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Chapter 1: Refresher course in key numerical skills The aim of this chapter is to provide the reader with a set of tools which can be used to study the topics in this book and will be useful to you throughout your studies. Author: Glyn Davis © Oxford University Press, 2012. All rights reserved. Chapter 1: Refresher course in numerical skills 1 Introduction (1/2) Numeracy is the ability to reason with numbers and other mathematical concepts. To be numerically literate, a person has to be comfortable with logic and reasoning. Some of the areas that are involved in numeracy include: basic number, algebra, probability and statistics. The importance of numeracy has been identified by the UK Commission for Employment and Skills (UKCES, 2009, http://www.ukces.org.uk/publications/employability-challenge-full-report, accessed 12/4/12) who defined employability skills as: " ..the skills almost everyone needs to do almost any job. They are the skills that make specific knowledge and technical skills fully productive. Without employability skills, UK PLC ceases to be a global economic force and individuals don't get and progress in rewarding jobs." Author: Glyn Davis © Oxford University Press, 2012. All rights reserved. Chapter 1: Refresher course in numerical skills 2 Introduction (2/2) Furthermore, the report identified literacy, language, numeracy and ICT skills which underpin employability skills as illustrated in Figure1.1 (UKCES, 2009, page 11) Author: Glyn Davis © Oxford University Press, 2012. All rights reserved. Chapter 1: Refresher course in numerical skills 3 Learning Objectives On successful completion of the module, you will be able to: • Understand terms such as square and square root, indices, and standard form (or scientific notation). • Identify a simple equation and solve. • Understand what a logarithm represents. • Identify both linear and non-linear equations. • Locate information on Excel 2010 mathematical functions. • Plot a graph of y against x where the relationship between y and x is linear or non-linear. • Describe change using calculus. Author: Glyn Davis © Oxford University Press, 2012. All rights reserved. Chapter 1: Refresher course in numerical skills 4 Basic Algebra (1/6) Squares The square of a number is that number multiplied by itself. For example, ‘three squared’ is written 32 which means 3 x 3 = 9 and ‘four squared' is written 42 which means 4 x 4 = 16. Square roots The square root of a number is a value that can be multiplied by itself to give the original number. The sign 4 means the square root of the number 4. For example, Indices 4 = 2. The index is the power of a number. The number 23 is said 'two to the power three' or 'two cubed' and means 2 * 2 * 2 = 8. The number 32 is said 'three to the power two' or 'three squared' and means 3 * 3 = 9. We could use the method describe earlier to calculate 23 or we could use the Excel symbol ^ or Excel function POWER () to undertake the calculation. Author: Glyn Davis © Oxford University Press, 2012. All rights reserved. Chapter 1: Refresher course in numerical skills 5 Basic Algebra (2/6) To solve simple equations Here are some examples of very simple equations: 2a = 8, a + 2 = 4, 3a + 6 = a + 10, 3(a + 2) = 9. In all these examples, there are letters and numbers on both sides of the equals sign and the letters have no powers higher than 1. (i.e. there are no a2 or a3 or b2 or b3 terms). Your answer must have a letter, which must be positive, on one side of the = sign, and a number on the other side. It does not matter which side of the equals sign the letter is! To solve simple equations you must follow a set of rules. For example, solve the equation 3(a + 2) = 9 for the unknown term. Step 1 Remove any brackets by multiplying them out: 3(a + 2) = 9, 3a + 6 = 9. Step 2 Put all the terms containing letters on one side and numbers on the other side: 3a + 6 = 9, 3a = 9 – 6, 3a = 3, a = 3/3 = 1. Check by substituting a=1 into equation 3(a + 2) = 3a + 6 = 9 = 3(1) + 6 = 3 + 6 = 9. Author: Glyn Davis © Oxford University Press, 2012. All rights reserved. Chapter 1: Refresher course in numerical skills 6 Basic Algebra (3/6) Standard form (or scientific notation) Very large and very small numbers must sometimes be expressed in standard form (also called scientific notation), A * 10n, where 1 < A < 10 and n is an integer. Translated this means that A must be a number between 1 and 10 and n is a positive or negative number. Here are some examples to clarify how we can write a number using standard form. For example, the number 87000 can be written in standard form as 8.7 x 104. Author: Glyn Davis © Oxford University Press, 2012. All rights reserved. Chapter 1: Refresher course in numerical skills 7 Basic Algebra (4/6) Logarithms and exponential functions The logarithm was invented by John Napier (c1614) and developed by Napier and Professor Briggs (c1624) to aid in the calculation process of index numbers. The term is an alternative word for an index or power of a given positive number base. For example, 32 = 9, we define the index (or exponent) 2 to be the logarithm of 9 to the base of 3 and write 2 = log3 9. For example, consider calculating the value of 104. We can solve this problem by multiplying 10 by itself 4 times: 104 = 10 x 10 x 10 x 10 = 10,000. This can be written in index form as 104 = 10,000. Then, log10 10,000 = 4. The logarithm of 10,000 to the base 10 is 4. The number 4 is the exponent to which 10 must be raised to produce 10,000. Author: Glyn Davis © Oxford University Press, 2012. All rights reserved. Chapter 1: Refresher course in numerical skills 8 Basic Algebra (5/6) Linear and non-linear equations An equation is a mathematical expression that allows a relationship to be written between one variable and another variable (or variables). For example, the relationship between the cost of 10 tins of baked beans can be written as C = 10p, where p is a term that represents the cost of one tin of baked beans. In this case we notice that the term labelled C is dependent upon the linear term p. In this case the power of p is equal to 1 and the relationship between C and p is called a linear relationship (or simple equation). Any relationship between variables which is not linear is called a non-linear relationship. For example, the relationship between two variables (x, y) might be of the form y = x2. In this example, we note that whatever the value of the variable x is equal to, the value of y is equal to x multiplied by x, and therefore y = x2 is a non-linear equation. Another way of describing this is to say that y is a function of x2. Author: Glyn Davis © Oxford University Press, 2012. All rights reserved. Chapter 1: Refresher course in numerical skills 9 Basic Algebra (6/6) Excel mathematical functions Excel provides a range of mathematical and statistical functions (see appendices) that can be useful in undertaking the analysis of data e.g. the Excel function ABS returns the absolute value of a number. For example, consider calculating the absolute value of -4. From Excel this is equal to ABS(-4) = 4. For a detailed list of Excel mathematical functions then please see the online refresher course in key numerical skills or the Microsoft support site for Excel 2010 at http://office.microsoft.com/en-us/excel-help/ Author: Glyn Davis © Oxford University Press, 2012. All rights reserved. Chapter 1: Refresher course in numerical skills 10 Drawing graphs (1/5) The Co-ordinate of a point In the next chapter Microsoft Excel will be used to create a range of graphs that can be used to help to visualize data e.g. bar charts, pie charts, histograms, and scatter plots. In this section we will explore the concept of co-ordinate geometry and how Excel can be used to plot algebraic relationships between two variables in the form y = mx + c. Before we do this we now need to understand the concept of a coordinate of a point and the Excel method we can use to plot a series of points. Author: Glyn Davis © Oxford University Press, 2012. All rights reserved. Chapter 1: Refresher course in numerical skills 11 Drawing graphs (2/5) Co-ordinates of a Point (X, Y) The coordinates of any point can be written as an ordered pair (x, y). Point P in the screenshot has coordinates (2, 3). Its horizontal distance along the x axis from the origin 0 is 2 units so the x coordinate is 2. Its vertical distance along the y axis from the origin 0 is 3 units so the y coordinate is 3. Remember the x coordinate is always written first. Author: Glyn Davis © Oxford University Press, 2012. All rights reserved. Chapter 1: Refresher course in numerical skills 12 Drawing graphs (3/5) Plotting straight line graphs Consider the equation of a line y = 2x + 4. We can plot this line if we know the data points that the line passes through. To do this we choose 3 values of x and calculate corresponding value of y. This will provide a set of (x, y) values which can then be plotted as illustrated below. Point A D B x 0 2 4 y = 2x + 4 4 8 12 If we plot these co-ordinate points onto a graph of y against x we will create Figure 1.3 Author: Glyn Davis © Oxford University Press, 2012. All rights reserved. Chapter 1: Refresher course in numerical skills 13 Drawing graphs (4/5) Linear equation parameters ‘m’ and ‘c’ Positive gradient, m>0 Negative gradient, m<0 No gradient, m=0 Infinite gradient, m=∞ Author: Glyn Davis © Oxford University Press, 2012. All rights reserved. Chapter 1: Refresher course in numerical skills 14 Drawing graphs (5/5) Plotting non-linear relationships when y is a polynomial of x In certain situations you may find that the relationship between two variables (say y and x) is not a linear (or line) relationship but a non-linear relationship. Figure 1.8 illustrates a graph of y against x for the non-linear equation y = 2x2 – 4x +2 We observe that this curve as a minimum value of the variable y when x is approximately equal to 1. If we know the theoretical relationship between y and x then we can use the concept of calculus to identify the value of x (and y) from this relationship (e.g. for y = 2x2 – 4x +2 we can show that the minimum value of y is 0 when the value of x equals 1). Author: Glyn Davis © Oxford University Press, 2012. All rights reserved. Chapter 1: Refresher course in numerical skills 15 Notes: Solving exactly y = 2x2 – 4x + 2 We can solve exactly the value of x where y = 0 by using the following equation. Given 0 = ax2 + bx + c, then x is given by Equation (1.2): b b 2 4ac x 2a For y = 2x2 – 4x + 2. This implies that a = +2, b = -4, c = + 2. Substituting these values into Equation (1.3) gives: b b 2 4ac 4 x 2a 42 422 4 22 16 16 4 0 1 4 4 This confirms that the value of x when y = 0 is x = 1. The curve meets the y = 0 axis at x = 1 and we say that we have one root (x = 1) to the quadratic equation solution of 2x2 – 4x + 2 = 0. Author: Glyn Davis © Oxford University Press, 2012. All rights reserved. Chapter 1: Refresher course in numerical skills 16 Describe change with calculus (1/5) In Section 1.2.4, we observed from Figure 1.8 that the quadratic equation as a minimum at approximately x = 1. y = 2x2 – 4x + 2. Figure 1.9 illustrates the shape of the curve in which we can observe that a minimum value of y occurs at x = - b/2a = -(- 4)/(2(2)) = 1 with a minimum value of y = 0. Author: Glyn Davis © Oxford University Press, 2012. All rights reserved. Chapter 1: Refresher course in numerical skills 17 Describe change with calculus (2/5) To calculate the gradient of the curve at a particular point on the curve we use the method of differentiation. Differentiation is a method that can be employed to solve problems where a variable is a function of another variable, for example, if y is a function of x then we can write this as y = f(x) e.g. y = x2, y = x3, y = x3 + 3x2 – 6. To differentiate this type of relationship we can use the following rule: If y = axn then the first derivative is given by the equation: dy anx n -1 dx The first derivative enables the calculation of the gradient of the curve y = axn at a particular point on the curve (tangent line). Now reconsider : y = 2x2 – 4x + 2. Now use the rules of differentiation to differentiate y = 2x2 – 4x + 2. Author: Glyn Davis © Oxford University Press, 2012. All rights reserved. Chapter 1: Refresher course in numerical skills 18 Describe change with calculus (3/5) y = 2x2 – 4x + 2 dy 2 * 2x 2 1 4(1)x 11 2 * 0x 01 dx dy 4x 1 4x 0 2 * 0 dx dy 4x 4 dx This equation represents the gradient of the curve y = 2x2 – 4x + 2 at a point on the curve (x, y). Note If x = 0, then the gradient of the curve at x = 0 is equal to 4(0) – 4 = -4. Therefore, the tangent of the line would slope downwards (gradient negative) with a unit increase in x producing a 4 unit decrease in y. If x = 1, then the gradient of the curve at x = 1 is 4(1) – 4 = 0. This tells me that we are now at the minimum value of y. If x = 3, then the gradient of the curve at x = 3 is equal to 4(3) – 4 = 8. Therefore, the tangent of the line would slope upwards (gradient positive) with a unit increase in x producing a 8 unit increase in y. Author: Glyn Davis © Oxford University Press, 2012. All rights reserved. Chapter 1: Refresher course in numerical skills 19 Describe change with calculus (4/5) Finding the minimum and maximum value of a function It can be shown that the minimum or maximum value will occur when the first derivative = 0: dy 0 dx Consider: y = 2x2 – 4x + 2 The minimum (or maximum) value of y occurs when dy/dx = 4x – 4 = 0. If we solve this equation we see that 4x – 4 = 0 only occurs when x = 1. Author: Glyn Davis © Oxford University Press, 2012. All rights reserved. Chapter 1: Refresher course in numerical skills 20 Describe change with calculus (5/5) How to decide if a minimum or maximum value? d2y It can be shown that the second derivative, 2 > 0 for a minimum and < 0 for a dx maximum value. y = 2x2 – 4x + 2 dy 4x 4 dx Minimum or maximum when dy/dx = 4x – 4 = 0. This occurs when x = 0. d2y The second derivative is > 0 at this point and given that it is > 0 then we have a minimum at x = 1. dx 2 4 Substituting this value of x into the equation for y gives a minimum value of y = 2(1)2 – 4(1) + 2 = 0. Therefore, minimum value of y at co-ordinate point (x, y) = (1, 0). Author: Glyn Davis © Oxford University Press, 2012. All rights reserved. Chapter 1: Refresher course in numerical skills 21 Conclusion In this presentation we explored using Excel in solving a range of numerical problems, including: Author: Glyn Davis © Oxford University Press, 2012. All rights reserved. Chapter 1: Refresher course in numerical skills 22