The Gradient Function
The Gradient Function Aim: To find the gradient function of curves of the form y=axn. To begin with, I should investigate how the gradient changes, in relation to the value of x. Following this, I plan to expand my investigation to see how the gradient changes, and as a result how a changes in relation to this. Method: At the very start of the investigation, I shall investigate the gradient at the values of y=xn. To start with, I shall put the results in a table, but later on, as I attempt to find the gradient through advanced methods, a table may be unnecessary. As I plot the values of y=x2, this should allow me to plot a line of best fit and analyze, and otherwise evaluate, the relationship between the gradient and x in this equation. I have begun with n=2. After analyzing this, I shall carry on using a constant value of "a" until further on in the investigation, and keep on increasing n by 1 each time. I shall plot on the graphs the relative x values and determine a gradient between n and the gradients. Perhaps further on in the investigation, I shall modify the value of a, and perhaps make n a fractional or negative power. Method to find the gradient: These methods would perhaps be better if I demonstrated them using an example, so I will illustrate this using y=x2. This is the graph of y=x2. I will find out the gradient of this curve, by using the three methods -
Solution of equations by numerical methods.
Solution of equations by numerical methods This investigation is to find equation solutions using three methods: . Change of sign using bisection, decimal search or linear interpolation. 2. Newton - Raphson. 3. Rearrangement of the equation f(x) = 0 into the form x = g(x). The change of sign methods are systematic searches, which use the positive and negative signs of the f(x) solutions to find the location of the root within the intervals found using the graph of the function curve. The change of sign method I have chosen to use within my investigation is the decimal search method. Fixed point iteration requires finding a single value or point as an estimation for the value of x, rather than establishing an interval as in the change of sign methods. The Newton - Raphson method and the rearrangement of equation f(x) = 0 into the form x = g(x) will be used to investigate this form of numerical equation solution. Once each of these methods have been investigated I will compare each of them, in order to find the easiest method for equation solution. This comparison will also include the negative and positive points of each method, such as problems that result in an inability to find the correct root and the speed of convergence to the correct root. Fixed Point Iteration - Newton - Raphson Method. To start off using the Newton - Raphson method, we must first take an
Solution of equations by Numerical Methods.
Ashley Hilsdon Div 3 Pure 2 Coursework Solution of equations by Numerical Methods Method 1: The Change of Sign method The simplest method for solving an f(x) function is to use a change of sign method; these include the methods of bisection, decimal search and linear interpolations. Unfortunately, as well as being the simplest methods, they are also relatively cumbersome. The bisection was employed below for the function of f(x)=x5+3x2+x+2. This was also displayed graphically in the graph below. The actual calculations for this method are summarised below in this table. I placed these with relevant diagrams The error bounds for this result are conveniently provided by the use of this method, and are -1.485<x<-1.475. The maximum error is therefore ± 0.005. However, this method is clearly not particularly easy to work with. In addition, as with many methods there are some functions for x, which do not work. An example is the function f(x)=x3-2x2-x. This is shown below. The problem in this situation is the fact that there is in fact two roots in a very short space. (Labelled (a.) and (b.) on the graph). Using the bisection method within the interval of [-1, 0], this is bisected into 0.5. The root of 0 is the one to which the calculations tend. However, the other root, of -0.4142 (to 5 s.f.) is missed. This is illustrated in more detail in the diagram below. Method
Solving Equations using Numerical Methods.
A2 Pure Maths Coursework Solving Equations using Numerical Methods There are three methods with which you can solve an equation. . Change of Sign Method 2. Newton-Raphson Method 3. Rearranging the equation f(x)=0 in the form x= g(x) Hardware and Software Used For the coursework I have used a computer for attaining more accurate results and to avoid errors. The software I have used are * Microsoft Word * GraphCalc (A software used for drawing graphs) Change of Sign Method x4- 4x +2 We see that the roots lie in between [0, 1] and [1, 2] We confirm this by looking for a change of sign x 0 2 f(x) 2 -1 0 f(x) is found by substituting the values of x in the equation. We look for roots in [0, 1] x 0 0.1 0.2 0.3 0.4 0.5 0.6 f(x) 2 .6001 .2016 0.8081 0.4256 0.0625 -0.2704 We look for roots in [0.5, 0.6] x 0.5 0.51 0.52 f(x) 0.0625 0.02765 -0.00688 We look for roots in [0.51, 0.52] x 0.51 0.511 0.512 0.513 0.514 0.515 0.516 0.517 0.518 f(x) 0.02765 0.02418 0.02071 0.01725 0.01380 0.01034 0.00689 0.00344 -0.000002 We look for roots in [0.517, 0.518] x 0.517 0.5171 0.5172 0.5173 0.5174 f(x) 0.00344 0.00309 0.00275 0.0024 0.00206 x 0.5175 0.5176 0.5177 0.5178 0.5179 0.518 f(x) 0.00172 0.00137 0.00103 0.00068 0.00034 -0.000002 Root is in between [0.5179, 0.5180] Failure Case There are some
Numerical Solutions of Equations
A Level Coursework - Numerical Solutions of Equations My coursework is based on finding roots of non-quadratic equations. I will be investigating 3 methods for finding these roots and comparing them. Change Of Sign Method Y = x4+4x-6 This method simply uses Microsoft Excel to tabulate points on a function in different stages. It 'narrows down' the desired root by getting progressively closer to it. The root is always in the interval between the 2 x values where the y values change sign. -5 599 -4 234 -3 63 -2 2 -1 -9 0 -6 -1 2 8 3 87 4 266 5 639 The roots of this equation lie in the intervals -2 < x < -1 & 1 < x < 2. To find the lower root, I tabulated -2 < x < -1. x y -2 2 -1.9 -0.5679 -1.8 -2.7024 -1.7 -4.4479 -1.6 -5.8464 -1.5 -6.9375 -1.4 -7.7584 -1.3 -8.3439 -1.2 -8.7264 -1.1 -8.9359 -1 -9 There is a change of sign between -2 & -1.9, therefore the boundaries of the root lie in this interval. I tabulated in steps of 0.01, and again in steps of 0.001 & 0.0001 x y -2 2 -1.99 .722392 -1.98 .449536 -1.97 .181385 -1.96 0.917891 -1.95 0.659006 -1.94 0.404685 -1.93 0.15488 -1.92 -0.09046 -1.91 -0.33137 -1.9 -0.5679 From this we have established the root lies between -1.9237 & -1.9238. I tabulated this one more time. x y -1.9238 0.002248669 -1.92379 0.002003871 -1.92378 0.001759078 -1.92377
Numerical Solutions of Equations.
P2 Maths Coursework Numerical Solutions of Equations HANNAN SHAH Introduction In this coursework I intend to use numerical methods to find the solutions of equations that cannot be solved algebraically. Many simple equations can be solved using methods such as factorising or the quadratic formula but such methods cannot be applied to more complicated equations. My intention is to use three methods of numerical analysis to solve some of these more complex equations, show problems associated with these methods and then to compare their relative merits. The methods I will be using are decimal search, fixed point iteration using Newton-Raphson and fixed point iteration after rearranging f(x) = 0 into the form x = g(x). Decimal Search Assuming that you are looking for the roots of the equation f(x) = 0 you want to find the values of x for which the graph of y = f(x) crosses the x-axis. As y = f(x) crosses the x-axis, f(x) changes sign (from + to -), provided that f(x) is continuous. Using decimal search I will attempt to find one root of the equation y = x5+4x²-1. The graph for this function is shown below. The graph clearly shows that there are three roots in the intervals [-2, -1]. [-1,0] and [0,1]. I am going to attempt to find the value of the root in the interval [0,1]. I will start at one decimal place by taking values of x between 0 and 1 in increments of
