# “Teach A Level Maths” Vol. 2: A2 Core Modules

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“Teach A Level Maths” Vol. 2: A2 Core Modules
17: Iteration using © Christine Crisp

Module C3 "Certain images and/or photos on this presentation are the copyrighted property of JupiterImages and are being used with permission under license. These images and/or photos may not be copied or downloaded without permission from JupiterImages"

There are some equations that we can’t solve.
e.g. However, we can find an approximate solution to some of these equations. The approximation can be very accurate, say to 6 or more decimal places. There are several methods of finding approximate solutions and in this presentation we will study one of them.

There are 2 stages to getting a solution:
Stage 1. Find a 1st estimate Stage 2. Find a formula to improve the estimate. Sometimes we can just spot an approximate solution to an equation. Can you spot the approximate value of the solution to Ans: It’s quite close to 0 as the l.h.s. is then 0 and the r.h.s. is 1. If we can’t quickly spot an approximation, we can use a method involving finding bounds for the solution.

Bounds are numbers which lie on either side of the solution.
If these are integers we call them integer bounds. e.g. To find integer bounds for we can sketch and This is the point . . . where and so the x coordinate gives the solution to

0 and 1 are the integer bounds.
Bounds are numbers which lie on either side of the solution. If these are integers we call them integer bounds. e.g. To find integer bounds for we can sketch and The solution lies between 0 and 1. 0 and 1 are the integer bounds. a We usually call the solution a, so

0 and 1 are the integer bounds.
Bounds are numbers which lie on either side of the solution. If these are integers we call them integer bounds. e.g. To find integer bounds for we can sketch and The solution lies between 0 and 1. 0 and 1 are the integer bounds. a We usually call the solution a, so Our first approximation to a, is any number between the bounds, say

If we can use Autograph or a graphical calculator, sketching is a good method of finding the integer bounds. Even without a graph plotter you may have been able to sketch these 2 graphs. However, if we can spot likely bounds, or if we are given values and want to show they are bounds, we can use the algebraic method that follows and avoid sketching.

Rearrange the equation to get zero on one side. e.g. For , get
To show how the method works I’m going to sketch ( but you won’t usually have to do this ). The solution, a, is now where At a, a To the left of a, e.g. at x = 0, To the right of a, e.g. at x = 1,

Rearrange the equation to get zero on one side. e.g. For , get
To show how the method works I’m going to sketch ( but you won’t usually have to do this ). has opposite signs on the left and right of a. a To the left of a, e.g. at x = 0, To the right of a, e.g. at x = 1,

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. Rearrange the equation to the form Let Define : 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. Rearrange the equation to the form Let Define : Find : Change of sign Find : The change of sign

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. Rearrange the equation to the form Let Define : Find : Change of sign Find : The change of sign Our 1st estimate of a is between these values, say You must always include this line.

It is possible to find bounds that are closer than the nearest integers.
For example, to find bounds accurate to 1 decimal place, we could use a decimal search. So if we had , a decimal search would calculate looking for a change of sign. However, integer bounds are good enough for the method of iteration we are studying.

e.g. 1 (a) Using a graphical calculator, or otherwise, sketch, on the same axes, the graphs of
and (b) From the sketch, find integer bounds for the solution, a, of the equation (c) Use an algebraic method to confirm these are correct and give a 1st approximate solution. Solution: (a) (b) a is the x-value at the point of intersection, so 0 and 1 are integer bounds. a

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

Not all functions that have a sign change between 2 numbers have a solution to between the numbers.
Look at this function: There is a change of sign . . . but no solution between 0 and 1. Do you notice anything that explains this? Ans: The function has an asymptote between 0 and 1. We couldn’t draw the curve without lifting the pencil off the paper. We say it is discontinuous.

Not all functions that have a sign change between 2 numbers have a solution to between the numbers.
Look at this function: There is a change of sign . . . but no solution between 0 and 1. Do you notice anything that explains this? You are unlikely to meet discontinuous functions in this work so just remember the effect on solutions.

An equation has a solution which may consist of one or more roots.
Exercise 1. Using a graphical calculator or otherwise, sketch suitable graphs to find integer bounds for the solution to Give a 1st approximation to the solution. Use the algebraic method to show that has a root between 2 and 3. Give an approximation to this root. An equation has a solution which may consist of one or more roots.

The integer bounds for a are 1 and 2. So,
1. Sketch and Solution: a The integer bounds for a are 1 and 2. So, Any number between 1 and 2 could be used as the 1st approximation.

Use the algebraic method to show that
has a root between 2 and 3. Solution: Rearrange to Define Change of sign (continuous function) Any number between 2 and 3 could be used as the 1st approximation.

The 1st approximate solution lies anywhere between the bounds
The 1st approximate solution lies anywhere between the bounds. The next stage is to improve this estimate. Rearrange the equation to the form You may spot lots of ways of doing this. e.g. e.g. For the equation :

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

Let’s take the 2nd arrangement:
Our 1st estimate of a we will call x0. We substitute x0 into the r.h.s. of the formula and the result gives the new estimate x1. ( Some people start with x1 which is just as good. ) We now have We will then keep repeating the process so we write the formula as This is called an iterative formula. to iterate means to repeat

So, Starting with we get ( 6 d.p. ) Because we are going to repeat the calculation, we use the ANS function on the calculator. Type and press ENTER Type the r.h.s. of the equation, replacing x with ANS, using the ANS button, giving Press ENTER and you get ( 6 d.p. ) Pressing ENTER again replaces with and gives the next estimate and so on.

We get Although I’ve only written down 6 decimal places, the calculator is using the greatest possible accuracy. If we continue to iterate we eventually get ( to 6 d.p. ) Error Bounds Since the answer is correct to 6 decimal places, the exact value of must be within of our answer. Tip: The index equals the number of d.ps. in the answer.

SUMMARY To find an approximation to a solution ( or root ) of an equation: Find a 1st approximation, often by finding integer bounds for the solution. Let this be x0 . Rearrange the equation into the form Write the arrangement as an iterative formula: Key x0 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.

e.g. 1(a) Show that the equation has a root a 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 Then, Change of sign

e.g. 1(a) Show that the equation has a root a 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: (b) gives Let It takes about 7 iterations to reach (4 d.p.)

Exercise 1. (a) Show that the equation has a solution a between 2 and 3. (b) Use the iterative formula with to find the solution, giving your answer correct to 4 d.p. 2. (a) Show that the equation has a solution between 1 and 2. (b) Use the iterative formula with to find the solution, giving your answer correct to 4 d.p.

Solutions 1. (a) Show that the equation has a solution between 2 and 3. Solution: (a) Let Change of sign (b) Use the iterative formula with to find the solution, giving your answer correct to 4 d.p. ( 4 d.p. )

Solutions 2. (a) Show that the equation has a solution between 1 and 2. Solution: Let Change of sign (b) Use the iterative formula with to find the solution, giving your answer correct to 4 d.p. Solution: We need ( 4 d.p. )

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

Now try the formula with
We get 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. The next presentation investigates convergence of iterative sequences.

The following slides contain repeats of information on earlier slides, shown without colour, so that they can be printed and photocopied. For most purposes the slides can be printed as “Handouts” with up to 6 slides per sheet.

There are some equations that we can’t solve.
However, we can find an approximate solution to some of these equations. e.g. There are several methods of finding approximate solutions and in this presentation we will study one of them. The approximation can be very accurate, say to 6 or more decimal places.

There are 2 stages to getting a solution:
Stage 1. Find a 1st estimate Stage 2. Find a formula to improve the estimate. Sometimes we can just spot an approximate solution to an equation. The solution to If we can’t quickly spot an approximation, we can use a method involving finding bounds for the solution. is quite close to 0 as the l.h.s. is then 0 and the r.h.s. is 1.

0 and 1 are the integer bounds.
The solution lies between 0 and 1. e.g. To find integer bounds for we can sketch and a 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 a, is any number between the bounds, say We usually call the solution a, so

Rearrange the equation to the form
Find : Change of sign The change of sign Our 1st estimate of a is between these values, say Define : Let 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.

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, a, of the equation (c) Use an algebraic method to confirm these are correct and give a 1st approximate solution. A 1st approximation is any number between 0 and 1. So,

You may spot lots of ways of doing this. e.g.
Rearrange the equation to the form The 1st 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 :

to iterate means to repeat
Let’s take the 2nd arrangement: Our 1st estimate of a we will call x0. We substitute x0 into the r.h.s. of the formula and the result gives the new estimate x1. 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 x1 which is just as good. )

SUMMARY To find an approximation to a solution ( or root ) of an equation: Find a 1st approximation, often by finding integer bounds for the solution. Let this be x0 . Rearrange the equation into the form Write the arrangement as an iterative formula: Key x0 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.

e.g. 1(a) Show that the equation has a root a 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

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

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.

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.