This method can fail if:
The graph has two roots that are close together,
The graph has a discontinuity,
The graph touches the x axis, but does not cross it.
E.g. A graph that has two roots close together:
y = 90x4 – 60x + 24.7
This would indicate that the graph does not cross the axis and therefore has no roots, but if the graph is sketched:
This method would therefore appear ineffective.
Newton-Raphson Method
The basic principle of this method involves making an estimate (x0) at the root of the equation. From that estimate, a line is drawn vertically up, until it meets the curve. A tangent to the curve is drawn at that point, and the point at which that tangent crosses the axis is the new estimate (x1). This process is repeated a few times until the value of x appears to have stopped changing.
The Newton-Raphson iterative formula is:
xn+1 = xn –
The equation I am going to solve is:
f(x) = 15x4 + √2 – x -
(x) = 60x3 -
x0 = 1
x1 =
x1 = 1 –
x1 = 1 – 0.2114
x1 = 0.7886
First Root:
x0 = 1
x = 0.5753
Second Root:
x0 = 0
x = 0.05231 ± 0.000005
f (0.052305) = 4.82872E-06
f (0.052315) = −2.65013E-05
The change of sign indicates a root.
This method will fail if x0 is on a turning point, or if the graph has a discontinuity:
E.g. A graph with a discontinuity:
y = ⅓ Ln (x – 2) +1
The tangent to the line crosses the axis after the root. The process cannot continue from here.
Rearrangement
y = 2x5 –
This equation shall be rearranged to find two separate equations in the form x=g(x). Instead of then finding the root of 0 = 2x5 – , we find the point of intersection between x=g(x) and the line y=x. To do this, the gradient near the point of intersection must be between -1 and 1, otherwise the method will diverge from the point of intersection.
= 2x5
2x =
x =
Solution: x = 0.748
As the gradient of this graph is between 1 and −1, this method can converge on an accurate answer for x.
0 = 2x5 –
2x5 =
2x =
x =
This method cannot work with this graph, as the gradient is smaller than −1, and the value for x consequently diverges.
Comparison
To compare the three methods, I shall use the equation used for the decimal search method:
y = x4+2x3−5x2+1
I shall solve the largest root of this equation using all three methods, and see which requires the fewest steps to converge on the answer.
Decimal Search:
Solution = 1.3315 ± 0.0005
f (1.331) = -0.00348
f (1.332) = 0.003284
Change of sign indicates a root somewhere between the two.
This took 4 steps to find an answer correct to five significant figures.
Newton-Raphson
This method took 4 steps. However, if it were to be done by hand, it would have taken much longer.
Rearrangement
This rearrangement cannot be solved by this method, as the gradient is not between 1 and -1 at any point.
y = x4+2x3−5x2+1
0 = x4+2x3−5x2+1
−2x3 = x4 − 5x2 + 1
x =
This method requires 7 steps, not including the work put in to rearranging the equation.
It is clear to see that decimal search is the simplest way to solve equations. It is however the slowest, as it a table of values has to be made each time, to improve accuracy and the equation has to be written into Excel, which can take time. Doing it by hand would also take a long time, as you would have to put numbers through the equation up to ten times before the change of sign is found. This will be very monotonous and time consuming.
The Newton-Raphson method is the fastest, especially if you have a computer with Autograph to draw the graphs and to the maths behind it for you. A few more steps are necessary if you use the formula each time to improve your answer.
The rearrangement method required the largest number of steps, and on top of that, the equation had to be rearranged twice.
Overall, I believe the Newton-Raphson method is the best method to use when solving equations, because it is the shortest by hand and without a doubt the shortest and easiest using a computer.
