1. Numerical methods
You have encountered equations that can be solved by a direct algebraic method – for example any linear or quadratic function can be easily solved.
Unfortunately, many other types of equations, often with real applications in fields as diverse as engineering and weather prediction, cannot directly be solved.
Mathematicians have developed numerical methods to find solutions to any desired degree of accuracy. These vary in sophistication and how fast they ‘find’ the root and in FP1 we consider 3 different methods:
Interval bisection
Very slow but fool-proof!
Linear interpolation
Faster but requires careful algebra
Newton-Raphson
Extremely fast with a standard rule to apply, but can go wrong.

2. From (a), function is increasing in interval [1,2]
Interval Bisection
WB10
(a) Show that the equation f (x) = 0 has a root  between x =1 and x = 2.
Establish root by showing change of sign in interval
Change of sign root in interval
(b) Starting with the interval [1, 2], use interval bisection twice to find an interval of width 0.25 which contains 
From (a), function is increasing in interval [1,2]

3. From (a), function is decreasing in interval [0,1]
Care must be taken establishing whether the function is increasing or decreasing Eg
(a) Show that the equation f (x) = 0 has a root  between x =0 and x = 1.
Establish root by showing change of sign in interval
Change of sign root in interval
From (a), function is decreasing in interval [0,1]
(b) Starting with the interval [0, 1], use interval bisection to find the correct to 1dp
A decreasing function reverses the way you interpret the results so use a diagram!

4. GCSE recap: similar triangles
Eg two similar triangles
The lengths of sides are multiplied by the scale factor k
x k
Eg if triangles RST and UVW are similar S V T
Hence R W U
This property can be used to solve problems

5. Similar shapes Eg PQ is parallel to ST
PQ = 4cm, ST = 10cm, QR = 3cm, RT = 9cm Q
4cm
3cm P
ΔSRT and ΔQRP are similar as: R
9cm
QPR and STR are alternate T
PQR and TSR are alternate
PRQ and SRT are vertically opposite
10cm
Work out the length of RS S
Make separate sketches
Similarity R R
Taking part of this statement,
3cm
9cm Q
4cm P S
10cm T

6. So function is increasing in interval [1 , 2]
This method can be used to approximate the roots of equations by a method called linear interpolation Eg
Use linear interpolation once on this interval to find an approximation to .
Give your answer to 3 decimal places.
The equation has a solution in the interval [1,2].
In this method, you imagine the curve to be a straight line between the limits of the initial interval, then use similar triangles to locate where this line crossed the x-axis
Similar triangles
So function is increasing in interval [1 , 2]

7. So function is increasing in interval [1.6 , 1.8]
WB11
The root  of the equation f (x) = 0 lies in the interval [1.6,1.8].
Use linear interpolation once on the interval [1.6, 1.8] to find an approximation to .
Give your answer to 3 decimal places.
As f(1.6) and f(1.8) are awkward, call their absolute values A and B to make the algebra easier and store them in your calculator’s memory
So function is increasing in interval [1.6 , 1.8]
Similar triangles
NB: the textbook has questions which ask for solutions correct to 1 or 2dp, which could take many iterations of this method and prove incredibly awkward and time-consuming. In reality you will only be asked to ‘use linear interpolation once’ in the exam.

8. The Newton-Raphson Method
There is a much faster method for finding roots of equations: Eg
Rearrange an equation to the form f(x) = 0
Differentiate to obtain f’(x)
Pick an initial solution x1
Evaluate to obtain
an improved solution x2
Repeatedly applying
will obtain increasingly accurate solutions
etc

9. WB12
(a) Write down, to 3 decimal places, the value of f(1.3) and the value of f(1.4).
(b) Taking 1.4 as a first approximation to α, apply the Newton-Raphson procedure once to f(x) to obtain a second approximation to α, giving your answer to 3 dp

10. WB13
(a) Show that f (x) = 0 has a root  between 1.4 and 1.5.
Change of sign root in interval
(b) Taking 1.45 as a first approximation to , apply the Newton-Raphson procedure
once to
to obtain a second approximation to , correct to 3dp

11. Making your calculator do the hard work
From the previous example, to solve , keep applying
A scientific calculator can do this for you if you tell it at a starting value and instruct it what to do with this value repeatedly:
Type ‘2 =‘ to make the Ans key x1
Then type and keep pressing =

12. Why does the Newton-Raphson method work?
Repeating this process will ‘home in’ on the root
Although much more powerful and ‘mechanical’ than the other methods, it does occasionally go wrong…

13. So function is increasing in interval [1 , 2]
Numerical methods
Eg The equation has a solution in the interval [1,2].
So function is increasing in interval [1 , 2]
Interval bisection
Linear interpolation
Similar triangles
Newton-Raphson using x1 = 2