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Numerical Method Course Work

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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)

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), 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.

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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.

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