• 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

Change of Sign Method

Extracts from this document...

Introduction

Numerical Methods Coursework

Change of Sign Method

To look at how this method I will find the roots of the equation x5-7x+5=0 by the change of sign method. Here is a graph of y=x5-7x+5 and an integer search showing where the roots of x5-7x+5=0:

image00.png

x

-5

-4

-3

-2

-1

0

1

2

3

4

5

y

-3085

-991

-217

-13

11

5

-1

23

227

1001

3095

This equation has 3 roots as shown in the table as the y value changes sign 3 times in the table, one between x=-2 and x=-1 another between x=0 and x=1 and another between x=1 and x=2. This is also shown in the graph. A root can be more accurately found using a decimal search or zooming in on a graph and identifying when a change of sign occurs. I will do this to find the root of this equation between x=-2 and x=-1:

x

-2

-1.9

-1.8

-1.7

y

-13

-6.461

-1.2957

2.7014

image11.png

As there is a change in sign between x=-1.8 and x=-1.7 there must be a root between x=-1.8 and x=-1.7 as shown on the graph above. Therefore a decimal search to two decimal places will be more accurate.

x

-1.8

-1.79

-1.78

-1.77

y

-1.2957

-0.8466

-0.409

0.0173

Therefore there is a root between x=-1.

...read more.

Middle

image01.png

image02.png

image03.png

image04.png

image05.png

image06.png

image07.png

image08.png

image09.png

1

1.25

1.27359

1.27389

1.27389

1.27389

1.27389

1.27389

1.27389

When x=1.27388 y=0.000034 and when x=1.27390 y=-0.000030. There is a change of sign between the two therefore:

x=1.2739 (5s.f.)

1.27388<x<1.27390

Now I will find the root between x=2 and x=3 and I will use x0=3 as this is close to the root and is most likely to find that root.

image01.png

image02.png

image03.png

image04.png

image05.png

image06.png

image07.png

image08.png

image09.png

3

2.58333

2.41243

2.37854

2.37720

2.37720

2.37720

2.37720

2.37720

When x=2.37719image12.pngand whenimage13.png. There is a change of sign between the two therefore:

x=2.3772 (5s.f.)

2.37719<x<2.37721

Failure of the Newton Raphson Method

When the gradient of the gradient of a curve is small the Newton Raphson method may give the root of the equation at a point you were not looking for. For example if I was looking for the root of the equation image15.pngbetweenimage16.png and image17.png. The iterative formula for image18.pngwould be:

image19.png

A sensibleimage20.pngto take would beimage21.pnghowever the result would look like this.

image01.png

image02.png

image03.png

image04.png

image05.png

image06.png

image07.png

image08.png

image09.png

image22.png

1

-1

-2.33333

-1.90831

-1.6162

-1.46103

-1.41736

-1.41423

-1.41421

-1.41421

image23.png

This shows that the Newton Raphson method has failed to find the required root between x=0 and x=1 but the root at x=-1.4142 therefore it is a failure.


Rearranging f(x)=0 in the form x=g(x)

To show how this method works, I will show how this method can be used to find all of the roots of the equation x³-3x²-4x+7=0. Firstly I will require values for x0 and I also need to know how many roots I am looking for. This can be done with an integer search and a graph of y=x³-3x²-4x+7.

image24.png

x

-5

-4

-3

-2

-1

0

1

2

3

4

5

y

-173

-89

-35

-5

7

7

1

-5

-5

7

37

...read more.

Conclusion

The Newton Raphson method was again fairly easy to perform and also gave reliable results although an accuracy check was required to ensure the root had been found but the method was not likely to fail. It was again also very easy to use with a spreadsheet program and graphing software although some time would have been spent in deriving the iterative formula. This method is slightly harder to perform with a graphical calculator as an iterative formula would need to be input, however this method is very fast.

The rearrangement to x=g(x) method was again easy to perform and gave reliable results although again an accuracy check would be required. However this method fails many times and more than one iterative formula was required to find all the roots of the equation. After calculating an iterative formula this method was again easy to perform with a spreadsheet program and graphing software. This method is similarly easy as the Newton raphson method however is more likely to take more iterations.

In conclusion this investigation has shown that the Newton Raphson method of solving equations has shown to be the better of the three methods. This is due to its speed, ease to use and likelihood not to fail.

Page

...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)

    0.5 2 1.414214 0.353535 3 1.732051 0.2887 4 2 0.25 So far, I cannot conclude anything from these data since the numbers are not whole and therefore hard to work with. However, this evidence will help support the formula which I will attempt to create by using binomial expansion.

  2. 2D and 3D Sequences Project Plan of Investigation

    c = 4 n = 3 27a + 9b + 3c = 26 n = 4 64a + 16b +4c = 64 = 16a + 4b + c = 16 4 To get rid of 'c' I will use this calculation; _16a + 4b + c = 16 4a +

  1. Change of Sign Method.

    It is therefore necessary to find the points of intersection. I will focus on the point of intersection that lies in the interval [-1,0]. I will use the previously evaluated iterative formula, using x0 as 1, for my starting value.

  2. newton raphson

    In dialog window I will see OVERFLOW. If I will choose the point, which is the one of the stationery points, this method will failure. Because in this case I will not have any chances to find another, closes to truth root point.

  1. Triminoes Investigation

    369a + 61b + 9c + d = 210 - equation 6 - 175a + 37b + 7c + d = 120 - equation 7 194a + 24b + 2c = 90 - equation 10 Equation 7 - Equation 8 I am doing this to eliminate d and to form

  2. Different methods of solving equations compared. From the Excel tables of each method, we ...

    For example in the following case: Equation A: � The graph above shows us that the roots of the equation are very close. Zoom in on the x and y axes, we can see that they all lie between x=0 and x=1.

  1. The method I am going to use to solve x&amp;amp;#8722;3x-1=0 is the Change ...

    shown below: The blue line is approaching to the root, and that is the answer of f(x)=0 Rearrangement B: 3x^5+5x�-1=0 x�=(1-3x^5)/5 x=V[(1-3x^5)/5] On Autograph software, I can draw the equation y=g(x)= V[(1-3x^5)/5]and also the line y=x. The graph is shown below: I want to find the intersection of the arrow

  2. Numerical integration can be described as set of algorithms for calculating the numerical value ...

    The use of polynomials: The mid-point rule is a second order polynomial which overestimates convex curves. The function I am integrating also follows a convex curve. As a result I expect the consecutive values of the mid-point to decrease as this polynomial overestimates.

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