- Level: AS and A Level
- Subject: Maths
- Word count: 2008
Solution of Equations by Numerical Methods
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Introduction
Damian Chandler 13MP 66555 0474 Maths Coursework
Solution of Equations by Numerical Methods
Change of Sign Method – Numerical Search
By using the computer program ‘Omnigraph’, I have produced this graph of the equation y = x^3-9x+7.
The roots of this equation are shown where the line crosses the X-axis. I can now see that the three roots of this equation lie between (-4,-3), (0,1) and (2,3). Due to the fact that I have only declared the roots to be within wide margins, I will need to use a different method to acquire a more accurate answer. Instead of simply saying “the root is about -3.3”, I can use a method called decimal search to find an answer to any degree of accuracy that I decide.
Decimal search works by finding two points on the x-axis of a graph, where between, the sign of Y changes, i.e. where the line crosses the x-axis. Without the graph I would be able to work out that the sign changes three times, meaning there are three roots.
This table shows the values of Y when X is between -5 and 4:
X | -5 | -4 | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 |
Y | - 73 | -21 | 7 | 17 | 15 | 7 | -1 | -3 | 7 | 35 |
From this I can now work out a more accurate answer to where the roots lie. I now know that a root lies between (-4,-3).
Middle
2
3
4
Y
-43.26
-10.58
4.096
6.776
3.456
0.136
2.816
17.496
It appears from the table that there is only a single root. However the curve of the equation looks like this:
Rearranging into the Form x = g(x)
The Equation y = x^3-14x-4, can be written in many different ways. By rearranging the equation I could write it as: ‘14x = x^3-4’ or ‘x = (x^3-4)/14.
The graph of the equation y = x^3-14x-4 looks like this:
If I now use the rearranged equation x = (x^3-4)/14 and draw the lines of both sides of the equation on a graph i.e. y = x and y = (x^3-4)/14, then the points where the two lines cross are the roots from the original equation.
This shows the graph of the two lines y=x and y = (x^3-4)/14.
x | -4 | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 |
f(x) | -12 | 11 | 16 | 9 | -4 | -17 | -24 | -19 | 4 |
The roots lie between (-4,-3), (-1,0) and (3,4).
Comparing this to my graph, I can see that it looks correct.
For my equation y = x^3-14x-4, to find the root in the interval (-1,0), I start at X1 = 1.
Xn^3 – 4 = Xn+1
14
X2 = (1^3 – 4)/14 = -0.21429
X3 = (-0.21429^3 – 4)/14 = -0.28642
X4 = (-0.28642^3 – 4)/14 = -0.28738
X5 = (-0.28738^3 – 4)/14 = -0.28741
X6 = (-0.28741^3 – 4)/14 = -0.28741
Root is -0.28741 to 5 d.p. This means the root is x = 0.28741 ± 0.000005
Failure of rearranging into the Form x = g(x)
Conclusion
The rearranging into the form g(x) is a very easy method as it simply requires just putting the new answer into another simple formula. This is still a tedious method as it requires a lot of repetition.
Although the Newton-Raphson method can be slightly confusing and a mistake can be made easily, it is by far the quickest. The main problem occurring, by an estimate being made near a turning point that doesn’t occur close to the x-axis can be eliminated by using the decimal search method to find the rough points at which the roots cross the x-axis. The other problem that can cause this method to take a long time is by the original estimate being made not close to the root. This will result in it taking a long time to reach the root. This can however be eliminated by combining the decimal search method with Newton-Raphson. This will allow decimal search to find the rough points at which the roots occur, then allowing Newton-Raphson to close in on the root without suffering the repetition of decimal search.
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