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Title: MEI (OCR) Maths - Core 3 - Coursework - Comparison of Methods
Description: 2nd year sixth form coursework core 3 module

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Ryan Guttridge

Comparison of Methods
Change of sign method – Decimal Search
This method will find an interval in which the root lies
...

It is possible to use the change of sign method on a table in order to identify and specify with
greater accuracy the interval in which the root lies
...

They are [-2,-1] [0, 1] and [1, 2]
...


The symbol

represents a change in sign
...
1
...


As is evident from the table; there is a sign change, and
therefore a root; in the interval [1
...
2]
...
1, 1
...
01
...

As seen from the table, there is a sign change and therefore a root in the interval [1
...
16]
...
15, 1
...
001, which is again shown in the table (above-right)
...
159, 1
...


In this particular instance the root is to be found to an accuracy of
Β±0
...
p
...
159,
1
...
However, to be able to find an actual value for the root
to 3 decimal places, the method must be repeated a final time to
determine the final decimal place i
...
whether the root is 1
...
160 to 3 d
...

It can be seen from the table that there is a sign change, and thus a
root in the interval [1
...
1599]
...
p
...
160 to 3 d
...
as all values are in the interval [1
...
1599]
...
1595 and the upper bound is 1
...
1595) = -0
...
1605) = 0
...
1595 and the upper bound of 1
...

There is very little difficulty in this method, it can be considered the easiest of the 3 methods,
but it is rather time consuming when great degrees of accuracy are required
...
The main fault with this method is that it
fails to find the root on many occasions – including when a repeated root is present or when
two roots are close together
...

The first intention is to find a root of the equation: 𝑓(π‘₯) = π‘₯ 9 βˆ’ 5π‘₯ + 2 = 0
This can be rearranged into the form π‘₯ = 𝑔(π‘₯) where

This gives the iterative formula:

π‘₯ 𝑛+1 =

π‘₯ 9 +2
𝑛
5

𝑔( π‘₯ ) =

π‘₯ 9 +2
5


...
40005 this is a root of the equation: π‘₯ 9 βˆ’ 5π‘₯ + 2 = 0
...
At which point, a downward vertical line is
drawn – the point at which this downward line crosses the x axis is the new prediction x1
...

This method is slightly more complex than the decimal search method – requiring a
rearrangement of the equation
...


Newton-Raphson Method
π‘₯ 𝑛+1 = π‘₯ 𝑛 βˆ’

𝑓(π‘₯ 𝑛 )
𝑓 β€² (π‘₯ 𝑛 )

The Newton-Raphson method is another fixed point estimation method, i
...
it
estimates the actual value of a root as opposed to the interval in which it lies
...

It is necessary to find f’(x) to use in the formula:
f(x) = π‘₯ 9 βˆ’ 5π‘₯ + 2
f’(x) = 9π‘₯ 8 βˆ’ 5
This gives the iterate formula:
π‘₯ 𝑛+1 = π‘₯ 𝑛 βˆ’

π‘₯ 9 βˆ’ 5π‘₯ 𝑛 + 2
𝑛
9π‘₯ 8 βˆ’ 5
𝑛

If the numbers in the sequence approach a limiting value then they are said to
β€˜converge’ to this value
...
Where this
tangent crosses the x-axis
will be the next prediction x2
...
5
Tangent at x=1
...
15988 and so the root lies in
the interval [1
...
160]
...
160 (to 3 d
...


The Newton-Raphson method is both time consuming and is probably the most
difficult of all 3 methods – requiring differentiation of the equation and then other
processes to obtain the desired result
...
e
...

In conclusion, these 3 methods all have their failings, the quickest is most likely the
rearrangement method, the easiest is the decimal search and the most reliable is the
Newton-Raphson method
...
Microsoft Excel provided an easy means of calculations – much faster
than using a calculator as once the formula is in place it simply needs to be β€˜dragged
down’ to replicate it
...

Omnigraph was very useful for this as it has a very versatile scale adjustment
system, also a tool that creates tangents at any point
...



Title: MEI (OCR) Maths - Core 3 - Coursework - Comparison of Methods
Description: 2nd year sixth form coursework core 3 module