Iteration

Some equations do not have “nice” solutions
- They are not integers (whole numbers), fractions or simple decimals
- Instead, they can be irrational decimal solutions that go on forever with no pattern
Iteration is a repeated process used to solve such equations
- the process starts with an initial value (starting value)
- after each stage of the process (after each "iteration"), a solution is produced
- the solutions get more and more accurate as more and more iterations are performed
  - these solutions are called estimates
Scientific calculators allow us to perform iterations very quickly using the ANS button
- Iteration questions will only be asked in the calculator exam

It Calc Notes fig1, downloadable IGCSE & GCSE Maths revision notes

Find the equation you would like to solve using iteration
- for example, x³ + x = 7
Rearrange this equation into the form x = f(x) by making any x the subject of the equation
- for example, $x = \sqrt[3]{7 - x}$
Replace the x on the left with x_n+1 (meaning the "next" value of x) and any x's on the right with x_n(meaning the "current" value of x)
- $x_{n + 1} = \sqrt[3]{7 - x_{n}}$
- This called the iterative formula
  - n and n+1 are just counters: n+1 is simply one more than n
  - n starts at 0 so the process starts with x₀_,the initial value
  - x₁ is your first estimate, x₂ is your second estimate, etc

Find a good initial (starting) value (x₀) near to the solution
- This is often given in the question, for example x₀ = 2
Store x₀ = 2 into your calculator (by typing 2 and pressing the "=" button)
- 2 is now stored under the "Ans" button
Type in the right-hand side of the iteration formula with "Ans" instead of x_n
- $\sqrt[3]{7 - Ans}$
Press "=" to find x₁ (be careful to only press "=" once)
- x₁ = 1.709975...
Without pressing any other button, press "=" again to find x₂
- x₂ = 1.742418...
Press "=" again to find x₃
- x₃ = 1.738849...
Repeat as many times as required
- the more you do, the closer the estimates get to the true solution

It Calc Notes fig2, downloadable IGCSE & GCSE Maths revision notes

x₁, x₂, x₃ ... etc are estimates to the solution of x = f(x)
- for example, x₁ = 1.709975..., x₂ = 1.742418..., x₃ = 1.738849... are estimates to $x = \sqrt[3]{7 - x}$
They are also estimates of solutions to any rearrangements of x = f(x)
- such as the original equation trying to be solved, x³ + x = 7
This makes x₁, x₂, x₃ ... estimates to the solution of the original equation
The more times you perform the iteration, the better the estimates get to the real solution

To find x₀(the initial / starting value), you are often asked to show that there is a solution in an interval
For example, show that there is a solution to x³ + x = 7 between 1 and 2
- Method 1: Leave a constant term (e.g. the 7) on the right, substitute x = 1 and x = 2 into the left and show that this gives values below and above 7
  - 1³ + 1 = 2 and 2³ + 2 = 10 which are below and above 7
  - A solution therefore lies between 1 and 2
- Method 2: Use "0" as your constant term on the right (by rearranging the equation into "... = 0"), then substitute in x = 1 and x = 2, showing this gives values below and above 0, i.e. negative and positive
- this is called a change of sign between 1 and 2
  - x³ + x - 7 = 0
  - Substitute x = 1 into the left-hand side: 1³ + 1 - 7 = -5 (negative)
  - Substitute x = 2 into the left-hand side: 2³ + 2 - 7 = 3 (positive)
  - A solution lies between 1 and 2 as there is a change of sign
Knowing an interval that contains the solution helps to find x₀
- If the solution is between 1 and 2 then you could choose either x₀ = 1 or x₀ = 2

Exam Tip

Be careful to not press =/EXE or "Ans" more than once at a time. If you do the best thing to do is to restart from the beginning.
Iteration questions always require working with a lot of decimal places, so write down all digits from your calculator display for x₁, x₂, etc. and round them at the end if necessary