- Level: AS and A Level
- Subject: Maths
- Word count: 1797
This coursework is about finding the roots of equations by numerical methods.
Extracts from this document...
Introduction
C3 Coursework - Numerical Solution to Equations
This coursework is about finding the roots of equations by numerical methods.
I am going to use three different methods to solve different equations. First of all, one root should be found successfully by using three different methods. Then, error bounds should be given and shown by graphic. A failure example is given and explained.
Change of Sign Method
I need to use an equation and use Autograph to get a rough interval and do a search to find intervals that show a change of sign.
Here is an example x³–2x²–3x+4=0
Solve x³–2x²–3x+4=0
Here is the table of values
x | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 |
F(x) | -38 | -6 | 4 | 4 | 0 | -2 | 4 | 24 |
This shows that there are three intervals containing roots:
(-2, -1), [1,2] and (2, 3)
Then carry out a decimal search in one of the identified intervals to find that root to the desired level of accuracy.
I use the interval (2, 3)
x | f(x) | |
2 | -2 | |
2.1 | -1.859 | |
2.2 | -1.632 | |
2.3 | -1.313 | |
2.4 | -0.896 | |
2.5 | -0.375 | |
2.6 | 0.256 | |
2.7 | 1.003 |
The change in sign tells us there is a root in the range (2.5, 2.6)
Now use decimal search and Excel within the interval [2.5, 2.6]
x | f(x) | |
2.51 | -0.31695 | |
2.52 | -0.25779 | |
2.53 | -0.19752 | |
2.54 | -0.13614 | |
2.55 | -0.07363 | |
2.56 | -0.00998 | |
2.57 | 0.054793 | |
2.58 | 0.120712 |
Middle
2.56150
-0.00034
2.56151
-0.00028
2.56152
-0.00021
2.56153
-0.00015
2.56154
-8.2E-05
2.56155
-1.8E-05
2.56156
4.63E-05
This graph shows us how the root is finally trapped in the range (2.56155, 2.56156)
What can cause failure?
One reason that the decimal search can fail is that there are several roots very close together in the same integer range.
For example, the equation y=54x³−225x²+309x−140
x | y | x | y | x | y | x | y |
1 | -2 | 1.61 | -0.37533 | 1.661 | -0.04956 | 1.6661 | -0.00509 |
1.1 | -0.476 | 1.62 | -0.32749 | 1.662 | -0.04103 | 1.6662 | -0.00419 |
1.2 | 0.112 | 1.63 | -0.27216 | 1.663 | -0.0324 | 1.6663 | -0.00329 |
1.3 | 0.088 | 1.64 | -0.20902 | 1.664 | -0.02368 | 1.6664 | -0.0024 |
1.4 | -0.224 | 1.65 | -0.13775 | 1.665 | -0.01488 | 1.6665 | -0.0015 |
1.5 | -0.5 | 1.66 | -0.05802 | 1.666 | -0.00598 | 1.6666 | -0.0006 |
1.6 | -0.416 | 1.67 | 0.030502 | 1.667 | 0.003005 | 1.6667 | 0.0003 |
1.7 | 0.352 | ||||||
There appears to be only one root in the interval (1, 2) but there are three.
After finding the first root we might miss the other two.
There are three roots close together.
We only get one change of sign when we do the integer search.
The search will find the first root 1.6666 in this interval
Unless we have drawn a graph there is a big danger we may miss the other two roots.
Newton Raphson
An example is given: y=x³−2x²+1
We can see there are roots in the intervals [-1, 0] and [1, 2]
We will find the root in the interval [-1, 0]
Xn+1= Xn–f(Xn)/f’(Xn)
If f(X)= x³−2x²+1 then f’(x)= 3x²-4x
Xn+1= Xn–( Xn ³−2 Xn ²+1)/ (3 Xn ²-4 Xn)
Using our example : y=x³−2x²+1
To find the root in the interval [-1, 0] we can take x1 =-1 or x1 = 0
I will use x1 = -1
x | y |
-1 | -0.71429 |
-0.71429 | -0.62658 |
-0.62658 | -0.61811 |
-0.61811 | -0.61803 |
-0.61803 | -0.61803 |
Can see that root is near to -0.6180 to 4 sf
Using Newton Raphson and Autograph Solver to find the root in the interval [-1, 0] Solution: x=-0.61803399, y=0
Then find the root in the interval [1, 2]
To find the root in the interval [-1, 0] we can take x1 =1 or x1 = 2
I will use x1 = 2
x | y |
2 | 1.75 |
1.75 | 1.642857 |
1.642857 | 1.619207 |
1.619207 | 1.618037 |
1.618037 | 1.618034 |
1.618034 | 1.618034 |
Can see that root is near to 1.6180 to 5 sf
x | y |
1 | 1 |
Conclusion
(-0, 1)
To find the root in the interval (0, 1)
Decimal search
X | y |
1 | -1 |
0.9 | -0.097 |
0.8 | 0.624 |
0.89 | -0.01701 |
0.88 | 0.061184 |
0.889 | -0.00911 |
0.888 | -0.00123 |
0.887 | 0.006639 |
0.8879 | -0.00044 |
0.8878 | 0.000349 |
0.88789 | -0.00036 |
0.88788 | -0.00028 |
0.88787 | -0.0002 |
0.88786 | -0.00012 |
0.88785 | -4.4E-05 |
0.88784 | 3.42E-05 |
The root is0.8878 to 4 s.f.
Newton-Raphson
- To find the root in the interval [0, 1] we can take x1 =0 or x1 =1
I will use x1 = 1
xn+1= xn - (3xn^3+xn^2-xn-2)/(9xn^2+2xn-1)
x | y |
1 | 0.9 |
0.9 | 0.88801 |
0.88801 | 0.887844 |
0.887844 | 0.887844 |
Can see that root is near to 0.8878 to 4 sf
x = g(x)
Take x1 = 1
Roots are given by −3x³−x²+x+2=0
Rearranges to:
3x³= −x²+x+2
x³ = 1/3(−x²+x+2)
x= [1/3(−x²+x+2)]1/3
- Iteration is Xn +1 = [1/3(−Xn ²+ Xn +2)]1/3
Using our example and finding the root in the interval [0, 1] , using X1=1
x | y |
0.0000 | 0.8736 |
0.8736 | 0.8736 |
Can see that root is near to 0.8736 to 4 sf
Each method achieved the same accuracy in:
16 calculations for change of sign
4 calculations for Newton Raphson
1 calculation for x = g(x)
The x=g(x) method was the fast one in this particular example
Change of sign was much slower than NR and x=g(x)
If I only had a scientific calculator – the change of sign method is easier to operate.
If I also have Excel - the x=g(x) method is easier because there was only one calculation.
If I also have Autograph – the x=g(x) method is easier because there was only one calculation..
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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