• 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

# 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)

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.

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.

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

# Related AS and A Level Core & Pure Mathematics essays

1. ## 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.

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

Due to the fact that we know T128 and T64 already, we can work out T256. The value of T256 can thus be extrapolated as [T128 - (1/22*(T64-T128))]. This method of calculating a quarter of the gap between T64 and T128 and adding (remember trapezium is an underestimate)

1. ## Numerical Method (Maths Investigation)

proceeds to zero. Upper Bound = 0.952545 Lower Bound = 0.952535 f(Upper Bound) = -0.0001612 f(Lower Bound) = 3.07510-5 Error Bound = 0.952545 - 0.95254 = 0.000005 Hence, Approximate Root Value = 0.95254 0.000005 HOW BASIS ITERATION FAIL TO CONVERGE TO THE REQUIRED ROOT To show the failure of the Basis Iteration, Xn+1

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

-0.3471 -3.18056 x 10^-4 -0.3472 -2.54210 x 10^-4 -0.3473 9.61518 x 10^-6 There is a change of sign between x=-0.3472 and x=-0.3473 x f(x) -0.34730 9.61518 x 10^-6 -0.34729 -1.67664 x 10^-5 There is change of sign between x=-0.34729 and x=-0.34730.

1. ## MEI numerical Methods

have to manually input a different values every iteration, this would also increase the chance of a mistake. Finally in order to make sure my approximation is correct I also made use of software called autograph. This piece of software allowed me to generate a graph of the equation y=+

2. ## Change of sign method.

0.3769531250 0.3771972656 0.3770751953 0.0001235805 -0.0003907864 -0.0001336152 0.3769531250 0.3770751953 0.3770141602 0.0001235805 -0.0001336152 -0.0000050205 0.3769531250 0.3770141602 0.3769836426 0.0001235805 -0.0000050205 0.0000592792 0.3769836426 0.3770141602 0.3769989014 0.0000592792 -0.0000050205 0.0000271292 0.3769989014 0.3770141602 0.3770065308 0.0000271292 -0.0000050205 0.0000110543 0.3770065308 0.3770141602 0.3770103455 0.0000110543 -0.0000050205 0.0000030169 0.3770103455 0.3770141602 0.3770122528 0.0000030169 -0.0000050205 -0.0000010018 0.3770103455 0.3770122528 0.3770112991 0.0000030169 -0.0000010018 0.0000010076 0.3770112991 0.3770122528

1. ## Repeated Differentiation

= a.b x(b-1) (b-1)! equation correct for n=1 (see previous work) [2] Assume true for n=k yk = a.b! x(b-k) (b-k)! [3] Differentiate both sides w.r.t. x yk+1 = a.b! . (b-k).xb-(k+1) (b-k)! = a.b! . xb-(k+1) (b-(k+1))! Therefore if equation true for n=k also true for n=(k+1).

2. ## Newton-Raphson Method: This is a fixed-point estimation method.

= -1.888888889 (-1 8/ 9) Method 2: Newton-Raphson method Fig1. Shows the function f(x) = 1/3x 3 - 5x + 1.4 Differentiating the function we get: f(x)` = x 2 - 5 So now, using the Newton-Raphson iteration equation we can find the roots x n+1 = x n - f(x n)/f`(x n)

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