Math review
Fundamental Algebraic Properties of Real Numbers
Definitions
1. The set of integers is {…,-4,-3,-2,-1,0,1,2,3,4,…}
2. Real numbers: All numbers that can be written in decimal form. Examples: 2, -2, 2.2, 2 , 1/7, etc.
Properties of Real Numbers
For real numbers a, b, c
Closure
a+b, ab are real
1+2 = 3, 1×2=2 are real
Commutative
a+b=b+a, ab = ba
1+2=2+1, 1×2=2×1
Associative
(a+b)+c=a+(b+c)
(ab)c=a (bc)
(1+2)+3=1+(2+3)
(1×2)×3=1×(2×3)
Distributive
a(b+c)= ab+ ac
(b+c) a= ba+ ca
3×(1+2)= 3×1+3×2
(1+2)×3= 1×3+2×3
Identity
a+0=a
a×1=a
5+0=5
5×1=5
Inverse
Additive: a + (-a) = 0
Multiplicative: a (1/a)=1
5+(-5) = 0
5×(1/5)=1
Cancellation
If a+b = a+c, then b=c
If ab = ac and a≠0, then b=c
Zero Factor
a×0=0×a=0
ab=0 if and only if a=0 or
b=0, or both
Negation
-1×(-a)= -(-a)=a
(-a)(-b)=ab
(-a)(b)= (a)(-b)=-(ab)
Exponents/Powers of a Number
Let a, b be a real number and m and n be integers: an=aa…a where a is multiplied by itself n
times.
Properties of Exponents
anam = an+m
e.g. 42×45=42+5=47=16,384.
(an)m = anm e.g. (42)5=42×5=410= a really big number.
anbn = (ab)n e.g. 22×32 =(2×3)2=62=36
a-1 = 1/a e.g. 2-1 = ½ = 0.5
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Note that these properties also hold if n and m are real numbers, but the solution can be what is
called a complex number, which we will never have to deal with in this class.
A set of two linear equations
Linear function all have a common form y = a + bx. a is called the intercept because it
the value of y for x = 0. b is called the slop, which tells us how fast y changes as x
changes: b = ∆y/∆x (can be seen the following way:
y1 = a + bx1
y 2 = a + bx 2
Subtract the second equation from the first one
y1 − y 2 = b ( x1 − x 2 )
Rearrange:
y1 − y 2
= ∆y
=b
x1 − x 2
∆x
1. When b > 0 x and y are directly related: when x increases, y increases.
2. When b < 0 x and y are inversly related: when x increases, y decreases.
3. When b = 0, y does not change at all.
1
y
3
2
x
Solution
A solution for two linear equations is a point in the plane that describes the intersection
of two lines.
Often, we are interested in solving for the intersection of two lines. Consider the
following equations:
1. y = a1 + b1x
2. y = a2 + b2x
to find the point (x,y) that satisfy the two equations, we use algebraic properties:
0 = (a1 − a 2 ) + (b1 −b2 ) x
subtracting 2. from 1.
− (a1 − a 2 ) = (b1 −b2 ) x
subtracting (a1 − a 2 ) from both sides
a − a1
dividing both sides by b1 − b2 .
x= 2
b1 − b2
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To find y substitute
a 2 − a1
for x in one of the equations.
b1 − b2
a 2 − a1
. We can find commen denominator by multiplying and dividing the
b1 − b2
first term by b1 − b 2, ,
a − a1 a1 (b1 − b2 ) + b1 (a 2 − a1 ) a 2 b1 − a1b2
b −b
y = a1 2 1 + b1 2
=
=
b1 − b2
b1 − b2
b1 − b2
b1 − b2
y = a1 + b1
The meaning of no solution
Note that if b1 = b2, then we can not find a solution for x. The two lones described by the
linear equations are parallel and do not intersect.
Example: market equilibrium
A demand for a good is an increasing line Q D = 120 − 2 P , the supply for the good is a
decreasing line Q S = 60 + P Market equailibrium is the intersection od demand and supply.
Solution: Q = 80, P = 20.
Elemantary Calculus
The definition of the derivative
Example:
An acquisition officer observed the cost of similar defense contracts for the past 10 years.
He is interested in the rate of change in the costs for contracts that last 2 years. The
following table depicts the values he obtained.
Duration in months
Cost in
$1,000
10
33
20
79
30
165
40
236
50
362
60
491
Using these values, he plotted the points on a graph of COST versus DURATION as
represented by the red dots in the illustration below.
COST
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DURATION
Please refer to the graphing functions station for methods for deriving a function from a
graph.
From the graph, he was able to find a function (illustrated by the blue line,
above) which represents the graph of cost c versus time t. The officer
found that the function which best fits the points on the graph is
c = 0.1 t2 + 2t
With this function, he was able to compute the rate of change in cost with respect to
duration by using the concept of derivatives. The acquisition officer took the derivative c'
of the function c
c' = 0.2 t + 2
With the derivative c', he can now determine the rate of change of cost of a contract as it
increases beyond two years or 24 months. The derivative c' at time t = 24 months given
in $1000's per month is then
c' = 0.2 × 24 + 2 = 2.48
Definition:
What is this derivative and how does it work? Let us see what the definition of derivative
is. A derivative f'(x) of a function f depicts how the function f is changing at point x. It is
necessary for the function f to be continuous at point x in order for there to be a
derivative at that point. A function which has a derivative is said to be differentiable.
The derivative is computed by using the concept of x. x is an arbitrary change or
increment in the value of x. You can see that if x in the example above is 10 months,
you can readily compute the rate of change in the function, but the officer wanted to
know the rate of change at a particular point x, not a span of 10 months. The derivative is
the limit approached by the rate of change in the function when x becomes arbitrarily
small.
To understand this idea of x consider the group of 10,000 marching men depicted
below:
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If we add one more man to the group you will probably not be able to notice the
difference, because the addition of one man is very small when compared to 10,000 men.
If we can measure the rate of change in a function is smaller and smaller increments, we
will approach the rate of change at a point. In general, the derivative of function f(x),
also called dy/dx can be defined as
This means that as the x gets very small, the difference between the value of the
function at x and the value of the function at x + x divided by x is the derivative.
How do we calculate a derivative?
Give x an arbitrary change,
f(x +
x, and compute the new value of the function.
Subtract the original value of the function, y = f(x), from the new value, y +
x), to obtain the change in the function.
Divide the change in the function,
rate of change.
y, by the change in x,
y=
x, to obtain the average
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Determine the limit, dy/dx, of the average rate of change of the function as
approaches zero
x
Derivatives of some common functions
Fortunately, we don't have to follow the steps on the previous page in order to
differentiate most functions since there are differentiation rules that can be used.
dy/dx c = 0
The derivative of a constant is zero.
Example: dy/dx 7 = 0
dy/dx c × x = c
The rate of change of a linear function is its slope.
Example: dy/dx 3 × x = 3
dy/dx (xn) = n × x(n-1)
Example: dy/dx (x4) = 4 × x 3
dy/dx (log x) = 1/x
The derivative of the log of x is its inverse.
Example: dy/dx (log (x + 1)) = 1 / (x + 1)
dy/dx (eax) = a eax
Example: dy/dx (e3x) = 3 e3x
dy/dx (sin x) = cos x
Example: dy/dx (sin y) = cos y
dy/dx (cos x) = -sin x
Example: dy/dx (cos ) = - sin
By using the derivative rules in combination, we can find the derivatives of many other functions.
Do you want to see the graphs of the derivatives of a few functions?
Graphing interactive workbench
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Determining Maximum and Minimum
Example:
George had a function of stresses on a particular airplane part with respect to time in
flight. He wished to determine the time when the part was under the most stress.
Given the function
f(t) = -t2 + 2 t + 3
George took the derivative of the function
f'(t) = -2t + 2
then determined that the derivative was equal to zero at t
The time of most stress was 1 minute into the flight.
= 1.
One of the convenient uses for the derivative of a function is finding the maximum or
minimum points of the function. The example on the definition page of this station shows that the
derivative indicates the rate of change of a function. Put another way, the derivative measures the
slope of the function at a particular point.
When the slope changes from positive to negative, the function is at its maximum
when the slope is zero. When the slope changes from negative to positive, it is at its
minimum when the slope is zero.
General laws of differentiation
Here are some basic laws which can be used to derive other differentiation rules.
The derivative of a constant is zero.
If a function does not vary (is constant), its rate of change is zero.
More formally, if y = c (a constant), then
Example:
y=5
dy/dx = d/dx (5) = 0
The derivative of the product of a constant and a function is equal to the
constant times the derivative of the function.
That is, if y = c u, then
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Example:
y=8×x
dy/dx = d/dx (8 × x)
dy/dx= 8 × d/dx (x)
dy/dx= 8 × (1) = 8
The derivative of the sum or difference of two functions is equal to the sum or
difference of the derivatives of the functions
Example:
y = 8 × x - x^2
dy/dx = d/dx (8 × x - x^2)
dy/dx= 8 × d/dx (x) - d/dx (x^2)
dy/dx= 8 - 2 × x
The derivative of the product of two functions is equal to the first function
times the derivative of the second, plus the second function times the derivative of
the first
That is, if y = u v, then
Example:
y = x ex
dy/dx = d/dx (x ex) = x d/dx (ex) + ex d/dx (x)
dy/dx = x ex + ex = ex (x + 1)
The derivative of a fraction, that is, the quotient of two functions, is equal to
the function in the denominator times the derivative of the function in the
numerator, minus the function in the numerator times the derivative of the function
in the denominator, all divided by the square of the function in the denominator
That is, if y = u/v, then
Actually, we can find the derivative of the quotient of two functions in an easier way. If
we consider the equation y = u/v as y = u v-1 and then apply the product rule of
derivatives (above) we get the same result.
Example:
y = ex/x
y = ex × x-1
Now applying the product rule, we have,
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dy/dx = ex d/dx (x-1) + x-1 d/dx (ex)
dy/dx = ex (-1)x-2 + x-1 ex
dy/dx = ex(- x-2 + x-1)
If the derivative of y = f(x) is dy/dx, then the derivative of the inverse function
which expresses x in terms of y is given by the formula
If y = f(u) and u = g(x), that is, if y is a function of a function, then
If y = f(t) and x = g(t), that is, if y and x are related parametrically, then
Example:
y = t3 and x = 2 t2
dy/dx = (dy/dt) / (dx/dt)
dy/dt = d/dt (t3) and dx/dt = d/dt (2 t2)
dy/dt = 3 t2 and dx/dt = 4 t
dy/dx = (3 t2) / 4 t
dy/dx = 3 t / 4
partial derivative
(Definition)
The partial derivative of a multivariable function is simply its derivative with respect
to only one variable, keeping all other variables constant (which are not functions of
the variable in question). The formal definition is
where
is the standard basis vector of the th variable. Since this only affects the th
variable, one can derive the function using common rules and tables, treating all other
variables (which are not functions of
) as constants. For example, if
, then
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Note that in equation
respect to
20
.
, we treated
as a constant, since we were differentiating with