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Krishan Mistry
Candidate number: xxxxx
Centre number: xxxxx

Core 3 Mathematics coursework – Numerical solutions of equations
...
The function is f(x) =x3+5x2-5x-4, illustrated as the graph
below
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positive

negative

Below is an image of the graph of the function zoomed in
...
This will enable me to approximately find where
the root is
...
The root lies
between the intervals [-6,-5],[-1,0] and [1,2], which means that there is a root between these two
values
...
The ‘change of
signs’ is highlighted in green
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748 and -5
...
This value can be
written as: -5
...
749, or as a value with error bounds: -5
...
0005
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An
example is the equation x³ + 5x² + 4
...
Below is the graph of the function f(x) = x³ + 5x² + 4
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positive

negative

Below is the table of values for this function
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There is a change of sign between x=-4 and x=-3
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Krishan Mistry
Candidate number: xxxxx
Centre number: xxxxx

However from the above graph, there are two other roots between x=-1 and x=0 or [-1,0]
...


positive

root

root

negative

I can see that the two other roots specified on the graph above have been missed by the Decimal
search method
...


Krishan Mistry
Candidate number: xxxxx
Centre number: xxxxx

Newton-Raphson Method
I will now carry out the Newton-Raphson Method on a new equation
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positive

negative

The Newton-Raphson Method relies on the general iterative formula:

Krishan Mistry
Candidate number: xxxxx
Centre number: xxxxx

For this function, it becomes:

xn  4 xn  7 xn  3
3

x n 1  x n 

2

3xn  8 xn  7
2

root

positive

negative

root
positive

negative

root

Krishan Mistry
Candidate number: xxxxx
Centre number: xxxxx

The table below shows the location of the three roots
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The graphs below show the Newton-Raphson iteration for the function
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4286 (as
shown in the
table )

Krishan Mistry
Candidate number: xxxxx
Centre number: xxxxx

positive

-0
...
3609 (as
shown in the
table )

Krishan Mistry
Candidate number: xxxxx
Centre number: xxxxx

This method continues and the root is 0
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This x value can be
written as: -0
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360875, or as a value with error bounds: -0
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000005
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The position of the root is evident
due to the change of signs occurring between the two error bounds
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An image of
the function f(x) = x³ + 2x²-2 is shown below
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positive

negative

Krishan Mistry
Candidate number: xxxxx
Centre number: xxxxx

If I use the Newton-Raphson iteration feature on Autograph when x n = 0, it appears as divergent
instead of the value where the tangent hits the x axis
...
The method will draw a
tangent at -2, and then try to find the x axis
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Krishan Mistry
Candidate number: xxxxx
Centre number: xxxxx

Rearrangement method
I will now carry out the Rearrangement of f(x) =0 to give x=g(x) on a new function and equation
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The equation I shall use is
x3+7x2+6x-3=0
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Krishan Mistry
Candidate number: xxxxx
Centre number: xxxxx

root

root

root

This equation could be rearranged in several ways to find the roots
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1st Rearrangement of the equation
1)
2)
3)
2nd rearrangement of this equation

1)
2)
3)
4)
3rd rearrangement of this equation
1)
2)
3)
4th rearrangement of this equation
1)
2)

Krishan Mistry
Candidate number: xxxxx
Centre number: xxxxx
3)
4)
5)

I can plot the four rearranged functions y=(−x³-7x²+3)/6 , y²=((-x³-6x+3)/7) , y²=(-6x+3)/(x+7) and y³=(-7x²6x+3) against y=x Autograph along with the I can see that when the two lines intersect, the root is
exactly the same as that of the original function y=x^3+7x^2+6x-3
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Root

negative
root
root

The line y=x and function g(x) = (−x^3-7x^2+3)/6 are stated on the graph below
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Another purple line is then horizontally drawn

Krishan Mistry
Candidate number: xxxxx
Centre number: xxxxx
from this meeting point to the line y=x
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The
graph below shows the state after 1 iteration using the x=g(x) feature on Autograph
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Again the purple line is drawn vertically to
meet the g(x) line , a and is then drawn from this point to the y=x line
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After a total of 72 iterations
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Krishan Mistry
Candidate number: xxxxx
Centre number: xxxxx

Below is a completed table on Excel showing the exact locations of the roots
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Krishan Mistry
Candidate number: xxxxx
Centre number: xxxxx

Krishan Mistry
Candidate number: xxxxx
Centre number: xxxxx

The root is therefore 0
...

The equation

can be rewritten as

Differentiating this can give can also


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34997 (5
...
p) as x and substitute this into the above equation
the gradient is -0
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However the same equation (−x^3-7x^2+3)/6=0 can also fail
...
The graph
below with the function g(x) = (−x^3-7x^2+3)/6 shows the same method with the x=g(x) feature on
autograph
...

The graph below shows the position after 10 iterations
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This is because the gradient is too steep and this method will only
work when the gradient is between -1 and 1
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The table below, done
on Excel, can show this because at x=-1, the root which the iteration didn’t expect to find is found
and the values diverge when x=-2
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I will start off with the Decimal search method
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34997
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root

root

Positive

Negative

root

Krishan Mistry
Candidate number: xxxxx
Centre number: xxxxx

Comparison (Decimal search)
I shall attempt to solve this equation (x³ + 7x² + 6x- 3 = 0) by the Decimal search method and hence
find the roots of the equation
...
This will enable me
to approximately find where the root is
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The root I found
for this equation was 0
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There is a change of sign between x=0
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34997
...
34996, 0
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This can be seen in
the tables above
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34996 ...
3499965 ± 0
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The rearrangement method found 0
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This means that both the
Decimal search and rearrangement method have successfully found the same root
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My iterative formula in this particular case would be x n 1 

x3  7x 2  6x  3

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Each root is highlighted in a different colour
...
This has been done on
Autograph, with x n = 0 (the starting value)

0
...
3636 (as
shown in the
table )

Krishan Mistry
Candidate number: xxxxx
Centre number: xxxxx

positive

negative
0
...
35 (as shown
in the table )

Krishan Mistry
Candidate number: xxxxx
Centre number: xxxxx

positive

negative

0
...
So the root is x = 0
...


Krishan Mistry
Candidate number: xxxxx
Centre number: xxxxx
Conclusion
I have now found the same root using the Decimal Search method, the Rearrangement method and
the Newton-Raphson method
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For the equation used for comparison, x³ + 7x² + 6x- 3 = 0,I thought that the Newton-Raphson
method was the most appropriate method to use
...
Although the Newton-Raphson and
Rearrangement methods were similar as they both had iterative processes, the Newton-Raphson
worked best because it had a general iterative formula

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The rearrangement method found the root after 106 calculations, but the
Newton-Raphson method found the root after 4 calculations
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However the Decimal search method took 5 calculations to get to this level of
accuracy, which was less than that on the rearrangement method
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However the
advantages of this method are that the method is quite easy to carry out
...
However the Newton-Raphson and rearrangement
method weren’t as ‘time-consuming’ as the Decimal search method
...

The software used to draw the graphs shown throughout my coursework was Autograph
...
Autograph was able to draw more advanced and sophisticated graphs of the functions in all
the methods, as Excel doesn’t have the capability and the facility to this
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However there
were some disadvantages of Autograph
...
,whereas Excel’s maximum level of accuracy was
to 15 decimal places, although formulas were quite tedious and complex to type in, and mistakes
could’ve been made easily, leading to incorrect values
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A simple representation is shown below
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This concludes that the Newton-Raphson method was the most appropriate
method to use

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Title: OCR MEI CORE 3 COURSEWORK ( COMPLETED W/ FULL MARKS)
Description: This is my OCR MEI Core 3 Mathematics coursework.This is the completed version and I scored 18/18 when I submitted this,which was in 2013.Please use as a reference guide only.

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