PRELIMINARY MATHEMATICS

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PRELIMINARY MATHEMATICS

• LECTURE 4
• Exponentials and Logarithmic Functions

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Reading

• Chiang, A. C. and K. Wainwright (2005)
  Fundamental Methods of Mathematical Economics,
  Forth Edition. McGraw-Hill. Chapter 10.

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Exponential functions
Exponential functions An exponential function is
defined as f (x) b x b gt 0 and b ? 1
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Exponential functions
Exponential functions An exponential function is
defined as f (x) b x b gt 0 and b ?1
Contrast this with the power function, f (x)
a xn . In power functions x was raised to a
exponent n. In exponential functions a is raised
to a exponent x.
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Exponential functions
For example let us consider the exponential
function y 10 x.
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Exponential functions
As can be seen from the graph, the exponential
curve is monotonically increasing, and y
increases at an increasing rate throughout. In
other words, constant increase in the independent
variable x lead to rapidly rising increase in y.
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Logarithmic functions
Inverse function When the roles of the dependent
and independent variables of the original
function is reversed, we have an inverse
function. Logarithmic (Log) functions are
inverse function of exponential functions.
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Logarithmic functions
Inverse function When the roles of the dependent
and independent variables of the original
function is reversed, we have an inverse
function. Logarithmic (Log) functions are
inverse function of exponential functions. For
example, if we have the exponential function
by bringing x to the left hand side
and equating it with a function of y,
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Logarithmic functions
The graph of
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Logarithmic functions
As with the exponential function, the curve is
monotonically increasing, but the log curve
increases at a decreasing rate.
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Exponential and logarithmic functions
Note that the graph of is
simply the graph of
re-plotted with the two axes transposed.
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Logarithmic
The meaning of logarithm (log) When we have two
numbers such as 4 and 16, which can be related to
each other by the equation 42 16, we define the
exponent 2 to be the logarithm (or log) of 16 to
the base of 4, and write if 42 16 log 416 2
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Logarithmic
The meaning of logarithm (log) When we have two
numbers such as 4 and 16, which can be related to
each other by the equation 42 16, we define the
exponent 2 to be the logarithm (or log) of 16 to
the base of 4, and write if 42 16 log 416 2
Note that the log of 16 to the base 4 is nothing
but the power to which a base (4) must be raised
to attain a particular number (16). In general,
if
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Logarithmic
Common log The base of the logarithm, bgt1 does
not have to be restricted to any particular
number, but in many cases, two numbers are widely
chosen as bases the number 10 and the number e.
When 10 is the base, the logarithm is known as
the common logarithm, symbolised by or
when the context is clear simply by log.
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Logarithmic
Common log A common logarithm is simply the
power or exponent to which 10 has to be raised to
equal any given number log 10 1000 3
because 10 3 1000 log 10 100 2 because 10
2 100 log 10 10 1 because 10 1 10 log
10 1 0 because 10 0 1 log 10 0.1 1
because 10 1 0.1 log 10 0.01 2 because
10 2 0.01
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Logarithmic
Natural log When e is the base, the logarithm is
known as the natural logarithm, denoted as ln or
log e. The illustration of natural logarithms
ln e3 log e e3 3 ln e2 log e e2 2 ln
e1 log e e1 1 ln 1 log e e0 0
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Logarithmic
What is e? The approximate value of e is
2.71828. This is a number which is defined as
the limit expression of
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Logarithmic
What is e? The approximate value of e is
2.71828. This is a number which is defined as
the limit expression of What do we mean by
this? As we make the number of m larger and
larger, the quantity of the expression
becomes smaller and smaller.
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Logarithmic
What is e? The approximate value of e is
2.71828. This is a number which is defined as
the limit expression of What do we mean by
this? However, because we are adding 1 to this,
the whole expression in the bracket
is greater than 1.
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Logarithmic
What is e? The approximate value of e is
2.71828. This is a number which is defined as
the limit expression of What do we mean by
this? And since we have the variable exponent,
as m becomes larger and larger, the value of
the whole of becomes
larger and larger as well.
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Logarithmic
What is e? The approximate value of e is
2.71828. This is a number which is defined as
the limit expression of What do we mean by
this? As m increases, and as m tends to
infinity, tends to a number e, approximated
by 2.71828.
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Logarithmic
What is e? The approximate value of e is
2.71828. This is a number which is defined as
the limit expression of What do we mean by
this? Mathematically, as or we can say that
e is the limit of as m
tends to infinity, in mathematical shorthand
written as
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Logarithmic
What is e? The approximate value of e is
2.71828. This is a number which is defined as
the limit expression of Why is this important?
Why is e widely used as the base of
logarithmic in Economics? Why is the limit of
important? ?
Formula for (continuous) interest compounding
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Rules of logarithmic
Rule 1 (log of a product) or (u, v
gt0) This corresponds to the product of
exponentials with the same base bm ? bn b m
n
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Rules of logarithmic
Rule 1 (log of a product) Example
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Rules of logarithmic
Rule 2 (log of a quotient) Again this
relates to the algebra rule whereby to divide
numbers with the same base, subtract the
exponents. (x?0)
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Rules of logarithmic
Rule 2 (log of a quotient) Example
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Rules of logarithmic
Rule 3 (log of a power) Contrast
with (b m ) n b mn
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Rules of logarithmic
Rule 3 (log of a power) For example
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Rules of differentiation
Differentiation of natural exponential function
Given
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Rules of differentiation
Differentiation of natural exponential function
Given where is a differentiable
function of x, the derivative is equal to the
original natural exponential function times the
derivative of the exponent.  
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Rules of differentiation
Differentiation of natural exponential function
Given where is a differentiable
function of x, the derivative is equal to the
original natural exponential function times the
derivative of the exponent.   For example,
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Rules of differentiation
Differentiation of natural exponential function
Differentiate By the product rule,
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Rules of differentiation
Differentiation of natural logarithmic function
Given
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Rules of differentiation
Differentiation of natural logarithmic function
Given Where is positive and
differentiable, the derivative is the derivative
of the original function divided by the original
function
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Rules of differentiation
Differentiation of natural logarithmic function
Given Where is positive and
differentiable, the derivative is the derivative
of the original function divided by the original
function For example,
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Basic compound interest
Under compound interest, interest payments are
made each year not on the original sum, but on
the current value of the original investment,
i.e. the original investment plus the accumulated
interest.
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Basic compound interest
Under compound interest, interest payments are
made each year not on the original sum, but on
the current value of the original investment,
i.e. the original investment plus the accumulated
interest. If the original/principal value is P0,
and the annual interest rate is i, after one
year, the value of the investment will be P1
P0 i P0 P0 (1 i)
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Basic compound interest
Under compound interest, interest payments are
made each year not on the original sum, but on
the current value of the original investment,
i.e. the original investment plus the accumulated
interest. If the original/principal value is P0,
and the annual interest rate is i, after one
year, the value of the investment will be P1
P0 i P0 P0 (1 i) After the second
year, P2 P1 i P1 P0 (1 i) i
P0 (1 i) P0 (1 i) (1 i) P0 (1
i)2
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Basic compound interest
A given principal P0 compounding annually at an
interest rate i for a given number of years t
will have a value Pt at the end of that time
given by the exponential function Pt P0 (1
i) t Formula for basic compound interest
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When interest is compounded more than once a year
If interest is compounded more than once a year
the value of the investment will rise even more
rapidly. Recall that if the interest was
compounded once a year, Pt P0 (1 i)
t For example, if interest is paid every 6
months, there will be two interest payments a
year, but at one half the annual rate. Hence the
compound interest formula becomes
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When interest is compounded more than once a year
This is because 6 months after the
investment,
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When interest is compounded more than once a year
This is because 6 months after the
investment, after one year,
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When interest is compounded more than once a year
This is because 6 months after the
investment, after one year,
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When interest is compounded more than once a year
This is because 6 months after the
investment, after one year,
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When interest is compounded more than once a year
This is because 6 months after the
investment, after one year,
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When interest is compounded more than once a year
after 18 months,
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When interest is compounded more than once a year
after 18 months,
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When interest is compounded more than once a year
after 18 months,
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When interest is compounded more than once a year
after 18 months,
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When interest is compounded more than once a year
after 2 years,
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When interest is compounded more than once a year
after 2 years,
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When interest is compounded more than once a year
after 2 years,
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When interest is compounded more than once a year
after 2 years,
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When interest is compounded more than once a year
so after t years, (when the principle value is Pt
and the interest rate i per annum is paid in 2
payments)
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When interest is compounded more than once a year
so after t years, (when the principle value is Pt
and the interest rate i per annum is paid in 2
payments)
In general, if interest is paid m times a year
for t years
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What is e?
Now consider the following problem. Suppose the
bank offers the interest rate of 100 per annum.
Assuming P0 is 1, if interest is to be
compounded once a year,
P11(1100) 2
If the interest is compounded semi-annually, P1
(150)(150) 2.25
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What is e?
By analogy,
P11(1100) 2
P1 (150)(150) 2.25
P1 2.37037037 P1 2.44140625 P1 2.
59374246 P1 2.704813829 P1 2.716923932
P1 2.718145927
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What is e?
If we keep on increasing the value of m, so that
the interest is compounded continuously, P1 tends
to a limit of 2.71828. This limiting value is
denoted by the letter e.
The discussion so far was for a special case
where the interest was compounded only for 1
year, the principal value P0 is set as 1, and the
annual interest rate is 100. To reach the
general formula for continuous compounding, we
need to allow for (1) more years of compounding
( t years ) (2) a principal/initial value other
than 1 (P0) (3) a nominal interest rate other
than 100 (i)
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What is e?
First, if 1 becomes e after 1 year of
continuous compounding and if we let e be the
new initial value in the second year, our asset
value at the end of 2 years will become e(e)
e2. After the third year, it will become e3.
In general, after t years, we have Pt e t.
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What is e?
Second, suppose that the initial value is P0. If
1 will grow into e t after t years of
continuous compounding at the nominal interest
rate of 100 per annum, P0 will grow into P0
et. In other words, the limit of will tend
to
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What is e?
Third, let us allow the nominal interest rate to
be i. The general formula for continuous interest
compounding becomes PtP0 e it.
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What is e?
Third, let us allow the nominal interest rate to
be i. The general formula for continuous interest
compounding becomes PtP0 eit.

• Recall that with
• an initial investment of P0
• to be invested for t years
• at a normal interest rate of i,
• the compound interest formula when compounding
  takes place m times is

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What is e?
Third, let us allow the nominal interest rate to
be i. The general formula for continuous interest
compounding becomes PtP0 eit.

• Recall that with
• an initial investment of P0
• to be invested for t years
• at a normal interest rate of i,
• the compound interest formula when compounding
  takes place m times is

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Third, let us allow the nominal interest rate to
be i. The general formula for continuous interest
compounding becomes PtP0 eit.

• Recall that with
• an initial investment of P0
• to be invested for t years
• at a normal interest rate of i,
• the compound interest formula when compounding
  takes place m times is

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Now let us consider what happens when the
frequency of compounding m increases
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Earlier we saw that
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Earlier we saw that
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Instantaneous rate of growth
As long as we can assume the general process of
exponential growth, we can apply the function
equally well to the growth of population, wealth,
or real capital etc. When applied to some other
context other than interest compounding, the
coefficient i no longer denotes the nominal
interest rate. The economic meaning of i can be
reinterpreted as the instantaneous rate of growth
of the function.
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Instantaneous rate of growth
Given the function Pt P0 e it. which gives
the value Pt at each point of time t , the rate
of change of Pt is found to be The
instantaneous rate of growth is
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Instantaneous rate of growth
The formula for the instantaneous rate of
growth is the ratio of the marginal function
divided by the total function. Recall that the
derivative of ln f ( t ) with respect to t is
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Instantaneous rate of growth
Hence an alternative method of finding the
instantaneous rate of growth Pt P0 e it.
Step 1. Take the natural log of the function ln
Pt ln P0 it ln e. ln P0 it Step 2.
Differentiate with respect to t
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Price elasticity of demand
Given a demand function Step 1. Taking logs on
both sides Step 2. Differentiate with respect to
lnP a is the price elasticity of demand
because
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PRELIMINARY MATHEMATICS

• LECTURE 5
• Optimization

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Reading

• Chiang, A. C. and K. Wainwright (2005) Chapters 9
  and 11.
• Thomas (1999) Chapters 10-12.

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Local maxima and local minima
In this diagram, A and B represent turning points
of the curve y f(x). A is a maximum turning
point and B is a minimum turning point. The value
of y at point A is known as the local maxima
(maximum) and the value of y at point B is called
the local minima (minimum).
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Local maxima and local minima
Note that the slope of the curve y f(x),
measured by is 0 at both turning points.
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Second order condition for local maxima and
minima
If we draw the derivative on a graph we can
confirm that
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Points of inflexion
The condition however is a necessary
condition for local maxima or minima but not a
sufficient condition. In the function below
the sign of at point A is changing from
positive through 0 to positive again, so this is
neither a maximum nor a minimum turning point,
but a point of inflection.
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Second order condition for local maxima and
minima
What distinguishes the two points is the
slope/gradient of the derivative function
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Second order condition for local maxima and
minima
At point A, which is the point of local maxima,
the slope of the curve is negative, while at
point B, which is the point of local minima, the
slope is positive.
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Second order condition for local maxima and
minima
The slope of is measured by the
derivative of known as the second
derivative, denoted as
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Second order condition for local maxima and
minima
we can see that at the local maxima on the
original curve, and
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Second order condition for local maxima and
minima
we can see that at the local maxima on the
original curve, and
and at the local minima,
and
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Second order condition for a point of inflection
A point of inflexion can be characterised by
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Second order condition for a point of inflection
A point of inflexion can be characterised by
However it is also possible to have a local
maxima or a local minima with
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Second order condition for local maxima and
minima
Hence, the obtained second order condition the
local maxima, and
and local minima,
and is a second order sufficient condition but
not necessary because it is also possible to have
a local maxima or a local minima with
and
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n th order condition for local maxima and minima
n th derivative test for relative extrema or
inflection point. If at a
stationary point, continue differentiating
until a non-zero higher-order derivative is
obtained.
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n th order condition for local maxima and minima
n th derivative test for relative extrema or
inflection point. If at a
stationary point, continue differentiating
until a non-zero higher-order derivative is
obtained. If when evaluated at
the stationary point then

• if n is an odd number , we have a point of
  inflection
• if n is an even number AND we have
  a local maxima
• if n is an even number AND we
  have a local minima

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nth derivative test an example.

• Let us consider y x 6.
• The first order necessary condition is obtained
  as
• and x 0,
• Next check the second order sufficient condition
• Evaluating at x 0 ,

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nth derivative test an example.
Because we cannot exclude the possibility that
the function is at a local maxima or local
minima, we carry out the nth derivative test to
check if this is a relative extrema or inflection
points. Third order derivative , at x 0,
Forth order derivative , at x 0,
Fifth order derivative , at x 0,
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nth derivative test an example.
Sixth order derivative Because the sixth
order derivative is non-zero, whereby the
non-zero is obtained at an order which is an even
number, and because the non-zero derivative is
positive, we have a local minima.
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nth derivative test an example.
This is confirmed by drawing a diagram
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Recap Differentiation of bivariate functions

• To recap How to evaluate stationary points
• Step 1. First order condition
• Equate the first derivative to zero and find
  all stationary points.

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Recap Differentiation of bivariate functions
To recap How to evaluate stationary
points Step 2. Second order sufficient
conditions Evaluate the second order derivative
at each stationary points. If we have a
local maxima If we have a local minima
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Recap Differentiation of bivariate functions
To recap How to evaluate stationary
points Step 3. If at any stationary point the
second order derivative is zero, carry on
differentiating until a non-zero higher order
derivative is found. If n is an odd number, we
have a point of inflection If n is an even
number AND we have a local maxima if n is
an even number AND we have a local minima
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Second order partial derivatives
Second order partial derivatives Given a function
z f (x, y), the second order (direct) partial
derivative signifies that the function has been
partially differentiated with respect to one of
the independent variables twice while the other
independent variable has been held
constant The notation with a double
subscript signifies that the primitive function f
is differentiated partially with respect to x
twice. The alternative notation resembles
that of but with the partial derivatives
symbol.
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Second order partial derivatives
Second order partial derivatives The cross
partial derivatives signifies that the function
has been partially differentiated with respect to
one of the independent variables and then in turn
partially differentiated with respect to another
independent variable. The two cross
partial derivatives are identical with each
other, as long as the two cross partial
derivatives are both continuous (Youngs
theorem).
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Unconstrained optimisation
Recall that for a function of a single variable
such as y f (x) any point for which and
was a local minima. Similarly, any point for
which and was a local maxima.
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Unconstrained optimisation
Now consider a function of two variables z f
(x, y) When does z have a local minima?
Extrapolating our arguments of local minima of a
single variable function, we may expect z to have
a local minima when it is not possible to obtain
a smaller value of z by making slight changes in
the values of x and y. There are three ways in
which we can make slight changes in x and
y (a) we can keep x constant and make y
vary (b) we can keep y constant and make x
vary (c) we can make both x and y vary at the
same time. If z has a local minima at point x
a and y b, then we should be able to change the
values of x and y by any of the methods (a), (b),
or (c), and yet not make the value of z smaller.
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Unconstrained optimisation
Let us first consider case (a). In this case, the
value of x is held constant. At a point where
and we cannot obtain a smaller z by varying
y alone.
Similarly, if we consider case (b), when and
we cannot obtain a smaller z by varying x
alone.
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Unconstrained optimisation
Case (c) is slightly complicated. The condition
that must be satisfied here is or
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Unconstrained optimisation
Hence, the second order conditions are as
follows
A point at which the first partial derivatives
are zero is a local maxima if
a local minima if
Note that if then we have
neither a maxima nor a minima but what is
called a saddle point
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Determinantal test

• In the two-variable case, the total differential
  is
• We can also derive the second-order total
  differential as

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Determinantal test
In the two-variable case, the total differential
is We can also derive the second-order total
differential as
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Determinantal test
If d 2 z gt 0 (positive definite), local
minima If d 2 z lt 0 (negative definite), local
maxima
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Determinantal test
Since the variables x and y appear only in
squares, this can be seen as a quadratic form. As
discussed in lecture 1, the condition for d 2 z
to be positive/ negative definite may be stated
by use of determinants. The quadratic form can be
expressed in matrix form as
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Determinantal test
The determinant with the second order partial
derivatives as its elements is called a Hessian
determinant.
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Determinantal test
The principal minors in this case are
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Determinantal test
Hence the second order conditions are d 2 z gt
0 (positive definite), iff and
, then local minima. d 2 z lt 0 (negative
definite), iff and , then local maxima.
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Determinantal test
If we consider a function of three choice
variables, The Hessian determinant is
113
Determinantal test
And the principal minors are
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Determinantal test
d 2 z gt 0 (positive definite), iff and
, then local minima. d 2 z lt 0 (negative
definite), iff and , then local
maxima.
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Unconstrained optimisation application
Application profit maximisation Let us consider
a firm that is a price taker, produces and sells
two goods, goods 1 and goods 2 in perfectly
competitive markets, and has a profit
function What would be the level of output
for goods 1 and 2 satisfying the first order
condition for maximum profit?
116
Unconstrained optimisation application
Application profit maximisation Let us consider
a firm that is a price taker, produces and sells
two goods, goods 1 and goods 2 in perfectly
competitive markets, and has a profit
function What would be the level of output
for goods 1 and 2 satisfying the first order
condition for maximum profit?
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Unconstrained optimisation application
Application profit maximisation Setting both
condition equal to zero,
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Unconstrained optimisation application
Application profit maximisation Setting both
condition equal to zero, Rewrite as
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Unconstrained optimisation application
Application profit maximisation Setting both
condition equal to zero, Rewrite
as Express this in terms of matrices
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Unconstrained optimisation application
Application profit maximisation Use Cramers
rule
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Unconstrained optimisation application
Application profit maximisation Let us consider
a firm that is a price taker, produces and sells
two goods, goods 1 and goods 2 in perfectly
competitive markets, and has a profit
function What would be the level of output
for goods 1 and 2 satisfying the first order
condition for maximum profit? The level of
output/supply for good 1 must be 40 and for good
2 be 24 in order to satisfy the first order
conditions.
122
Unconstrained optimisation application
Application profit maximisation Let us consider
a firm that is a price taker, produces and sells
two goods, goods 1 and goods 2 in perfectly
competitive markets, and has a profit
function What would be the level of output
for goods 1 and 2 satisfying the first order
condition for maximum profit? The level of
output/supply for good 1 must be 40 and for good
2 be 24 in order to satisfy the first order
conditions. Next, check the second order
conditions to verify if we have local maxima.
123
Unconstrained optimisation application
Application profit maximisation Let us consider
a firm that is a price taker, produces and sells
two goods, goods 1 and goods 2 in perfectly
competitive markets, and has a profit
function Let us consider whether the
stationary point we found at Q1 40 and Q2 24
is a local maxima or a local minima.
124
Unconstrained optimisation application
Application profit maximisation The Hessian
matrix can be written as Hence the
profit is at local maxima at Q1 40 and Q2 24