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

Solution of equations by Numerical Methods.

Extracts from this document...


Ashley Hilsdon Div 3

Pure 2 Coursework

Solution of equations by Numerical Methods

Method 1: The Change of Sign method

The simplest method for solving an f(x) function is to use a change of sign method; these include the methods of bisection, decimal search and linear interpolations.

Unfortunately, as well as being the simplest methods, they are also relatively cumbersome. The bisection was employed below for the function of f(x)=x5+3x2+x+2. This was also displayed graphically in the graph below.


The actual calculations for this method are summarised below in this table. I placed these with relevant diagrams

The error bounds for this result are conveniently provided by the use of this method, and are -1.485<x<-1.475. The maximum error is therefore ± 0.005.

However, this method is clearly not particularly easy to work with. In addition, as with many methods there are some functions for x, which do not work. An example is the function f(x)=x3-2x2-x. This is shown below.


...read more.


Using an estimate for (a.) as 0, the Newton-Raphson iterative formula was then used.

This next diagram gives some indication as to how the Newton- Raphson method actually works. What it basically does is to ‘slide’ the tangent across and is indeed sometimes known as ‘tangent sliding’.


For this method, the root can be calculated much more quickly then using the Change of Sign method. In accordance with the question the root was calculated to five significant figures, this means that the real value of x exists between 0.16655 and 0.16665, which give an error value of ± 0.00005.

This method, despite its speed in calculation a root, does not work for every function; this is illustrated below.

Method 3: Fixed Point Estimation- Rearranging the function

The third and final method for numerically solving a function of x is by rearranging an existing function of x and then using fixed-point iteration. This is illustrated for the function f(x)=x3+x-3 below.

Diagram of f(x)=x3+x-3

As is displayed graphically, the function with regards to (g)

...read more.


Rearranging the function is another useful technique, especially when a ‘cobweb’ effect is produced, as this gives natural error bounds for the root. However, it relies on an important rearrangement of the function, in conjunction with the plotting of at least two graphs. Once this has been done, the iterative formula is used, though within my calculations did not prove as fast as the Newton-Raphson method.

As I have already mentioned, the rearrangement method, relies more heavily on suitable graphical generating software or hardware; this is therefore a clear disadvantage that it faces. However, the change of sign method requires an interval for the root and the Newton-Raphson method requires an approximation to the result. Though, as this investigation has proved clear, regardless of the method, it has been necessary to have some idea of the function before attempting calculations. This is particularly important in the case of turning points and roots close to each other.

...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

    C3 Coursework - different methods of solving equations.

    5 star(s)

    -1.96180 -0.004775254 -1.96179 -0.004280068 -1.96178 -0.003784896 -1.96177 -0.003289738 -1.96176 -0.002794593 -1.96175 -0.002299463 -1.96174 -0.001804347 -1.96173 -0.001309244 -1.96172 -0.000814155 -1.96171 -0.000319081 -1.96170 0.00017598 X f(x) -2.0 -2 -1.99 -1.45736 -1.98 -0.92928 -1.97 -0.41553 -1.96 0.084135 -1.95 0.569938 -1.94 1.042111 -1.93 1.500882 -1.92 1.946474 -1.91 2.37911 -1.90 2.79901 The root of the function f(x)

  2. Marked by a teacher

    Estimate a consumption function for the UK economy explaining the economic theory and statistical ...

    3 star(s)

    There are 3 parameters in this equation. They are LC_1 , LY and constant. T-value are 6.11, 1.92 and 3.63. Where 1.92 is smaller than 2, which means that the coefficient of constant (0.206148) is insignificant in this equation.

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

    Then the process is repeated, as shown in graph 2 The gradient of the tangent at (x1, f(x1)) is f'(x1). Since the equation of a straight line can be written y-y1 = m(x-x1), The equation of the tangent is y-f (x1)

  2. Solving equations by numerical methods - The Interval Bisection method

    The first step, with an equation f(x)=0, is to rearrange it into form x=g(x). Any value of x for which x=g(x) is clearly a root of the original equation. For the next method I decided to take this equation: The first step is to rearrange the equation into the form of x=g(x).

  1. Numerical solutions of equations

    0.680 -0.005007 0.681 0.001510 The change of sign is between x = 0.680 and x = 0.681 x f(x) 0.6801 -4.355665x10-3 0.6802 -3.704544x10-3 0.6803 -3.053279x10-3 0.6804 -2.401872x10-3 0.6805 -1.750322x10-3 0.6806 -1.098629x10-3 0.6807 -4.467927x10-4 0.6808 2.051866x10-4 The change of sign is between x = 0.6807 and x = 0.6808 x f(x)

  2. Triminoes Investigation

    a new equation which will be equation 11, when equation 7 - equation 8. 175a + 37b + 7c + d = 120 - equation 7 - 65a + 19b + 5c + d = 60 - equation 8 110a + 18b + 2c = 60 - equation 11 Equation

  1. C3 COURSEWORK - comparing methods of solving functions

    sf Using Newton Raphson and Autograph to find the root in the interval [-4,-3]: We now are using the Newton Raphson and Autograph to find the other two roots. I can see that the other roots lie within these intervals [-1, 0] and [0, 1]: From the table below, I

  2. Evaluating Three Methods of Solving Equations.

    of x, a tangent is drawn and extended up to the x-axis. This tangent should intercept the x-axis at a point nearer to the root of our equation. We then draw another tangent to the curve at the point corresponding to this new value of x and continue this process

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