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# Solutions of equations

Extracts from this document...

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:

1. Change of sign method
2. Newton-Raphson method
3. 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

=-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.)

Finding Root C

X0 = 1

X1 = 2.12979

X2 = 1.62013

X3 = 1.40784 X6 = 1.34816

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 =  -143(-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

1 or g2. This was also a repeated process where I pressed execute has many times as required until I found the root to the degree of accuracy I required.

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.

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