C3 Coursework

7/25/2019 C3 Coursework

1/20

Joseph Wheatland

C3 Coursework

In this coursework I will be using three diferent methods to solve an

equation. These are the:

Change o sign method which involves a decimal search ewton!"aphson method

"earrangement o #$% & ' to $ & g#$%

I will use each o these methods to (nd a success and a ailure and at the

end will identi) the practicalities o each method in diferent situations

compared to each other.

Change o sign method

I we were to graph an equation with roots to the equation #$% & '* it

would be seen that the equation will cross or touch the line. When it

crosses the $ a$is* the value o #$% & ' will change rom a positive to a

negative number between two intervals either side o the actual root. We

can keep narrowing the intervals to keep getting closer to the actual root

value but the value will never actuall) be met* which is wh) I am going to

test this method to three decimal places. I will then repeat this between

this two integers until a change o sign occurs again.

+) using a decimal search where integers are chosen to put into the

equation #$% & '* I can (nd two values that la) either side o the actual

route.

,$ample

I will use this method to solve the equation '.-$/0$123'$/04.0 & '.

To do this I will (nd the values o #$% & '.-$/0$123'$/04.0 and detail

where the value o the #$% changes rom a positive value to a negative. 5s

seen in the graph plotted in autograph* there are three roots or this

equation. I will be searching or the second root.

Consecutive integers take the place o $ in the unction. I the unction o

' is used we arrive at the answer 04.0:

#'% & '.-$'/0$'123'$'/04.0 & 04.0

Equation 1

6


2/20

Joseph Wheatland

With the unction o 6 we can see that the answer is now negative#,quation 0%. This means there is a root in this interval. We can now

narrow down this search to locate the root in the '!6 range.

#6% & '.-$6/0$6123'$6/04.0 & !3.3

Equation 2

5s the #'% is positive and #6% is negative this tells us the graph must

cross the $ a$is between ' and 6. We can now narrow down one o the

roots to being in the range #'* 6%

Figure 1: f(x) = 0.5x+2x!30x+2".2

It is now possible to do a decimal search in which I will split the interval

into ten and will again calculate the values o #$% or the values o $ in the

'!6 interval until a change in sign occurs. This step is repeated until the

accurac) comes to our decimal places and the number can be accuratel)

rounded to 3 decimal places.

0

#6% & !3.3

#'% &


3/20

Joseph Wheatland

#te$ 1 #te$ 2

$ #$%

!7 68'.7

!8 689.0

!- 686.7

!4 644.0

!3 669.7

!0 99.0

!6 --.7' 04.0

6 !3.3

0 !03.9

3 !34.3

4 !36.9

- !63.3

3

"oot in interval #'*6%as there is a change osignwhere the line crosses$

#te$ " #te$

$ #$%

6 !3.3

'.!

'.96--

'.96.738

"oot in interval#'.9*'.%

$ #$%

'.98'.670

09

'.986'.6767

96

'.980'.6483

4

'.983'.60'

'8

'.984'.'-4

79

'.98-'.'7''

-7

'.988'.'448

43

'.987'.'60

3-"oot in interval#'.987*'.989%

'.989

!'.''86

7

; #$%

'.9 6.738

'.966.477

06'.90

6.00'494

'.93'.838

4

'.94'.7'7-

-0

'.9-'.4-0'

83

'.98'.670

09"oot in interval#'.98*'.97%

'.97 !'.'-8

-


4/20

Joseph Wheatland

$ #$%

'.987'.'60

3-

'.9876'.'688

-

'.9870'.'646

-4'.9873 '.'668

64'.9874 '.'''

74

'.987-'.''8-

34

'.9878'.''3

4

'.9877'.''64

-4"oot in interval#'.9877*'.9879%

'.9879

!'.''6'

This data has split the values o $ into equal intervals so as to get the

degree o accurac) as desired #4 decimal places%. We can now diagnose

that #$% & ' or '.9877 < $ < '.9879. This means that the value o the

root is '.989 #3 d.p% and the ormer value o #$% describes the error

bounds o the answer. I we were to again search or a change o sign

between '.9877 and '.9879 then it would )ield us a more accurateanswer. =ur error bound proves our answer is not the de(nitive one.

4

#te$ 5


5/20

Joseph Wheatland

Figure 2: f(x) = 0.5x+2x!30x+2".2

%0.&'0.&'&*

The decimal search method I undertook was a success as the unction

crosses the $ a$is between the values o '.9877 and '.9879 #(gure 0%.When the $ a$is is intersected* #$% & '. Thereore* as the value o

#'.9877% is positive and the value o #'.9879% is negative* we know that

there is an intersection trapped between these two points and hence a

root.

Faiure of t,e -,ange of sign et,o/

In some cases this method will ail when there are two or more roots in

close pro$imit) in an integer range. I this is the case* then when we do

the integer search onl) one change o sign ma) be noticed and roots

could be overlooked. Without the help o a graph we could miss them

altogether and the method would be a ailure as not all the roots would

have been ound.

I>m using the equation '.-$/0$123'$/- & ' to (nd a ailure using the

change o sign method. To do this I will (nd the values o the unction #$%

& '.-$/0$123'$/- which have an intersection between them and

thereore a change o sign. With an integer search it appears as i there

are no roots between these values as the signremains positive throughout and doesn>t change

-

Interval #'.9877* '.9879%

#'.9879%

is

,?5TI@,

#'.9877%

isA=BITI@,

$ #$%' -

6 36.-

0 66

3 '.-

4 3

- 06.-


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Joseph Wheatland

#(gure 3%. +ut as we can see on the graph between three and our there

are two roots. It is not picked up b) this method as the sign changes back

to positive in the same interval.

Figure 3: f(x) = 0.5x+2x!30x+5

5s the two roots haven>t been picked up with the decimal search and a

success is onl) a success when all the roots can be ound with the

method* this method has ailed in this e$ample.

ewton a$,son

The newton "aphson method uses an iterative ormula to calculate new

values o $ which become more accurate to the actual value o the root.

This iteration works b) (nding the point at which a tangent to the curve

crosses the $ a$is. This new value o $ is inputted into the iterative

ormula and will (nd a tangent which crosses the $ a$is closer to the

actual root each time.

The iterative ormula used can be generalised to $n/6 & $n !f(xn)

f'(xn)

The starting point or the ormula is known as $'which is an integer whichla)s near the root )ou are attempting to (nd. ?iven this initial integer

estimate* we can use this numerical method to (nd an estimate to the

root.

et #$% & $/07$1/-$20.4. The equation #$% & ' has two roots #(gure 4%

between D!6*'E and D'*6E. We will (nd both o the roots using the ewton

"aphson method starting with the right root #D'*6E%.

8

Integer search

doesn>t pick up

these two roots

5t #4% the sign

is ositive too

5t #3% the sign

is ositive

The table o integers or $

below indicates a change

o sign twice which leads

us to the conclusion that

$ #$%!3 69.8

!0 97.8

!6 69.8

' !0.4

6 3'.8

0 603.8


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Joseph Wheatland

Figure ": f(x) = x+2x+5x!2."

To begin this method it (rst requires an estimate or the root. This value

#$'% will be ' as it lies reasonabl) close to the root. Thereore:

$'& '

#$% & $/07$1/-$20.4

>#$% & 3$0/-4$/-.

This can be substituted into the general ormula in place o $nas ollows:

;n/6& $n !f(x0)

f

'

(x0)

& ;n/6& $n !xn

3+27xn2+5xn2.4

3xn

2

+54

xn+5

;6& $'!x0

3+27x02+5x02.4

3x0

2+54x0+5

& ' !032702+502.4

302+540+5

& '.49

ow that we have a value or $6this can be substituted into the iterative

ormula again to give out another answer closer to that o the root.

Instead o $'we now substitute it with $6* and the $6at the beginning o

the ormula becomes an $0as it will (nd the second iteration.

$0& $6!x1

3+27x12+5x12.4

3x1

2+54x1+5

& '.49 !0.48

3270.482+50.482.4

30.482+540.48+5

&

'.077

While this process can be repeated to achieve a greater accurac) it is

much easier to use sotware to calculate the values o $3etc. as it will (nd

better values quickl) with much less rounding error.

7


8/20

Joseph Wheatland

Figure -:

ewton

"aphson

tangent

diagram or #$% & x+2x+5x!2."

The graph illustrates that the line converges

towards the root ver) rapidl). It has onl) taken

eight iterations to (nd the e$act route and ater

(ve the digits remained ($ed to si$ decimal places.

Btarting at ' #$o%in(gure -* this graph illustrates

how a line is created which shoots towards the

curve up or down and creates a tangent which then

shoots of to the $ a$is. This point now becomes

$6. The line shoots to the curve o )& #$%

again and another tangent orms which shoots

o to the $ a$is. When this crosses the $ a$is it is closer to the root and

more accurate to the roots real value. This point is known as $0. Thisprocess repeats and will get more accurate each time. 5s the iterations

occur* the change in $nbecomes less and less as it reaches closer towards

the root* until it is miniscule or reaches Gero. This means the method has

ound when the #$% & ' and hence the root. The line in this case is

converging to a value near '.06973.

The root between D'*6E is '.0697 to (ve signi(cant (gures in this case

with the solution bounds o '.0698-


9/20

Joseph Wheatland

F#'.0698-% & !'.'''6434-3'0-F#'.0698-% & 0.80006'8,!''-

The change o sign means there is de(nitel) a root in between the values

and both o these numbers round to '.0697 so we can sa) or certain this

is the answer to this root to (ve signi(cant (gures.

There is one more root that lies between D!6*'E that still needs to be ound.

5gain the same method will be used as beore but the value o $'will be

changed to !6 as it is closer to this root and will (nd it in ewer iterations.

We can see that the method has ound the root with ease in onl) (ve

iterations once again. The iteration gives the

root as !'.4'996 to (ve signi(cant (gures.

ewton a$,son et,o/ faiure

5 ailure in the case o the newton "aphson

method can be de(ned when either the line

diverges awa) rom the closest root rom the

starting integer or when a diferent root is ound

'.0697 H '.'''''-

;

6

Curve o

;

0

;

'

($ed

(gures

;n

change in$n

!'.--8-067

4'.4'4347

908!

'.44079677

-

'.6-097'

3!

'.46'409608

'.'303-384

!'.4'9960'

'.''686-

068!

'.4'99'99-3 4.'8,!'8!

'.4'99'99-3 0.-8,!66


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Joseph Wheatland

than the one we are searching or. This could happen due to the gradient o the

tangent being ver) close to Gero and not reaching the $ a$is or a long time as it

has a steep crossing point or the unction is discontinuous and the tangent is

ormed but the new value o $ndoesn>t hit the curve again.

et #$% & #0$3!4$0/6%63 #Figure -.6% I will attempt to (nd the root on the ar right

in the interval D6*0E in this e$ample or the equation #$% & '.

Figure 5.1: f(x) = (2x3"x2+1)143

I we take $'to equal 0 then the e$pected outcome would be or the root on the

let o this number to be ound. owever* due to a steep $ a$is where the line

crosses the turning point the tangent is thrown of rom where it should go andthe $6value is now on the let o the root #(gure -.0%. When the line goes rom

the $0 number to $3the intersection o the $ a$is is urther awa) rom the root we

are searching or too. 5s there is a ver) low gradient when the line shoots

vertical towards the line rom $3* the line now diverges into in(nit) in the

opposite direction rom the local root.

Figure

5.2$ change

6'

;4;'

;6;0;3

;4

The line (rst goes onto the otherside o the root we are looking or.

The negative value o the

intersection at the $ a$is show that

a large divergence awa) rom our

root has occurred


11/20

Joseph Wheatland

in $n6.80-'

'6'.374

'0.8-3

336.'3433

69

6.6-009

6.4846'4

!0.74'-

93.3-9'

30

= g(x) et,o/

This third method is known at the $ & g#$% method which involves

rearranging the equation #$% & ' to $& g#$%. This new unction can then

be used as an iterative ormula. When using this method* the unction ) &g#$% is plotted against )&$ and the points at which these points intersect

will now satis) the equation #$% & '.

I am going to use the $ & g#$% method to solve 0$ !3$1!0$ /'.7&' so let

#$% & 0$ !3$1!0$ /'.7&' #(gure 8%.

Figure ': f(x) = 2x 3x2x +0.

To do this I will use the iteration $n/6 & g#$n% where g#$% is a

rearrangement o the equation 0$ !3$1!0$ /'.7&' into the orm $ & g#$%.

0$ & 0$3!3$0/'.7

; & #0$3!3$0/'.7%0

;n/6& #0$n3!3$n

0/'.7%0

66

5dding the 0$ onto one side

allows us two then divide b)

two to leave an $ on one side.

This leaves the ormula in the

new $ & g#$% orm.

The $ & g#$% iterative ormula now becomes ;n/6& #0$n

3!3$n0/'.7%0in which a new number is

created with each iteration and substituted

back into the ormula or a more accurate

answer each time.


12/20

Joseph Wheatland

5n integer search can be used now to see which number to use or the

starting value o this ormula #$'%. In 6gure there are three turningpoints which signi(es there are three roots. I am going to ocus on the root

between D'*6E so I will make $'& '

$ #$%

!6 !0.3

' '.7

6 !0.3

0 '.7

3 06.7

This method uses an iterative ormula Kust like the newton

"aphson method. Lsing $'& ' the (rst iteration can

be ound b) substituting into the ormula

;6& #0$'3! 3$'0/'.7%0 & #0$'3!3$'0/'.7%0 & '.3-

;0& #0$63!3$60/'.7%0 & #0$'.3-3!3$'.3-0/'.7%0 & '.0'60-

5s this is an iterative ormula that repeats its steps it is once again muchquicker to use sotware to (gure out urther iterations. Figure &shows

the equations ) & $ and ) & g#$% plotted against each other with the

original ) $% also in view. The two latter unctions cross at the same $

value as the middle root in the equation #$% & '. This is because the

unctions are Kust rearrangements o

each other.

Figure &: 7 = f(x) = 2x 3x2x +0. 7 =g(x) = (2x33x2+0.)42 7 = x

60

Figure

M&

$

M &

#$%

M &

g#$%


13/20

Joseph Wheatland

ow that we>ve continued with several more iterations

there are si$ ($ed (gures or the value o the root. The

value can be rounded to '.0836 #- sig. (gures%. We cancheck that there actuall) is a root near this value b)

creating solution bounds and checking or a change o

sign between '.083'- and '.0836-.

F#'.083'-% & 6.000949,!''-F#'.0836-% & !6.407669,!''-

There is a change o sign indicating a root lies between

these numbers and both o these numbers round to

'.0836 so this is the correct root to (ve signi(cant(gures.

I we now go back to our graph displa)ing the cobweb convergence the

pattern is relativel) simple to spot the pattern the line takes. The initial

integer estimate o ' #bottom let o 6gure % is ollowed up the curve to

the curve ) & g#$% and then a horiGontal line leads right to the ) & $ line

as part o the (rst iteration. The $ coordinate o this point is now $6. The

line goes vertical towards the ) & g#$% line again and repeats the same

pattern in the ne$t iteration. When the horiGontal line drawn reaches the

)&$ curve again* this point can be known as $0. Lltimatel) the line

converges towards the point where ) & $ and ) & g#$% cross which is the

63

There are si$

($ed signi(cant

(gures

$ change in $n

'.3- '.3-

'.0'60- '.64'97-

'.03-4-906 '.'9440'906

'.048'4'97 '.'47-'4-6

'.074''60 '.'09'4300

'.0-7'04 '.'6869769

'.087393068 '.''49'000

'.08697--86 '.''--'78--

'.08-'' '.''306-339

'.08306''4 '.''697698

'.08436'-84 '.''6'6--

'.08387484- '.'''83-6

'.084'4-303 '.'''37'879

'.08390303 '.'''06-

'.083--063 '.'''60-99

'.0839969- 7.34,!'-

'.083048'- 4.09,!'-

'.0839899 0.4,!'-

'.0836406 6.4-,!'-

'.083'-747 9.48,!'8

'.0836'87 4.3,!'8

'.083'79'- 0.97,!'8

'.083'49 6.89,!'8

'.083'9-'4 .78,!'7

'.083''70 -.8,!'7

'.083'9746 3.30,!'7

'.083'934 6.3,!'7


14/20

Joseph Wheatland

same $ coordinate as where the equation #$% & ' as the root we are

looking or.

Fig ure

The reason or the

cobweb like shape is due to the

magnitude o the root or g>#$% when calculated at the root.

When the value o g>#$% lies in the bounds !6 < gN#$% t reach a point on a curve. Failure would also

constitute (nding the wrong root which is not the closest root to the

starting integer.

I am going to once again rearrange m) equation #$% & ' into the orm $ &

g#$%.

64

"oot o #$% &

'

M&g#$% line

steeper than )& $

M & $ has

gradient o 6

;3

;0

;6

;'


15/20

Joseph Wheatland

5s the roots are ound b) the equation ) & 0$ !3$1!0$ /'.7* I will

rearrange this into the ormx

x=(3x+0.35)1 /2

. The iteration or this

equation is $n/6& #$n3O $n/'.3-%60.

I will choose a diferent root to work on this time between the integers D!

0*!6E. The intersection is nearest to the integer !0 so I will make the value

o $'equal !0. 5t (rst glance #(gure 6'% it is clear to see that the gradient

o the equation ) & g#$% is much higher than that o ) & $ which has a

gradient o 6. This means that g>#$%P6Q we would e$pect the line to

diverge awa) rom the intersection in a staircase like pattern. This isn>t

too clear to see however as ater the line hits ) & $ or the second time* it

tries to go vertical but shoots to in(nit). The line visualising $ & g#$% is

ver) steep so the line created will not reach it. The graph does in act

diverge awa) rom the intersection as e$pected #in a staircase pattern%

thereore. The table below shows that the graph is not converging at all

towards the desired too and is in act moving urther awa) with each

iteration in the interval D!0*!6E.

Figure 6'

n $

2x 3x2x +0.

' !0 !03.36 !0.90- !80.890

0 .89- 6-68.90

3 !44.0

!69699468.

9

We know that the g#$% & #$2$/'.3-%60so thereore:

g>#$% &

3x21

2

x

3

x+

7

30

6-

M&$

;0

;'

M&

g#$%

M

$%

;6

The value or $ is the root weound earlier.


16/20

Joseph Wheatland

g>#$% &

3(0.26391)21

2

(0.26391)3(0.26391)+

7

30

The value o g#$% is below !6 so a cobweb diverging pattern is ormed as

we e$pected.

Co$arison of t,e et,o/s

,ach method has its own bene(ts in diferent situations and now I>ll

compare them b) selecting a previousl) used equation and completing it

using all three methods. In (nding a success or the $ & g#$% method I

used the equation 0$ !3$1!0$ /'.7&'. I>m going to use each method to

solve the root which lies between the integers D'*6E.

8e-ia sear-,

n $#$% & 0$ !3$1!0$

/'.7

' ' '.7

6 '.6 '.470

0 '.0 '.68

3 '.3 !'.668

4 '.06 '.688000

- '.00 '.638'88 '.03 '.6'-834

7 '.04 '.'74949

9 '.0- '.'437-

'.08 '.'603-0

6' '.07 !'.'6334

66 '.086 '.''68680

60 '.080 '.''8'374-8

63 '.083 '.''097-94

64 '.084 !'.'''099-60

6- '.0836 '.''0---9668 '.0830 '.''004304

67 '.0833 '.''60897

69 '.0834 '.''686'470

6 '.083- '.''604'48

0' '.0838 '.'''77-6

06 '.0837 '.'''8866'9

00 '.0839 '.'''344-8

03 '.083 0.9'-80,!'-

04 '.084 !'.'''099-60

0- '.0836 !3.-36,!'8

68


17/20

Joseph Wheatland

08'.083

'6 0.49'7,!'-

07'.083

'0 0.670-0,!'-

09

'.083

'3 6.9--8,!'-

0'.083

'4 6.-346,!'-

3''.083

'- 6.0009-,!'-

36'.083

'8 .'804,!'8

30'.083

'7 -.9739,!'8

33'.083

'9 0.73690,!'8

34

'.083

' !4.33743,!'7

ewton a$,son

n $$n/6 & 0$ !3$1!0$ /'.7

8$R0!8$!0

' ' '.3-''''338

6 '.3- '.08807'39

0 '.08807 '.0836633-

3'.0836

6 '.083'9983

4'.083'

0.84,!'6

-'.083'

0.84,!'6

The root is close to '.0836 so we will check to see i it is trapped

between '.083' and '.0836 in (ve iterations.

F#'.083'% & 0.9'-8039,!''-

F#'.0836% & !3.-366'8,!''

The value o the root is between these two numbers so we can sa) that the root is'.0836 to (ve signi(cant (gures.

= g(x)

n$

change in$n

' '.3- '.3-

6'.0'6

0-'.64'97-

0'.03-

48'.'9440'

906

3'.048'

46'.'47-'4

-6

67

The change o sign method

has taken 34 iterations to trap

the root between '.083'9

and '.083'. This means

that the value o the root in

the interval D'.6E is '.0836 to

The root has been ound as

'.0836 to (ve signi(cant

(gures ater 03 iterations. The

root is trapped between

'.083' and '.0836.


18/20

Joseph Wheatland

4'.074'

'.'09'4

300

-'.0-7

'3'.'68697

69

8'.0873

93'.''49'

000

7'.0869

78'.''--'7

8--

9'.08-'

6'.''306-

339

'.0830

6'.''6976

98

6''.0843

66'.''6'6

--

66'.0838

7-'.'''83-

6

60'.084'

4-'.'''37'

879

63'.0839

0'.'''06-

64'.083

--'.'''60-

99

6-'.0839

907.34,!'-

68'.083

0-4.09,!'-

67 '.083 0.4,!'-

69'.083

646.4-,!'-

6

'.083

'8 9.48,!'8

0''.083

664.3,!'8

06'.083

'90.97,!'8

00'.083

'6.89,!'8

03'.083

'.78,!'7

,ach method ound the root to (e signi(cant (gures but

with a var)ing number o steps. The accurac) was

achieved in:

34 iterations or the change in sign method

- iterations or the newton "aphson method

03 iterations or the $ & g#$% method

Comparing all three methods* the newton "aphson

method worked a lot aster to (nd out the root to the

69


19/20

Joseph Wheatland

same accurac) as the other methods in this particular

e$ample. The change o sign method was the slowest

but the easiest to understand while $ & g#$% was in the

middle o the other two or speed and is also diScult todo compared to the change o sign method. The $ &

g#$% and newton "aphson methods are more tasking

but were much quicker in getting the required

accurac)* but the $ & g#$% method was still a deal

slower than the newton "aphson.

The easiest method b) ar was the change in sign* as itdidn>t require sotware or ease o use and )ou could

keep narrowing it down to a high accurac). This method

requires a lot o work to do in spreadsheet sotware

involving ormula manipulation but the ease o use and

little understanding required* give this method good

merit.

The newton "aphson method was rather slow to do b)

hand and rounding errors occurred. owever with the

help o the autograph sotware and e$cel the results

were ver) eas) to achieve as the computer could

repeat the iterations with ease. The method does

require being able to diferentiate though so it is one othe more diScult methods. The general ormula means

all that is required is a reasonable estimate or the root

which la)s near the intersection o the $ a$is )ou are

searching or. 5s the answers can be substituted each

time it doesn>t var) too much* although it is time

consuming i sotware isn>t available to )ou.

6


20/20

Joseph Wheatland

The $ & g#$% method is also not too diScult as it onl)

requires the abilit) to rearrange #$% & '. This method

converges aster than the decimal search method but

slower than the newton "aphson. It can take a longtime to converge though as not all rearrangements o

the equation into g#$% orm would be suitable. I

spreadsheet sotware was available then the time

consuming convergence issue would not occur.

I onl) spreadsheet sotware was available then the

easiest method to use would be the change in sign asthe rows can be auto completed once the general

ormula is interpreted. It is easier to spot i an error has

occurred too. +ut without sotware it is possible or the

user to overlook the possibilit) o more than one sign

change in an interval. 5utograph is a helpul actor in

the $ & g#$% method especiall) as Kust rom looking at

the steepness o the rearranged curve compared to ) &$ suggests i the outcome will converge* diverge*

staircase or cobweb. The visualisations that graphing

sotware provide or the newton "aphson method too

means that )ou can tell i the particular equation is

going to work earl) on.

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