• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

Numerical Method Course Work

Extracts from this document...

Introduction

Numerical Method Course Work

  1. Problem Specification

Finding all the roots of equation (x+1) sin (1/x)=0 for 0.05<x<1.The reason for why I use numerical methods is I can’t solve this equation by any analytical method. And it lies beyond my mathematics knowledge to solve by an algebraic method.

  1. Strategy

I am not sure which method I will use for finding all roots of the equation. So I want to see which method is most efficient method for finding a root, efficient method means can find the roots most quickly. The methods are shown below,

 Bisection Method with the formula: If a root lies between X = X0 and X = X1, then X2 = 0.5 ( X0 + X1 ) will give an approximation to the roots;

 Secant Method     with the formula: If X = X0 and X1 are approximation to a root of X = X0. A better approximation to the root will usually be given by X = X1 – (X1 - X0 )*f (X1 )/[f(X1 )-f(X0 )];

 Linear Iteration    If x = a and x = b are approximations to a root of f (x)

...read more.

Middle

), X0is given. This will usually converge to a root of f (x) = 0 near to x = x0. The method is called the Newton-Raphson method and has second order convergence. (Appendix 5)

In my course work, the equation is  (x+1) sin (1/x) = 0.

So differentiate this function by using chain rule, product rule, and trig. Differentiation.

Y =F (x) = (x+1) sin (1/x))

Y= u * v

        U’ = 1

        V’ = -(1/x^2)* cos. (1/x)   (Chain rule and Trig. Differentiation)

        F ‘(x) = u v’ + v u’  (Product Rule)

                  = Sin (1/x) –  (x+1)/x^2* cos. (1/x)

Then I put the function f (x) = (x+1) sin (1/x) into the graphic calculator, using the table to store the information with the range 0.05 as starting point, 1 as the end point, 0.01 as the pitch. I use the calculator to see how many roots are there roughly. The calculator lists a table  and draws a graph for me. (Appendix 3 and 4)

From the table and graph, we can see that there may be 6 roots in the interval (0.05, 1). There are between 0.05 and 0.06, 0.06 and 0.07, etc. Then I use the same method to check whether or not there is exactly only 1 root in the interval. Instead, I use 0.05 as starting value, 0.

...read more.

Conclusion

0 is correct to 1 decimal place, X1 will probably be correct to 2 decimal places, X2 to 4 decimal places, and so on.

Because Ms Excel spreadsheet just can go to the accuracy as far as 9 decimal places, so my solution just can achieve 8 decimal places, because of rounding in the computer program

  1. Interpretation

The roots of the equation (x+1) sin (1/x) =0 are

* 0.053051648,

* 0.063661977,

* 0.106103295,

* 0.159154943,

* 0.318309886,

* 0.079577472.  (8 decimal places).

Because the Successive errors of all the roots become smaller and turn to 0 at the end, and the En/ En-1^2 ratio becomes constant, I can say my solution is valid.

So the limitation of Newton-Raphson method is I need to choose the roots very carefully otherwise I can’t find the roots I want to find. Particularly to find roots at 0.0795774472, it is very important to choose the correct value.

 And the disadvantage of Secant method is that, like the Newton-Raphson method, can lead to divergence in some cases. (Appendix 7)

And for the fix-iteration, this method seem not so good to the equation (x+1)sin (1/X) =0

The bisection method is good, reliable method for solving an equation f (x) = 0, it always go to the convergence but it reduces the rate of going to convergence, because it halves the interval width.

...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)

    5 625 1250 1000 My theory was correct. I shall now try and binomially prove this... 2(x+h)4 - 2x4 = 2(x4 + h4 +4hx� + 6x�h� + 4xh�) - 2x4 = x + h - x h 2x4 + 2h4 +8hx� + 12x�h� + 8xh� - 2x4 = 2h4 +8hx�

  2. The open box problem

    then I will know that this is right. x x 24 24-2x x 24-2x 24 Once again I will use the same equation but change the 15 to a 24 this time so the equation will be V=x(24-2x)^2. Again I will construct tables and graphs to find out the maximum volume and the value of x.

  1. C3 Coursework: Numerical Methods

    Using the Newton Raphson Method was also made simpler using Microsoft Excel. As with change of sign, I used Microsoft Excel to replicate the formulae I used. This was also made simpler using the Auto fill function. As the formulae for the Newton Raphson method was more complex than the

  2. MEI numerical Methods

    multiple formulas in order to satisfy the methods, for more information on the formulas read the sections below where it deeply talks about them and how they were created. Solution to the problem: Method of bisection: To use the method of bisection, we must rearrange the equation so it = 0.

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

    from this fix point and eventually it will converge to the roots. This method requires us to start with a point which is close to the target root and we call it. Then we will find the gradient - at Cartesian co-ordinate and we can find the better estimation -

  2. Change of Sign Method.

    To check that this is correct, substitute x= -0.91725595 into f(x) = -0.00000031423 and x= -0.91725605 into f(x) = 0.000000200258 The above calculations illustrate that there is a change of sign. Therefore the root is -0.9172560�0.00000005. Failure of the Fixed Point Iteration Method I will now attempt to find the

  1. Although everyone who gambles at all probably tries to make a quick mental marginal ...

    However, on any one trial, it is much more likely to lose money instead of making infinite profit. If it is considered that when P? E, one has lost money, one has the infinite sum of the geometric sequence odds of making money with this game.

  2. 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

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