Looking at Indices
Maths AS GURU P1 ALGEBRA Looking at Indices The manner in which we count is based on the number of fingers (digits) that we have. Our number system is the product of centuries of development . The symbols originated with the Hindus and the name from the Romans. Decimal, means tenth or tithe. The numbers are put together so that the position of any particular digit in a whole number represents its value multiplied by 10, 100, 1000, etc. 6 × 00 + 5 × 0 + 7 × = 657 These multiples of ten can often be written more conveniently as 10, 102, 103, etc. Ten is called the base and the small number above and to the right is called the index (when there are several, the word is indices). 2873 = 2 × 1000 + 8 × 100 + 7 × 10 + 3 × 1 2 × 103 + 8 × 102 + 7 × 101 + 3 × 100 Top This notation logically extends to include indices of zero and of negative numbers. 00 = 1 Using negative indices allows the position of digits after the decimal point to represent fractions of whole numbers. 0-1 = 1/10 and 10-2 = 1/100 873 59/100 is the same as 8 × 102 + 7 × 10 + 3 × 100 + 5 × 10-1 + 9 × 10-2 The idea of bases and indices can be extended to algebra. a is the base and m is the index. This whole expression should be read as: "a to the power of m" am In this lesson we will look at how index notation works. We will develop rules for simplifying
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 -
Numerical Method (Maths Investigation)
NUMERICAL METHOD INTRODUCTION It is very useful to use Numerical Method to find the roots of an equation that cannot be solved ALGEBRAICALLY. Quadratic equations in the format of can be solved by Quadratic Formula: . However, for polynomial equations, that have highest power more than 2, has to be solved through Trial and Error, which is very hard and tedious to determine their roots. Moreover, some roots of polynomial equations may be not Integer or Large numbers, which make things harder. Therefore NUMERICAL METHOD is developed to help to solve polynomial equations. In P2, we learned that there are two types of Numerical Method: * Change of Sign Method o Decimal Search o Interval Bisection o Linear Interpolation * Fixed Point Iteration o Newton-Raphson Method o Rearrangement Method About Coursework Things have to do for Coursework: * One kind of Change of Sign Method; * Newton-Raphson Method; * Rearrangement Method; * Failures of each methods; * Error Bound of each methods; * Comparison made with the three methods above; and * Ease of use and Availability of Hardware and Software to do coursework. CONTENTS: PAGE No. Description 2 Change of Sign Method and its Failure 8 Newton-Raphson Method and its Failure 2 Rearrangement Method and its Failure 7 Comparison made onto one roots of an equation with the 3 Methods above 22 Ease of use and
Functions Coursework - A2 Maths
C3 Coursework: Numerical Analysis Change of sign method I want to find the roots of the equation x3-3x-1=0 to five decimal places. Consider the function f(x)=x3-3x-1; hence I want to find the roots of the equation f(x)=0. If the function is continuous in the interval [a,b], where a and b are values x could take, and either: f(a)<0 and f(b)>0 or: f(a)>0 and f(b)<0 then there must be at least one root of the equation f(x)=0 in the interval [a,b]. Therefore we are looking for a change of sign, which will then indicate the interval in which there lies a root of the equation f(x)=0. The table shows values of f(x) at integer values of x from -5 to 5. x -5 -4 -3 -2 -1 0 2 3 4 5 f(x) -111 -53 -19 -3 -1 -3 7 51 09 We can identify 3 intervals which have a change of sign: [-2,-1],[-1,0] and [1,2] If we sketch the graph of the function, i.e. y=f(x), the x values at the point where y=0 are the roots to the equation f(x)=0, since y=f(x). Three roots are confirmed in these intervals i.e. each interval contains one root. The gradient formula can be calculated: f´(x)=3x2-3 f´(5)=72, and after x=5, the gradient formula shows that the gradient just continues to increase with increasing x therefore the curve y=f(x) never crosses the x-axis after x=5. f´(-5)=72, and for any x value lower than x=-5, the gradient formula shows that the gradient just continues
Pure Mathematics 2: Solution of equation by Numerical Methods
Pure Mathematics 2: Solution of equation by Numerical Methods Introduction: In this coursework, I am going to solve equations by using the Numerical Methods. Numerical methods are used to solve equation that cannot be solved algebraically e.g. quadratic equations ax²+bx+c=0 can be solved using this formula: x= -b± V b² - 4ac 2a Therefore numerical methods would not be used for quadratic equations. I will be working with cubic equation because there is no formula to solve it. There are three methods, which I will be using: * Change of sign method * Newton-Raphson method * Rearranging f(x) = 0 in the form x = g(x) Change of sign method: This method is concerned with when a function crosses the x-axis, and by definition changes sign (+ and -). If we are looking the root of equation f(x) = 0. The point at which the curve crosses x-axis is the root. Once an interval where f(x) changes sign then the root must be in the interval. f(a) > 0 f(b) < 0 Therefore root must be between [a,b] f(a) < 0 f(b) > 0 Root is between the interval [a,b] To find the interval of each root for the equation, I'll be doing a decimal search first. Lets take the equation y = x³ - 12x + 5 x -4 -3 -2 -1 0 2 3 4 f(x) -11 4 21 6 5 -6 -11 -4 21 There are 3 roots in this equation and they are in these intervals: [-4, -3] [ 0 , 1 ] [ 3 , 4 ] Now, I am going to use
Numerical solution of equations, Interval bisection---change of sign methods, Fixed point iteration ---the Newton-Raphson method
Numerical solution of equations Introduction What is numerical method? We may occasionally encounter some equations which cannot be solved by algebraic or analytical method. In order to obtain the answers, numerical methods can be used for solving those equations. In this coursework, I am going to use three numerical methods to solve the cubic and non-trivial equations. The methods are shown as follows: * Interval bisection---change of sign methods * The Newton-Raphson method---fixed point iteration * Rearranging the equation f(x)=0 into the form x=g(x)---fixed point iteration Each method shown above will be applied to one different cubic and non-trivial equation. In addition to the application, I will compare the other two methods by applying one of the three equations, thereby comparing their ease of use and speed of convergence. *in this coursework, programme including Microsoft Excel (used for calculations) and Autograph 2.10 (used for drawing graphs) will be used so as to enhance the process of the coursework. Interval bisection---change of sign methods Given that the equation y=f(x) is a continuous function in which no asymptotes or other breaks occur, we look for the roots of the equation f(x) = 0. In the usual case, all we need to do is just to factorize the formula and then solve the equation easily. In this case, a formula y = (2x-3)(x+1)(x-2)- which is
Estimate a consumption function for the UK economy explaining the economic theory and statistical techniques you have used.
Estimate a consumption function for the UK economy explaining the economic theory and statistical techniques you have used. Consumption has been considered as the most important single element in aggregate demand, accounting for almost 66% of GDP in 1989. Therefore, it is essential that the level of consumption be predicted accurately, for even a small percentage error may lead to a large absolute error. Another reason why consumption such important is that the marginal propensity of consume is one of the factor which is used to determine the size of the multiplier. This will influence the changes of investment and government spending. Moreover, as saving ratio, another factor that used to predict the behavior of consumers, has fluctuated rapidly during these years, so the consumption become more important. Thus, many economists attempt to develop many theories and equations to predict consumers' expenditure. Therefore, in this project, there are two main theories and their equations will be given. Some data from Economic Trends Annual Supplement and graphs will be used to estimate the performance of the model on the basis of some econometric tests. Keynes, John Maynard introduced consumption function in 1936. In The General Theory of Employment Interest and Money, Keynes pointed out that " We should therefore define what shall call the propensity to consume as the
The open box problem
The open box problem An open box is to be made from a sheet of card. Identical squares are to be cut off at the corners so that the card can be folded into the open box. The diagram below shows the sheet of card and the four corners, which are to be cut off. There are two objectives I shall be investigating: . To determine the size of the four square cuts that will make the volume of the box as large as possible with any given square sheet of card. 2. To determine the size of the four square cuts that will make the volume of the box as large as possible with any given rectangular piece of card. First I shall investigate objective 1 as I think it will be easier to do. Objective 1: The square sheet of card. There are 2 ways to solve this problem: I can use algebra, or I can use trial and improvement. I think I will start by using trial and improvement; I will construct a series of tables and graphs and see if I can find any patterns. If so, then I will be able to come up with a hypothesis, which I can then test to see if I can solve this first objective. Then I will attempt to solve the same problem using algebra. Method 1: Trial and improvement To do this I need to make up some dimensions and then apply them to the square. I will choose the dimensions 6x6 to start with. I will call the length of the square cuttings x. The second drawing shows the open box with the
Solving Equations. Three numerical methods are discussed in this investigation. There are advantages and also disadvantage in these methods and they will be discussing throughout this investigation.
Numerical Solution of Equation INTRODUCTION In many situations, equation can be solving by algebra and by graphical analysis. However when equation involves terms which had a higher order than two then it became very difficult to solve by algebra. In this investigation we are focus of cubic functions - functions which involves term. It is possible to solve algebraically but it had certain level of difficulties because it involves imaginary number and this increase the difficulties of solving cubic equations algebraically. A problem also occur because was not introduce before early 16 century therefore mathematicians use numerical analysis instead of solving cubic equation algebraically. Three numerical methods are discussed in this investigation. There are advantages and also disadvantage in these methods and they will be discussing throughout this investigation. In cubic functions, there is 1 solution which constitute by 3 or less distinct roots. All roots of the equations are laid on the x axis, in other words it is intercepting linear equation x = 0. Three methods we are discussing in this investigation is base on this concept and derive to find one target root in an equation. ) Method of Bisection This is an example of method of interval estimation; it was used in continuous function. In continuous function f(x) = 0, if it intercepts liner equation x = 0 then this
I am going to solve equations by using three different numerical methods in this coursework and compare the methods which are Bisection, Newton-Raphson, and Rearranging method.
Introduction-Solving equations by numerical methods Numerical Methods are used for solving equations which solutions are not possible to be solved by algebraic or analytical methods. Now I am going to solve equations by using three different numerical methods in this coursework and compare the methods which are Bisection, Newton-Raphson, and Rearranging method. The comparison is about their ease of use and speed of convergence. In this coursework, I will use some software which are Microsoft Excel (used for calculations) and Autograph 3.0 (used for drawing graphs) to help me complete the coursework. Bisection method Bisection Method is looking for a sign change of a continuous function. Actually, we couldn't use some normal methods which are algebraic or analytical methods to solve some particular cases. I have chosen an equation y=2x³-3x²-8x+7 which is a non-trivial equation. By using the autograph, we can see the graph (Below) cross the x axis and there are three roots. Now take the root in the interval [-2,-1] and start by taking the mid-point of the interval,-1.5. f(-1.5)=5.5,so f(-1.5)>0.Since f(-2)<0,the root is in[-2,-1.5]. Now take the mid-point of this second interval,-1.75. f(-1.75)=1.09375,so f(-1.75)>0.Since f(-2)<0,the root is in[-2,-1.75]. And if we continue this method, than we can find out the root that I have shown on the graph. I am going to combine
