# Change of Sign Method - Mathematical Essay

# Change of Sign Method - Mathematical Essay

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More ↓In order to find the roots of an equation that cannot be solved

algebraically, I can use numerical methods to do this instead. One of

these methods is the change of sign method. From looking at a graph of

my equation I can find two integers that my root lies between, then

from there, using spreadsheets, I can use the change of sign method to

discover where the root lies to five decimal places.

I have chosen to try to solve the equation: 5x3-7x+1=0

First, I drew the graph of y=5x3-7x+1 in autograph to find where the

roots roughly lie by looking where the graph cuts the x-axis.

Y=5x3-7x+1

[IMAGE]

From this graph I can see that there are three roots, in the intervals

[-2, -1], [0, 1] and [1, 2]. Looking at the root in the interval [1,

2], we bisect this interval and find the midpoint, 1.5.

f(1.5) = 7.375, so f(1.5)> 0. Since f(1) <0, the root is in [1,

1.5].

Now I am going to take the midpoint of this second interval, 1.25.

f(1.25) = 2.016, so f(1.25)> 0. Since f(1) <0, the root is in [1,

1.25].

The midpoint of this reduced interval is 1.125.

f(1.125) = 0.244, so f(1.125)> 0. Since f(1) <0, the root is in [1,

1.125].

The method then continues in this manner until the required degree of

accuracy is achieved. However, this takes a long time to converge the

root, and can be solved more rapidly using a spreadsheet.

To put my data into a spreadsheet, I first needed to design one that

achieved the desired purpose: finding the roots of the equation using

the change of sign method. Below, the formula which I typed into my

spreadsheet are shown:

Change of sign method

y=5x^3-7x+1

a

b

(a+b)/2

f(a)

f(b)

f((a+b)/2)

1

2

=(A5+B5)/2

=5*A5^3-7*A5+1

=5*B5^3-7*B5+1

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### Related Searches

=IF(D5*F5<0,A5,C5)

=IF(D5*F5<0,C5,B5)

=(A6+B6)/2

=5*A6^3-7*A6+1

=5*B6^3-7*B6+1

=5*C6^3-7*C6+1

=IF(D6*F6<0,A6,C6)

=IF(D6*F6<0,C6,B6)

=(A7+B7)/2

=5*A7^3-7*A7+1

=5*B7^3-7*B7+1

=5*C7^3-7*C7+1

=IF(D7*F7<0,A7,C7)

=IF(D7*F7<0,C7,B7)

=(A8+B8)/2

=5*A8^3-7*A8+1

=5*B8^3-7*B8+1

=5*C8^3-7*C8+1

=IF(D8*F8<0,A8,C8)

=IF(D8*F8<0,C8,B8)

=(A9+B9)/2

=5*A9^3-7*A9+1

=5*B9^3-7*B9+1

=5*C9^3-7*C9+1

=IF(D9*F9<0,A9,C9)

=IF(D9*F9<0,C9,B9)

=(A10+B10)/2

=5*A10^3-7*A10+1

=5*B10^3-7*B10+1

=5*C10^3-7*C10+1

If this number is positive, it replaces f(b) as the upper bound. If it

is negative it replaces f(a) as the lower bound of the interval.

[IMAGE]

[IMAGE]

[IMAGE]

[IMAGE]

[IMAGE]

If this number is negative, it is the lowest bound for the interval.

1. These numbers are the integers that I have discovered my roots to

be between.

From this spread sheet I can find the upper and lower bounds of the

intervals so that I can narrow down the accuracy of the root of the

equation. Shown above are only the first few rows of the spreadsheet,

to find the roots to a high enough accuracy, the formulas are filled

down for more rows. If I look at the numbers that are present in the

spreadsheet once the equation has been entered, we can find the root

to a give number of decimal places.

Change of sign method

y=5x^3-7x+1

a

b

(a+b)/2

f(a)

f(b)

f((a+b)/2)

1.0000000000

2.0000000000

1.5000000000

-1.0000000000

27.0000000000

7.3750000000

1.0000000000

1.5000000000

1.2500000000

-1.0000000000

7.3750000000

2.0156250000

1.0000000000

1.2500000000

1.1250000000

-1.0000000000

2.0156250000

0.2441406250

1.0000000000

1.1250000000

1.0625000000

-1.0000000000

0.2441406250

-0.4401855469

1.0625000000

1.1250000000

1.0937500000

-0.4401855469

0.2441406250

-0.1140441895

1.0937500000

1.1250000000

1.1093750000

-0.1140441895

0.2441406250

0.0609855652

1.0937500000

1.1093750000

1.1015625000

-0.1140441895

0.0609855652

-0.0275378227

1.1015625000

1.1093750000

1.1054687500

-0.0275378227

0.0609855652

0.0164708495

1.1015625000

1.1054687500

1.1035156250

-0.0275378227

0.0164708495

-0.0055966303

1.1035156250

1.1054687500

1.1044921875

-0.0055966303

0.0164708495

0.0054213097

1.1035156250

1.1044921875

1.1040039063

-0.0055966303

0.0054213097

-0.0000916085

1.1040039063

1.1044921875

1.1042480469

-0.0000916085

0.0054213097

0.0026638633

1.1040039063

1.1042480469

1.1041259766

-0.0000916085

0.0026638633

0.0012858806

1.1040039063

1.1041259766

1.1040649414

-0.0000916085

0.0012858806

0.0005970744

1.1040039063

1.1040649414

1.1040344238

-0.0000916085

0.0005970744

0.0002527175

1.1040039063

1.1040344238

1.1040191650

-0.0000916085

0.0002527175

0.0000805507

1.1040039063

1.1040191650

1.1040115356

-0.0000916085

0.0000805507

-0.0000055299

1.1040115356

1.1040191650

1.1040153503

-0.0000055299

0.0000805507

0.0000375101

1.1040115356

1.1040153503

1.1040134430

-0.0000055299

0.0000375101

0.0000159901

.1040115356

1.1040134430

1.1040124893

-0.0000055299

0.0000159901

0.0000052301

[IMAGE][IMAGE]1.1040115356

1.1040124893

1.1040120125

-0.0000055299

0.0000052301

-0.0000001499

1.1040120125

1.1040124893

1.1040122509

-0.0000001499

0.0000052301

0.0000025401

1.1040120125

1.1040122509

1.1040121317

-0.0000001499

0.0000025401

0.0000011951

Here I can see that the final root is consistent to five decimal

places and is therefore is correct to five decimal places

Error bounds need to be established for the roots. To make sure that

the final answer that I have is correct I can have a look at the error

bounds. For my answer here I can see that to five decimal places, my

root between 1 and 2 is 1.10401. So I can use the change of sign

method to check my answer.

I can take my final answer one degree of accuracy further and look at

the change of sign of f(x) so when:

f(1.104005) = -7.9268 <0

f(1.104015) = 3.3557> 0

Therefore because my answer lies between the positive and negative, it

must be accurate.

However, there are some instances where this method does not succeed

and we are unable to find the root of the equation using the change of

sign method. For example if there are three roots between two

integers, the spreadsheets can only find one of those roots and so it

appears as though the other two does not exist unless a graph is also

drawn. These roots are unable to be found, and therefore the method

fails for some equations.

Y=9(5x-1)5-4(5x-1)2+0.5

[IMAGE]

From this graph of: 9(5x-1)5-4(5x-1)2+0.5=0

we can see that there are three roots between 0 and 1. However, if we

look at the spreadsheet for the change of sign method, we can see that

in fact, only one root is found between these two values.

The formulae that we enter in to the spreadsheet are in fact the same

as with the successful change of sign, because in both instances we

want to find the roots of the equation.

As we can see from the spreadsheet shown below, a root is actually

found for the equation, but only one of them, not all three of the

ones between [0, 1]. Therefore for this equation the method fails.

a

b

(a+b)/2

f(a)

f(b)

f((a+b)/2)

0.0000000000

1.0000000000

0.5000000000

0.5000000000

5.5000000000

-0.2187500000

0.0000000000

0.5000000000

0.2500000000

0.5000000000

-0.2187500000

0.2587890625

0.2500000000

0.5000000000

0.3750000000

0.2587890625

-0.2187500000

0.0042419434

0.3750000000

0.5000000000

0.4375000000

0.0042419434

-0.2187500000

-0.1213693619

0.3750000000

0.4375000000

0.4062500000

0.0042419434

-0.1213693619

-0.0605677068

0.3750000000

0.4062500000

0.3906250000

0.0042419434

-0.0605677068

-0.0284970393

0.3750000000

0.3906250000

0.3828125000

0.0042419434

-0.0284970393

-0.0121916348

0.3750000000

0.3828125000

0.3789062500

0.0042419434

-0.0121916348

-0.0039885210

0.3750000000

0.3789062500

0.3769531250

0.0042419434

-0.0039885210

0.0001235805

0.3769531250

0.3789062500

0.3779296875

0.0001235805

-0.0039885210

-0.0019332887

0.3769531250

0.3779296875

0.3774414063

0.0001235805

-0.0019332887

-0.0009050543

0.3769531250

0.3774414063

0.3771972656

0.0001235805

-0.0009050543

-0.0003907864

0.3769531250

0.3771972656

0.3770751953

0.0001235805

-0.0003907864

-0.0001336152

0.3769531250

0.3770751953

0.3770141602

0.0001235805

-0.0001336152

-0.0000050205

0.3769531250

0.3770141602

0.3769836426

0.0001235805

-0.0000050205

0.0000592792

0.3769836426

0.3770141602

0.3769989014

0.0000592792

-0.0000050205

0.0000271292

0.3769989014

0.3770141602

0.3770065308

0.0000271292

-0.0000050205

0.0000110543

0.3770065308

0.3770141602

0.3770103455

0.0000110543

-0.0000050205

0.0000030169

0.3770103455

0.3770141602

0.3770122528

0.0000030169

-0.0000050205

-0.0000010018

0.3770103455

0.3770122528

0.3770112991

0.0000030169

-0.0000010018

0.0000010076

0.3770112991

0.3770122528

0.3770117760

0.0000010076

-0.0000010018

0.0000000029

0.3770117760

0.3770122528

0.3770120144

0.0000000029

-0.0000010018

-0.0000004994

0.3770117760

0.3770120144

0.3770118952

0.0000000029

-0.0000004994

-0.0000002483

From this chart we can see that when the formulas look to find the

root between [0,1], however when it sees the root between [0, 0.5], it

does not try to look for the other two, between [0.5, 1]. Hence,

failure.

Newton - Raphson method

This is another fixed-point estimation method, and as for the previous

method, it is necessary to use an estimate of the root as a starting

point.

I start with an estimate, x, for a root of f(x) = 0. Next, I must draw

a tangent to the curve y = f(x) at the point (x, f(x)). The tangent

will cut the x axis at the next approximation for the root, then

tangent will then be found at that new point and so on.

I chose to start looking trying to find the roots for the equation:

3(x/2)3-4(x/2)+1.5=0

so I plotted the graph of y=3(x/2)3 -4(x/2)+1.5 in autograph.

Y=3(x/2)3-4(x/2)+1.5

[IMAGE]

I can see that there are three roots to my equation, in the intervals

[0, 1], [1, 2], [-2, -3]. I can use a spreadsheets and close ups of my

graph to find the roots to a given number of decimal places. I have

chosen to look closely at the root in the interval [0, 1].

Shown below are the formulae that I entered into the spreadsheet to

obtain my required results.

Newton Raphson

y=3(x/2)^3-4(x/2)+1.5

0

[IMAGE]=A4-((3*(A4/2)^3-4*(A4/2)+1.5)/(((9*A4^2)/8)-2))

=B4

=A5-((3*(A5/2)^3-4*(A5/2)+1.5)/(((9*A5^2)/8)-2))

=B5

=A6-((3*(A6/2)^3-4*(A6/2)+1.5)/(((9*A6^2)/8)-2))

=B6

=A7-((3*(A7/2)^3-4*(A7/2)+1.5)/(((9*A7^2)/8)-2))

=B7

=A8-((3*(A8/2)^3-4*(A8/2)+1.5)/(((9*A8^2)/8)-2))

=B8

=A9-((3*(A9/2)^3-4*(A9/2)+1.5)/(((9*A9^2)/8)-2))

[IMAGE]

This is to find out the point where the tangent of my equation crosses

the x-axis. (see below)

[IMAGE]

The reason the formula A4-((3*(A4/2)^3-4*(A4/2)+1.5)/(((9*A4^2)/8)-2))

is used is because we are trying to find out the new point where the

tangent of my equation at my given integer point crosses the x-axis.

As we can see from the close up of my graph below, tangents are used

to get an increasingly more accurate value for my root.

[IMAGE]

[IMAGE]

The gradient of the tangent at (x1, f(x1)) is f'(x1). Since the

equation of a straight line can be written:

y-y1=m(x-x1)

The equation of the tangent is:

y-f(x1)=f'(x1)[x-x1]

The tangent cuts the x-axis at (x2, 0), so

0-f(x1)=f'(x1)[x2-x1]

-f(x1)= f'(x1)x2-f'(x1)x1

f'(x1)x1-f(x1)=f'(x1)x 2

(f(x1)/f'(x1))x1-(f(x1)/ f'(x1)= x2

x2= x1-(f(x1)/ f'(x1))

Hence the equation is re-arranged to give a closer value for my root

on the x-axis. This is the equation that is then typed into my

spreadsheet, shown above.

The final value for my root is shown in the numerical spreadsheet

below, for the root between [0, 1].

Newton Raphson

y=3(x/2)^3-4(x/2)+1.5

0.000000000000000

0.750000000000000

0.750000000000000

0.865714285714286

0.865714285714286

0.875982340295208

0.875982340295208

0.876073029827270

0.876073029827270

0.876073036958846

0.876073036958846

0.876073036958846

0.876073036958846

0.876073036958846

0.876073036958846

0.876073036958846

0.876073036958846

0.876073036958846

0.876073036958846

0.876073036958846

0.876073036958846

0.876073036958846

So to five decimal places, the root in the interval [0,1], is 0.87607.

To find the other two roots of the equation, I can enter in different

starting integers into my spreadsheet to find the root to a given

number of decimal places, I have chosen to look at it to five. So for

the root in the interval [1, 2], the value is 1.74318 and for the root

in the interval [-2, -3], the value is -2.61925.

Error bounds need to be established for the roots however, if we look

at our final answer to five decimal places, then take that accuracy

one decimal place further, we can see whether our answer is correct.

So for 0.87607:

f(0.876065) = 0.00000913453385

f(0.876075) = -0.0000022311064

Because, the function of one of these numbers is positive and the

other is negative, the root that I am searching for must be between

the two, hence my answer of 0.87607 is a sensible one.

However, there are some equations that this method fails to find the

roots of. If the gradient of the tangent is more than 1 or less than

-1, the tangent crosses the x-axis at a point that is further away

from the root than the original integer, and hence an increasingly

less accurate answer is obtained.

To demonstrate this failure, I have chosen to use the equation:

y=ln(2x2-4)+x and to solve it using the equation ln(2x2-4)+x=0. First,

I shall look at the graph, as shown below.

[IMAGE]

This equation fails not only because the gradient of the graph is too

steep but also because the graph is discontinuous and there are

actually no values for x between [-2, 2] anyway.

When we apply the Newton Raphson tangents, as shown in the graph

below, we can see that the tangent moves off away from the root, and

an overflow is produced, hence spreadsheets also fail.

[IMAGE]

The tangent moves into the discontinuous area, due to its high

gradient

[IMAGE]

Neither integers either side of the root are able to move towards a

value for it, and I am also able to see this with the use of a

spreadsheet. The formulae that are entered into this spreadsheet are

the same that would be applied in a successful Newton Raphson

investigation (see above). However, when the numbers are entered, the

spreadsheet looks like this:

2.000000000000000

0.871235212960036

0.871235212960036

#NUM!

#NUM!

#NUM!

#NUM!

#NUM!

And for when we start with 1 as our integer, the spreadsheet simply

looks like this:

1.000000000000000

#NUM!

#NUM!

#NUM!

#NUM!

#NUM!

#NUM!

#NUM!

It is not possible to use the Newton Raphson method to solve this

equation.

Rearranging f(x)=0 in the form x=g(x)

The first step I must take to find a root for the equation f(x)=0, is

to rearrange it in the form x=g(x). To set up the spreadsheet we must

look at the line of y=x in conjunction with the rearrangement of my

equation.

There are some rearrangements that can be solved using this method and

others that this method fails to find the root for. I am going to look

at the equation: y=3x3-2x2+4x+7.

I am first going to rearrange it so that: y=3x3-2x2+4x+7

3x3=4x-2x2+7

x3=(4x-2x2+7)

3

x=((4x-2x2+7)/3)(1/3) then I will solve it so that ((4x-2x2+7)/3)(1/3)=0.

[IMAGE]

This is the graph of y=((4x-2x2+7)/3)(1/3). I can see here that the

line y=x and my g(x) cross at one point between the coordinates ([1,

2],[1, 2]). I can now use a spreadsheet to find this root to a given

number of decimal places. The root is found in the fixed point

iteration method by taking a line from my chosen integer to my g(x)

line, then from that point to the y=x line, then back to the g(x) line

and so on until the root is found to the highest possible degree of

accuracy. As shown on the graph below, this equation creates a cobweb

effect when these lines are drawn. Here we can see how the method

converges to the root by moving between the two lines until a suitable

value is found.

[IMAGE]

Text Box: This is the formula of my g(x), the rearrangement of my original f(x) equation.

To find these roots on a spreadsheet I must enter these formulae, as

explained below:

Text Box: The new value that my g(x) produced is then used again to find the g(x) of that value, eventually to get closer to the value of my root.

[IMAGE]

[IMAGE]

x

g(x)

1

=((4*A2-2*A2^2+7)/3)^(1/3)

=B2

=((4*A3-2*A3^2+7)/3)^(1/3)

=B3

=((4*A4-2*A4^2+7)/3)^(1/3)

=B4

=((4*A5-2*A5^2+7)/3)^(1/3)

=B5

=((4*A6-2*A6^2+7)/3)^(1/3)

=B6

=((4*A7-2*A7^2+7)/3)^(1/3)

The reason this equation succeeds is because the gradient of my line

is between 1 and -1. This means that the lines of iteration can move

between the y=x line and the y=((4x-2x2+7)/3)(1/3) line, without there

being a divergent.

Here is the spreadsheet of the success of iteration, and hence the

root to five decimal places, as I have chosen to find.

x

g(x)

1

1.44224957

1.44224957

1.421044364

1.421044364

1.423056043

1.423056043

1.422869683

1.422869683

1.422886986

1.422886986

1.42288538

1.42288538

1.422885529

1.422885529

1.422885515

1.422885515

1.422885517

1.422885517

1.422885516

1.422885516

1.422885516

1.422885516

1.422885516

1.422885516

1.422885516

However, my original equation of y=3x3-2x2+4x+7 can be rearranged to

give: y=3x3-2x2+4x+7

-4x=3x3-2x2+7

4x=2x2-3x3-7

x=2x2-3x3-7

4

I can try to use this using the iteration method to find the roots of

this equation. Below is the graph of this equation, with the y=x line.

[IMAGE]

Here we can see that there is a root in the coordinates ([0, -1], [0,

-1]). However, when we look at the graph with the iteration cobweb on,

we can see that in fact the cobweb spirals the wrong way, and

continuously moves outwards, away from the root, regardless of which

integer we chose to start from.

[IMAGE]

For the spreadsheet, because we have used a different rearrangement

for the equation, the formulae on the spreadsheet are slightly

different:

x

g(x)

-1

=(2*A2^2-3*A2^3-7)/4

=B2

=(2*A3^2-3*A3^3-7)/4

=B3

=(2*A4^2-3*A4^3-7)/4

=B4

=(2*A5^2-3*A5^3-7)/4

=B5

=(2*A6^2-3*A6^3-7)/4

=B6

=(2*A7^2-3*A7^3-7)/4

Text Box: This is the new arrangement of my equation, so this is now entered into my spreadsheet.

This equation fails because the gradient of the line is greater than

1, hence steeper than the y=x line.

Comparison of methods

In order to draw a conclusion about the easiest numerical method to

use, it is important to ensure that the same equation is used for each

method, and then the ease and speed of convergence to the root

assessed.

I decided to use the same equation as I did for my successful change

of sign method, which was y=5x3-7x+1. I then drew the graph of this

again, as shown below.

[IMAGE]

Next, I had to ensure the root of the equation 5x3-7x+1=0 could be

found using the Newton Raphson method and the rearrangement method,

both graphs respectively are found below

[IMAGE]

[IMAGE]

I can see that a root in the interval [0, 1] can be found using each

method. We need to look at the spreadsheets however to make sure that

the same root is found exactly to a given number of decimal places, I

have chosen to round to five. All of the formulae entered into the

spreadsheets are the same as shown in the individual analyses

previously.

Change of sign method

y=5x^3-7x+1

a

b

(a+b)/2

f(a)

f(b)

f((a+b)/2)

0.0000000000

1.0000000000

0.5000000000

1.0000000000

-1.0000000000

-1.8750000000

0.0000000000

0.5000000000

0.2500000000

1.0000000000

-1.8750000000

-0.6718750000

0.0000000000

0.2500000000

0.1250000000

1.0000000000

-0.6718750000

0.1347656250

0.1250000000

0.2500000000

0.1875000000

0.1347656250

-0.6718750000

-0.2795410156

0.1250000000

0.1875000000

0.1562500000

0.1347656250

-0.2795410156

-0.0746765137

0.1250000000

0.1562500000

0.1406250000

0.1347656250

-0.0746765137

0.0295295715

0.1406250000

0.1562500000

0.1484375000

0.0295295715

-0.0746765137

-0.0227093697

0.1406250000

0.1484375000

0.1445312500

0.0295295715

-0.0227093697

0.0033770204

0.1445312500

0.1484375000

0.1464843750

0.0033770204

-0.0227093697

-0.0096745566

0.1445312500

0.1464843750

0.1455078125

0.0033770204

-0.0096745566

-0.0031508496

0.1445312500

0.1455078125

0.1450195313

0.0033770204

-0.0031508496

0.0001125667

0.1450195313

0.1455078125

0.1452636719

0.0001125667

-0.0031508496

-0.0015192713

0.1450195313

0.1452636719

0.1451416016

0.0001125667

-0.0015192713

-0.0007033847

0.1450195313

0.1451416016

0.1450805664

0.0001125667

-0.0007033847

-0.0002954171

0.1450195313

0.1450805664

0.1450500488

0.0001125667

-0.0002954171

-0.0000914272

0.1450195313

0.1450500488

0.1450347900

0.0001125667

-0.0000914272

0.0000105693

0.1450347900

0.1450500488

0.1450424194

0.0000105693

-0.0000914272

-0.0000404291

0.1450347900

0.1450424194

0.1450386047

0.0000105693

-0.0000404291

-0.0000149299

0.1450347900

0.1450386047

0.1450366974

0.0000105693

-0.0000149299

-0.0000021803

0.1450347900

0.1450366974

0.1450357437

0.0000105693

-0.0000021803

0.0000041945

0.1450357437

0.1450366974

0.1450362206

0.0000041945

-0.0000021803

0.0000010071

0.1450362206

0.1450366974

0.1450364590

0.0000010071

-0.0000021803

-0.0000005866

0.1450362206

0.1450364590

0.1450363398

0.0000010071

-0.0000005866

0.0000002102

Newton Raphson

y=5x^3-7x+1

0.000000000000000

0.142857142857143

0.142857142857143

0.145034843205575

0.145034843205575

0.145036371205923

0.145036371205923

0.145036371206683

0.145036371206683

0.145036371206683

0.145036371206683

0.145036371206683

0.145036371206683

0.145036371206683

Rearrangement:(5x3+1)/7

x

g(x)

0

0.142857143

0.142857143

0.144939608

0.144939608

0.145032012

0.145032012

0.145036175

0.145036175

0.145036362

0.145036362

0.145036371

0.145036371

0.145036371

0.145036371

0.145036371

0.145036371

0.145036371

0.145036371

0.145036371

0.145036371

0.145036371

0.145036371

0.145036371

From the spreadsheets we can see that the root of this equation to

five decimal places is 0.14504. The change of sign takes the longest

to converge to the root, taking 21 rows to find it. The Newton Raphson

and rearrangement methods were both very quick to converge, however

the Newton Raphson found it in 4 rows, whereas the rearrangement took

5.

The change of sign method is also very complicated to set up a

spreadsheet for because it has many different columns and it would be

easy to make a mistake when entering a formula, which would make the

whole spreadsheet useless. The rearrangement method can also be

difficult because of errors made when rearranging the equations; this

can take time and be difficult to find one that is able to find the

root. Also two graphs have to be drawn, which did not prove to be

difficult for me using autograph, but by hand, there is more chance of

error. However, the spreadsheet is easy to set up and use. On the

whole I conclude that the Newton Raphson method is the best method to

use because the spreadsheet is relatively easy to set up, only one

graph has to be drawn, no rearrangements have to be found and for this

equation, and most others, the Newton Raphson method found the root

the quickest.

For this investigation I used several different types of software on

my computer. For the spreadsheets I use edexcel, which was easy to use

and much more efficient than trying to work out f(x) or g(x) for each

line. Autograph was another very useful program because the graphs are

drawn neatly and accurately, with no chance of error, and is also very

quick to use. The other benefit of autograph is that there are certain

functions that can be used, such as the Newton Raphson lines and g(x)

iteration lines can be drawn on the graph in autograph.

I used two different types of hardware, my computer with all the

programmes listed above and my graphics calculator. The computer was

much easier to use because it has a larger screen which I can zoom in

on, and more than one thing can be looked at at one time. The graphics

calculator was also unreliable and often difficult to pinpoint the

exact root that I was trying to look for.