I can take my final answer one degree of accuracy further and look at the change of sign of f(x) so when:
f(1.104005) = -7.9268 < 0
f(1.104015) = 3.3557 > 0
Therefore because my answer lies between the positive and negative, it must be accurate.
However, there are some instances where this method does not succeed and we are unable to find the root of the equation using the change of sign method. For example if there are three roots between two integers, the spreadsheets can only find one of those roots and so it appears as though the other two does not exist unless a graph is also drawn. These roots are unable to be found, and therefore the method fails for some equations.
From this graph of: 9(5x-1)5-4(5x-1)2+0.5=0
we can see that there are three roots between 0 and 1. However, if we look at the spreadsheet for the change of sign method, we can see that in fact, only one root is found between these two values.
The formulae that we enter in to the spreadsheet are in fact the same as with the successful change of sign, because in both instances we want to find the roots of the equation.
As we can see from the spreadsheet shown below, a root is actually found for the equation, but only one of them, not all three of the ones between [0, 1]. Therefore for this equation the method fails.
From this chart we can see that when the formulas look to find the root between [0,1], however when it sees the root between [0, 0.5], it does not try to look for the other two, between [0.5, 1]. Hence, failure.
Newton – Raphson method
This is another fixed-point estimation method, and as for the previous method, it is necessary to use an estimate of the root as a starting point.
I start with an estimate, x, for a root of f(x) = 0. Next, I must draw a tangent to the curve y = f(x) at the point (x, f(x)). The tangent will cut the x axis at the next approximation for the root, then tangent will then be found at that new point and so on.
I chose to start looking trying to find the roots for the equation: 3(x/2)3-4(x/2)+1.5=0
so I plotted the graph of y=3(x/2)3 -4(x/2)+1.5 in autograph.
I can see that there are three roots to my equation, in the intervals [0, 1], [1, 2], [-2, -3]. I can use a spreadsheets and close ups of my graph to find the roots to a given number of decimal places. I have chosen to look closely at the root in the interval [0, 1].
Shown below are the formulae that I entered into the spreadsheet to obtain my required results.
The reason the formula A4-((3*(A4/2)^3-4*(A4/2)+1.5)/(((9*A4^2)/8)-2)) is used is because we are trying to find out the new point where the tangent of my equation at my given integer point crosses the x-axis. As we can see from the close up of my graph below, tangents are used to get an increasingly more accurate value for my root.
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)=f’(x1)[x-x1]
The tangent cuts the x-axis at (x2, 0), so
0-f(x1)=f’(x1)[x2-x1]
-f(x1)= f’(x1)x2-f’(x1)x1
f’(x1)x1-f(x1)=f’(x1)x2
(f(x1)/f’(x1))x1-(f(x1)/ f’(x1)= x2
x2= x1-(f(x1)/ f’(x1))
Hence the equation is re-arranged to give a closer value for my root on the x-axis. This is the equation that is then typed into my spreadsheet, shown above.
The final value for my root is shown in the numerical spreadsheet below, for the root between [0, 1].
So to five decimal places, the root in the interval [0,1], is 0.87607. To find the other two roots of the equation, I can enter in different starting integers into my spreadsheet to find the root to a given number of decimal places, I have chosen to look at it to five. So for the root in the interval [1, 2], the value is 1.74318 and for the root in the interval [-2, -3], the value is –2.61925.
Error bounds need to be established for the roots however, if we look at our final answer to five decimal places, then take that accuracy one decimal place further, we can see whether our answer is correct.
So for 0.87607:
f(0.876065) = 0.00000913453385
f(0.876075) = -0.0000022311064
Because, the function of one of these numbers is positive and the other is negative, the root that I am searching for must be between the two, hence my answer of 0.87607 is a sensible one.
However, there are some equations that this method fails to find the roots of. If the gradient of the tangent is more than 1 or less than –1, the tangent crosses the x-axis at a point that is further away from the root than the original integer, and hence an increasingly less accurate answer is obtained.
To demonstrate this failure, I have chosen to use the equation: y=ln(2x2-4)+x and to solve it using the equation ln(2x2-4)+x=0. First, I shall look at the graph, as shown below.
This equation fails not only because the gradient of the graph is too steep but also because the graph is discontinuous and there are actually no values for x between [-2, 2] anyway.
When we apply the Newton Raphson tangents, as shown in the graph below, we can see that the tangent moves off away from the root, and an overflow is produced, hence spreadsheets also fail.
Neither integers either side of the root are able to move towards a value for it, and I am also able to see this with the use of a spreadsheet. The formulae that are entered into this spreadsheet are the same that would be applied in a successful Newton Raphson investigation (see above). However, when the numbers are entered, the spreadsheet looks like this:
And for when we start with 1 as our integer, the spreadsheet simply looks like this:
It is not possible to use the Newton Raphson method to solve this equation.
Rearranging f(x)=0 in the form x=g(x)
The first step I must take to find a root for the equation f(x)=0, is to rearrange it in the form x=g(x). To set up the spreadsheet we must look at the line of y=x in conjunction with the rearrangement of my equation.
There are some rearrangements that can be solved using this method and others that this method fails to find the root for. I am going to look at the equation: y=3x3-2x2+4x+7.
I am first going to rearrange it so that: y=3x3-2x2+4x+7
3x3=4x-2x2+7
x3=(4x-2x2+7)
3
x=((4x-2x2+7)/3)(1/3) then I will solve it so that ((4x-2x2+7)/3)(1/3)=0.
This is the graph of y=((4x-2x2+7)/3)(1/3). I can see here that the line y=x and my g(x) cross at one point between the coordinates ([1, 2],[1, 2]). I can now use a spreadsheet to find this root to a given number of decimal places. The root is found in the fixed point iteration method by taking a line from my chosen integer to my g(x) line, then from that point to the y=x line, then back to the g(x) line and so on until the root is found to the highest possible degree of accuracy. As shown on the graph below, this equation creates a cobweb effect when these lines are drawn. Here we can see how the method converges to the root by moving between the two lines until a suitable value is found.
To find these roots on a spreadsheet I must enter these formulae, as explained below:
The reason this equation succeeds is because the gradient of my line is between 1 and –1. This means that the lines of iteration can move between the y=x line and the y=((4x-2x2+7)/3)(1/3) line, without there being a divergent.
Here is the spreadsheet of the success of iteration, and hence the root to five decimal places, as I have chosen to find.
However, my original equation of y=3x3-2x2+4x+7 can be rearranged to give: y=3x3-2x2+4x+7
-4x=3x3-2x2+7
4x=2x2-3x3-7
x=2x2-3x3-7
4
I can try to use this using the iteration method to find the roots of this equation. Below is the graph of this equation, with the y=x line.
Here we can see that there is a root in the coordinates ([0, -1], [0, -1]). However, when we look at the graph with the iteration cobweb on, we can see that in fact the cobweb spirals the wrong way, and continuously moves outwards, away from the root, regardless of which integer we chose to start from.
For the spreadsheet, because we have used a different rearrangement for the equation, the formulae on the spreadsheet are slightly different:
This equation fails because the gradient of the line is greater than 1, hence steeper than the y=x line.
Comparison of methods
In order to draw a conclusion about the easiest numerical method to use, it is important to ensure that the same equation is used for each method, and then the ease and speed of convergence to the root assessed.
I decided to use the same equation as I did for my successful change of sign method, which was y=5x3-7x+1. I then drew the graph of this again, as shown below.
Next, I had to ensure the root of the equation 5x3-7x+1=0 could be found using the Newton Raphson method and the rearrangement method, both graphs respectively are found below
I can see that a root in the interval [0, 1] can be found using each method. We need to look at the spreadsheets however to make sure that the same root is found exactly to a given number of decimal places, I have chosen to round to five. All of the formulae entered into the spreadsheets are the same as shown in the individual analyses previously.
From the spreadsheets we can see that the root of this equation to five decimal places is 0.14504. The change of sign takes the longest to converge to the root, taking 21 rows to find it. The Newton Raphson and rearrangement methods were both very quick to converge, however the Newton Raphson found it in 4 rows, whereas the rearrangement took 5.
The change of sign method is also very complicated to set up a spreadsheet for because it has many different columns and it would be easy to make a mistake when entering a formula, which would make the whole spreadsheet useless. The rearrangement method can also be difficult because of errors made when rearranging the equations; this can take time and be difficult to find one that is able to find the root. Also two graphs have to be drawn, which did not prove to be difficult for me using autograph, but by hand, there is more chance of error. However, the spreadsheet is easy to set up and use. On the whole I conclude that the Newton Raphson method is the best method to use because the spreadsheet is relatively easy to set up, only one graph has to be drawn, no rearrangements have to be found and for this equation, and most others, the Newton Raphson method found the root the quickest.
For this investigation I used several different types of software on my computer. For the spreadsheets I use edexcel, which was easy to use and much more efficient than trying to work out f(x) or g(x) for each line. Autograph was another very useful program because the graphs are drawn neatly and accurately, with no chance of error, and is also very quick to use. The other benefit of autograph is that there are certain functions that can be used, such as the Newton Raphson lines and g(x) iteration lines can be drawn on the graph in autograph.
I used two different types of hardware, my computer with all the programmes listed above and my graphics calculator. The computer was much easier to use because it has a larger screen which I can zoom in on, and more than one thing can be looked at at one time. The graphics calculator was also unreliable and often difficult to pinpoint the exact root that I was trying to look for.
