As you can see I have found only the first route and without a graph I wouldn’t be aware of the two routes actually present between this interval. The reason for the failure is the fact of two routes being in one interval. If the intervals were smaller then it would have worked.
Newton Raphson Method
In this method an estimate of the route is taken a line equal to x is taken up until it hits the curve and then a tangent is drawn down to the x-axis. This new position on the x-axis is treated in the same way as the first and so another tangent is drawn and taken back down to the x –axis. This is repeated until a sufficient value for the route is found. I began by using the change of sign method. This gave me an Idea of where my solutions would be. I then found the respective figures to substitute into the Newton-Raphson iterative formula:
Xn+1=xn- (f (xn)/(f’ (xn)
For this my equation will be f (x)=2x^5-5x^3+1
This was the beginning stages of this method and shows the change of signs I had to investigate. There were three routes as shown by the highlighted cells.
After taking an estimate of the roots to be –1.5, 0.5 and 1.5 I started to use the iterative formula. On the first route this idea becomes clear as we can see by the highlighted red cells that for the first value of the root drawn out by the equation becomes the starting point for the next set of figures to go through the iterative equation
The three routes I have found using this method are all highlighted in green. They can also be seen on the below graph
In this method I also looked at the error bounds of my solutions. I took the final figure up by 0.0000000005 and down by the same. I then put these two separate solutions into the original function to see if the would still force a change of sign which they did.
Green corresponds with route 1, Red with route 2 and yellow with route three. All of them showing a change of signs meaning that the error bounds have been established to a suitable degree of accuracy.
Like change of sign method it is also possible for Newton-Raphson to fail. To illustrate this I am going to use the equation f(x)=2x^3-5x^2+2.
Again I started the method by looking for a change of sign. I decided to try and get a solution for the route between 0 and 1, so I placed my first guess at 0.1. I then put this through the iterative formula.
However this seems to have picked up a different solution. From the change of sign we can see there is one between 2 and 3 so it therefore must be this. It can also be seen on the graph below.
It happens because the tangent is diverged away from the suspected route on the first iteration but then homes in on another route as shown below.
Rearranging
This method is also an iterative process, where like the last method it uses the last answer to get closer to the route until it eventually converges infinitely close to it. It requires a particular rearrangement of the equation f(x)=0 into a form of x=g(x) which will allow convergence to a route of the equation. However this will only happen if –1 ∠ g’(a) ∠ 1for values of x close to the route. For my investigation into this method I used the equation f(x)=x^5-4x+2 represented graphically like this
From this I can see all 3 possible routes. I am going to look at the route between –2 and –1 particularly. To do this I need to rearrange my formula into the form x=g(x)
F(x)=x^5-4x+2
⇒ x^5-4x+2=0
⇒x^5=4x-2
⇒x=(4x-2)^0.2
if I plot the graphs of y=x and of y=(4x-2)^0.2 then these graphs will intersect at the points of the routes.
I can now find the routes by using the rearrangement of the formula to give the iterative formula, so xn+1=(4x-2)^0.2
I first took an estimate of –1 and then found g(x) by substitueing in. I then used the g(x) of –1 as x and found g(x) of this.I repeated this until I had an accurate value of the route. This came out to be –1.51851. The value of g(X) was less than one enabling the route to be found but if it was grater than one or less than one then the route would fail as I have already mentioned.
If I were to rearrange the equation into the form of y=(x^5+4)/2 to try and find the same route the iterations seem to diverge to other routes because of the constants of the method I have mentioned above.
We can also see this failure when I try to calculate the route….
…as the route comes out to be 0.508499. This is because the gradient at the route( g(x) ) is greater than one, causing it to fail and diverge to the middle route.
Comparison of Methods
Using the equation I used for the change of sign method y=x^3-5x+3 I will find the same route using the two other methods giving me a chance to compare them in terms of speed of convergence and in terms of ease of use with available hardware and software. The root I found using the said above method was one between –3 and –2 as shown on the graph below.
Newton-Raphson
The Newton raphson method shows the route to be –2.490863615 ± 0.0000000005 as shown in red above.
Rearranging method
I found with this equation that the Newton-raphson method was the fastest to converge to the route. It was very reliable and it was difficult to find an equation for which it didn’t work. It was so fast to converge to a route because o fits logarithmic method of tending to a route, meaning to begin with it tends towards the solution very quickly and as the steps get smaller it becomes even more so accurate. The method was very easy to use due to the power of excel and the ability to copy, paste and fill down within the cells. If it were to be done manually it would be much more difficult due to the equations involved and types of polynomial expressions I was using.
The change of sign method was the simplest available to me and was very quick and easy to use. Unlike the two other methods there were no formula required. Each time I repeated the method the solution became more accurate and the route's accuracy increased to another significant figure was found. The method is reliable in most cases but if a turning point is discovered to be at y=0 or there are two routes within one interval on a graphical projection of the equation then it fails due there being no change of sign in the equation. One disadvantage of the method is that an approximate value of the route needs to be known, so a graph must first be plotted. The convergence time to the routes is relatively fast but it does require a great deal of work. If this method was to be done manually it would take a great deal of time more so than if it is done on excel due to the benefits of excel I have already mentioned.
The final method of rearrangement was relatively easy to use and there were no major calculations that needed to be done apart from the rearrangement at the beginning that did not prove to be too difficult. I did find however that it seemed to tend to the strongest route and would ignore the weaker routes when they are required and still tend to the stronger ones. Also as we already know when a route has a gradient lager than 1 or less than –1 the formula did not work because it couldn’t find the route and so failed. I found it the least successful of the methods although it could be used to find an accurate route quickly it often failed. The time in which it takes to use the method is greatly increased by the use of excel like both other methods, if it were to be used manually, again like both other methods, it would be very time consuming.