Change of Sign Method - Mathematical Essay

Change of Sign Method - Mathematical Essay

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Change of Sign Method - Mathematical Essay

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

=5*C5^3-7*C5+1

=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.
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