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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
 42  422  4 

22
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 11  2 * 0x 01
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