula

  • The gradient between $( x_{1} , y_{1} )$ and $( x_{2} , y_{2} )$ is $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

How do I apply linear interpolation using gradients?

• This is best explained with an example

• If the root of $\mathrm{f} ( x ) = 0$ lies in the interval $[ 1 , 2 ]$ and $\mathrm{f} ( 1 ) = - 5$ and $\mathrm{f} ( 2 ) = 10$, then:

  • A is the point $( 1 , - 5 )$

  • B is the point $( 2 , 10 )$

  • X is the point $( x , 0 )$

    • This is the x-intercept of the line AB
    • A, X and B lie on the same straight line

  • The gradient of AX equals the gradient of XB

    • The order matters (not BX)

  • Use the gradient formula to set these equal

    • $\frac{0 - ( - 5 )}{x - 1} = \frac{10 - 0}{2 - x}$ giving $\frac{5}{x-1}=\frac{10}{2-x}$

  • Solve the equation to find $x$

    • $5 ( 2 - x ) = 10 ( x - 1 )$
    • expand and solve to get $x=\frac{4}{3}$

• This process can be **repeated **to get the next approximation

• This gradient method does** not** require a sketch

What other methods can I use to do linear interpolation?

• You can sketch the points A, B and X on a diagram and use similar triangles either side of the x-axis to find an equation in $x$

  • This leads to the same equation as the gradients above

• Or you can use coordinate geometry to find the equation of the line AB

  • Use $y - y_{1} = m ( x - x_{1} )$

  • Then find its x-intercept

    • Substitute in $y=0$ and solve for $x$

Exam Hint If the calculations involve lengthy decimals, you can use $\mathrm{f} ( a )$ and $\mathrm{f} ( b )$ to stand for their numbers in your working!

Worked Example The equation $\mathrm{f} ( x ) = 0$ where $\mathrm{f} ( x ) = x^{3} - \sqrt{x} - 2$ has a root in the interval $[ 1.4 , 1.5 ]$.

Use linear interpolation once to find an approximation to the root, giving your answer to 3 decimal places.

Find $\mathrm{f} ( 1.4 )$ and $\mathrm{f} ( 1.5 )$

$\begin{gathered} \mathrm{f} ( 1.4 ) & = 1.4^{3} - \sqrt{1.4} - 2 = - 0.439215... \\ \mathrm{f} ( 1.5 ) & = 1.5^{3} - \sqrt{1.5} - 2 = 0.150255... \end{gathered}$

It is easier to write them as $\mathrm{f} ( 1.4 )$ and $\mathrm{f} ( 1.5 )$ for now Let A, B and X be the points $( 1.4 , \mathrm{f} ( 1.4 ) )$, $( 1.5 , \mathrm{f} ( 1.5 ) )$ and $( x , 0 )$ Use the gradient formula $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$* to find the gradient of *AX

gradient of* AX is *$\frac{0 - \mathrm{f} ( 1.4 )}{x - 1.4}$

Use the gradient formula $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$* to find the gradient of *XB

gradient of* XB is *$\frac{ \mathrm{f} ( 1.5 ) - 0}{1.5 - x}$

Set the two gradients equal to each other

$\frac{0 - \mathrm{f} ( 1.4 )}{x - 1.4} = \frac{ \mathrm{f} ( 1.5 ) - 0}{1.5 - x}$

Simplify and cross multiply

$\begin{gathered} \frac{- \mathrm{f} ( 1.4 )}{x - 1.4} & = \frac{ \mathrm{f} ( 1.5 )}{1.5 - x} \\ - \mathrm{f} ( 1.4 ) ( 1.5 - x ) & = \mathrm{f} ( 1.5 ) ( x - 1.4 ) \end{gathered}$

Expand and collect the $x$ terms on the right-hand side

$\begin{gathered} - 1.5 \mathrm{f} ( 1.4 ) + \mathrm{f} ( 1.4 ) x & = \mathrm{f} ( 1.5 ) x - 1.4 \mathrm{f} ( 1.5 ) \\ 1.4 \mathrm{f} ( 1.5 ) - 1.5 \mathrm{f} ( 1.4 ) & = \mathrm{f} ( 1.5 ) x - \mathrm{f} ( 1.4 ) x \end{gathered}$

Factorise out the $x$, then make $x$ the subject

$\begin{gathered} 1.4 \mathrm{f} ( 1.5 ) - 1.5 \mathrm{f} ( 1.4 ) & = x [ \mathrm{f} ( 1.5 ) - \mathrm{f} ( 1.4 ) ] \\ \frac{1.4 \mathrm{f} ( 1.5 ) - 1.5 \mathrm{f} ( 1.4 )}{ \mathrm{f} ( 1.5 ) - \mathrm{f} ( 1.4 )} & = x \end{gathered}$

Substitute in $\mathrm{f} ( 1.4 )$ and $\mathrm{f} ( 1.5 )$ from above

$\begin{gathered} \frac{1.4 \times 0.150255... - 1.5 \times ( - 0.439215... )}{0.150255... - ( - 0.439215... )} & = x \\ 1.47451... & = x \end{gathered}$

Give your final answer to 3 decimal places

Final answer: $1.475$to 3 decimal places

If you did the working without$\mathrm{f} ( 1.4 )$and$\mathrm{f} ( 1.5 )$, keep lots of decimal places