Pure 2 Coursework

Task I) Systematic search for a change of sign.

We are looking for solutions to f (x)=0, i.e. where the curve crosses the x-axis. The first thing to do is to draw a graph. For this method, initial boundary values are needed, and a graph is a good way to do this, and to check that the particular equation can be solved using this method.

It is convenient if the boundary values c & d are a whole unit apart e.g. 2 and 3.It will be noted that c and d have opposite signs, in this example c is -ve and d +ve. Now, you have to calculate f (x) for x, which is incremented in steps of 0.1 from c to d. The resulting table might look like this:

We now know that the root we are looking for lies between: c+0.6 and c+0.7
because this is where there is a change of sign and the graph obviously crosses the x-axis. Also note that you only need to calculate values until there is a change of sign, one need not go further. These are the limits for the root, which must lie between them. We can find the next decimal value of the root by repeating this method using an interval of 0.01 . Once further bounds are found you can repeat the decimal search using an interval of 0.001 and so on to the required accuracy.

Unlike the change of Sign method, this requires only one starting value, which should be an estimate of the root we wish to find for f(x)=0. First, a graph should be drawn, if only a rough sketch.

Now, you find the tangent to the curve at x1, and find where it hits the x-axis. This point is your new estimate, x2, and you repeat this technique to a required accuracy.

We can see how this method gives us a good approximation to the root, as each step gets us a lot closer to the root.

The gradient of the tangent at (x1, f(x1) ) is given by:

y - f(x1)    =   f ' (x1)

       x  - x1                      

Where (x,y) is a point on the tangent and f ' (x1) is dy/dx of the curve at x1. Where the tangent cuts the x-axis, x = x2 and y = 0:

0  -  f(x1)  =  f ' (x1)

      x2 - x1

-  f(x1)  =  x2  -  x1

      f ' (x1)

      x2  =  x1  -   f(x1)

                   f ' (x1)

This is a form of the Newton-Raphson Formula:

     Xn+1 = Xn -  f(Xn)

                 f'(Xn)

Using this formula iteratively, one can find an accurate approximate root fairly fast.

An Example of its use:

I will use this method to find all roots of y = f(x) = 2x^4.5 - 4x^3.5 + x
First, a graph

There are three roots, the first of which is 0 because the intercept is 0 and other parts are positive powers of x.

In the terms of which method was easier and quicker I would say that the change of sign method was the most convenient. Alternatively the Newton-raphson method required you to differentiate first then work out the values.

The use of Microsoft excel and omni graph was far more beneficial for all three methods in the sense that the values for which x lay between were much easier to obtain, as was the use of tangents.

Conclusion

I have found that the Change of sign method is a very simple method to use, and only requires some simple checks before using it. The built in error-bounds and steady rate of convergence at 1 decimal place per cycle make it useful. Once programmed in to a computer, the many simple calculations become even easier.

The rearrangement of equations was complicated and required a certain amount of thinking before a calculator/computer could take over, including checking that g’ (x) was between -1 and 1. Its rate of convergence also varied a lot, depending on g’ (x). It also often requires more than one rearrangement to find all roots to an equation.

The Newton Raphson method converged the fastest of them all and was very easy to use once programmed in to my computer. It always tries to converge to a real root, but you may find the wrong one if your initial estimate was near a turning point.

I think that the Newton Raphson method is the best because of its speed of convergence and because a good starting estimates will always find the required root.