Solutions of equations
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Introduction
Gareth Brown Solutions of Equations
C3
Solutions of equations
By Gareth Brown
Introduction
I will be investigating the solutions of equations using the following three methods:
- Change of sign method
- Newton-Raphson method
- Fixed point iteration after rearranging the equation from form f(x) =0 into the form x = g(x)
I will solve the equations with these methods as I cannot solve them by any algebra I have met so far. The equations I am going to use are the following:
- y = 243x3-378x2+192x-32 = 0
- y = ex-x3-1.4 = 0
- y = 43x3-60x2-250x-115 = 0
- y = x3-5x+2.6 = 0
In my investigation I will give illustrated examples of the methods failing for the root I am looking for. All failures will be explained.
In my investigation I will use a Casio calculator to help draw graphs and to help identify where the rots are. I will also have computer software such as Autograph to help draw graphs and to show certain methods graphically.
Change of Sign Method
I will be using the “decimal search” method to check for a change of sign in the equation:
y = 243x3-378x2+192x-32
The graph of this function is given below.
I then decided to confirm the shape of the graph by finding how many turning points there is.
Dy = 729x2-756x+192 = 0
Dx
Quadratic Equation: D= b2-4ac
=-7562-(4*729*192)
=11664 > 0
=> 2 Real Turning Points (Graph Shape Confirmed)
By putting in Values of X into the equation I found some values of Y.
Table 1
x | F(x) |
-4 | 22400 |
-3 | -10571 |
-2 | -3872 |
-1 | -845 |
0 | -32 |
1 | 25 |
2 | 784 |
3 | 3703 |
4 | 10240 |
Middle
To find root A.
X0 = -2
X1 = -2-(e-2-(-2)3-1.4) = -1.43232
(e-2 -3-22)
X2 = -1.13190
X3 = -1.02608
X4 = -1.01226
X5 (root) = -1.01203 Answer (5.d.p.)
Answer Root A: -1.01203 (5 .d.p)
Error: -1.01203 +0.000005
Error bounds: [-1.012025, -1.012035]
F(x) = ex-x3-1.4
F(-1.012025) = -0.0000073
F(-1.012035) = 0.0000198
A change of sign occurs and so confirms there is a root.
Finding Root B
X0 = 0.8
X1 = -0.22618
X2 = 0.69116
X3 = 0.21887
X6 = 0.37282 ( 5.d.p.)
Answer Root B: 0.37282 (5.d.p.)
Finding Root C
X0 = 1
X1 = 2.12979
X2 = 1.62013
X3 = 1.40784
X6 = 1.34816
Answer Root C: 1.34846(5.dp.)
Newton-Raphson Failure
For the Newton-Raphson method to fail the X0 starting position must be between a turning point and the root. This will therefore not find the required root.
For example with the equation:
F(x) = 43x3-60x2-250x-115
The graph for this function is shown below.
The table of values is given below:
x | F(x) |
-3 | -1066 |
-2 | -199 |
-1 | 32 |
0 | -115 |
1 | -382 |
2 | -511 |
3 | -244 |
4 | 677 |
5 | 2510 |
Using the Newton-Raphson formula of: X1=X0 - f(X0)
f’(X0)
F(x) = 43x3-60x2-250x-115
F’(x) = 129x2–120x-250
So my formula is: X1=X0 - 43x03-60x02-250x0-115
129x02–120x0-250
Finding Root A
X0 = -1
X1 = -1– 43(-1)3-60(-1)2-250(-1)-115 = 31
129(-1)2–120(-1)-250
X2 = 20.87085
X3 = 14.14418
X11(root)= 3.36156 This is actually ROOT C.
The graph 1 shows the iteration process. It shows the first iteration more clearly than my graph 2: Only the first 7 or 8 iterations are shown clearly here
The graph 2 below is showing the same iteration process but the y-axis goes up to a million.
Graph 3 shows the iteration process from X7 between the x values of 3.
Conclusion
Problems with Starting Points
There were no problems with starting points in the decimal search method. In the failure of decimal search I had to recognise the f(x) lowering and then raising. For the rearrangement method the starting point mattered depending also on which g(x) to use. I had to view my graph to see which derivative of g(x) was near 1 or -1. I noticed if I selected the other g(x) then I would find another root. When using the Newton-Raphson method the starting value mattered greatly because I noticed if I started with a different value then a different root could have been found. Also with the Newton-Raphson method I could never use the starting value of “0” as the solution of the formula would be trivial as in we would be dividing by zeros. Autograph helped me solve problems with starting points as it showed where about the solutions were, without autograph I would have to try more than one value.
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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