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

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                                      A2 Mathematics---Core 3 Coursework                  Felix Cheng

Numerical solution of equations


What is numerical method?

We may occasionally encounter some equations which cannot be solved by algebraic or analytical method. In order to obtain the answers, numerical methods can be used for solving those equations. In this coursework, I am going to use three numerical methods to solve the cubic and non-trivial equations. The methods are shown as follows:

  • Interval bisection---change of sign methods
  • The Newton-Raphson method---fixed point iteration
  • Rearranging the equation f(x)=0 into the form x=g(x)---fixed point iteration

Each method shown above will be applied to one different cubic and non-trivial equation. In addition to the application, I will compare the other two methods by applying one of the three equations, thereby comparing their ease of use and speed of convergence.

*in this coursework, programme including Microsoft Excel (used for calculations) and Autograph 2.10 (used for drawing graphs) will be used so as to enhance the process of the coursework.

Interval bisection---change of sign methods

Given that the equation y=f(x) is a continuous function in which no asymptotes or other breaks occur, we look for the roots of the equation f(x) = 0. In the usual case, all we need to do is just to factorize the formula and then solve the equation easily. In this case, a formula y = (2x-3)(x+1)(x-2)- which is non-trivial equation, is chosen for showing how the roots of the equation will be obtained.

 But in my equation the product of the factor is not equal to 0. It implies that neither using algebraic nor analytical method is possible. Thus, an alternative way to find out the solution has to be sought.

The solution I would like to introduce first is the change of sign method.

When we are solving the equation f(x) =0, the values of x for which the graph of y=f(x) crosses the x axis is expected to achieve. A characteristic is also needed to be mentioned is that when the curve crossed the x axis, which means y =0, the f (x) sign change, it indicate that interval must contain a root. After we spot a root in the interval, the mid-point of the interval will be taken.

Graph 1.1 ---y = (2x-3) (x+1) (x-2)-1

When the graph of y = f(x) cross the x axis, f(x) change the sign of its value.

To solve the equation by bisection iteration, the values of f(x) has to be worked out so as to find the intervals where there is a sign change within them. The location of the interval can be achieved by substituting the value of x in to the equation f(x) = 0.


From the table1.2 above, it can be seen that the roots lie in the interval of [-1, 0], [1, 2] and [2, 3]

Here, I can take a=1 and b=2, whereas the mid-value is taken as (a+b) /2

Graph 1.3

We can achieve the value of the root as about 1.213, to three decimal places.

Graph 1.4---bisection of f(x) = (2x-3) (x+1) (x-2)-1=0

To solve f(x) = (2x-3) (x+1) (x-2)-1=0 by using bisection method in Excel spreadsheet, the root can be obtained by inserting the formulae as follows,

Spreadsheet 1.5

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In order to make sure there are always sign changes between the 2 values of interval [a, b], two columns showing  f (a) < 0 and f (b) > 0 will be added on the spreadsheet.

Spreadsheet 1.6

By using Excel spreadsheet 1.6, I discovered that in the 20th term, the root 1.21293 of 5 decimal places with error bounds  ±is got. This gives the solution bounds as 1.212929046< x < 1.212930954. To ensure the root is correct, it is essential to check the presence of the sign change.

f (1.212929046) =

f (1.212930954) =

Since sign change can be ...

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