1. P2 Chapter 8 CIE Centre A-level Pure Maths © Adam Gibson

2. In Chapter 8, we will cover: Finding roots numerically Sign change rule and the decimal search The iterative method

3. The Numerical Solution of Equations Any equation can be rearranged into the form f(x)=0. Example: can be rearranged to: i.e. we set The solutions of this equation are the roots of the function These roots are the values of x where the graph of f(x) crosses the x-axis. In general, there may be any number of these roots, including zero.

4. Why “numerical”? What is a “numerical” solution and why do we have a whole chapter for it? Because most equations are very difficult to solve! Here are a few examples of “normal” equations: Can you solve these? In many cases, you do not know how many roots there are, if any. We say that these equations “cannot be solved analytically”. We can still try to find an approximate solution, using a calculator or a computer.

5. The sign change rule Let’s start with a very simple equation: First, rearrange it into the form f(x)=0: Try to sketch the graph of this function. Use what you know!

6. Did you get the plot right? We need to find the approximate value of this root

7. Sign change rule and decimal search We know that the derivative is always positive. We know that the function is continuous. Therefore, there is only one root and we can use the sign change rule to find it. Read the sign change rule on page 108. Now we’re going to use the process of decimal search to find the root accurate to 3 decimal places. First, between which two integers does the root lie? Answer: 0 and -1.

8. Decimal search continued What are the values of f(0) and f(-1)? f(0)=1, f(-1)=-0.63 This suggests that the root is about 3/8 of the distance from -1 to 0. Look at this diagram: 0.63 1.0 ~3/8 ~5/8 “Chord approximation”

9. Decimal search continued So: we next try the value x = -0.6 f(-0.6)=-0.05 This is negative; what should we do next? We try the next higher value at this accuracy:- f(-0.5)=0.107 This is positive; what do you conclude? Apply the sign change rule. The root is between -0.6 and -0.5. The root should be about 1/3 of the distance from -0.6 to -0.5. So let’s try -0.57.

10. Decimal search continued f(-0.57)=-0.0044 This is negative; what should we do next? We try the next higher value at this accuracy:- f(-0.56)=0.0112 Apply the sign change rule. The root is between -0.57 and -0.56. The root should be about 4/15 of the distance from -0.57 to -0.56. You can use 1/3 instead of 4/15. There is no point in trying to be very accurate with this measurement! Which value of x should we try next?

11. Decimal search continued f(-0.567)=0.00023 This is positive; what should we do next? We try the next lower value at this accuracy:- f(-0.568)=-0.0013 Apply the sign change rule. The root is between -0.568 and -0.567. Remember: we want an answer to 3 decimal places, so we just need to choose one of these answers. To decide between -0.568 and -0.567, which single value of x should we use?

12. Decimal search continued f(-0.5675) = -0.0006 This is negative; what is our final answer? The only solution to the equation is, accurate to 3 d.p. This process can be used up to any number of decimal places, as required. It is laborious *, but can easily be programmed into a computer. * adj. = involving a lot of work, from Latin labor, laborare

13. Finding roots by iteration As discussed, the decimal search technique is laborious and may, in some difficult cases, be too slow. There is a more elegant technique, based on the concept of sequences. The idea is based on this trick: Rewrite as Choose a starting value such as Then rewrite the equation as an inductively defined sequence:

14. Finding roots by iteration - continued Let’s return to our original equation: As a sequence: Set Find Answer: -0.13534,-0.87342,-0.41752 (Store intermediate values in the calculator)

15. Finding roots by iteration The iterations converge to the value -0.567 (3 d.p.) We chose a starting value of -2

16. Finding roots by iteration - continued When and how does this process work? A difficult and complex question, which we will not study! Read the box on p. 113 Will the iterative process work for any function? Certainly not!

17. Convergence and divergence A quick review from Chapter 14 P1: A geometric series is convergent if …?

18. Convergence and divergence - continued Sometimes the sequence does not converge to the root of f(x), instead it diverges. Going back to our example: Instead of writing: we will write: and we will guess a starting value of The following table illustrates what happens:

19. Convergence and divergence - continued Have we made a mistake? There is no mathematical error, it’s just that this technique does not always work. Iteration r = 1-0.56-0.57982 r = 2-0.579818-0.54504 r = 3-0.54504-0.6069 r = 4-0.606896-0.4994 r = 4-0.499398-0.69435 r = 5-0.694352-0.36478 r = 6 -0.364777-1.00847 r = 7-1.0084690.008434 r = 80.008434ERROR!

20. Convergence and divergence - continued The iterations diverge away from the root Outside the domain of ln(-x)