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

In this investigation I aim to investigate three methods of finding the roots to equations and then compare them. I aim to be able to use fully and understand all three techniques.

Extracts from this document...

Introduction

Mathematics Coursework Aim In this investigation I aim to investigate three methods of finding the roots to equations and then compare them. I aim to be able to use fully and understand all three techniques. The three methods that I am going to examine are: ?Decimal Search/Change of Sign Method ?The Newton-Raphson Method ?The Re-arrangement Method. Initial Exploration of Methods When I started this project I did not have a deep understanding of these methods I was using. In order to develop my understanding I experimented with the methods and also with the equations that I was going to use so that I knew what they would look like, making it easier for me to create equations to my own specifications. Decimal Search/Change of Sign Method In the piece of course work we are asked to find (or not to find) the roots of equations that we choose. But what are the roots. To put it simply they are the values that can be given to x so that f(x) is equal to zero. This can be displayed graphically by plotting the values of x against values of f(x). This will produce a line that will be continuous (will not be using any discontinuous equations in this project). ...read more.

Middle

If you look at my TABLES section you will see that I have easily found the root at roughly 0.14. But if we look at the two charts in green we can see that this method does not find any negative roots even when the resolution is increased. It is only when, with information, not from the search method but from omnigraph, that we can find the solutions between -1.1 -1. The solutions can be seen closer in a graphical from here. I have not gone further with the decimal search method to find them, because I have already proved that I can find roots and all I needed to do here was to prove that first it seemed as is they didn't exist and then that they did. The Newton-Raphson method The Newton-Raphson method is much more sophisticated than the Decimal Search method. It involves a very clever method to find out where the roots of an equation are. It works on the principle that a curve will be pointing (it's tangent) at the root of an equation. To find the roots it takes the derivative of the equation and then calculates where the tangent to the line at a given point will cross the x-axis. ...read more.

Conclusion

I have demonstrated four failures with the Newton-Raphson Method, but there I still one more even in this equation! The Newton-Raphson method can in fact find a root that doesn't even exist. If you look in the tables section you will see one root that is found where I haven't tested form accuracy. This root does not exist. This is because the root almost exists, it is where it looks like a double root on the large scale picture But as you can see from this graph it is about a 1trillionth away from the x-axis, but does not touch and then due to software and/or hardware limitations the spreadsheet program thinks that there is a double root at this point and creates a non-existent root, with is a far worse fault than missing out on a root that does exist. Software/Hardware limitations are a large problem in situations like this. Most programs will only use a certain number of bits to determine a point and that number may further be limited if it being used on a low quality computer. If you look on the graph above you can see the resolution. The values all seem to be distinct trillionths away from each other and this can be problem in scenarios like this. The Re-arrangement Method Original y=(x^6-2*x^4+4*x^2-8x+3.141592654)/20 Success y=(x^6-2*x^4+4*x^2+3.141592654)/8 Failure y=((x{^{6}+4x{^{2}-8x+3.141592654)/2){^{0.25} Comparison of Methods ...read more.

The above preview is unformatted text

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

    C3 Coursework - different methods of solving equations.

    5 star(s)

    in the f(x) : (-1)3-6(-x)2+2(-1) + 2 = -7 This is the tangent drawn: From the equation, we can work out where the line crosses the X axis by making equal to 0: -0.58824 So now, our X1 value is -0.58824 instead of 0.

  2. Marked by a teacher

    The Gradient Function

    5 star(s)

    + n (n-1) (xn-2) h� 2! + n (n-1)(n-2) (x^n-3)(h)� + n(n-1)(n-2)(n-3) (xn-4)(h4) + ...hn 3! 4! = (nx^n-1)(h) + n(n-1) (x^n-2)(h�) + ... +h^n 2! Therefore, (x+h)n - xn = nxn-1 + n(n-1) (xn-2)(h) + ... + hn-1 h 2!

  1. Marked by a teacher

    Estimate a consumption function for the UK economy explaining the economic theory and statistical ...

    3 star(s)

    Here R^2, the coefficient of determination, is the most commonly used of the quality of the fit of the equation to a data sample. The equation of R^2 is: R^2=(?(Y'i-Y*)2)/(?(Yi-Y*)^2), where Y'i are the predicted value from the derived equation, the Yi are the actually recorded values, Y* is the value of the mean.

  2. The open box problem

    Using this equation I will again draw graphs and tables to find out the maximum volume and the value for x. X 0.5 1.0 1.5 2.0 v 10.5 12 7.5 0 Here we can see the maximum volume lies between 0.5 and 1.5 so I will now draw another table

  1. Numerical Method (Maths Investigation)

    FAIL to solve such equation. n Xn f(Xn) f / (x1) Xn+1 1 0.5000000000 0.467106781 0.353553391 -0.82118 2 -0.82117749 #NUM! #NUM! #NUM! 3 #NUM! #NUM! #NUM! #NUM! REARRANGEMENT METHOD Another method of Fixed Iteration Method. Any equation f(X) = 0 can be arranged in the form X = g(X)

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

    <1, X=-0.34500825, g'(x) =-0.271173069, and that is a success example. Rearranging equation failure: To show that failure I chose the root in the interval (+1 to +2), and apply the same rearrangement: g(x) = (2x�-3x�-2)/5.Because I think there are still some cases that this method fails.

  1. MEI numerical Methods

    is: As k-> ? then ? -> 0 In other words as the value of K increases, the value of ?(the root) decreases. The graph appears to have some form of geometric progression if we look at the graph. In order for it to be a geometric progression it must have a common multiple.

  2. C3 COURSEWORK - comparing methods of solving functions

    < -1 then the iteration will diverge away from the root in a cobweb fashion and it will become a failure case Evaluating g'(x) using calculus Since g(x) = 0.8x³+2x²–1 g'(x) = 2.4x2+4x The roots were x = -2.7876 ([-3, -2])

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