## Solution of equations by numerical methods

This investigation is to find equation solutions using three methods:

- Change of sign using bisection, decimal search or linear interpolation.
- Newton – Raphson.
- Rearrangement of the equation f(x) = 0 into the form x = g(x).

The change of sign methods are systematic searches, which use the positive and negative signs of the f(x) solutions to find the location of the root within the intervals found using the graph of the function curve. The change of sign method I have chosen to use within my investigation is the decimal search method.

Fixed point iteration requires finding a single value or point as an estimation for the value of x, rather than establishing an interval as in the change of sign methods. The Newton – Raphson method and the rearrangement of equation f(x) = 0 into the form x = g(x) will be used to investigate this form of numerical equation solution.

Once each of these methods have been investigated I will compare each of them, in order to find the easiest method for equation solution. This comparison will also include the negative and positive points of each method, such as problems that result in an inability to find the correct root and the speed of convergence to the correct root.

Fixed Point Iteration – Newton – Raphson Method.

To start off using the Newton – Raphson method, we must first take an estimate of the root as a starting point. For a root f(x) we shall start with estimation x1 –we then draw a tangent to the curve y = f(x) at the point (x1, f(x1)). The point at which the tangent cuts the x-axis gives the second approximation for the root, and so the process is repeated.

The gradient of the tangent at (x1, f(x1)) is f’(x1), and as the equation of a straight line can be written y – y1 = m(x – x1), the equation of the tangent can be said to be y - f(x1) = f’(x1)[x – x1]. This tangent cuts the x –axis at (x2, 0) so we can say that 0 – f(x1) = f’(x1)[x2 – x1].

Rearranging 0 – f(x1) = f’(x1)[x2 – x1] gives x2 = x1 – f(x1) which gives us the Newton -

F’(x1)

Raphson iterative formula:

Xn+1 = xn – f(xn) . f’(xn)

The formula that I have chosen to find the roots of using the Newton – Raphson method is y = f(x) = x^7-6x+4 which gives the graph shown below.

Returning to the initial function f(x) = x^7 – 6x + 4, we must find f’(x) in order to use the iterative formula. Using this initial function, we find f’(x) = 7x^6 – 6.

As the iterative formula states: xn+1 = xn – f(xn)

f’(xn)

- xn+1 = xn – xn^7 – 6xn + 4

7xn^6 – 6

which gives us the following convergence table: