Dr J Frost (jfrost@tiffin.kingston.sch.uk) GCSE Iteration Dr J Frost (jfrost@tiffin.kingston.sch.uk) www.drfrostmaths.com @DrFrostMaths Last modified: 20th June 2017

Term to term formulae A sequence is generated using the following term-to-term formula. 𝒙 𝟏 is the first term. 𝑥 1 =3, 𝑥 𝑛+1 = 𝑥 𝑛 +2 What would be the first 4 terms of the sequence? 𝒙 𝟏 =𝟑 𝒙 𝟐 = 𝒙 𝟏 +𝟐=𝟓 𝒙 𝟑 = 𝒙 𝟐 +𝟐=𝟕 𝒙 𝟒 = 𝒙 𝟑 +𝟐=𝟗 ? ? ? ? How therefore do we interpret 𝑥 𝑛+1 = 𝑥 𝑛 +2? “Each term is the previous term plus 2” ?

Test Your Understanding 1 Why do you think the sequence refers to the first term as 𝑝 0 rather than 𝑝 1 ? Because 𝒑 𝟎 indicates the number of slugs ‘after 0 days’, i.e. the initial value. ? 119 ? 2 A sequence is defined as: 𝑥 1 =5, 𝑥 2 =2 𝑥 𝑛+2 =2 𝑥 𝑛+1 − 𝑥 𝑛 Determine 𝑥 5 . 𝒙 𝟑 =𝟐 𝒙 𝟐 − 𝒙 𝟏 =−𝟏 𝒙 𝟒 =𝟐 𝒙 𝟑 − 𝒙 𝟐 =−𝟑 𝒙 𝟓 =𝟐 𝒙 𝟒 − 𝒙 𝟑 =−𝟒 ?

Iteration ? ? a ? ? ? ? Solve 𝑥= 𝑥+1 using: The quadratic formula (hint: square both sides first and make one side 0) Iteration We turn the equation into a term-to-term formula, by making the left-hand-side 𝑥 𝑛+1 and any instance of 𝑥 on the RHS 𝑥 𝑛 . Don’t worry yet about why it works… ? a ? 𝑥 2 =𝑥+1 𝑥 2 −𝑥−1=0 𝑎=1, 𝑏=−1, 𝑐=−1 𝑥= 1± 1+4 2 𝑥= 1+ 5 2 =1.618… 𝑥 𝑛+1 = 𝑥 𝑛 +1 𝑥 0 =1 You will always be given a starting value in the exam. The 0 means we’ve done “0 iterations”. Use your term-to-term formula to get successively better solutions to the original equation. 𝑥 1 = 𝑥 0 +1 =1.41421… 𝑥 2 = 𝑥 1 +1 =1.553777… 𝑥 3 = 𝑥 2 +1 =1.598053… 𝑥 4 = 𝑥 3 +1 =1.611847… ? Use the ANS key to reuse the previous answer. If you type in 1=, then type 𝐴𝑁𝑆+1 , you can then spam the equals key! Bro Note: We only want the positive solution because in the original equation, the RHS is positive (square rooting can’t give a negative value) and thus the LHS, i.e. 𝑥 must be positive. ? ? ? We could keep going for ever to keep getting more accurate solutions. The exam will usually tell you how many iterations it wants (here we did 4).

Why use iteration then? In the previous example, we could have found out the ‘exact’ answer using the quadratic formula. However, for some functions, the exact solution is either complicated and difficult to calculate: ? 𝒙 𝟑 +𝟐 𝒙 𝟐 −𝟑𝒙+𝟒=𝟎 or there’s no ‘exact’ expression at all! (involving roots, sin, cos, etc.) ? 𝒙− 𝐜𝐨𝐬 𝒙 =𝟎 Exact solution not expressible  Iteration therefore is an intelligent way of doing ‘trial and improvement’ that gets us the solution correct to progressively more and more decimal places on each iteration, even if there is no way to write the answer in a 100% accurate way.

Another Example Bro-Exam Note: Most exam questions have two parts: the first will get you to rearrange the equation so you have 𝑥 on its own on the LHS [Edexcel New SAMs Paper 3H Q14c] a) Show that the equation 𝑥 3 +4𝑥=1 can be arranged to give 𝑥= 1 4 − 𝑥 3 4 ? 𝟒𝒙=𝟏− 𝒙 𝟑 𝒙= 𝟏 𝟒 − 𝒙 𝟑 𝟒 Bro Tip: Use the target expression as clues for how to rearrange. Each term is over 4, which suggests we need to divide by 4 at some point. b) Starting with 𝑥 0 =0, use the iteration formula 𝑥 𝑛+1 = 1 4 − 𝑥 𝑛 3 4 twice, to find an estimate for the solution of 𝑥 3 +4𝑥=1. Write all the digits on your calculator display. 𝒙 𝟎 =𝟎 𝒙 𝟏 = 𝟏 𝟒 − 𝟎 𝟑 𝟒 =𝟎.𝟐𝟓 𝒙 𝟐 = 𝟏 𝟒 − 𝟎.𝟐 𝟓 𝟑 𝟒 =𝟎.𝟐𝟒𝟔𝟎𝟗𝟑𝟕𝟓 For iterative methods, give your answer as a decimal NOT as a fraction. Why? Because fractions imply an exact answer, but this method only gets an approximation of the solution) ? ?

Test Your Understanding a) Show that the equation 𝑥 3 −𝑥−19=0 can be arranged to give 𝑥= 3 𝑥+19 ? 𝑥 3 =𝑥+19 𝑥= 3 𝑥+19 b) Starting with 𝑥 0 =0, use the iteration formula 𝑥 𝑛+1 = 3 𝑥 𝑛 +19 four times to find an approximate solution to 𝑥 3 −𝑥−19=0 Write all the digits on your calculator display. ? 𝑥 0 =0 𝑥 1 =2.6680401… 𝑥 2 =2.7878899… 𝑥 3 =2.7930050… 𝑥 4 =2.7932236…

Roots and Showing Roots Lie in a Range Terminology: The roots are the solutions of an equation in the form 𝑓 𝑥 =0. If 𝑦=𝑓(𝑥) then the roots are the 𝑥-intercepts of the graph. 2 −𝑥 −𝑥=0 Show that the equation has a root between 0 and 1. 𝟐 −𝟎 −𝟎=𝟏>𝟎 𝟐 −𝟏 −𝟏=− 𝟏 𝟐 <𝟎 So 𝟐 −𝒙 −𝒙 must be equal to 0 for 𝟎<𝒙<𝟏 a ? The key is to show that substituting into the left gives <0 for one of them and >0 for the other (make sure you write “<0” and “>0”. Therefore it must pass 0 somewhere in between. Rearrange to the form 𝑥=𝑓 𝑥 to set up an iterative formula. 𝟐 −𝒙 =𝒙 𝒙 𝒏+𝟏 = 𝟐 − 𝒙 𝒏 b ? Choose a starting value (from step (a)) and generate the sequence. 𝒙 𝟎 =𝟎.𝟔 𝒙 𝟏 =𝟎.𝟔𝟔𝟎, 𝒙 𝟐 =𝟎.𝟔𝟑𝟑, 𝒙 𝟑 =𝟎.𝟔𝟒𝟓, 𝒙 𝟒 =𝟎.𝟔𝟒𝟎, 𝒙 𝟓 =𝟎.𝟔𝟒𝟐 c ? We could have chosen any starting value between 0 and 1. Preferably near the middle!

Exercise For each iterative formula, find 𝑥 1 , 𝑥 2 , 𝑥 3 and 𝑥 4 correct to three decimal places 𝑥 𝑛+1 = 1 𝑥 𝑛 +1, 𝑥 0 =1.5 𝒙 𝟏 =𝟏.𝟔𝟔𝟕, 𝒙 𝟐 =𝟏.𝟔, 𝒙 𝟑 =𝟏.𝟔𝟐𝟓, 𝒙 𝟒 =𝟏.𝟔𝟏𝟓 𝑥 𝑛+1 =2− 𝑥 𝑛 3 8 , 𝑥 0 =1.5 𝒙 𝟏 =𝟏.𝟓𝟕𝟖, 𝒙 𝟐 =𝟏.𝟓𝟎𝟗, 𝒙 𝟑 =𝟏.𝟓𝟕𝟏, 𝒙 𝟒 =𝟏.𝟓𝟏𝟔 𝑥 𝑛+1 = 5 𝑥 𝑛 −1 2 , 𝑥 0 =2 𝒙 𝟏 =𝟐.𝟏𝟐𝟏, 𝒙 𝟐 =𝟐.𝟏𝟗𝟐, 𝒙 𝟑 =𝟐.𝟐𝟑𝟏, 𝒙 𝟒 =𝟐.𝟐𝟓𝟒 𝑥 𝑛+1 = 1 5−2 𝑥 𝑛 , 𝑥 0 =0.5 𝒙 𝟏 =𝟎.𝟐𝟓, 𝒙 𝟐 =𝟎.𝟐𝟐𝟐, 𝒙 𝟑 =𝟎.𝟐𝟐𝟎, 𝒙 𝟒 =𝟎.𝟐𝟏𝟗 1 3 Show that 𝑥 2 +4𝑥−29=0 can be written in the form 𝑥= 29−4𝑥 . 𝒙 𝟐 =𝟐𝟗−𝟒𝒙 𝒙= 𝟐𝟗−𝟒𝒙 Use the iteration formula 𝑥 𝑛+1 = 3 𝑥 𝑛 +19 to find 𝑥 4 to 3 decimal places. Start with 𝑥 0 =0. 𝒙 𝟏 =𝟐.𝟔𝟔𝟖 𝒙 𝟐 =𝟐.𝟕𝟖𝟖 𝒙 𝟑 =𝟐.𝟕𝟗𝟑 𝒙 𝟒 =𝟐.𝟕𝟗𝟑 Show that 𝑥− 𝑥 −1=0 has a root between 2.6 and 2.7 𝟐.𝟔− 𝟐.𝟔 −𝟏=−𝟎.𝟎𝟏𝟐<𝟎 𝟐.𝟕− 𝟐.𝟕 −𝟏=𝟎.𝟎𝟓𝟕>𝟎 Therefore 𝟐.𝟔<𝒙<𝟐.𝟕 Using 𝑥 𝑛+1 = 𝑥 𝑛 +1 and 𝑥 0 =2.6, find 𝑥 3 to 3 decimal places. 𝒙 𝟏 =𝟐.𝟔𝟏𝟐, 𝒙 𝟐 =𝟐.𝟔𝟏𝟔 𝒙 𝟑 =𝟐.𝟔𝟏𝟖 ? ? ? ? ? 4 ? ? For the equation 𝑥− 𝑥+3 =0, we can use the iterative formula 𝑥 𝑛+1 = 𝑥 𝑛 +3 and 𝑥 0 =2. Explain the relationship between the values of 𝑥 1 , 𝑥 2 , 𝑥 3 and the equation 𝑥− 𝑥+3 =0 𝒙 𝟏 , 𝒙 𝟐 , 𝒙 𝟑 are successfully closer approximations to a root/solution of 𝒙− 𝒙+𝟑 =𝟎 2 ? ?

Why does this method work? Note: You are not expected to know this for your GCSE exams. Solve 𝑥= 𝑥+1 This is known as a staircase diagram (given the shape!) Recall we used the recurrence 𝑥 𝑛+1 = 𝑥 𝑛 +1 𝒚=𝒙 𝒚= 𝒙+𝟏 Finding the solution to 𝑥= 𝑥+1 is the same as sketching 𝑦=𝑥 and 𝑦= 𝑥+1 and seeing the point at which they intersect. This value of 𝑥 is the solution to 𝑥= 𝑥+1 This gives 𝑥 1 =1.414… This is then fed back into 𝑥 𝑛 +1 for the next iteration, i.e. the 𝑦 value becomes the new 𝑥 value! This is equivalent to moving to the line 𝑦=𝑥. 𝒙 𝟏 𝒙 ∗ 𝒙 𝟎 When 𝑥 0 =1, we would find 𝑥 0 +1 . This is the 𝑦 value on the 𝑦= 𝑥+1 graph. We can repeat this process using 𝑥 1 =1.414… to get 𝑥 2 .

There are questions on iteration available on the DFM Homework Platform. www.drfrostmaths.com/homework