• 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...


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


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.




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


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.


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. Numerical solutions of equations

    the point of intersection between y = x and y = g(x) = (-4x2+2)1/5 is between x = 0 and x = 1. Like before, I will take my starting value of x (x1) to be 0.5. The same process of substituting values of x into the iterative formula, as in the last Rearrangement is carried out here.

  2. Numerical Solutions of Equations

    x y = x5-3x+1 0.3 0.10243 0.31 0.072863 0.32 0.043355 0.33 0.013914 0.34 -0.01546 It has taken another five steps to find the root to an accuracy of 1 decimal place.

  1. C3 Coursework: Numerical Methods

    Using the Newton Raphson Method was also made simpler using Microsoft Excel. As with change of sign, I used Microsoft Excel to replicate the formulae I used. This was also made simpler using the Auto fill function. As the formulae for the Newton Raphson method was more complex than the

  2. C3 Numerical Solutions to Equations

    This is the root of the equation f(x)=0 At the root x = 1.75650874, g'(x) = 0.479 (3sf). �g'(x)�<1 for the iterations to converge. If �g'(x)�>1 the iterations will not converge on the correct root. For example sing the same rearrangement of the equation to find the root between

  1. 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.

  2. Solving Equations. Three numerical methods are discussed in this investigation. There are advantages and ...

    Close roots In some cubic function there are 2 close roots and this method will also fail in this situation. Our target root is between 0 and 1 and if we use 0 and 1 as the upper and lower bound then we will find another root instead of our target root.

  1. Numerical Solutions of Equations.

    = 0.49269, y is +0.000005. When x = (0.492685 - 0.000005) = 0.49268, y is -0.000037. This change of sign means the root lies between this two values for x. Problems with decimal search Although decimal search was successful in finding this particular root there are problems with the method in certain situations that result in no root being detected.

  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