• 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

Methods for Advanced Mathematic

Extracts from this document...

Introduction

Methods for Advanced Mathematics (C3) Coursework

Introduction:

In this coursework, I am going to solve equations by using the Numerical Methods. The Numerical methods are used to solve equation that cannot be solved by using algebra.

Here are the 3 Numerical methods:

  1. Change of sign method
  2. Newton-Raphson method
  3. Rearranging f(x) = 0 in the form of x=g(x)

Change of sign method

This method is based on how an equation has a change of sign either from positive to negative or negative to positive on either side of a root when crossing the x-axis. It is best to sketch a diagram of the equation before starting with the method, so we would have a clear structure of what the equation looks like, the number of roots the equation has and the roots approximate position.

To investigate this method I would use the following equation

0 = 3x3+11x2+2x-5

The graphs below show the function f(x) = 3x3+11x2+2x-5

image00.png

image01.png

x

f(x)

6

1051

5

655

4

371

3

181

2

67

1

11

0

-5

-1

1

-2

11

-3

7

-4

-29

-5

-115

-6

-269

x (1dp)

f(x)

-1

1

-0.9

-0.077

As we can see from the results table, there is a change of sign between [-1, -0.9]. This indicates the root is in the interval. To make sure the result is as accurate as possible, I will repeat the process until a sufficient number of decimal places are achieved. I will calculate this to 3 decimal places.

A zoomed in of the root between -1 & 0

image02.png

x (2dp)

f(x)

X (3dp)

f(x)

-1

1

-0.91

0.028387

-0.99

0.890203

-0.909

0.017823

-0.98

0.780824

-0.908

0.007264

-0.97

0.671881

-0.907

-0.00329

-0.96

0.563392

-0.95

0.455375

-0.94

0.347848

-0.93

0.240829

-0.92

0.134336

-0.91

0.028387

-0.9

-0.077

x=-0.9075 +/- 0.0005

Error Bounds

The calculations above show that the root lies in the interval 0.908 to 0.907. I can now conclude that the root to the equation 0 = 3x3+11x2+2x-5 has a root of -0.9075 +/- 0.0005

(3 d.p) being half the interval between the error bounds.

Limitations of Change of Sign Method

This method works for most equations of x, however, there are cases when this method fails to converge. An example for this is when two roots are within the same interval and the method shows only one change of sign. This would indicate that it has only one root within the interval and the second root is over looked.

I will use the equation 0 = x4+5x3-7x+2 to demonstrate how this method would fail.

Here is a graph of the function f(x) = x4+5x3-7x+2

image03.png

The graph shows the equation 0 = x4+5x3-7x+2 has 3 roots clearly distinguished. However, when I try to find the solutions by using the change of sign method it only display one change of sign and the second change of sign is over looked.

x

f(x)

-3

-31

-2

-8

-1

5

0

2

1

1

2

44

...read more.

Middle

Xn

f(xn)

f'(xn)

xn+1

-2

1

2

-2.5

-2.5

-2

10.5

-2.30952381

-2.30952381

-0.34899039

6.908163265

-2.25900526

-2.25900526

-0.02234667

6.028575984

-2.25529847

-2.25529847

-0.00011743

5.965242424

-2.255278783

-2.255278783

-3.3065E-09

5.964906509

-2.255278783

x = -2.25528 (5 d.p.)

Root 2

Xn

f(xn)

f'(xn)

xn+1

-1

-2

-6

-1.333333333

-1.333333333

-0.18518519

-4.666666667

-1.373015873

-1.373015873

-0.00484909

-4.419123205

-1.37411317

-1.37411317

-3.9015E-06

-4.412009679

-1.374114054

x = -1.37411 (5 d.p.)

Root 3

Xn

f(xn)

f'(xn)

xn+1

1

-2

14

1.142857143

1.142857143

0.2303207

17.26530612

1.129517055

1.129517055

0.00210532

16.95002322

1.129392847

1.129392847

1.8169E-07

16.94709769

1.129392837

x = 1.12939 (5 d.p.)

Limitation of Newton-Raphson Method

Although Newton-Raphson is a great method of finding roots to an equation, but when the value is taken close to or at a minimum or maximum point, the method would fail to find the point to a root. I will use the equation 0 = x3-3x+1 to demonstrate how Newton-Raphson method fails to coverage.

Here is a graph of the function f(x) = x3-3x+1


image05.png

I have chosen 0.9 as a starting point to find the root between 0 and 1.

Newton Raphson Formula

Xn/x=1

f(Xn)= x3-3x+1

f’(Xn)=3x2-3

Result

Xn

f(xn)

f'(xn)

xn+1

0.9

-0.971

-0.57

-0.80351

-0.80351

2.89176

-1.06312

1.916558

1.916558

2.290216

8.019582

1.63098

1.63098

0.445623

4.980287

1.541503

1.541503

0.038457

4.128691

1.532188

1.532188

0.0004

4.0428

1.532089

1.532089

4.51E-08

4.041889

1.532089

As we can see from the results above, the answer is showing the root between interval [1,2] rather than the root between [0,1]. Here is a graph used to demonstrate how the method fails.


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

Rearranging method is also an iterative process. It is similar to the Newton Raphson method as it uses the last answer to get closer to the value of the root.

...read more.

Conclusion

Although the change of sign method has the fastest technique to calculate the roots of an equation, the accuracy of the answer compare to the other two methods is far from beyond. Both the Newton-Raphson method and the Rearranging method had given a much accurate results by giving the answer up to nine or more decimal places.

In terms of accuracy via speed, the Newton-Raphson method reaches a high degree of accuracy in a relatively short amount of time compare to the “Rearranging” method. The Newton-Raphson method requires a differential of the equation first before the process can begin, but once is set up highly accurate results can be achieved in a short amount of time. In the “Rearranging” method, the equation has to be rearranged in the form of x=g(x) and once the equation is rearranged, it has to be checked each time by the differential of x=g(x). The steps required from the “Rearranging” method are more than the Newton-Raphson method and therefore the Newton-Raphson method is quicker than the “Rearranging” method.

The use of software Omnigraph, Coypu and Microsoft Excel has benefited for all three methods. The graphs were easily plotted by using Omnigraph and Coypu, so the approximate interval values can be easily spotted. Microsoft Excel has helped me to calculate and display all of my calculations. However, the required differentiation from the ‘Newton Raphson’ and ‘Rearranging’ cannot be done by any program/software, therefore it had to be done by hand.

...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)

    Therefore the gradient function of x5 = 5x4, which corresponds to my results earlier in the tables, proving my predictions of the gradient function was correct. After investigating the gradient function where a is a constant 1, and n is always a whole number, I have concluded this section with

  2. Change of Sign Method.

    When x = -0.9165, f(-0.9165) = -0.00389006712 When x = -0.9175, f(-0.9175) = 0.00125526563 The error bounds of the root -0.917 are -0.917 � 0.0005. Newton-Raphson Method The Newton Raphson method will now be applied to find the same root to the equation. It is first necessary to look for a change of sign between integer values so as to choose a suitable starting value.

  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. MEI numerical Methods

    the upper bounds and lower are two opposite signs the root must lie within the range of values of the interval. As K varies As we proved above, if K = 1 the root is 0.479731007. However what if K, was 2, 3 etc, could a pattern be distinguished, the

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

    key call 'bisection iteration' then we can enter upper and lower bound and this program will be able to help us to find the root automatically. However this method is not always working properly. There are two situations which lead to the failure of the method 1)

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

    By checking sign change: f (0.635085) = +4.57111 x 10-5, which is positive f (0.635095) = -7.72189 x 10-5 , which is negative This tells that my answer is correct and it is in the bounds. Differentiating the formula of --> I would get Inserting the solution x = of 0.63509 into g'(x), I would get 0.093078.

  1. The Gradient Fraction

    Results x Gradient 2 7.82 3 12.41 4 16.8 -2.5 -9.23 -4 -17.77 Values at point: 2: x= 1.15, y=9 [9/1.15] 3: x= 1.45, y= 18 [18/1.45] 4: x= 1.9, y= 32 [32/1.9] -2.5: x= 1.3, y = -12 [-12/1.3] -4: x= 1.8, y=-32 [-32/1.8] Above are the differences in the 'x' and 'y' scales.

  2. In this investigation I am going to investigate three methods of finding the roots ...

    This is the process whereby a table of values is constructed which represents whether the value of f(x) with certain values of x is positive or negative. By zooming in on the places where the value of f(x) switches from positive to negative (or vice versa), values for the root of this function can be found.

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