There is a change of sign and therefore a root in the region between -0.261 and -0.260.
The next part is to establish where, between -0.261 and -0.26, a change of sign occurs.
There is a change of sign and therefore a root in the region between -0.2606 and -0.2605.
The next part of the investigation is establishing where, between -0.2606 and -0.2605, a change of sign occurs.
There is a change of sign and therefore a root between -0.26053 and -0.26052.
As shown visibly, and as can be proved mathematically, the root of the equation which lies between -1 and 0 is closer to -0.26052 than it is to -0.26053. Therefore, the root of the equation which lies between -1 and 0 is -0.26052 to 5 significant figures.
Because of the nature of the answer when given to five significant figures, the amount of error is +0.000005. This makes the percentage error of the answer is 0.0019% which is very small and in turn makes the answer very accurate.
The change of sign method, however, isn’t always the best way of finding the value of a root. There are two ways which it can go wrong. The first of which involves a tangent.
This is the graph y=5x3+7x2-x-3. Visibly, we can see that there is a root between -1 and 0 here. The root here is at -1. But it is in the form of a tangent. If we apply the method in this case, the change of sign isn’t picked up.
The other way it can go wrong is if there are two changes of sign between one whole number.
This is the graph y=5x3+3x2-2x-1. There are two changes of sign between -1 and 0 as we can see. If, however the ability to see the graph before the method was applied is not available, problems could occur as there would be two changes of sign and choosing which root to go for becomes an issue. Here, the two changes of sign are between -0.8 and -0.7 as well as -0.5 and -0.4.
Newton-Rapshon method
This is the graph y=3x3-2x2-5x-1. The root under investigation is the one which lies furthest left between -1 and -2. The Newton-Rapshon method involves using a starting value. A line is drawn from the starting value on the x-axis directly vertically upwards until it intersects the line created by the graph. At the intersection, the tangent is drawn which passes through the x-axis. At the new intersection, between the tangent and the x-axis, another line is directly vertically drawn upwards until it intersects the line the graph created. The process continues until the root is found to a satisfactory degree of accuracy.
The starting value which will be used is -1.
As shown, the first Newton-Rapshon iteration shows the tangent will intercept between -0.9 and -0.8.
The equation is y=3x3-2x2-5x-1. When x=-1, y=-1
dy/dx=9x2-4x-5. Therefore the gradient at -1 is 8.
y=mx+c; y=-1, m=8, x=-1, c=?
y-mx=c; -1-(8x-1)=7; Equation of tangent; y=8x+7
When y=0, x=-7/8=-0.875
Δx=-0.125
The first Newton-Rapshon iteration has resulted in the answer -0.875.
The equation is y=3x3-2x2-5x-1. When x=-0.875, y=-0.1660156
dy/dx=9x2-4x-5. Therefore the gradient at -0.1660156 is 5.3906.
y=mx+c; y=-0.1660156, m=5.3906, x=-0.875, c=?
y-mx=c; -0.1660156-(5.3906x-0.875)=4.55076; Equation of tangent; y=5.3096x+4.55076
When y=0, x=-4.55076/5.3096=-0.857081512
Δx=-0.017918488
The second Newton-Rapshon iteration has resulted in a more accurate answer -0.857081512. As the number of iterations increase, the change in x (Δx) decreases creating a much more accurate answer.
As shown, the Newton-Rapshon iterations have further narrowed down the value of the root creating the final answer of -0.84236 to 5.s.f. Because of the nature of the final value, the accuracy is to a +0.000005 standard which is very accurate. In this instance, the percentage error is 0.00059%.
The value of the other two roots had to established also using the method.
This shows the process of Newton-Rapshon iterations aiming to find out the value of the root between -1 and 0 which lies closer to zero. The starting value chosen in this case in 0. This has resulted in a value of -0.22708 to 5.s.f. being obtained for the root.
The other root lies between 1 and 2.
The starting value for this root was 2. This root has been found to be 1.7368 to 5.s.f.
However, there are occasions where finding the root using the Newton-Rapshon method doesn’t always work.
This is the graph y=2x3+2x2-2x-1. If the root between -2 and -1 was under investigation and the starting value of -1 was used, the gradient of the tangent at this point is zero. Therefore, no tangent can be drawn from this point to intercept the x-axis because the tangent at this point is horizontal.
y=2x3+2x2-2x-1; When y=-1, x=1
dy/dx=6x2+4x-2; When x=-1, dy/dx=6-4-2=0
y=mx+c; y=-1, m=0, x=1, c=?
y-mx=c; -1-0=-1 Therefore, equation of tangent is y=-1
When y=0, y cannot =0. Therefore the Newton-Rapshon method cannot be applied here.
