Sign Change Re-cap Show that each of the following have a solution for x: 1.The intersection of (between 1.8 and 1.9) and 2.The intersection of(between.

Sign Change Re-cap

Show that each of the following have a solution for x:1. The intersection of (between 1.8 and 1.9)

and2. The intersection of (between 3.6 and 3.7)

and

3. The equation (between 0.4 and 0.5)

𝑦=𝑥4−6 𝑥 𝑦=1/𝑥

𝑦=𝑠𝑖𝑛𝑥+1 𝑦=log10𝑥

√𝑥=4 𝑥−1

Rearranging equations

Find as many ways as you can of rearranging to make x the subject:

𝑥4−5 𝑥2+1=0

𝑥2−10 𝑥+1=01

√𝑥+3=1−5𝑥

2𝑥+𝑥=4𝑥

Iteration

.Use iteration to approximate solutions

to difficult equations.

Begin to understand convergence and staircase and cobweb diagrams.

starter

The 1st approximate solution lies anywhere between the bounds. The next stage is to improve this estimate.

)(xg• Rearrange the equation to the form .xYou may spot lots of ways of doing this. e.g.

31 xx e.g. For the equation :

31

1 x x

2

1

x

x x

You may spot lots of ways of doing this. e.g.

x )(xg• Rearrange the equation to the form .

x 23 )1( x 31 xx

The 1st approximate solution lies anywhere between the bounds. The next stage is to improve this estimate.

(i) Square:or (ii) Rearrange:

Cube root:or (iii) Rearrange:

2xDivide by :

xx 13 31 xx

xx 13 31 xx

31 xx e.g. For the equation :

to iterate means to repeat

Let’s take the 2nd arrangement: 31

1 xx 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

31

01 1 xx

31

1 1 nn xx

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. )

31

131

01 5011 xxx

Starting with we get500 x

Because we are going to repeat the calculation, we use the ANS function on the calculator.

31

1 1 nn xx So,

• Type and press ENTER

50• Type the r.h.s. of the equation, replacing

x with ANS, using the ANS button, giving 31

1 ANS• Press ENTER and you get ( 6

d.p. ) 6641050

• Pressing ENTER again replaces with and gives the next estimate and so on.

664105050

66410501 x( 6

d.p. )

Casio emulator

500 x66410501 x

56987802 x

If we continue to iterate we eventually get

6054230 ( to 6

d.p. )

Although I’ve only written down 6 decimal places, the calculator is using the greatest possible accuracy.

We get

6105000000050 orof our answer.

Since the answer is correct to 6 decimal places, the exact value of must be within

Error Bounds

Tip: The index equals the number of d.ps. in the answer.

Exercise

3 21 25 nn xx

1. (a) Show that the equation has a solution a between 2 and 3.

2523 xx

(b) Use the iterative formulawith to find the solution, giving your answer correct to 4 d.p.

520 x

Solutions 1. (a) Show that the equation has

a solution between 2 and 3.2523 xx

Solution: (a) Let

025)( 23 xxxf

0132522)2( 23 f

0112533)3( 23 fChange of sign (continuous

function) 32

3 21 25 nn xx (b) Use the iterative formula

with to find the solution, giving your answer correct to 4 d.p.

520 x

62582...,65662,52 10 xx( 4

d.p. )

Domino trail

In table groups

Convergence diagrams

Some arrangements of an equation give formulae which do not give a solution.

31 xx We earlier met 3 arrangements of

6054230 ( to 6

d.p. )

We used (ii) with to find the solution

500 x

Now try (i) with 500 x

We get ...,940,300,770 321 xxx

and after a while the sequence just oscillates between 1 and 0.

This iterative sequence does not converge.

x 23 )1( x(i) 3

1

1 xx(ii)2

1

x

xx

(iii)

We get ,060,171 21 xx

21

1

n

nn

x

xx

Now try the formula with .500 x

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.

)(xgy

xy

31

1 xy xy 3

1

1 xy xy

0x

Exercise B

Page 136

Convergence diagrams

Iteration

31

1 x x

2

1

x

x x

You may spot lots of ways of doing this. e.g.

x )(xg• Rearrange the equation to the form .

x 23 )1( x 31 xx

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: 2xDivide by :

xx 13 31 xx

xx 13 31 xx

31 xx e.g. For the equation :

Iteration

to iterate means to repeat

Let’s take the 2nd arrangement: 31

1 xx 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

31

01 1 xx

31

1 1 nn xx

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. )

IterationSUMMARY 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 )(xgx Write the arrangement as an iterative

formula:)(1 nn xgx

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.

Iteration

xx 13(ii) xx 13(iii)23 )1( xx (i)

31

1 xx 2

1

x

xx

Some arrangements of an equation give formulae which do not give a solution.

31 xx We earlier met 3 arrangements of

6054230 ( to 6

d.p. )

We used (ii) with to find the solution

500 x

Trying (i) with 500 x

gives ...,940,300,770 321 xxxand after a while the sequence just oscillates between 1 and 0.

The iterative sequence does not converge.

Iteration

gives ,060,171 21 xx

2

1

x

xx

Trying the arrangement with 500 x

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.

Iteration

Solution: (b) gives3 2 xx 3

1 2 nxnx

Let 510x 3 5131 22 0 xx

4141386512 x377613 x

37351It takes about 7 iterations to reach (4 d.p.)