- Level: AS and A Level
- Subject: Maths
- Word count: 1498
Numerical Solutions of Equations
Extracts from this document...
Introduction
A2 Level Maths Coursework
A Level Coursework – Numerical Solutions of Equations
My coursework is based on finding roots of non-quadratic equations. I will be investigating 3 methods for finding these roots and comparing them.
Change Of Sign Method
Y = x4+4x-6
This method simply uses Microsoft Excel to tabulate points on a function in different stages. It ‘narrows down’ the desired root by getting progressively closer to it. The root is always in the interval between the 2 x values where the y values change sign.
-5 | 599 |
-4 | 234 |
-3 | 63 |
-2 | 2 |
-1 | -9 |
0 | -6 |
1 | -1 |
2 | 18 |
3 | 87 |
4 | 266 |
5 | 639 |
The roots of this equation lie in the intervals -2 < x < -1 & 1 < x < 2.
To find the lower root, I tabulated -2 < x < -1.
x | y |
-2 | 2 |
-1.9 | -0.5679 |
-1.8 | -2.7024 |
-1.7 | -4.4479 |
-1.6 | -5.8464 |
-1.5 | -6.9375 |
-1.4 | -7.7584 |
-1.3 | -8.3439 |
-1.2 | -8.7264 |
-1.1 | -8.9359 |
-1 | -9 |
There is a change of sign between -2 & -1.9, therefore the boundaries of the root lie in this interval.
I tabulated in steps of 0.01, and again in steps of 0.001 & 0.0001
x | y |
-1.924 | 0.007146 |
-1.9239 | 0.004697 |
-1.9238 | 0.002249 |
-1.9237 | -0.0002 |
-1.9236 | -0.00265 |
-1.9235 | -0.00509 |
-1.9234 | -0.00754 |
-1.9233 | -0.00999 |
-1.9232 | -0.01243 |
-1.9231 | -0.01488 |
-1.923 | -0.01732 |
x | y |
-1.93 | 0.15488 |
-1.929 | 0.130146 |
-1.928 | 0.105457 |
-1.927 | 0.080812 |
-1.926 | 0.056212 |
-1.925 | 0.031657 |
-1.924 | 0.007146 |
-1.923 | -0.01732 |
-1.922 | -0.04174 |
-1.921 | -0.06612 |
-1.92 | -0.09046 |
x | y |
-2 | 2 |
-1.99 | 1.722392 |
-1.98 | 1.449536 |
-1.97 | 1.181385 |
-1.96 | 0.917891 |
-1.95 | 0.659006 |
-1.94 | 0.404685 |
-1.93 | 0.15488 |
-1.92 | -0.09046 |
-1.91 | -0.33137 |
-1.9 | -0.5679 |
Middle
-1.92376
0.001269505
-1.92375
0.001024725
-1.92374
0.00077995
-1.92373
0.000535179
-1.92372
0.000290412
-1.92371
4.56502E-05
-1.9237
-0.000199108
The other root of the equation can be found using this method. It is, to 5 significant figures, 1.1144.
Failure of the Change of Sign Method
The Change of Sign method only works for certain equations, e.g. where there is a change of sign or where a change of sign is shown in the first table of values. One equation where the Change of Sign method does not find the roots is y = 4x4-4x+1.
x | y |
-5 | 2521 |
-4 | 1041 |
-3 | 337 |
-2 | 73 |
-1 | 9 |
0 | 1 |
1 | 1 |
2 | 57 |
3 | 313 |
4 | 1009 |
5 | 2481 |
x | y |
0 | 1 |
0.1 | 0.6004 |
0.2 | 0.2064 |
0.3 | -0.1676 |
0.4 | -0.4976 |
0.5 | -0.75 |
0.6 | -0.8816 |
0.7 | -0.8396 |
0.8 | -0.5616 |
0.9 | 0.0244 |
1 | 1 |
Newton Raphson Method
x3 – 10x + 6 = 0
The Newton Raphson Iteration works as follows:
It is based on the formula
F’(Xn) is F(Xn) differentiatied, so if F(Xn) = x3 – 10x + 6, than F’(xn) = 3x2 - 10. Iterations work by repeating a formula that eventually converges toward a root.
Xo is the original estimation of a root. Therefore we replace Xn with Xo.
My original estimate for the middle root is 3, so substitute 3 for Xo.
Therefore: f(Xn) = 33 – 10x3 + 6 = 3
And f’(Xn) = 3*32 – 10 = 17
X1 = 3 – (3/17) = 2.823529412 or 48/17
X2 = 2.803785064
X3 = 2.803542459
X4 = 2.803542496
X5 = 2.803542496
The other two roots of this equation are:
0.62434 & -3.4279, each to 5 significant figures.
This method can be illustrated graphically:
Failure of the Newton-Raphson Method
However, the Newton-Raphson method does not work for all roots. One example of an equation where the Newton Raphson method does not work is y = 2/X3+6.
x | y |
-5 | 5.99936 |
-4 | 5.998047 |
-3 | 5.99177 |
-2 | 5.9375 |
-1 | 4 |
0 | #DIV/0! |
1 | 8 |
2 | 6.0625 |
3 | 6.00823 |
4 | 6.001953 |
5 | 6.00064 |
Conclusion
However, the time it takes to set up and apply a method also affects it’s efficiency.
The Newton Raphson method took the least amount of steps to find the root, but also took the most time to substitute into the formula as the function had to be differentiated. Although this took a while, once the formula was in the calculator, it could easily be repeated for each Xn.
The Change of Sign method took 6 steps to converge to the root (up to 5 s.f.) but took by far the most time to work out. This was because the table on excel for each step had to be changed, while using the other methods a calculator could just re-iterate the chosen formula.
The Fixed Point iteration took more steps to converge to the root, twice that of Change of Sign and three times that of the Newton Raphson method. However, it was the quickest to find using hardware and software as the formula was easily re-arranged and could then be quickly written into and then re-iterated on a graphical calculator for each Xn.
This student written piece of work is one of many that can be found in our AS and A Level Core & Pure Mathematics section.
Found what you're looking for?
- Start learning 29% faster today
- 150,000+ documents available
- Just £6.99 a month