1. Iteration The solution lies between 0 and 1. e.g. To find integer bounds for we can sketch and  0 and 1 are the integer bounds. We often start by finding numbers ( bounds ) that lie on either side of the solution. If these are integers we call them integer bounds. Our first approximation to , is any number between the bounds, say We usually call the solution , so. - Sign Change

2. Iteration Rearrange the equation to the form Find : Change of sign The change of sign Our 1 st estimate of  is between these values, say Define : Let Find : The Algebraic Method:  If we want to show that 0 and 1 are integer bounds, we show that has different signs at 0 and 1. - Sign Change

3. Iteration Solution: (c) ( Confirm bounds are 0 and 1 ) e.g. 1 (a) Using a graphical calculator, or otherwise, sketch, on the same axes, the graphs of and Rearrange equation to : Define : The change of sign (b) From the sketch, find integer bounds for the solution, , of the equation (c) Use an algebraic method to confirm these are correct and give a 1 st approximate solution. A 1 st approximation is any number between 0 and 1. So, - Sign Change

4. Iteration You may spot lots of ways of doing this. e.g. Rearrange the equation to the form. The 1 st approximate solution lies anywhere between the bounds. The next stage is to improve this estimate. (i) Square: (ii) Rearrange: Cube root: (iii) Rearrange: Divide by : e.g. For the equation : - Calculator Method

5. Iteration to iterate means to repeat Let’s take the 2 nd arrangement: Our 1 st estimate of  we will call x 0. We substitute x 0 into the r.h.s. of the formula and the result gives the new estimate x 1. We now have We will then keep repeating the process so we write the formula as This is called an iterative formula. ( Some people start with x 1 which is just as good. ) - Calculator Method

6. Iteration SUMMARY To find an approximation to a solution ( or root ) of an equation:  Find a 1 st approximation, often by finding integer bounds for the solution. Let this be x 0.  Rearrange the equation into the form  Write the arrangement as an iterative formula:  Key x 0 into a calculator and ENTER.  Key the r.h.s. of the formula into the calculator, replacing x with ANS.  Press ENTER as many times as required to get the solution to the specified accuracy. - Calculator Method

7. Iteration e.g. 1(a) Show that the equation has a root  in the interval. (b) Using the arrangement write down an iterative formula and use it to find the root correct to 4 decimal places. Solution: (a) Let Change of sign

8. Iteration Solution: (b) gives Let It takes about 7 iterations to reach (4 d.p.)

9. Iteration Some arrangements of an equation give formulae which do not give a solution. We earlier met 3 arrangements of ( to 6 d.p. ) We used (ii) with to find the solution Trying (i) with gives and after a while the sequence just oscillates between 1 and 0. The iterative sequence does not converge.

10. Iteration gives Trying the arrangement with The iteration then fails because we are trying to square root a negative number. Some arrangements of an equation give an iterative sequence which converges to a solution; others do not converge.

11. Iteration SUMMARY To draw a diagram illustrating iteration:  Draw and on the same axes.  Mark on the x -axis and draw a line parallel to the y -axis from to ( the curve ).  Continue the line, going parallel to the x -axis to meet  Continue the line, going parallel to the y -axis to meet Repeat 4 different diagrams are possible. - Convergence Diagrams

12. Iteration The gradients of for the converging sequences are between  1 and +1 Staircase: diverging Cobweb: converging Staircase: converging Cobweb: diverging - Convergence Diagrams