REVISION SHEET – PURE MATHEMATICS 4furthermaths.org.uk/manager_area/files/teaching/NM/NMmei_rev.pdfthe Further Mathematics network – V 07 1 1 REVISION SHEET – NUMERICAL METHODS

the Further Mathematics network – www.fmnetwork.org.uk V 07 1 1

REVISION SHEET – NUMERICAL METHODS (MEI)

APPROXIMATING FUNCTIONS

(Finite) Difference Tables. Suppose that you are given a collection of data points for some function, for example:

Note: we must have equally spaced x-values to use any of the methods in this chapter which relate to looking at difference tables. xi fi Δfi Δ2fi Δ3fi Δ4fi Δ5fi0 0.2 1.6 2.7 1.1 0.1 −0.1 1 1.8 4.3 3.8 1.2 0 2 6.1 8.1 5 1.2 3 14.2 13.1 6.2 4 27.3 19.3 5 46.6

x 0 1 2 3 4 5 f(x) 0.2 1.8 6.1 14.2 27.3 46.6

The main ideas are: • Constructing and interpreting

difference tables • Using a difference table to

calculate Newton’s interpolating polynomial

• Lagrange’s

Polynomials/Method

Before the exam you should know: • How to construct and interpret a difference table, and the

notation used • The formula for the Newton Interpolating polynomial

(also known as Newton Forward Difference Method)

00 0

20 102

30 1 203

f ( ) f f

( )( ) f2!

( )( )( ) f ..3!

x xxh

x x x xh

x x x x x xh

.

−= + Δ +

− −Δ +

− − −Δ +

• Lagrange’s Polynomials which fit points without equally

spaced x-values. For example ( )( ) ( )(( )

( )( ) ( )( )( )( )

( )( )

A x b x c B x c x af xa b a c b c b a

C x a x bc a c b

)− − −= +

− − − −− −

+− −

−

goes through (a,A), (b, B) and (c, C)

Column of Second Differences

Column of Third Differences

Column of Fourth Differences

Column of Fifth Differences

Column of First Differences

This entry is calculated as

27.3 – 14.2 = 13.1

This entry is calculated as

8.1 – 4.3 = 3.8

As a rule of thumb, if the variation in a column is less than about 10% of the average value for that column. we would say that it is nearly constant

The third differences column is nearly constant so in Newton’s Interpolation formula and Newton’s Interpolating polynomial use terms up to and including Δ3f0 for a good approximation, unless you require an exact fit.


Newton’s Interpolating Polynomial (a.k.a Newton’s Forward Difference Method) Example The cubic 3f ( ) 2 1x x x= − + passes through the points (−1, 2), (0, 1), (1, 0) and (2, 5). Show that Newton’s Interpolating Polynomial for these four points is the original cubic polynomial. Solution

-1 2-1

0 1 0-1 6

1 0 65

2 5

fiΔ 2fiΔ 3fiΔfiix0x

1x

2x

0f0fΔ2

0fΔ

30fΔ

The four points, (−1, 2), (0, 1), (1, 0) and (2, 5) give the xi and the fi values

So, substituting in the formula and with h = 1, gives

[ ]

2 30 0 1 0 1 20 0 0 02 3

3

3

( )( ) ( )( )( )f ( ) f f f f2! 3!

( ( 1))( 0) ( ( 1))( 0)( 1)2 ( ( 1)) ( 1) 0 62 6

2 ( 1) 0 ( 1) ( 1)2 1

2 1

x x x x x x x x x x x xxh h h

x x x x xx

x x x xx x x

x x

− − − − − −= + Δ + Δ + Δ

− − − − − − −⎡ ⎤ ⎡ ⎤= + − − × − + × + ×⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦= + − − + + + −

= − − + −

= − +

Langrange’s Technique Example For a certain function f, f(1) = 4.5, f(2) = 4.9, f(4) = 5.8 and f(7) = 8.4. Use Lagrange’s Method to estimate f(5) by fitting a cubic to the graph of f at the points given. Solution Note that because the x-value we have are not evenly spaced, we have no choice but to use Lagrange’s method here. The curve passing through (a,A), (b, B), (c, C) and (d, D) is

( )( )( ) ( )( )( ) ( )( )( ) ( )( )(( )( )( ) ( )( )( ) ( )( )( ) ( )( )(A x b x c x d B x a x c x d C x a x b x d D x a x b x cy

a b a c a d b a b c b d c a c b c d d a d b d c− − − − − − − − − − − −

= + + +− − − − − − − − − − − −

))

Using this with the points (1, 4.5), (2, 4.9), (4, 5.8), (7, 8.4) gives

4.5( 2)( 4)( 7) 4.9( 1)( 4)( 7) 5.8( 1)( 2)( 7) 8.4( 1)( 2)( 4)(1 2)(1 4)(1 7) (2 1)(2 4)(2 7) (4 1)(4 2)(4 7) (7 1)(7 2)(7 4)

x x x x x x x x x x x xy − − − − − − − − − − − −= + + +

− − − − − − − − − − − −

Now you must let x = 6 to get an approximation to f(6), notice that here there is no need to simplify the interpolating polynomial itself!

4.5(6 2)(6 4)(6 7) 4.9(6 1)(6 4)(6 7) 5.8(6 1)(6 2)(6 7) 8.4(6 1)(6 2)(6 4)f (6)(1 2)(1 4)(1 7) (2 1)(2 4)(2 7) (4 1)(4 2)(4 7) (7 1)(7 2)(7 4)

2 4.9 6.44 3.73 7.28 (to 2 d.p.)

− − − − − − − − − − − −≈ + + +

− − − − − − − − − − − −= − + + =



ERRORS AND APPROXIMATION

The main ideas are: • Significant Figures and

Decimal Places • Measures of error • Propagation of Errors

Before the exam you should know: • The formula for measures of error. These are error,

relative error, absolute error and absolute relative error. • The rules for propagation of relative errors addition,

multiplication and division. • That care must be taken when subtracting approximations

to nearly equal quantities, many significant figures can be lost in accuracy.

Measures of Error

When a given value x is approximated by X : • the errorε is given by X xε = − .

• the relative error is defined by, errorrelative error = exact value x

ε= .

• the absolute error is defined as the modulus of the error. In other words,

absolute error X xε= = − . • the absolute relative error is defined as

errorabsolute relative errorexact value

X xx xε −

= = = .

Note: usage of these terms on exam papers has not been completely consistent to date, e.g. an approximation of 0.6 to 0.612 may be described as having an error of 0.012 instead of an error of –0.012.

Subtraction of Nearly Equal Quantities Example The values X = 123.453 and Y = 120.351 are approximations to values x and y which are both correct to 6 significant figures. Give x y− correct to as many significant figures as is possible with this information. Solution Since X = 123.453 is correct to 6 significant figures

123.4525 123.4535x≤ < . Similarly, since Y = 123.351 is correct to 6 significant figures

120.3505 120.3515y≤ < . Therefore

123.4525 120.3515 123.4535 120.3505 or

3.101 3.103

x y

x y

− < − < −

< − <

So x – y must be 3.10 to 3 significant figures, but it cannot be given to 4 significant figures from the information available.


i)

ii)

iii)

Propagation of relative error upon multiplying or dividing approximations Example Suppose that X = 1.3 is used as an approximation to x = 1.3478 and Y = 0.03 is used an approximation to y = 0.0297,

What is the relative error in each of these approximations?

What is the relative error when XY is used an approximation of xy?

What is the relative error when XY is used as an approximation of

xy ?

Find any relationships between the values you have calculated? Solution

i) 1

1.3 1.3478 0.03547 (to 5 d.p.)1.3478

X xrx− −

= = = −.

20.03 0.0297 0.01010 (to 5 d.p.)

0.0297Y yr

y− −

= = =.

ii) The relative error when XY is used to approximate xy is:

3(1.3 0.03) (1.3478 0.0297)

1.3478 0.02970.039 0.4002966 0.02572 (to 5 d.p.)

0.4002966

XY xyrxy− × − ×

= =×

−= = −

a. A possible relationship is that 1 2 0.02537r r+ = − is approximately equal to this value.

iii) The relative error when XY is used to approximate

xy is

a.

4

1.3 1.347843.33333 45.380471380.03 0.0297 0.04511 (to 5 d.p.)1.3478 45.38047138

0.0297

X xY yr x

y

− − −= = = = −

.

b. A possible relationship is that 1 2 0.03547 0.01010 0.04557 (to 5 d.p.)r r− = − − = − is approximately equal to this value.

In fact these observations hold in general: If X is an approximation of x with relative error r1 and Y is an approximation of y with relative error r2 then:

• the relative error in XY as an approximation of xy is approximately r1 + r2

• the relative error in XY as an approximation of

xy is approximately r1 − r2.



NUMERICAL DIFFERENTIATION

The main ideas are: • Using the forward and

central difference formulae • Knowing about the order of

convergence for each method and being able to extrapolate from approximations you have calculated

• An appreciation of how the gradient of a function influences the error in f(x) when there is an error in x.

Before the exam you should know: • Know the formulae for the forward and central

difference approximations to the derivative (or gradient) of a function.

• The forward difference approximation to the derivative of a function f at a value x is given by

f ( ) f ( )f ( ) x hxh

x+ −′ ≈ where h is small.

• The central difference approximation to the derivative of a function f at a value x is given by

f ( ) f ( )f ( )2

x h x hxh

+ − −′ ≈ where h is small.

• If X is used as an approximation to x and the error is h (so that X – x = h) then the error when f(X) is used as an approximation to f(x) is approximately f ( )x h′

• You should know that the central difference method is a second method and the forward difference method is a first order method (and what this means!). Using the forward and central

difference formulae. Example For the function f(x) = x2sinx, calculate the forward difference approximations to with

. f (2)′

0.1,0.05,0.025,0.0125h = Solution The forward difference approximation to f ( )x′ is given by

f ( ) f ( )f ( ) x h xxh

+ −′ ≈ .

In each case below x = 2. With h = 0.1, this gives

2 2f (2 0.1) f (2) 2.1 sin 2.1 2 sin 2f (2) 1.69561 0.1

+ − = −′ ≈ = (to 4 d.p.)

With h = 0.05,

2 2f (2 0.05) f (2) 2.05 sin 2.05 2 sin 2f (2) 1.83901 0.05

+ − = −′ ≈ = (to 4 d.p.)

These approximations appear to be approaching a value slightly less than 2.


Example Values of a function, f, for various values of x are given in the following table x 1 1.1 1.2 1.3 1.4 f(x) 10 13 11 12 13 Find an approximation to f using forward difference approximations h = 0.2. (1.2)′Solution: The central difference approximation with h = 0.2 is,

f (1.3) f (1.1) 12 13f (1.2) 50.2 0.2− −′ ≈ = = −

Remembering the formulae Probably the easiest way to remember the formula for forward difference approximation to the derivative and central difference approximation to the derivative is to commit the following diagrams to memory. Forward

Difference Approximation

Central Difference

Approximation

y-axis

x-ax

y-axis

x x+hf(x)

f(x+h)h

f(x+h)-f(x)

How the gradient of a function influences the error in f(x) when there is an error in x. Example Calculate the error incurred if the approximations X = 6.0 and X = 5.9 are used in the function 3f ( )x x= to obtain approximations to f(5.92). Relate these to the derivative of f. Solution The exact value of f(5.92) is . Using X = 6.0 gives f(X) = 216. The error in this approximation of f(5.92) is

35.92 207.474688=

f (6) f (5.92) 216 207.474688 8.525312− = − = .

The relationship with the derivative of f is that f ( ) f ( ) f ( )X x x h′− ≈ where h = X – x. In fact

. ( )2f ( ) 3 5.92 6 5.92 8.411136x h′ = × × − = Using X = 5.9 gives f(X) = 205.379. The error in this approximation of f(5.92) is

f (5.9) f (5.92) 205.379 207.474688 2.095688− = − = − . This time h = 5.9 – 5.92 and ( )2f . ( ) 3 5.92 5.9 5.92 2.102784x h′ = × × − = −

gradient of chord

change in coord

change in coord

f ( ) - f ( )

y

x

x h x

h

=

+=

x-axx x+hx-hf(x-h)

f(x+h)

f(x)

f(x+h)-f(x-h)

gradient of chord

change in -coordinate=

change in -coordinate

f ( ) f ( )

2

y

x

x h x h

h

+ − −=

2h



NUMERICAL INTEGRATION

The main ideas are: • Using the midpoint rule,

trapezium rule and Simpson’s rule to approximate integrals.

• Knowing how accurate each

of these methods is so that you are able to estimate error.

Before the exam you should know: • You should be familiar with all the standard notation

used in Numerical Integration. • You should know the formulae for Mn, the midpoint

rule with n strips, Tn the trapezium rule with n strips and Sn, Simpson’s rule.

• You should know the following formulae, and use

them to your advantage in the exam.

21 ( )2

1 ( 23

n n

n n

T T M

S T M

= +

= + )

n

n

• You should know that the midpoint and trapezium

rule are second order methods. This means that halving the strip width gives you a roughly four times more accurate approximation. You should know that Simpson’s rule is a fourth order method. This means that halving the strip width gives you a sixteen times better approximation. You should be able to display some knowledge of how these facts can allow you to estimate the error in a given approximation by looking at the difference between that and a previous approximation, see the last section on this sheet for more details.

• Remember that each time you apply these rules, h

changes and the values of f1, f2,… change according to the particular strip width you are using.

The Midpoint Rule

x

y

bam1 m2 m3 m4 m5 m6 m7 m8

The composite form of the mid-point rule, using n strips, each of width h gives

1 2f ( ) (f ( ) f ( ) ... f ( ))b

n na

x dx h m m m M≈ + + + =∫ ,

where are the values of x at the midpoints of the strips. 1 2, ,..., nm m m


The Trapezium Law Simpson’s Rule In general, over any interval (a, b) divided into an even number, 2n, of strips, of width h, the composite Simpson’s rule gives,

[ ]0 1 3 5 2 1 2 4 2 2 2f ( ) S f 4(f f f ... f ) 2(f f ... f ) f3

b

n na

hx dx − −≈ = + + + + + + + + + +∫ n n

Simpson’s rule approximates the function by a series of quadratics, one for each pair of neighbouring strips.

Note that the above approximation can be calculated as 1 ( 23n nS T M= + )n

How accurate are my estimates?

x

y

ba

The composite form of the trapezium law using n strips, each

of width h gives [ ]0 1 2 3 1f ( ) f 2(f f f ... f ) f2

b

n na

hx dx −≈ + + + + + +∫ .

Note the notation here, f0 is the value of the function at the left hand end of the first strip, f1 is the value of the function at the left hand end of the second strip (or the right hand end of the first strip)…and so on. Finally, fn is the value of the function at the right hand end of the nth strip.

x

y

21

0.8

In the table below and to the right. we have a series of trapezium

rule estimates to 2

1

sin xdx∫ . Immediately to the right is the graph

of this function How accurate can we give an estimate to this integral based on these results?

Trapezium Differences Ratio ofEstimates Ti+1-Ti Differences

T1= 0.9354441T2= 0.9670954 0.03165131T4= 0.9750775 0.00798207 0.25218769T8= 0.9770783 0.00200085 0.25066821

T16= 0.9775789 0.00050057 0.25017771

0.2The first thing to say is that T8 and T16 agree to two decimal places, they both round to 0.98. To three decimal places they disagree, T8 rounds to 0.977 whereas T16 rounds to 0.978. As the trapezium rule is a second order method we can safely say that to

two decimal places 2

1

sin( ) 0.98x dx =∫ . We cannot safely say what

the estimate is to three decimal places. Estimates get about four times better (four times closer to the true value) upon each halving of the strip width. This means that the ratio of differences will be about 0.25 as is shown in the table on the right. We may wish to extrapolate to obtain further estimates of

16 832 16" "

3T TT T −

= + , 16 8 16 864 16 2" " etc etc.

T2

4 4T T T TT T − −

= + +

In fact you may wish to look at the number 16 8 1

True ValueT8

6 8 816 2 3 ...

4 4 4T T T TT − − −

+ + + +T4 T16

16T T which

can be found by summing an appropriate geometric progression.



RATE OF CONVERGENCE IN NUMERICAL PROCESSES

Detecting First Order Convergence by looking at Ratio of Differences Example Show, by considering ratios of differences, that the following sequence has first order convergence.

The main ideas are: • Using the forward and

central difference formulae. • Knowing about the order of

convergence for each method and being able to extrapolate from approximations you have calculated

• An appreciation of how the gradient of a function influences the error in f(x) when there is an error in x.

Before the examination you should know:

• A converging sequence is said to have first order convergence if, for some fixed positive number k,

1 2

0 1

absolute error in absolute error in , ,..absolute error in absolute error in

x xk kx x

≈ ≈

• A converging sequence is said to have second order convergence if, for some fixed positive number k,

( ) ( )

1 22 2

0 1

absolute error in absolute error in , ,...absolute error in absolute error in

x xk kx x

≈ ≈

• For methods depending on h, like those found in

Numerical Integration and Differentiation, a method is said to be an nth order method if the absolute error in an approximation is proportional to hn.

• The forward difference approximation is a first order

method. The central difference approximation, trapezium rule and midpoint rule are second order methods. Simpson’s Rule is a fourth order method.

x0 x1 x2 x3 x4 x5 x60.38 0.381672 0.38192295 0.38195972 0.38196509 0.38196587 0.38196599 Solution The ratios of differences are given in the table below.

n x n x n+1-x n (x n+1-x n)/(x n+2-x n+1)0 0.38 0.0016720000000 0.1500928761 0.381672 0.0002509552880 0.1465230922 0.381922955 0.0000367707447 0.1459895193 0.381959726 0.0000053681433 0.1459113884 0.381965094 0.00000078327325 0.381965877

Ratios of differences

You can calculate the ratios of differences very quickly using a spreadsheet program.

The ratio of differences column, being nearly constant, provides evidence of first order convergence. Detecting Second Order Convergence Second or higher order convergence is much faster than first order and so you can get a very good approximation to the limit with only a few iterations. Since such an approximation is very close to the actual value of the limit you can use it to estimate closely the absolute error in each term.


You should know the following facts:

Method Effect of halving h or doubling n

Ratio of differences

between such estimates

Forward Difference Approximation The estimate gets twice as close 12

Central Difference Approximation The estimate gets four times as close 14

Midpoint Rule The estimate gets four times as close 14

Trapezium Rule The estimate gets four times as close 14

Simpson’s Rule The estimate gets sixteen times as close 116

Example Below are Trapezium rule estimates to some integral. By considering the ratio between differences in successive estimates, give the value of the integral to the number of decimal places you feel is justified.

T2 T4 T8

2 1.81 1.7652 Solution T2T4

0.0448 0.19T8

1.7652 1.81 2 The ratio of differences is

This is very close to the value of

0.00448 0.2358 (to 4 d.p.)

0.19=

14

predicted by the theory.

.

The extrapolated value of T16 is

0.04481.7652 1.7544

− = .

Notice how we need to subtract here because clearly T16 is expected to be less than T8. The extrapolated value of T32 is

2

0.0448 0.04481.7652 1.75124 4

− − = .

This sequence of values converges to

2 3

2 3 4

0.0448 0.0448 0.04481.7652 ...4 4 4

1 1 1 11.7652 0.0448 ....4 4 4 4

10.044841.7652 0.0448 1.76521 31

41.750267 (to 6 d.p.)

− + + +

⎡ ⎤= − + + + +⎢ ⎥⎣ ⎦⎛ ⎞⎜ ⎟

= − = −⎜ ⎟⎜ ⎟−⎝ ⎠

=

Comparing these values, it looks as though the estimate of 1.75 is correct to 2 d.p.



THE SOLUTION OF EQUATIONS

The main ideas are:

The Bisection Method This is probably the easiest of all the methods to apply. The key points are:

• Using the bisection method, the

method of false position (also known as linear interpolation), fixed point iteration, the Newton Raphson Method, and the secant method to approximate the solution of an equation.

• Knowing how accurate each of these methods is so that you are able to estimate error.

Before the exam you should know: • And be totally familiar with using the five methods

of approximating solutions to equations used in this chapter. These are, the bisection method, the method of false position, fixed point iteration, the Newton Raphson Method, and the secant method.

• For every one of the methods above, you should

know how to judge when you have the approximation to the required degree of accuracy and then check that this is indeed the case.

• You should know the order of convergence of the

Newton-Raphson method. It has second order convergence – you should know what this means.

• You should know the requirements on the function g

to find a fixed point using fixed point iteration.

• You must begin with two points that straddle the solution of f(x) = 0. These are usually called a1 and b1

The reason that they are chosen is that the sign of f(a1) is the opposite of the sign of f(b1). This means that we expect the function to cross the x-axis somewhere between them.

• The sign of 1 1f2

a b+⎛⎜⎝ ⎠

⎞⎟ is calculated. Depending on this we know that the change of sign has to occur

either between a1 and 1 1

2a b+ or between 1 1

2a b+ and b1. The appropriate pair become our a2 and b2 and

the whole process is repeated. • Note that at every stage in this process you can say that the solution x satisfies . This means

that if ana x b< < n

n and bn are close enough you can give the solution to any degree of accuracy.

The Method of False Position (also known as Linear Interpolation)

Given two values, a and b that straddle a root of f(x) = 0 an approximation to the root is given by

f ( ) f ( )f ( ) f ( )

a b b acb a−

=−

This is the point where the straight line between and ( , crosses the x-axis. This method

can be iterated by testing the sign of f(c) to find whether the root is between a and c or between c and b.

( , f ( ))a a f ( ))b b

x

x)

ba

f(

f(a)

f(b)

α

x1

c

the Further Mathematics network – www.fmnetwork.org.uk V 07 1 1 Fixed Point Iteration

However we need a little more than this if we are to approximate the solution of f(x) = 0. We need the gradient of the g(x) to be between –1 and 1 around the fixed point.

One of the main skills you need to acquire for fixed point iteration is to get from an equation like f(x) = 0 to an equation of the form x = g(x). Here are some examples

2 22 0x xx

− = ⇔ = ,

2 14 sin 1 04sin

x x x xx x

+ − = ⇔ =+

x

y

x

y

Gradient of red line negative but >- 1 around the fixed point results in a converging series of values, illustrated by the inward moving cobweb in this diagram.

Gradient of red line negative and <- 1 around the fixed point results in a non-converging series of values, illustrated by the outward moving cobweb in this diagram.

y = g(x)

If the fixed point iteration method produces values 1 2 3, , ,...x x x where

1 g( )n nx x+ = ,converging to the fixed point x of g, then we expect

1error in g ( )error in

n

n

x xx+ ′≈

Newton Raphson-Method To generate a sequence of values converging to a root of f(x) = 0, near to x = x0, use the following iterative

formulae: 1f ( )f ( )

rr r

r

xx xx+ = −

′. This method has second order

convergence.

Below the Newton Raphson method is being used to approximate a solution of x3 + 2x - 1 = 0. So we have

3

1 2

23 2

r rr r

r

x xx xx+

+ −= +

+1 in this case. Using a starting value of

x0 = 1 gives 3

0 01 0 2

0

2 1 1 2 11 03 2 3 2

x xx xx+ − + −

= − = − =+ +

.6

3 31 1

2 1 2 21

2 1 0.6 2 0.6 10.6 0.46493506343 2 3 0.6 2

x xx xx+ − + × −

= − = − =+ × +

Then x3 = 0.453467173, x4 =0.453397654, x5 = 0.453397651, x5 = 0.453397651. Notice how quickly the sequence converges. You can get your calculator to perform these calculations very quickly using the ‘ANS’ feature. This also applies to Fixed Point Iteration so make sure you know how to do this.

Secant Method If x = x0 and x = x1 are approximations to a root of f(x) = 0, a better approximation to the root will usually be given by

0 1 1 02

1 0

f ( ) f ( )f ( ) f ( )

x x x xxx x

−=

−. This can be

repeated with x1 and x2 replacing x0 and x1 to obtain a value x3 and so on.. These calculations can take quite a while on a pocket calculator! The main difference between the secant method and the method of false position are that in the secant method x0 and x1 need not straddle the solution whereas in the method of false position the first two values, a and b, should straddle the root. When x0 and x1 do straddle the root the point generated, x2 is exactly the same as c in the method of false position.

y x=

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REVISION SHEET – PURE MATHEMATICS 4furthermaths.org.uk/manager_area/files/teaching/NM/NMmei_rev.pdfthe Further Mathematics network – V 07 1 1 REVISION SHEET – NUMERICAL METHODS