Maurice Yap 6946 – Core 3 Mathematics Coursework – 4752/02 Methods for Ada!ced Mathematics
Using numerical methods to find roots of and solve polynomial equations
This report will explore and compare the advantages and disadvantages of three different numerical methods used to solve polynomial equations, where analytical methods cannot easily be used. It will explore instances where, for some reason, they fail and also examine their ease, efficiency and usefulness in solving polynomial equations.
Change of sign decimal search method
The change of sign! method can be used to find an approximation of a root to an equation to a specified accuracy, using a decimal search."olynomial equations can be illustrated graphically as the function y # f$x%, as shown below in figure &. The points where the curve intersects the x'axis are the real roots of the equation f$x% #(, because the x'axis is where y # (. If the curve crosses this line, the values for f$x% when x is slightly larger and smaller than the root will be positive and negative, either way round $given that the values chosen for x are not beyond any other roots%.) logical and systematic way to use this to solve an equation to a certain degree of accuracy is a decimal search, where having already identified integer intervals where roots occur, the interval is divided into ten, and f$x% for each of the ten new values for x is found. ) search for a change of sign $* or '% is conducted and the process is repeated in the interval where the change of sign occurs until the level of accuracy desired is achieved. )fter this, the same technique is applied to find the other roots and thereby solving the equation.
+xample of an application of the change of sign method
or example, consider solving the following equation, by first finding the greatest root to five significant figures-
It is shown in figure & that there are three roots to this equation. That which is labelled root c! will be attempted to be found.
The task, ‘Aeroplane Landing’, requires the modelling of the velocity of a landing aeroplane from thetime it touches down to coming to a rest in the following situation:An aeroplane of mass 120,000kg comes in to land at a runway. After touchdown it initiallyslows down from air resistance. When it has slowed down enough this is augmented by aconstant force from the wheel brakes.We are to investigate a suitable mathematical model using Di
erential Equations to explain the natureof the forces acting on the aeroplane during these 26 seconds.The task has already provided us with data for the velocity each second from 0 to 26 to aid with themodelling cycle, which will be discussed later in order to test our models. (see
Verifying the Model
2 Simplifying and Setting up the Model
2.1 Assumptions Made
There are a number of assumptions made in the model, listed in descending order of relative importance:1.
The aerodynamics of the plane do not change.
That is, the plane does not deploy any ﬂaps,ailerons or elevators during landing. This is the most signﬁcant since it has a direct consequencefor the air resistance experienced by the plane.2.
The aeroplane acts as a point.
This simpliﬁes the complex aerodynamics and drag by suggestingthat we can consider the plane as a single point at its centre of mass rather than as a whole body.3.
There is no wind speed on the ground.
We don’t know if there is any wind going againstthe plane - which would act against motion and contribute to its deceleration, or with the plane,which would work against deceleration. However we don’t know what the conditions are, so it isbest to assume there is no headwind - which in reality is unlikely.4.
The runway is completely level.
Even with very small slopes, the component of the 120,000gweight would still be very large in magnitude to be able to a
ect the deceleration largely.5.
The landing is smooth.
If the landing was rough, it would cause the normal reaction force tovary as the plane landed, which would a
ect friction and hence the constant braking force.6.
The surface of the runway is uniform.
Since we’re considering a constant braking force, itdoesn’t matter exactly what the surface is, so long as it doesn’t change - in reality there could bedebris or oil on sections which a
ect the coe
cient, though it would be minor.7.
The landing is completely straight.
If the plane were to be turning while landing, it wouldincrease the friction of the tires and we would need to consider the force needed to change thedirection component of its velocity.8.
The fuel burnt during landing is negligible.
This would mean that the mass of the planedecreased, hence its inertia and resistance to deceleration decreased. However in 26 seconds, thefuel burnt will be a tiny proportion of its 120,000kg mass.9.
The weather conditions do not change.
This would e
ect the surface of the runway, windspeeds (see assumptions 2 & 5) as well as possibily the density of air, which would a
ect airresistance, however any changes to the weather within 26 seconds would be very minimal.10.
ects are negligible.
Even at the maximum velocity of 96m/s, it is very small incomparison to the speed of light, 3
m/s, so changes to mass and the e
ect of time dilationwould be incredibly small, infact it is likley immeasurable.2