- Level: AS and A Level
- Subject: Maths
- Word count: 741
Solving Equations using Numerical Methods.
Extracts from this document...
Introduction
A2 Pure Maths Coursework
Solving Equations using Numerical Methods
There are three methods with which you can solve an equation.
- Change of Sign Method
- Newton-Raphson Method
- Rearranging the equation f(x)=0 in the form x= g(x)
Hardware and Software Used
For the coursework I have used a computer for attaining more accurate results and to avoid errors.
The software I have used are
- Microsoft Word
- GraphCalc (A software used for drawing graphs)
Change of Sign Method
x4-4x +2
We see that the roots lie in between [0, 1] and [1, 2]
We confirm this by looking for a change of sign
x | 0 | 1 | 2 |
f(x) | 2 | -1 | 10 |
Middle
0.0024
0.00206
x | 0.5175 | 0.5176 | 0.5177 | 0.5178 | 0.5179 | 0.518 |
f(x) | 0.00172 | 0.00137 | 0.00103 | 0.00068 | 0.00034 | -0.000002 |
Root is in between [0.5179, 0.5180]
Failure Case
There are some cases where this method cannot be used to solve the equation.
Newton-Raphson Method
f `(x) = ∫f(x)
= ∫ (x5 – 4x +1)
= 5x4 -4
Roots are in between [0,1] and [1, 2].
We look for the roots in [0,1]
xn+1 = xn – f (xn)
f ` (xn)
xn+1= xn - (xn5 – 4xn+1)
(5xn4 -4)
When x1= 0
x2 = 0 – ((0)5 - 4(0) + 1)
5 (0)4 - 4
= 0 - 1
-4
= 0.25
x3 = 0.25 - ((0.25)5- 4(0.25) + 1)
5(0.25) 4
= 0.25024533856722
x4 = 0.20524 - ((0.20524)5- 4(0.20524) + 1)
5(0.20524) 4
= 0.25024534093233
Root will be in between 0.250235 and 0.250245
Failure Case
Rearranging the Equation in the form x=g(x)
We start this method by putting f(x) = 0.
x4
Conclusion
x
xn+1 = (4xn - 2)1/2
xn
x1 = 1 x1 = 2
x2 = 1.414x2 = 1.2247
x3 = 1.352x3 = 1.3901
x4 = 1.3654x4 = 1.3573
x5 = 1.3626x5 = 1.3643
x6= 1.3632x6 = 1.3628
x7 = 1.3630x7 = 1.3631
x8 = 1.3631 x8 = 1.3631
x9 = 1.3631
x = x4 + 2
4
is equivalent to solving
y = x
y = x4 + 2 is y = g(x)
4
The solution is where the line crosses the graph
Rearrangement will work if its gradient g’(x) is less than the gradient of y = x otherwise it fails. The gradient of y = x is 1.
| g’(x) | < 1 method works
| g’(x) | > 1 method fails
g(x) = x4 + 2
4 4
g’(x) = 4x3
4
g’(x) = x3
In [0, 1] , say x= 0.5
g’(x) = (0.5)3
= 0.625 < 1 . The method works
In [1, 2] , say x= 1.5
g’(x) = (1.5) 3
= 5.0625 > 1. The method fails.
This student written piece of work is one of many that can be found in our AS and A Level Core & Pure Mathematics section.
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