| BJØRN EINARSEN, 2006: | | |
| FIRST AID KIT IN MATHEMATICS
FOR Faculty of Business, Public Administration and Social Work, Oslo University College |
MESSAGES | TIME TABLES | 
 |
| CONTENTS: | More printerfriendly issue |   |
Lectures Autumn 2006 | Spring 2006 | Curriculum | Motto |  Problems
to solve |
Spice |
This text has been used as reading in a previous introductory
course in mathematics at Local Government Administration program at Oslo
University College.
The text might be used as a toolbox or first aid kit
in the present course in Introductory Matematics
READINGS IN INTRODUCTORY COURSE
IN MATHEMATICS AUTUMN 2005:
*Bjørnestad, Olsson,
Søyland and Tolcsiner: Matematikk for økonomi og samfunnsfag.
Høyskoleforlaget, latest edition with solutions.
*This text in paper or as an internet site, part 1 & 2 Download First Aid Kit part 2
OTHER TEXT AND LITERATURE:
*Risnes, Martin: Matematikk for økonomer, Univ.forlaget,
siste utgave.
*Sydsæter, Knut: Elementær algebra og funksjonslære,
Univ.forl., siste utg.
*Sydsæter, Knut: Matematisk analyse, bind 1, Univ.forlaget
1994
Roman numerals: I, II, III, IV, V, VI, VII, VIII, IX, X,...
binary digits: 0000, 0001, 0010, 0011, 0100, 0101,..(for computers)
Arabic numerals: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, ...
N = set of natural numbers i.e whole positive numbers N = { 1, 2, 3, ....}
Z = set of whole numbers, positive and negative Z = { ...., -2, -1, 0, 1, 2, ...}
Q = set of rational numbers
(numbers that might be written as fractions with whole
numbers in nominator and denominator, see 2a.)
R = set of real numbers for instance 1, 1.5, Ö 2
R - Q = set of irrational numbers
(numbers that might not be written as fractions with
whole numbers in nominator and denominator, for example Ö
2 = 1,41421356237.....).
N Í Z Í Q Í R
(N is a subset in Z that is a subset of Q, that is a subset of R)
intervals
<-2, 4> the interval between -2 and 4, but except -2 and 4.
[-2, 4] the interval from -2 to 4 including -2 and 4.
SOME EXPLANATIONS
= means 'equals'
:= could mean 'make equal to'.
(You see, in computer programming it is necessary to
distinguish between the static 'equals' and the dynamic 'make equal to')
==> means 'leads to' or 'implies'
<== means 'is a consequence of'
<==> means 'is equivalent to'
example: a <== b means 'a is a consequence of b', but could also mean 'b implies a'. This is an example that shows that mathematics is an international language and that we can choose whether we read from left to right in a Western way or from right to left in Arabic way!
OPERATIONS ON NUMBERS
In Norway we usually follow continental European conventions of decimal comma , but in this English version of our First Aid Kit in Mathemathics we will try to follow English/American conventions and use decimal point.
Addition 1 + 2 = 3
Subtraction 3 - 2 = 1
Multiplication 4 * 3 or 4 x 3 = 12
Division 4 / 3 or 4 : 3 = 1.3333
POWERS FOR STARTERS
Numbers are multiplied by itself a certain number of times: a*a = a2, a*a*a = a3 and so on
SQUARE OF BINOMIALS
(a+b)2 = a2 + 2ab + b2
(a-b)2 = a2 - 2ab + b2
(a+b)*(a-b) = a2 - b2
| 2a. FRACTIONS |  |
 |
|
|
40 40 is nominator
fraction line signifies the same as division sign
5 5 is denominator
TO EXPAND a fraction is to multiply nominator and denominator by the same number, for instance:
TO REDUCE a fraction is to divide nominator and denominator by the same number, for instance:
Percent or " per cent" means "per hundred". Thus
20 percent or 20% means
20
Thus 20 percent of 250 means that
100.
20
* 250 = 50.
100
20
you may
write as 0.2 so the expression above you may write as 0.2 * 250 = 50.
100
10 percent you may write as 0.1
20 percent you may write as 0.2
30 percent you may write as 0.3
40 percent you may write as 0.4
50 percent you may write as 0.5
100 percent you may write as 1 and so on
TO COMPUTE A PERCENTAGE CHANGE IS TO COMPUTE THE DIFFERENCE AS PERCENTAGE OF THE INITIAL VALUE
For instance, the population in a municipality increases by 250 from 5000 to 5250. This is
250
*100 = 5 % increase, and not 4.76 %.
5000
TO ADD A CERTAIN PERCENTAGE TO A NUMBER
To add 10 percent means that we want 100 percent
+ 10 percent = 110 percent.
110 percent is the same as
110
= 1.1.
100
This means that if you want to add 10 percent to a number, you may simply multiply the number by 1.1.
250 * 1.1 = 275
IF YOU WANT TO ADD - YOU MUST MULTIPLY BY:
5 percent
1.05
10 percent
1.1
15 percent
1.15
50 percent
1.5
75 percent
1.75
100 percent
2 and so on
TO SUBTRACT A CERTAIN PERCENTAGE FROM A NUMBER
To subract 10 percent from a number means that we want
100 percent - 10 percent = 90 percent.
90 percent is the same as
90
= 0.9.
100
This means that if you want to subtract 10 percent from a number, you simply multiply the number by 0.9.
250 * 0.9 = 225.
IF YOU WANT TO SUBTRACT - YOU MUST MULTIPLY BY
5 percent
0.95
10 percent
0.9
15 percent
0.85
20 percent
0.8
30 percent
0.7
50 percent
0.5
75 percent
0.25
90 percent
0.1 and so on
a to the fifth power is a multiplied by itself 5 times;
a*a*a*a*a = a5. a is the base, 5 is the exponent.
COMPUTATION RULES FOR POWERS 
am*an = am+n
(a*b)m = am * bm
(am)n = am*n
Let a be a positive number and let n be a natural number. The equation
Xn = a has the solution X = ±
,
i.e. plus/minus nth root of a. If the root index n = 2, we usually
write 
instead of 
.
QUADRATIC ROOT
X2 = a ==> X = ± 
The solution to the equation X2 = 4 is X =±
= ± 2 because (+2)*(+2)= 4 and
(-2)*(-2)= 4. The quadratic root of 4 is +2.
CUBIC ROOT
X3 = a ==> X =
example: The solution to the equation X3 =
8 is x = 2 because 2*2*2 = 8.
2 is the cubic root or 3rd root of 8.
nTH ROOT.
Xn = a has the solution X = ±
If the root index is an even number there might be two solutions. If the root index is an uneven number the radicand might be a negative number.
Xn = a ==> X = ±
ROOT EXPRESSIONS MIGHT BE WRITTEN AS POWERS:
± = 
(This is useful if you want to find the 5.th root of 3125 on your calculator)
POWER COMPUTATION AND ROOT ARITHMETIC ARE TWO OPPOSITE OPERATIONS
= a
REMEMBER: COMPUTATION RULES FOR ROOTS ARE THE SAME AS
FOR POWERS
| ONE UNKNOWN | ONE UNKNOWN WITH FRACTION | TWO UNKNOWN | THREE UNKNOWN | |
x*3 = x + 8
To solve an equation in one unknown is to find the value(s) for the unknown (for instance x) that satisfies the equation.
A condition for solving the equation is to isolate the unknown on the left side. The two sides of an equation is often referred to as left hand side (L) and right hand side (R).
In order to isolate the unknown in an equation in one unknown you may use the addition rule and/or the multiplication rule
The addition rule says that we may add or subract the same number (or value) on both sides.
3 = 3 ==> 3+4 = 3+4
3 = 3 ==> 3-2 = 3-2
The multiplication rule says that we may multiply or divide by the same number on both sides.
3 = 3 ==> 3*4 = 3*4
3 = 3 ==> 3:2 = 3:2
Let us find x in this equation
x*3 = x + 8 ==>
Vi subtract x from both sides:
x*3 - x = x - x + 8 ==>
x*3 - x = 8 ==>
2*x = 8 ==>
Vi divide by 2 on both sides:
2*x:2 = 8:2 ==>
x = 4
L = { 4 }
==============
EQUATIONS WITH TWO UNKNOWN, X AND Y
In order to solve an equation in two unknown, we need two equations I and II, for instance
I 2*X + 3*Y = 19
II X + Y = 7
A solution method is called the substitution method. We find X expressed by Y from equation II:
X = (7 - Y) and we can substitute (7-Y) for X in equation I.
I 2*(7-Y) + 3*Y = 19 ==>
14 - 2*Y + 3*Y = 19 ==>
Y = 19 - 14 = 5 ==>
X = (7 - 5) = 2.
EQUATIONS WITH THREE UNKNOWN, X, Y AND Z
Use the substitution method, i.e you express either
X, Y or Z as function of the two other variables. Substitute the
variable you just found into the two other equations, and the problem is
reduced to two equations in two unknown. Maybe an even better
approach is to use a method to eliminate one of the variables X,
Y or Z. Then you can use the multiplication rule on one or more of
the three equations. You should also know that equations in a set
of equations might be added together.
Example.:
I) x +2y +
3z = 140
II) -6x +7y + 8z = 320
III) x + y -
z = 0
Here we might like to eliminate the variable x by multipling
equation I and III by 6.
I) 6x +12y + 18z =
840
II) -6x +7y + 8z = 320
III) 6x + 6y - 6z
= 0
I +II give 19y + 26z =1160 (IV)
II +III give 13y + 2z =320 (V)
Here we might eliminate z by multipling equation
V with 13.
IV) 19y + 26z =1160
V) 169y + 26z =4160
V -IV give 150y = 3000 ==> y=3000/150=20
Equation IV gives us 19(20) +26z =1160 ==> 26z=1160-380
==> z=780/26=30
Equation I gives us x +2(20) +3(30) = 140 ==>x
= 140 -40 -90 = 10
If you think that it was a little strange just to add two equations together, just remember that this is the same as using the addition rule saying that we may add or subtract the same number (or value) on both sides of an equation!
COMPUTING WITH LETTER CONSTANTS
In some equations you may see that letters are used as constants. In such cases you must first find out what variable is the unknown, for instance X and treat all other numbers and letters as constants. For instance: Solve the equation with respect to X, i.e Find X when
X*3 = X + q ==>
X*3 -X = q ==>
2*X = q ==>
X = (q/2)
3*x -7 < 7 - x
> means greater than
< means less than
>= means greater than or equal to
<= means less than or equal to
3 < 5 means 3 is less than 5
A condition for solving an inequality is to isolate the unknown on the left side. The two sides of an inequality are often referred to as left hand side (L) and right hand side (R).
In order to isolate the unknown in an inequality you may use the addition rule. You may also use the multiplication rule, but remember to reverse the inequality sign, if you multiply or divide by a negative number.
The addition rule says that we may add or subtract the
same number or value on both sides.
3 < 4 ==> 3+4 < 4+4
3 < 4 ==> 3-2 < 4-2
The multiplication rule says that we may multiply or divide by the same number on both sides.
3 < 4 ==> 3*4 < 4*4
3 < 4 ==> 3:2 < 4:2
Look, if you multiply or divide by a negative number, the inequality sign must be reversed.
3 < 4 ==> 3*(-4) > 4*(-4)
3 < 4 ==> 3:(-2) > 4:(-2)
The solution of
3*x -7 < 7 - x ==>
We add x on both sides:
3*x + x - 7 < 7 - x + x ==>
4*x -7 < 7 ==>
We add 7 on both sides:
4*x - 7 + 7 < 7 + 7 ==>
4*x < 14 ==>
Vi divide both sides by 4:
4*x : 4 < 14 : 4 ==>
x < 3.5
L = < ¬ , 3.5 >
If you have a quadratic equation like this:
a*x2 + b*x + c = 0
you may find the solution in the following manner:
| 6. FUNCTIONS | Draw f(x) | Draw f(x) and g(x) | Function shapes | |
Y = f(x) x is independent variable, Y is dependent variable
In economics we might use Q for quantity rather than the more anonymous X and measure the dependent variable in pounds, dollars, euros or kroners .
LINEAR FUNCTIONS (FIRST DEGREE FUNCTIONS)
The simplest functions we know are of the form Y = a
+b*x. for instance, Y = 0 +3x i.e. Y = 3x. This
means that Y= 0 when x=0, that Y=3*1 = 3 when x=1, that Y = 3*2 =
6 when x = 2 and so forth.
What can this be? Assume that one apple costs 3 kroners.
Then Y =3x may be a function showing what x apples cost if one apple costs
3 kroners. To find out what 7 apples cost, we use the formula
Y = 3*7 = 14, or we may read the graph of the function
at x=7.
 |
Df = domain,
i.e. where x is defined, maybe between 0 and 100. Rf = range,
|
FUNCTIONS OF ZERO DEGREE.
Functions of zero degree are the same as constant functions,
i.e. they are of the form Y= c, for instance Y=100. The fact that
the function is parallell to the x-axis shows that x doesn't influence
Y.
The function Y=100 may express fixed costs. For
instance, we do not expect the housing costs for a hot dog stand to rise
if additional hot dogs are sold.
SECOND DEGREE FUNCTIONS (QUADRATIC FUNCTIONS)
Second degree functions are of the form Y = a*x2
+b*x + c, i.e a function where the highest power is 2,
for instance Y = 3 +x2:
FUNCTIONS OF THIRD AND HIGHER DEGREES
Functions where the highest power is 3, we call third
degree functions. Functions where the highest power is 4, we call forth
degree functions and so forth.
WHEN THE FUNCTION IS A FRACTION
We may experience functions that are fractions, for instance
Y = (x+6)/(2x-1). Notice that the graph of such functions split in two where the
denominator equals zero. Remember that we should avoid dividing by
zero. If we forget this, we get function values equal to + or - eternity
(+
or -).
The very fact that functions often are models of reality,
it may be very interesting to study functions further. We therefore
like to do function analysis.
More about that in FIRST AID KIT IN MATHEMATHICS, part II.
LOGARITMS
Assume the function f(x) = 10x.
This function shows the base 10 powered by the exponent x. If this function is drawn, you might see that f(x) = 5 if x is ca. 0.699. The number 5 might be expressed as the base 10 in the power of 0.699. The exponent 0.699 is called the LOGARITM of 5, or LOG(5) = 0.699, because 100.699 = 5.
Maybe you can't find the function LOG on your calculator
, but instead the function LN. This is the natural logaritm. (The natural)
logaritm of 5 is 1,609437912 that is the number e = 2,718281828 must be
powered by to make the number 5. I.e. 2.7182818281.609437912
= 5.
If you want to find out why this number e was chosen,
see Risnes(1989).
COMPUTING RULES FOR LOGARITMS:
LN(X*Y) = LN(X) + LN(Y)
One advantage of logaritms is the possibility to solve equations of this type:
In a municipality the population decreases by 6% per year. Should this decrease go on, how many years would it take till the population is just half of what is is today?
Let us assume that present population is 5000. We must find x = number of years in this equation:
5000 * (0.94)x = 5000/2 ==>
5000 * (0.94)x = 2500 ==>
(0.94)x = 0.5 ==>
LN( (0.94)x ) = LN(0.5) ==>
LN(0.5)
x * LN(0.94) = LN(0.5) ==> x = ¾¾¾¾
= 11.2
LN(0.94)
It will take little more than 11years before the population
in this municipality is half of what it is today.
an = a1 + (n-1)*d
(Find nth term in an arithmetic succession)
(a1
+ an)
Sn =
* n (Find the sum of the
n first terms in an arithmetic succession)
2
The succession 1, 3, 5, 7, 9, 11 is an arithmetic succession with a1 = 1 and d = 2.
Usage: interest on serial loan
an = a1 * kn-1 (Find
nth term in a geometric succession)
kn - 1
Sn = a1 *
(Find the sum of the n first terms in a geometric succession)
k - 1
a1
= (Find
the sum of an eternal geometric succession)
1 - k
(The succession must converge. -1 < k < +1)
The succession 1, 2, 4, 8, 16, 32, 64 is a geometric succession with a1 = 1 and k = 2.
Usage: interest, discount, annuity, investment
analysis, depreciation.
9. DIFFERENTIATION - FIND
THE NEWTON QUOTIENT FOR FUNCTIONS
This is a tool that is very useful if you want to study or analyse functions. (Functions may be seen as models of the real world) For instance we might want to see how functions increase or decrease and we might want to find the maximum and the minimum of a function.
DIFFERENTIATION OF A CONSTANT
f(x) = c ==> f'(x) = 0
example: f(x) = 4 ==> f'(x) = 0.
(The derivative of a constant is 0)
DIFFERENTIATION OF A POWER
f(x) = xn ==> f'(x) = n * x n-1
example: f(x) = x4 ==> f'(x) = 4*x3
f(x) = c * xn ==> = n * c * x n-1
example: f(x) = 5*x3 ==> f'(x) = 3*5*x3-1
= 15*x2
TERM BY TERM RULE:
f(x) = g(x) + h(x) ==> f'(x) = g'(x) + h'(x)
example: f(x) = g(x) + h(x) = 7*x3 + 5*x9 ==>
f'(x) = 3*7*x3-1 + 9*5*x9-1 = 21*x2+
45*x8
DIFFERENTIATION OF A PRODUCT
f(x)= g(x) * h(x) ==> f'(x) = g'(x)*h(x) + h'(x)*g(x)
example: f(x) = g(x) * h(x) = (6 + 7*x3) * (3 - 5*x9) ==>
f'(x) = 3*7*x3-1 * (3 - 5*x9) + (6 + 7*x3) * (-9*5*x9-1) ==>
f'(x) = 21*x2 * (3 - 5*x9) + (6 + 7*x3) * (-45*x8) ==>
f'(x) = 63*x2 - 105x11 -270*x8 - 315*x11
f'(x) = 63*x2 - 420x11 -270*x8
DIFFERENTIATION OF A FUNCTION-FUNCTION ( CHAIN RULE)
f(x) = g(h(x)) ==> f'(x) = g'(h)*h'(x), f.eks:
f(x) =
==>
f'(x) =
First you differrentiate the most exterior function with respect to h, then multipy by the derivative of the inner function h(x).
DIFFERENTIATION OF A QUOTIENT
g(x)
g'(x)*h(x) - h'(x)*g(x)
f(x) =
==> f'(x) =
h(x)
(h(x))2
In principle this is an redundant rule as we may use the product rule and the chain rule. Remember that
g(x)
= g(x) * (h(x))-1
h(x)
| 10. FUNCTIONS IN ECONOMICS | |
|
STK-funk.xls | Abbrevations we should know about | |
Suppose we have total cost TC(Q) = 0.0002*Q3
- 0.15*Q2 + 240*Q + 24000
Then
marginal cost MC = TC'(Q) = 0.0006*Q2 - 0.3*Q + 240
total variable cost TVC = 0.0002*Q3 - 0.15*Q2 + 240*Q
total fixed cost TFC = 24000
average cost ATC = TC/Q = 0.0002*Q2 - 0.15*Q + 240 + (24000/M)
average variable cost AVC = TVC/Q = 0.0002*Q2 - 0.15*Q + 240
These functions are all continous in the actual intervals.
A function's maximun or minimum may be found where the
1st derivative of the function equals zero. If the second derivative
of the function is less than zero we have a maximum.If the second derivative
of the function is above zero we have a minimum. If the second derivative
of the function is equal to zero we have an inflextion point. 
To find Q for mimimum AVC, differentiate the AVC-function and make it equal to 0. The solution for Q is the quantity that gives minimum AVC. If this Q-value is used in the AVC function, we find minimum AVC
To find Q for mimimum ATC, differentiate the ATC-function and make it equal to 0. The solution for Q is the quantity that gives minimum ATC. If this Q-value is used in the ATC function, we find minimum ATC
Suppose that total revenue TR(Q) = 300*Q, i.e. we have a constant market price = 300 kr.
Then
marginal revenue MR = 300.
Q for max profit is found where MC = MR
<==> 0.0006*Q2 - 0.3*Q + 240 = 300 ==> Q for max profit = 653
You may also comute a profit function,
(Q) = R(Q) - TC(Q)
==>
(Q) = 300*Q - 0.0002*Q3
+ 0.15*Q2 - 240*Q - 24000
If you differentiate this function and set the first derivative
equal to zero, you may find Q for max profit. The maximum profit
is found by putting this value for Q in the profit function.
Remember that max profit is no guarantee for profit being a positive
number. It could mean a situation with minimum loss!
11. SIMPLE INTEGRATION
OR ANTI-DIFFERENTIATION (WAITING LIST)
For a given function f(x) we wish to find a function F(x)
that has this characteristic: F'(x) = f(x).
In other words we want to find a function F(x) that when
differentiated gives us the f(x) function that we started with. We
call F(x) for the
anti-derivative, or the integral of
f(x). Integration and differentiation are two opposite or reverse
computing operations.
ò f or ò f(x) dx you read as the undecided integral of f(x).
"dx" in the expression above does not have any meaning by itself, it is simply part of the notation for an integral.
(n =/= -1)
Example:
Suppose that the marginal cost MC = 10 + 5*Q,
What do we know about the total variable cost TVC?
Control: F(Q) = 10*Q +(5/2)*Q2 ==> F'(Q)
= 10 + 5*Q = f(Q).
APPLICATION OF INTEGRALS: SURFACE AREA
If we draw the marginal cost MC we may find that the TVC for a certain quantity is the area under the MC-curve between 0 and the quantity.
Let us compute TVC at Q=100 units as the area under the
MC-curve betwen 0 and 100 units:
(q.e.d)
