• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
Page
  1. 1
    1
  2. 2
    2
  3. 3
    3
  4. 4
    4
  5. 5
    5
  6. 6
    6
  7. 7
    7
  8. 8
    8
  9. 9
    9
  10. 10
    10
  11. 11
    11
  12. 12
    12
  13. 13
    13
  14. 14
    14
  15. 15
    15
  16. 16
    16
  17. 17
    17
  18. 18
    18
  19. 19
    19
  20. 20
    20
  21. 21
    21
  22. 22
    22
  23. 23
    23
  24. 24
    24
  25. 25
    25
  26. 26
    26
  27. 27
    27
  28. 28
    28

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.

image74.pngimage00.pngimage01.png

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.

image75.jpg

Table 1

x

F(x)

-4

22400

-3

-10571

-2

-3872

-1

-845

0

-32image20.pngimage22.png

1

25

 2

784

3

3703

4

10240

...read more.

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.)

image78.pngimage26.pngimage25.pngimage24.pngimage27.pngimage28.png

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

image29.png

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

image29.png

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.

image79.pngimage30.pngimage28.pngimage31.png

The table of values is given below:

x

F(x)

-3

-1066image33.pngimage32.png

-2

-199image33.png

-1

32image34.png

0

-115

1

-382

2

-511image33.png

3

-244image36.png

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

image80.png

X0 = -1

X1 =  -143(-1)3-60(-1)2-250(-1)-115 = 31

                     129(-1)2–120(-1)-250

X2 = 20.87085

X3 = 14.14418image29.png

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

image81.pngimage44.pngimage43.pngimage42.pngimage41.pngimage40.pngimage39.pngimage38.pngimage37.pngimage45.png

The graph 2 below is showing the same iteration process but the y-axis goes up to a million.

image82.pngimage41.pngimage43.pngimage42.pngimage46.pngimage44.pngimage40.png

Graph 3 shows the iteration process from X7 between the x values of 3.

...read more.

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.

...read more.

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

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related AS and A Level Core & Pure Mathematics essays

  1. Marked by a teacher

    The Gradient Function

    5 star(s)

    The general pattern here between the two values of x and the gradient, the value of the gradient function here is 4x�. Using this formula, the next 5 values of x and the gradient are: x x� 4x� 5 125 500 6 216 864 7 343 1372 8 512 2048

  2. Numerical solutions of equations

    The graph of this function and the calculations are below: I can see that the root lies between x = -2 and x = -3. x f(x) -2 0.00810000 -2.1 0.00044205 -2.2 0.00000001 -2.3 0.00074120 -2.4 0.01048576 -2.5 0.05090664 -2.6 0.15752961 -2.7 0.37973325 -2.8 0.78074896 -2.9 1.43766095 -3.0 2.44140625 As

  1. Solving Equations. Three numerical methods are discussed in this investigation. There are advantages and ...

    Method of Bisection: n a f(a) b f(b) c f(c) error 1 0 1 1 -0.5 0.5 0.3125 0.5 2 0.5 0.3125 1 -0.5 0.75 -0.1015625 0.25 3 0.5 0.3125 0.75 -0.1015625 0.625 0.106445313 0.125 4 0.625 0.106445313 0.75 -0.1015625 0.6875 0.002319336 0.0625 5 0.6875 0.002319336 0.75 -0.1015625 0.71875 -0.049697876

  2. Numerical solution of equations, Interval bisection---change of sign methods, Fixed point iteration ---the Newton-Raphson ...

    They all lie in the same interval [-1, 0]. Bisectional method has failed to find all the roots when several roots are too close with each other. Fixed point iteration ---the Newton-Raphson method An estimate of the root as a starting point has to be set.

  1. I am going to solve equations by using three different numerical methods in this ...

    As the figures show blow: 2 5.2 0 7.2 5.2 72.0672 1 38.688 72.0672 152833.4368 2 6318.898 152833.4368 1.42797E+15 3 2.8E+10 1.42797E+15 1.16471E+45 4 2.45E+30 1.16471E+45 6.32E+134 5 1.63E+90 6.32E+134 #NUM! 6 4.8E+269 #NUM! #NUM! 7 #NUM! #NUM! #NUM! 8 #NUM!

  2. Using Decimal search

    -0.68847 -0.61736 -0.60540 -0.60507 -0.60507 These results show that for a positive value of X, using the Newton-Raphson method on eq'n 5x^3-1.5x+0.2=0, it finds the root in the interval (-0.6,-0.7). As shown in Graph 12, a tangent is taken at 0.3 to find the root between he interval (0.2,0.3).

  1. Change of Sign Method.

    = -0.00000260721 and x = 1.546825 into f(x) =0.00003192987 The above calculations illustrate that there is a change of sign. Therefore the root is 1.54682�0.000005. Failure of the Newton-Raphson Method There are situations when the root you wish to find cannot be determined using the Newton-Raphson method. This will be illustrated using the equation y=3x5-5x4.

  2. In my coursework I will be using three equations to investigate their solutions using ...

    and my solution bounds are (-0.796925? x ? -0.796935) This is the equation used for this graph I am going to demonstrate that the root that lies between (1,5) can not be solved using the change of sign method. 0 27.53373021 0.5 25.57575346 1 18.13454671 1.5 8.96010996 2 1.80244321 2.5 0.41154646 3 8.53741971 3.5 29.93006296 4

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work