• 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

MEI numerical Methods

Extracts from this document...

Introduction

Numerical methods coursework

Introduction:

Consider the problem x + ktanx = 1, the value of x lies between 0 and π/2 for different values of K, it is measured in radians. Normally a solution of a mathematical problem of this nature requires the root of an equation. If we attempt to find the root algebraically, we begin by rearranging the equation to give 0, hence the equation is x + ktanx – 1 = 0. Now lets assume the value of K is 1, we are left with x + tanx – 1 = 0. The only information we are provided is that the value of x is between 0 and π/2. Since x appears twice in the equation, the problem cannot be solved algebraically. In the sense that there are equations which cannot be integrated and require numerical methods to approximate the answer, eg the midpoint rule. In the same way there are equations which cannot be solved algebraically and require numerical methods to approximate the root.

To solve this solution I shall attempt to use numerical methods. I will be using the method of bisection, fixed point iteration, secant method and false position method. All these methods require the use of an excel spreadsheet since all methods require iterations which will get closer and closer to the real answer (converge). This is why the Newton raphson method isn’t being used as it requires differentiation which can be hard to use on computers. The other methods can be converted into formulas and inputted into excel which can easily perform multiple iterations, simply by dragging down the cells.

Strategy:

Method of bisection:

An approximation of the root can be obtained by this method; in order to use the method of bisection the first thing which must be done is to find an interval estimate of the root.

...read more.

Middle

Formulas used:

image10.png

Fixed point iteration:

To use fixed point iteration we need to rearrange the equation so that it equals x. There are two possible rearranged forms are:

  • X= 1 – ktanx
  • X= arctan(1-x)/k

Fixed point iteration uses recurrence relation so by finding the answer of one iteration we produce X(1), we then use X(1) as x in the formula to produce X(2), this is how we get closer to the root by using this method.

For example the first iteration for x = 1 – ktanx assuming that k = 1 is:

X = 1 – tan(π/4)

The value of tan(π/4) = 1

1-1 = 0, hence the value of the first iteration is 0, for the second iteration we would do:

X = 1 – tan(0)

The value of tan(0) = 0

1-0 = 1 hence the value of the second iteration is 1 and we would carry on. This task can be tedious to carry on doing it again and again especially for different values of K and if it has a lot of iterations, hence we use a spreadsheet.

Formulas used for spreadsheet:

image11.png

Secant method:

The secant method requires two approximations of the root in order to work. Fortunately we’ve been provided with two roots, 0 and π/2. We can now call x(0) as π/2 and x(1) as 0, we can then substitute these values into the following equation.

image12.png

Formula used for spreadsheet:

image13.png

False position method:

Similar to the secant method, the false position method also requires two

approximations of the root. I will use one root a as 0 and the other root b as π/4. Bearing this in mind we can substitute the values into the formula:

image14.png

The formula is the same as the secant method however remember the false position method roots approximation must support the sign change argument.

Formula used for spreadsheet:

image15.png

Proof of root:

image16.png

...read more.

Conclusion

image08.png

0.4797310065 < x < 0.4797310075, to test the bounds are correct we can do the f of both of these values and it would have to support the sign change argument, as proved above.

This means the root is 0.479731007 ± 0.0000000005. We can however lower the boundaries and thereby create a better approximation of the root. In order to lower the boundaries we merely continue reducing/increasing them until we hit the boundary of sign changes.

f(0.4797310073) = 0.0000000000444833

f(0.4797310072= -0.0000000002

Conclusion

The root to the equation is, 0.4797310065 < x < 0.4797310075 hence 0.479731007 to 9 D.P.

image09.png

This is my approximation of the root compared to the actual root, bear in mind this is zoomed in by a lot. However how valid is this approximation of the root?

If we look back we found out there were problems with the formulas themselves however that was merely for the purpose of finding the best method, this cannot be used as a limitation because it doesn’t apply to this equation.  

However earlier I did mention that excel is accurate to 15 D.P, my answer is only valid to 9 D.P, this is definitely a limitation, if I had more time I would change my figures and calculate it to 15 D.P.

Another way to test its validity we merely insert it back into the original equation,

 x + tanx – 1 =0. Notice the =0 part, if this approximation is valid then by using our approximation (x) in this equation we should end up with an answer close to 0. The result of the equation is -6.37 x 10^-10. Notice the degree of the polynomial is a high minus number meaning this number is tiny and extremely close to 0. This means my approximation of the root is valid. In order to improve on this approximation we could find the root to more D.P as explained earlier.

...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. OCR MEI C3 Coursework - Numerical Methods

    found 1.00000 0.00000 The value diverges, despite the starting value being close to the root. This is because the tangent crosses the asymptote, as shown below: Fixed point iteration with x=g(x) Finding a root f(x)=x5-4x+3 The graph of y=f(x): We shall rearrange the equation into the form x=g(x)

  2. Am going to use numerical methods to solve equations that can't be solved algebraically

    Y= g(x) = Making the iterative formula Xn+1= X1= 2 X2= 1.584893192 X3= 1.496651121 X4= 1.474924955 X5=1.469374182 X6=1.467942484 X7=1.467572301 X8=1.467476526 X9=1.467451742 X10=1.467445329 X11=1.467443669 X12=1.467443239 To 3.d.p = 1.4675 + 0.0005 This rearrangement was successful due to the gradient between [1, 2] being smaller than 1.

  1. The open box problem

    x x 12 15-2x x 15-2x 12 Again I use the same equation as before but this time I replace 12 with 15. So the equation will be V= x(15-2x)^2 . I will now construct tables and two graphs to again show the maximum volume and x.

  2. C3 Coursework: Numerical Methods

    As the Newton Raphson method relies on the gradient of the points it is not possible to use the Newton Raphson method the find the roots of the equation y=log(x+3)-x. I shall now attempt to use the Newton Raphson formula to find the roots of the equation.

  1. Change of Sign Method.

    I will continue the process using the increment of 0.001 within the interval [-0.92, -0.91]. x -0.920 -0.919 -0.918 -0.917 f(x) 0.014112 0.008970441 0.003827368 -0.001317213 The root therefore lies in the interval [-0.918, -0.917]. I shall now use increments of 0.0001 in the interval [-0.918, -0.917].

  2. In this coursework, I am going to solve equations by using the Numerical Methods. ...

    When we look at the graph we can see 2 roots, we stated in our objective that "if any method fails to find all the roots then the method has failed". The reason why this method failed is because the function just touches the x-axis but doesn't cross it, so

  1. I am going to solve equations by using three different numerical methods in this ...

    -0.345008342 10 -0.27117 -0.34500834 -0.345008188 11 -0.27117 -0.34500819 -0.34500823 12 -0.27117 -0.34500823 -0.345008218 13 -0.27117 -0.34500822 -0.345008221 14 -0.27117 -0.34500822 -0.345008221 15 -0.27117 Root is -0.3450082 to 7d.p. Error bounds is -0.3450082�0.00000005. Root bounds is -0.34500825<x<-0.34500815. Check X Y -0.34500825 -0.345008213 -0.34500815 -0.34500824 X g'(x)

  2. Fractals. In order to create a fractal, you will need to be acquainted ...

    - 1303.22i 1358.8 z8 = -1.5504*106 + 1.0026*106 1.8463*106 As you can see, the fourth iteration does not stay inside the circle. Here is a better illustration of why this point is not in the Julia set: Here is another example of a point that is in the Julia set

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