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

numerical solutions-Comparison of the three methods and Newton Raphson

Extracts from this document...

Introduction

Numerical Solutions of equations

  1. Newton Raphson method

Equation to be solved is x³−5x−1=0  The function f(x)= x³−5x−1is shown below

image00.png

There are 3 roots. I will first find the root in the interval [2, 3]

I will do the first few lines of calculation manually.

The formula to use is:  xn+1= xn –f(xn)/ f’(xn)

Therefore I must first differentiate x³−5x−1 which is 3x2-5

Using x1=3

X2:   3 – [(33-5 x 3-1)/(3 x 32 -5)] = 2.5

X3:   2.5 – [(2.53-5 x 2.5-1)/(3 x 2.52 -5)] = 2.3455

I will now work out all 3 roots using autograph until 5 significant figures are guaranteedimage01.pngimage06.pngimage07.png

The 3 boxes above show how I obtained the 3 roots of the equation x³−5x−1=0    

Which are -0.20164, -2.1284 and 2.3301

Below is the function f(x)= x³−5x−1 showing where I applied the

...read more.

Middle

image10.png

image11.png

Error bounds for interval [2, 3]

2.3301 is the root in this interval to 5 significant figures.

Therefore the error is 2.3301 ± 0.00005 I will now perform the change of sign test to confirm it is within these limits.

Lower limit is 2.33005 then f (2.33005) = -0.000098648

Upper limit is 2.33015 then f (2.33015) = 0.0010302

There is a change of sign which confirms root in interval is 2.3301 ± 0.00005

When does this method fail?

The Newton Raphson method does not always work, I will show this Using the equation (5x+4)1/7 =0

This function is shown below with autographs attempt to find the root using tangents to the curve starting at x= -1

The function f(x) = (5x+4)1/7

image12.png

As you can see from the graph and the Values of x in the table, it was unable to find the root in the interval [-1, 0].image13.png

This is because of the gradient of the curve, where my starting value is very close to a turning point. Then rather than converging towards the root, it is diverging further and further away on the x axis, therefore overflows and is unable to find the root.

Comparison of methods

...read more.

Conclusion

Comparing Ease of use of Hardware and software

It’s very obvious that the hardware and software speeds up the process of working out the solutions dramatically, as manually is very time consuming and easier to make mistakes.

Autograph is very easy to use and therefore Newton Raphson and rearrangement methods are very easy to perform, as you simply have to add the equation and perform the functions required which only require very few steps and choosing the starting point.  However change of sign requires more knowledge of Excel, having to input the formula yourself and choosing appropriate values for x to converge to the root. However once you know how to use excel you are able to drag down and it copies the formula for each value of x without having to retype it. For the manual calculations I was able to use a calculator, although this was time consuming, setting the starting point to “ans” allowed me to quickly get all the values of x by simply pressing the equals button. Autograph also allowed me to zoom in on roots, changing axes, etc, this helped make the process of finding roots and showing failure much easier.

...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)

    I shall try and investigate this. 3x� x y second value x second value y gradient 4 48 4.1 50.43 24.6 4 48 4.01 48.2403 24.06 4 48 4.001 48.024003 24.006 3 27 3.1 28.83 18.6 3 27 3.01 27.1803 18.06 3 27 3.001 27.018003 18.006 2 12 2.1 13.23

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

    reduces is the same as the 'ratio of differences' between successive estimates. One way of getting around this problem is by using the Simpson's rule. One particular method of the Simpson's rule is explained on page 4; however it requires an excess of calculation.

  1. Solving Equations Using Numerical Methods

    Using the table of iterates we can see that the root is at 0.1255. To confirm this, I will look for a change is sign. F(x) = y=x3-4x+0.5 F(0.126) = -0.001996 F(0.125) = 0.001953 As a change of sign has been found and confirmed here, this is where the root lies.

  2. MEI numerical Methods

    This creates the secant method, this requires two approximations, x(0) and x(1). Secant method: Evaluating this formula means, the first approximation x0 multiplied by the function of the second approximation x1, minus the function of the first approximation x0 multiplied by the second approximation x1.

  1. Math Portfolio Type II - Applications of Sinusoidal Functions

    Explain how the latitude of a location is related to the hours of daylight, and explain how this relationship is illustrated by the differences in the parameters in the two equations. The latitude of a location is related to the hours of daylight, since there are more hours of

  2. Numerical solutions of equations

    f'(xn) For my equation, f'(xn) = 4x3+1 when my function is differentiated to find the gradient function. The iterative formula I will use to find the roots of this function is: xn+1 = xn- xn4+xn-1 4xn3+1 I will start by finding out the negative root first.

  1. C3 Numerical Solutions to Equations

    This process is then repeated on the new x values until they converge on the root to the required level of accuracy. Taking x0 as the first guess at the root, the tangent to the curve at (x0,f(x0)) crosses the x axis at x1, the second guess.

  2. C3 COURSEWORK - comparing methods of solving functions

    Also, if the too roots are too close together, we might miss the other two after finding the first one. Consequently, we would be unlikely to seach further in the intervals. Despite the fact that x=g(x) method also involved some calculations, it?s not the quickest way to find all the

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