Numerical solutions of equations

Numerical solutions of equations

Introduction• This chapter gives you several

methods which can be used to solve complicated equations to given levels of accuracy

• These are similar to methods which computers and calculators will use, and hence can be used in computer programming

Teachings for Exercise 2A

Numerical solutions of equationsYou can solve equations of the

form f(x) = 0 using interval bisection

Interval bisection is a variation on Trial and Improvement which you will have

seen at GCSE level

Interval Bisection is an iterative process which allows us to find a root to

whatever degree of accuracy we wish (usually 1-2 decimal places!)

An iterative process is one which is a short set of instructions which are then

repeated as many times as needed

As a result such processes can be used in computers and calculators so

they can solve equations

2A



Use Interval Bisection to find √11 to 1 decimal place

Set this up as an equation:

2A

𝑥=√11𝑥2=11

𝑥2−11=0

Square both sides

Subtract 11

𝑓 (𝑥 )=𝑥2−11

𝑓 (3 )=(3)2−11

¿−2

𝑓 (4 )=(4)2−11

¿5

Sub in integers until we find a change of sign

So an solution lies between 3 and 4

Numerical solutions of equationsYou can solve equations of the form

f(x) = 0 using interval bisection

Use Interval Bisection to find √11 to 1 decimal place

So a solution lies between 3 and 4.

Now we set up a table, subbing these 2 values into f(x), as well as the

midpoint of these

When you have found the midpoint and substituted it in, choose the positive and negative answers

closest to 0

The answer will be between these. Now repeat the process for these 2

numbers

2A

𝑓 (𝑥 )=𝑥2−11

𝒇 (𝒂+𝒃𝟐 )𝒂+𝒃

𝟐𝒂 𝒃𝒇 (𝒂) 𝒇 (𝒃)

3 4 3.5−2 5 1.25

𝑓 (𝑥 )=𝑥2−11

3 3.5 1.25−2 3.25 −0.438

3.25 −0.438 3.5 1.25 3.375 0.391

3.25 −0.438 3.375 0.3913.3125−0.027

3.3125−0.027 3.375 0.3913.343750.181

Our answer must be between 3.3125 and 3.34375

To one decimal place, the answer therefore must be 3.3!



Show that a root of the equation:

lies between 0 and 1

Use interval bisection 4 times to find an approximation for this root

2A

𝑥3−5 𝑥+3=0

𝑓 (𝑥 )=𝑥3−5 𝑥+3

𝑓 (0 )=(0)3−5(0)+3

¿3

𝑓 (1 )=(1)3−5(1)+3

¿−1

Sub in 0 and 1 to show the sign of

the answer changes

As the sign has changed, a solution must lie between 0 and 1…



Show that a root of the equation:

lies between 0 and 1

Use interval bisection 4 times to find an approximation for this root

2A

𝑥3−5 𝑥+3=0

𝒇 (𝒂+𝒃𝟐 )𝒂+𝒃

𝟐𝒂 𝒃𝒇 (𝒂) 𝒇 (𝒃)

0 1 0.53 −1 0.625

𝑓 (𝑥 )=𝑥3−5 𝑥+3

0.5 0.625 1 −1 0.75−0.328

0.75−0.3280.5 0.625 0.6250.119

0.6250.1190.75−0.3280.6875−0.113

Our approximation is the final bisection

0.6875 (or round if necessary)

Teachings for Exercise 2B


form f(x) = 0 using linear interpolation

In linear interpolation, you first draw a sketch of the function between 2

intervals

Then, you draw a straight line between the interval coordinates (this will be a rough approximation to the

curve

You can then use similar triangles to find the place the straight line crosses

the x-axis (the ‘root’ as it were)

You then update the interval and repeat the process…

2B



A solution of the equation:

lies in the interval [1,2]. Use linear interpolation to find this root, correct

to one decimal place.

2B

𝑥3+4 𝑥−9=0

𝑓 (𝑥 )=𝑥3+4 𝑥−9

𝑓 (1 )=(1)3+4 (1 )−9

¿−4

𝑓 (2 )=(2)3+4 (2 )−9

¿7

Sub in 1 and 2 to show the sign of

the answer changes

As the sign has changed, a solution must lie between 1 and 2…






2B

𝑥3+4 𝑥−9=0

𝑓 (𝑥 )=𝑥3+4 𝑥−9𝑓 (1 )=−4𝑓 (2 )=7

Now sketch the graph between x = 1 and x = 2 (the limits you were given)

It does not have to be really accurate!

(1,-4)

(2,7)

After sketching the graph between the limits, draw a straight line between them

The place this crosses the x-axis is an approximation for the root

You can call it x and then use similar triangles to find its value

x x

y






2B

𝑥3+4 𝑥−9=0

(1,-4)

(2,7)

x x

y

4

7

x-1 2-x

Imagine creating triangles using the x-axis and the coordinates marked

Label the sides, using x as the place the straight line crosses the x-axis

These two triangles are similar – ie) They have the same angles (both have a right angle and two other pairs that are the same – you can see this from the vertically opposite angles at the centre and the ‘alternate’ angles ta the top and bottom!)

In similar triangles, a long side divided by a shorter side will give the same answer (provided that equivalent sides are used!)






2B

𝑥3+4 𝑥−9=0

(1,-4)

(2,7)

x x

y

4

7

x-1 2-x

𝑥−14 =

2− 𝑥7

7𝑥−728 =

8−4 𝑥28

7 𝑥−7=8−4 𝑥11𝑥=15

𝑥=1511 (1.3636…)

Short side ÷ longer side in each triangle gives the same answer…

Multiply the whole left by 7 and the whole right by 4Multiply by 28 to

leave the numerators

Add 4x, Add 7

Divide by 11So the value of x is

15/11 or 1.3636…






2B

𝑥3+4 𝑥−9=0

(1,-4)

(2,7)

x x

y

4

7

x-1 2-x

So the value of x is 15/11 or 1.3636…

15/11

𝑓 (𝑥 )=𝑥3+4 𝑥−9

𝑓 ( 1511 )=( 1511 )3

+4 ( 1511 )−9𝑓 ( 1511 )=−1.009767……

Sub in 15/11 to find the value on the curve at this

pointCalculate

(15/11,-1.009….)

We now repeat the process from this new point…






2B

𝑥3+4 𝑥−9=0

(2,7)

x

y

(15/11,-1.009….)

As you can see, the new estimate for the root is closer than the first approximation

Repeat the process using these new values (strictly speaking the original estimate was x1, and this one is x2 – use this notation when you solve these problems!)

x






2B

𝑥3+4 𝑥−9=0

(2,7)

x

y

(15/11,-1.009….)

x

7

1.009…

x-15/11 2-x

𝑥− 15111.009…

=2− 𝑥7

7 (𝑥− 1511 )=1.009 (2−𝑥 )

7 𝑥− 10511 =2.018−1.009 𝑥

8.009 𝑥=11.563…

𝑥=1.4438

Cross-multiply

Expand brackets

Rearrange

Solve

Repeat this process several times until your answer is accurate to the requested

degree Try to be as accurate as possible at

each stage, avoiding rounding too much

Draw a new sketch at each stage Use x1, x2, x3 to represent each

approximation!

Teachings for Exercise 2C


form f(x) = 0 using the Newton-Raphson process

The Newton-Raphson formula is as follows:

2C

𝑥𝑛+1=𝑥𝑛−𝑓 (𝑥𝑛 )𝑓 ′ (𝑥𝑛)

Our current approximation for

the root

Our next approximation for

the root

The function we are solving, with our current

approximation substituted in

The derivative of the function we are solving,

with our current approximation substituted

in




Use the Newton-Raphson process to find the root of the equation:

Use x0 = 3 and give your answer to 2 decimal places.

Find the function and its derivative first…

2C


𝑥4+𝑥2=80

𝑥4+𝑥2=80𝑥4+𝑥2−80=0

𝑓 (𝑥 )=𝑥4+𝑥2−80

𝑓 ′ (𝑥 )=4 𝑥3+2𝑥

Rearrange

Differentiate

𝑓 (𝑥 )=𝑥4+𝑥2−80𝑓 ′ (𝑥 )=4 𝑥3+2𝑥𝑥0=3





2C


𝑥4+𝑥2=80

𝑓 (𝑥 )=𝑥4+𝑥2−80𝑓 ′ (𝑥 )=4 𝑥3+2𝑥𝑥0=3


𝑥1=𝑥0−(𝑥0 )4+(𝑥0 )2−80 4 (𝑥0 )3+2 (𝑥0 )

𝑥1=3−(3 )4+(3 )2−80 4 (3 )3+2 (3 )

𝑥1=2.912

𝑥1=2.912

Our current approximation is x0, replace the fraction with equivalent expressions

Sub in x0 = 3

Calculate





2C


𝑥4+𝑥2=80

𝑓 (𝑥 )=𝑥4+𝑥2−80𝑓 ′ (𝑥 )=4 𝑥3+2𝑥𝑥0=3

𝑥2=𝑥1−(𝑥1 )4+(𝑥1 )2−80 4 (𝑥1 )3+2 (𝑥1 )

𝑥2=2.912−(2.912 )4+ (2.912 )2−80 4 (2.912 )3+2 (2.912 )

𝑥2=2.908…

𝑥1=2.912

Sub in x1 = 2.912

Calculate

Repeat the process, but now we use the value of x1 to find x2

𝑥2=2.908…

As both x1 and x2 round to 2.91, then this is the solution to 2 decimal

places!



Use the Newton-Raphson process twice to find approximate a root of

the equation:

Use x0 = 2 as your first approximation and give your answer to 3 decimal places.

2C


𝑥3+2𝑥2−5𝑥−4=0

𝑓 (𝑥 )=𝑥3+2𝑥2−5 𝑥−4

𝑓 ′ (𝑥 )=3 𝑥2+4 𝑥−5Differentiate

𝑓 (𝑥 )=𝑥3+2𝑥2−5 𝑥−4𝑓 ′ (𝑥 )=3 𝑥2+4 𝑥−5𝑥0=2




the equation:


2C


𝑥3+2𝑥2−5𝑥−4=0

𝑓 (𝑥 )=𝑥3+2𝑥2−5 𝑥−4𝑓 ′ (𝑥 )=3 𝑥2+4 𝑥−5𝑥0=2


𝑥1=𝑥0−(𝑥0)

3+2 (𝑥0 )2−5 (𝑥0 )−4 3 (𝑥0 )2+4 (𝑥0 )−5

𝑥1=2−(2)3+2 (2 )2−5 (2 )−4 3 (2 )2+4 (2 )−5

𝑥1=1.866…

𝑥1=1.866…

Sub in x0 as our first

approximation and replace

the parts of the fraction

Sub in x0 = 2

Calculate




the equation:


2C


𝑥3+2𝑥2−5𝑥−4=0

𝑓 (𝑥 )=𝑥3+2𝑥2−5 𝑥−4𝑓 ′ (𝑥 )=3 𝑥2+4 𝑥−5𝑥0=2

𝑥2=𝑥1−(𝑥1)

3+2 (𝑥1 )2−5 (𝑥1 )−4 3 (𝑥1 )2+4 (𝑥1 )−5

𝑥2=1.866−(1.866)3+2 (1.866 )2−5 (1.866 )−4

3 (1.866 )2+4 (1.866 )−5

𝑥2=1.8558 . .

𝑥1=1.866…

Sub in x1 =

1.866

Calculate

𝑥2=1.856 (3𝑑𝑝 )

It is important to note that the Newton-Raphson method will not always work, sometimes tending

away from a root rather than towards it

Usually, choosing a different first approximation will correct this!

Summary• You have learnt several iterative

methods for solving equations

• Although these may seem cumbersome, the ideas involved in these are used in computers and calculators

• They will also work when other methods fail to find answers!

2C

Documents

Numerical solutions of equations