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

C3 Numerical Solutions to Equations

Extracts from this document...


Adam Arstall

Numerical Solutions to Equations Coursework

Change of sign method

The change of sign method involves finding the interval in which a root of an equation lies by taking two values of x and showing that the root lies between them as the value of f(x) for each case has a different sign. A change of sign will always indicate a root if the function is continuous.

This method will be used to find a root of the equation f(x)=x⁵+x⁴−2x³+5x²−7x−2=0. As the function f(x) is continuous, a change of sign will always indicate a root.


The method will be used to find the root which lies between -2 and -3


As the root lies in the interval [-2.7211575,-2.7211574], x=-2.72115745 ± 0.000000005


Here it is shown that f(x) changes from negative to positive between -2.7211575 and-2.7211574 and the root to f(x)=0 is between these x values.

This method can fail to find roots in some cases. For example if the equation has a repeated root as shown below for the equation x³−0.96x²−5.

...read more.


Therefore x=-1.53407020 ± 0.000000005

To confirm this root there must be a change of sign. f(-1.5340703) = 1.16*10^-6.

f(-1.5340701) = -1.08*10^-6. Therefore there is a root as the function is continuous.



These two diagrams show the convergence of the iterations at different magnifications. It is shown that the root lies between -1.53407035 and -1.53407025.

Taking 1 as the first guess gives the following results


Therefore x=0.48269595 ± 0.000000005

f( 0.48269594) = -1.08*10^-7. f(0.48269596) = 0.0113

Therefore there is a root as the function is continuous.

Taking 4 as the first guess gives

Therefore x=4.05137424 ± 0.000000005image16.png

f(4.05137423) = 2.34*10^-7  f(4.05137425) = -1.65*10^-7

Therefore there is a root as the function is continuous.

This method can fail for some starting points such as ones where the gradient of the curve is very small which can lead to the iterations converging on the wrong root. For the equation ¼x³−½x²−2x+3=0, an initial guess of 2.35 to find the root between 0 and 3 makes the iteration converge on the root between -4 and -2 as shown below.


Rearranging f(x)=0 in the form x=g(x)

In this method the equation f(x)

...read more.


The change of sign method was fairly easy to use as it requires little extra hardware or software and can be carried out manually if necessary using only a calculator. This makes it particularly useful if computer software is not available. However the need for much manual computation can make the process quite laborious and time consuming. The Newton-Raphson and x=g(x) methods are relatively similar in terms of ease of use with hardware and software. Both make good use of Autograph software visually interpret equations before using an Excel spreadsheet to carry out the calculations to find each root. Compared to the change of sign method both are generally more easy to use as once the initial formula has been entered it is very quick and simple to do the iterations many times.

...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)

    +4x = 4x Therefore the gradient function is correct. Now, I will change the value of n by 1 to see whether the same rule applies with different values of "a" and n. y = 2x3 x y second value x second value y gradient 4 128 4.1 137.842 100.86

  2. MEI numerical Methods

    The secant method does not require two approximations of the root to be used which are of opposite signs, unlike method of bisection and method of false position; as a result secant method may not always converge. However to carry on this way is impractical and doesn't relate to my

  1. Numerical solutions of equations

    My negative root is x = -1.220744 (6 decimal places) Now I will find the positive root of this function. The starting value (x1) is approximately 1.5 when f(x)=0. x1 = 1.5 x2 = 1.116379 x3 = 0.862068 x4 = 0.745761 x5 = 0.725049 x6 = 0.724492 x7 = 0.724491959 x8 = 0.724491959 I can see some convergence from x6.

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

    Another rearrangement is done: A graph is then drawn in computer If we out starting value as 0 then is converge to another root but not target root, therefore we use starting value 1 and it also converges to another unexpected root therefore this rearrangement fails to find target root.

  1. Numerical solution of equations, Interval bisection---change of sign methods, Fixed point iteration ---the Newton-Raphson ...

    -1.328125 0.625 0.1191406 0.125 4 0.625 0.11914063 0.75 -1.328125 0.6875 -0.612549 0.0625 5 0.625 0.11914063 0.6875 -0.6125488 0.65625 -0.248627 0.03125 6 0.625 0.11914063 0.65625 -0.2486267 0.640625 -0.065212 0.015625 7 0.625 0.11914063 0.640625 -0.0652122 0.6328125 0.0268483 0.0078125 8 0.6328125 0.02684832 0.640625 -0.0652122 0.63671875 -0.019211 0.00390625 9 0.6328125 0.02684832 0.6367188 -0.0192111 0.634765625

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

    Raphson Method, I can't get the point I want, but the further one. Therefore, it is failure. Rearranging equation method: This method is rearranging the equation f(x) =0 into form x=g(x).Thus y=x and y=g(x) can cross together, and then we can get a single value which can be estimated for the root.

  1. OCR MEI C3 Coursework - Numerical Methods

    -0.00339 2.64527 x6 -1.76245 -0.00001 2.62496 x7 -1.76244 0.00000 2.62488 x8 -1.76244 0.00000 2.62488 x=2.62488 (5d.p.) Failure of the Newton-Raphson method The method fails for the same function if x1=0 xn f(xn) f'(xn) x1 0 0.60317 -0.14172 x2 4.25600 1.05148 -0.02495 x3 46.40573 1.00046 -0.00002 x4 Value could not be

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

    Simpson's rule is different to the trapezium rule in that it fits a parabola between successive triples of points, whereas the trapezium rule fits a straight line between successive pairs of points. As a result the number of strips must not be even.

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