1. C3 Chapter 4: Numerical Methods
Dr J Frost
Last modified: 30th August 2015

2. Overview ? ? ? Finding the root of a function 𝑓 𝑥 is to:
solve the equation 𝒇 𝒙 =𝟎
However, for some functions, the ‘exact’ root is either complicated and difficult to calculate: ? ?
𝒙 𝟑 +𝟐 𝒙 𝟐 −𝟑𝒙+𝟒=𝟎
or there’s no ‘algebraic’ expression at all! (involving roots, logs, sin, cos, etc.) ?
𝒙− 𝐜𝐨𝐬 𝒙 =𝟎
Exact solution not expressible 
In an exam you are expected to:
Be able to prove that a root lies within a range.
Use ‘numerical methods’ to find approximations for a root.
Prove a solution is correct to a certain degree of accuracy.
𝒚=𝒙− 𝐜𝐨𝐬 𝒙
Root

3. #1: Proving a solution lies in a range
Show that 𝑓 𝑥 =𝑒 𝑥 +2𝑥−3 has a root between 𝑥=0.5 and 𝑥=0.6 ?
𝑓 0.5 =−0.351 𝑓 0.6 =0.022…
There is a change in sign, so root must lie between 0.5 and 0.6
Bro Exam Tip: In the mark scheme they’re looking for:
Finding the function output for the two values.
Referring to a ‘change in sign’.
0.5,−0.351
0.6, 0.022
If the 𝑦 value goes from negative to positive or vice versa, then clearly the 𝑦 values must pass 0 somewhere in between.

4. #2: Using iteration to approximate a root
𝑒 is just a constant like 𝜋 is. We’ll learn about it in a later chapter, but for now you can find it to the right of your calculator.
Edexcel C3 Jan 2013 a ?
0= 𝑒 𝑥−1 +𝑥−6 6−𝑥= 𝑒 𝑥−1 ln 6−𝑥 =𝑥−1 ln 6−𝑥 +1=𝑥
! To solve 𝑓 𝑥 =0 by an iterative method, rearrange into a form 𝑥=𝑔 𝑥 and use the iterative formula 𝑥 𝑛+1 =𝑔 𝑥 𝑛
We’ll see why it works later.

5. #2: Using iteration to approximate a root
Edexcel C3 Jan 2013
𝑥 0 , 𝑥 1 , 𝑥 2 represent successively better approximations of the root, where 𝑥 0 is the starting value.
Bro Tip: Initially type 𝑥 0 (i.e. 2) onto your calculator.
Now just type:
ln 6−𝐴𝑁𝑆 +1
And then spam your = key to get successive iterations. b ?
𝑥 0 =2 𝑥 1 = ln 6−2 +1=2.386… 𝑥 2 = ln 6−2.386… +1=2.2847… 𝑥 3 =2.3125

6. #3: Proving a solution to a given accuracy
Edexcel C3 Jan 2013
Thinking back to Year 8, if the root is correct to 3dp, what’s the smallest and greatest value it could be?
Smallest:
Greatest:
Clearly if we can prove the solution lies in this range, then it would round to How can we prove the solution lies in this range?
As we did before!
𝒇 𝟐.𝟑𝟎𝟔𝟓 =−𝟎.𝟎𝟎𝟎𝟐𝟕𝟓 𝒇 𝟐.𝟑𝟎𝟕𝟓 =𝟎.𝟎𝟎𝟒𝟒𝟐
Change of sign so solution is in range [𝟐.𝟑𝟎𝟔𝟓, 𝟐.𝟑𝟎𝟕𝟓]. ? ? ?

7. Why do it work?
Why do we rearrange to the form 𝑥=𝑔(𝑥) before using 𝑥 𝑛+1 =𝑔 𝑥 𝑛 . Why does it (usually) converge to the correct root?
𝑥= 2 𝑥 2 +2
𝑥 𝑛+1 = 2 𝑥 𝑛 2 +2
𝑓 𝑥 =− 𝑥 3 +2 𝑥 2 −2
1: Since 𝑥= 2 𝑥 we’re looking for the point where the lines 𝑦=𝑥 and 𝑦= 2 𝑥 intersect.
4: And we repeat this two-step process of calculating 𝑦 and then making the new 𝑥 that 𝑦.
We can see we’re spiralling inwards towards the root.
3: But that output gets used as the new input to the iterative formula, i.e. the 𝑦 value 𝑥 𝑛+1 gets used as the new 𝑥. This is equivalent to moving horizontally to the red 𝑦=𝑥 line.
2: Suppose we started with the approximation 𝑥 0 =1.5. Then would give us the 𝑦 value on the blue line.
𝑥 0

8. Complete the past paper questions provided (solutions on next slides)
Exercises
Complete the past paper questions provided
(solutions on next slides)
File Ref: C3-Chp4-NumericalMethodsExamQuestion

9. June 2009 Q1

10. Jan 2010 Q2

11. June 2012 Q2

12. June 2011 Q2

13. June 2008 Q7