Mathematical Methods for
Economics- II
course taught by
DR. R AKESH N IGAM
Edited by:
Guru Prasad S
Kavinaya K
Rohit Kumar
Thiruvarul T S
Notes scribed by BA batch of 2022-25
Preface
“If you can’t explain it simply, you don’t understand it well enough.”
- ALBERT EINSTEIN
With the above thoughts, this book has been compiled with the lecture notes from the Mathematical Methods for Economics- II course module, a
part of the Madras School of Economics BA 2022-2025 curriculum, taught by Dr.
Rakesh Nigam. This book is an attempt to put this course on paper.
Mathematical Methods for Economics- II course at Madras School of Economics is
deeply focused on the Mathematical foundations behind Economics. This books
attempts to develop the understanding of mathematics needed to understand
courses in Economics and Mathematics. These topics have become very important and are virtually used in many disciplines.
This book is in no way an attempt to duplicate other popular Mathematics books
available in the market but it’s an attempt to write a book that develops concepts
from elementary level and we hope that every readers find interest in Mathematical Methods, be able to find relevant applications in real world as well as motivates
them to study further in details. We encourage students and working professionals
alike, to use these notes as reference material in order to sharpen one’s conceptual
understanding and intuitive learning.
2
Contents
1 Continuity and Differentiability
1.1 Introduction . . . . . . . . . . . . . . . . . . . .
1.2 Continuity . . . . . . . . . . . . . . . . . . . . .
1.2.1 Real life applications of Continuity . . .
1.2.2 Squeeze theorem . . . . . . . . . . . . .
1.3 Differentiability . . . . . . . . . . . . . . . . . .
1.3.1 Real life applications of Differentiability
1.4 Practice problems . . . . . . . . . . . . . . . . .
1.4.1 Problems on Continuity . . . . . . . . .
1.4.2 Problems on Differentibility . . . . . . .
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2 Integral Calculus
2.1 Integral calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 Indefinite integrals . . . . . . . . . . . . . . . . . . . . . . . . . . .
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3 Indefinite Integrals
3.1 Introduction to indefinite integrals:
3.2 Integration by Parts . . . . . . . .
3.3 Integration by Partial Fractions . .
3.4 Integration in quadratic forms: . .
3.4.1 Type 1 . . . . . . . . . . . .
3.4.2 Type 2 . . . . . . . . . . . .
3.4.3 Practice questions . . . . .
3.5 Application of integrals: . . . . . .
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31
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39
Area Under Curves
4.1 Introduction . . . . . . . . . . . . . . . .
4.2 Meaning and Concept . . . . . . . . . .
4.2.1 Area with respect to the X axis .
4.2.2 Area with respect to Y axis . . . .
4.2.3 Area between 2 curves . . . . .
4.2.4 Area between a curve and a line
4.3 Applications of This Concept . . . . . . .
4.3.1 Applications in Architecture . . .
4.3.2 Applications in AI . . . . . . . . .
4.3.3 Applications in Economics . . . .
4.4 Practice Questions . . . . . . . . . . . .
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41
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5
Riemann Integral
5.1 Definition . . . . . .
5.2 Properties . . . . . .
5.2.1 Linearity . . .
5.2.2 Monotonicity
5.2.3 Additivity . .
5.3 Questions . . . . . .
5.3.1 Theorem . .
5.4 Application . . . . .
5.5 Summary . . . . . .
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48
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6 Intermediate Value Theorem
6.1 Definition: . . . . . . . . .
6.2 Application 1 . . . . . . .
6.3 Application 2 . . . . . . .
6.4 Application 3 . . . . . . .
6.5 proof . . . . . . . . . . . .
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54
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7 Newton Leibnitz Formula
7.1 Proof of NLF: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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8 Improper Integrals
8.1 Introduction . . . . . . . . . . . . .
8.2 Types of Improper Integrals . . . .
8.2.1 Type 1 . . . . . . . . . . . .
8.2.2 Type 2 . . . . . . . . . . . .
8.3 Applications of Improper Integrals
8.3.1 Present Value . . . . . . . .
8.3.2 Radioactive waste . . . . .
8.3.3 Capital Value . . . . . . . .
8.4 Practice Problems . . . . . . . . . .
8.5 Conclusion . . . . . . . . . . . . .
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10 Logarithmic Function
10.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.1.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.2 Properties of Logarithm . . . . . . . . . . . . . . . . . . . . . . . .
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9 Theorem
9.1 Mean Value Theorem . . . . . . . . . . . . . . . . . . . . .
9.1.1 Proof: . . . . . . . . . . . . . . . . . . . . . . . . .
9.2 Applications of Mean Value Theorem . . . . . . . . . . . .
9.3 Fundamental Theorem of Calculus . . . . . . . . . . . . .
9.3.1 First Fundamental Theorem of Integral Calculus .
9.3.2 Second Fundamental Theorem of Integral Calculus
9.4 Applications . . . . . . . . . . . . . . . . . . . . . . . . . .
9.5 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . .
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10.3 Types of logarithm . . . . . . . .
10.4 Power Law Behaviour . . . . . .
10.4.1 Log-Log Plot . . . . . . .
10.4.2 semi-log plot . . . . . . .
10.5 Applications of Logarithm . . . .
10.5.1 COMPOUND INTEREST :
10.5.2 Economic growth . . . . .
10.5.3 Financial analysis . . . . .
10.5.4 Elasticity of demand . . .
10.6 Practice Questions . . . . . . . .
10.7 Conclusion . . . . . . . . . . . .
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11 Trigonometry
11.1 Degrees and Radians . . . . . . . . . . .
11.2 Trigonometric Functions . . . . . . . . .
11.3 Trigonometric Properties . . . . . . . . .
11.4 Odd and Even functions . . . . . . . . .
11.5 Sum Formulae in Trigonometry . . . . .
11.6 Product Formulae in Trigonometry . . .
11.7 Sum Product Formulae in Trigonometry
11.8 Polar Co-ordinates . . . . . . . . . . . .
11.9 Cartesian Co-ordinates . . . . . . . . . .
11.10Applications of Trigonometry . . . . . .
11.10.1Measuring Heights . . . . . . . .
11.10.2Aviation . . . . . . . . . . . . . .
11.10.3Criminology . . . . . . . . . . . .
11.10.4Video Game Development . . . .
11.10.5Economics and Finance . . . . .
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12 Binomial Theorem
12.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . .
12.2 What is the special formula? . . . . . . . . . . . . . . . . .
12.3 The Art of Counting – Permutation and Combination . . .
12.3.1 Deriving the Special Formula . . . . . . . . . . . .
12.4 Pascal’s Triangle . . . . . . . . . . . . . . . . . . . . . . . .
12.5 Applications of Binomial Theorem . . . . . . . . . . . . .
12.5.1 In the Distribution of The Internet Protocol Address
12.5.2 Used to forecast the economy . . . . . . . . . . . .
12.5.3 For ranking . . . . . . . . . . . . . . . . . . . . . .
13 PERMUTATION and COMBINATION
13.1 introduction . . . . . . . . . . . . . . . . . . . . . . .
13.2 Fundamental Principle of Counting . . . . . . . . . .
13.2.1 Multiplication Principle . . . . . . . . . . . .
13.2.2 Addition . . . . . . . . . . . . . . . . . . . . .
13.3 Permutation . . . . . . . . . . . . . . . . . . . . . . .
13.3.1 Permutations when all the objects are distinct
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13.4 Combination . . . . . . .
13.5 Some solved Questions . .
13.6 Applications . . . . . . . .
13.6.1 Probability . . . .
13.6.2 Binomial theorem
13.6.3 Business . . . . .
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14 Hyperbolic Functions
14.1 Introduction . . . . . . . . . . . . . .
14.2 Features of Hyperbolic Functions . .
14.2.1 Cosh(x) . . . . . . . . . . . .
14.2.2 Sinh(x) . . . . . . . . . . . .
14.2.3 tanh(x) . . . . . . . . . . . .
14.3 Hyperbolic Identities and Rules . . .
14.3.1 Identities . . . . . . . . . . .
14.3.2 Osborn’s Rule . . . . . . . . .
14.4 Hyperbolic Equations . . . . . . . . .
14.5 Applications of Hyperbolic Functions
14.6 Practice Problems . . . . . . . . . . .
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127
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134
135
136
15 Maxima and Minima and L’Hopital Rule
15.1 maxima and minima . . . . . . . . . . . . .
15.2 APPLICATIONS OF MAXIMA AND MINIMA .
15.3 L’Hopital’s Rule . . . . . . . . . . . . . . . .
15.3.1 Proof of L’Hopital’s Rule . . . . . . .
15.4 Taylor’s series . . . . . . . . . . . . . . . . .
15.5 Error in Approximation . . . . . . . . . . . .
15.6 Exercise . . . . . . . . . . . . . . . . . . . .
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138
138
140
141
141
141
142
143
16 taylor series
16.1 Introduction . . . . . . . . . . . .
16.2 application of taylor series . . . .
16.2.1 Taylor’s series about x = 0
16.2.2 Taylor’s series about x = a
16.2.3 Taylor’s series about z = 0
16.3 inverse function . . . . . . . . . .
16.4 Tangent equation . . . . . . . . .
16.5 Maxima and Minima . . . . . . .
16.5.1 Other scenarioes . . . . .
16.6 critical point . . . . . . . . . . . .
16.6.1 example: . . . . . . . . .
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144
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17 taylor series in 2D
151
17.1 Taylor Series in 2D (x,y) . . . . . . . . . . . . . . . . . . . . . . . . 151
17.1.1 Taylor series about t = 0 . . . . . . . . . . . . . . . . . . . 151
17.2 Gradient Vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153
6
Lecture 1
Continuity and Differentiability
20 january 2023 | Aarcha Mariam, Anisha Srivastava, Purva Pandit, S Megana
Prabha
Lectures on Mathematical Methods for Economics II by Dr Rakesh Nigam
1.1
Introduction
Continuity and differentiability are fundamental concepts in calculus that are used
to describe the behavior of functions. Continuity is a property of a function that
means that it has no abrupt changes or breaks in its graph. Differentiability, on
the other hand, is a property of a function that means that it has a well-defined
derivative at a point. The derivative is a measure of how quickly the function is
changing at that point.
Functions that are both continuous and differentiable are particularly important
in calculus because they allow us to use powerful tools like the Mean Value Theorem, the Fundamental Theorem of Calculus, and the Chain Rule to analyze their
behavior and solve problems.
1.2
Continuity
In mathematics, continuity is a fundamental concept that refers to the absence of
any sudden changes or breaks in a function. A function is said to be continuous if,
for every point in its domain, the value of the function approaches the limit of the
function as the input approaches that point. This means that the function can be
drawn without lifting the pen from the paper at that point.
More formally, a function f(x) is continuous at a point c in its domain if and only
if the following three conditions are met:
1. f(c) is defined (i.e., c is in the domain of f).
2. The limit of f(x) as x approaches c exists.
3. The limit of f(x) as x approaches c is equal to f(c).
If a function is continuous at every point in its domain, it is said to be a continuous
7
function.
limx→ c− f (x) = limx→ c+ f (x) = f (c)
Continuity is a fundamental concept in calculus and is used to define important
properties such as differentiability and integrability. It allows us to study the behavior of functions in a neighborhood around a particular point, and to make
conclusions about the function’s behavior on a larger scale.
Example - 1
Let’s consider the function f(x) = 2x + 1.
To prove continuity at any point, we will take the limit as x approaches a from
both the left and the right, and show that the limit exists and is equal to f(a).
Let’s first consider the left-hand limit:
limx→ a− f (x)
= limx→ a− (2x + 1)
= 2a + 1
Now, let’s consider the right-hand limit:
limx→ a+ f (x)
= limx→ a+ (2x + 1)
= 2a + 1
Since the left-hand limit and the right-hand limit both exist and are equal to 2a
+ 1, we can conclude that the limit of f(x) as x approaches a exists and is equal
to f(a) = 2a + 1. Therefore, we can say that f(x) = 2x + 1 is continuous for all
values of x in its domain.
Left Continuity
A function f(x) is left-continuous at a point a if the following condition holds:
limx→ a− f (x) = f (a)
Here, the symbol (x → a− ) means that x approaches a from the left-hand side
(i.e., from smaller values of x), and f(a) is the value of the function at the point a.
Intuitively, a left-continuous function is one that ”jumps up” to its value at a point
a, rather than having a ”jump down” or a ”jump across” at that point. Leftcontinuous functions are commonly used in analysis, probability theory, and other
areas of mathematics.
8
Right Continuity
Formally, a function f(x) is right continuous at a if:
limx→ a+ (f (x)) = f (a)
For example, the function g(x) = x2 is right continuous at every point, including
x=0, because limx→ 0+ x2 = 0 = x2 .
1.2.1
Real life applications of Continuity
Continuity has a wide range of real-life applications in various fields, including
mathematics, physics, engineering, economics, and more. Here are a few examples:
Calculus: Continuity is a fundamental concept in calculus, and it is used to prove
the existence of limits and derivatives of functions. Calculus is used in a wide
range of applications, including physics, engineering, economics, and more.
Physics: Continuity principles are also applied in physics, particularly in the study
of fluid mechanics. The principle of continuity in fluid mechanics is used to analyze the behavior of fluids in pipes, channels, and other flow systems. Continuity
is also used in electromagnetism to describe the flow of electric charge.
Engineering: Continuity is used in engineering to ensure that systems function
correctly and without interruption. It is also important in engineering applications
such as structural analysis and design. For example, in electrical engineering, continuity testing is used to determine whether an electrical circuit is complete.
Economics: Continuity is used in economics to describe the continuity of a consumer’s preferences over different goods and services. This concept is essential in
the theory of consumer choice.
Overall, continuity is an essential concept that has many real-life applications. It
is used in a wide range of fields and helps to ensure the smooth functioning of
systems and processes.
1.2.2
Squeeze theorem
The squeeze theorem, also known as the sandwich theorem or the pinching theorem, is a fundamental theorem in calculus that is used to evaluate limits. In
mathematical notation, the squeeze theorem can be expressed as:
If h(x) ≥ f (x) ≥ g(x) for all x in some neighborhood of a (except possibly at a),
and if
9
limx→ a h(x) = L and limx→ a g(x) = L, then limx→ a f (x) = L as x approaches a.
Figure 1.1: Squeeze theorem
This theorem is particularly useful when evaluating limits involving trigonometric
functions or exponential functions. It helps us to identify the limit of a function by
”squeezing” it between two other functions whose limits are known.
Example - 2
Here is an example of how the squeeze theorem can be used to evaluate a limit:
1
Let’s consider the limit of the function f (x) = x2 sin( )
x
as x approaches 0. To evaluate this limit, we can use the squeeze theorem by finding two other functions g(x) and h(x) that ”squeeze” f(x) between them.
1
First, we note that -1 ≤ sin( ) ≤ 1
x
for all x, so we can set g(x) = −x2 and h(x) = x2 .
Then, we have:
1
g(x) = −x2 ≤ x2 sin( ) = f (x) ≤ x2 = h(x)
x
Now, we take the limits of g(x) and h(x) as x approaches 0:
limx→ 0 g(x) = limx→ 0 (−x2 ) = 0
limx→ 0 h(x) = limx→ 0 (x2 ) = 0
10
Since both limits are equal, we can apply the squeeze theorem and conclude that:
1
limx→ 0 f (x) = limx→ 0 x2 sin( ) = 0
x
Therefore, the limit of f(x) as x approaches 0 is 0.
1.3
Differentiability
Differentiability is a property of a function that describes how smoothly the function changes as its input variable changes. A function is said to be differentiable
at a point if its derivative exists at that point.
Geometrically, the derivative of a function at a point represents the slope of the
tangent line to the curve at that point. The derivative measures how much the
function changes as its input variable changes by a small amount.
A function is said to be differentiable over an interval if it is differentiable at every
point in the interval.
The formal definition of differentiability for a function f(x) at a point x=a is as
follows:
f(x) is differentiable at x = a if the following limit exists:
limh→ 0
f (a + h) − f (a)
h
Figure 1.2: Differentiability
If the limit exists, it is denoted by f’(a), which is the derivative of f(x) at x = a.
The function f(x) is said to be differentiable on an interval if it is differentiable at
11
every point within that interval.
Slope of AB = tanθ =
f (a + h) − f (a)
h
As h → 0 ; B → A
limh→ 0
limh→ 0
∆y
=
∆x
f (a + h) − f (a)
→
h
Slope of the tangent to f(x) at x=a
Example - 3
Let’s consider the function f (x) = x3 − 2x2 + x.
To prove that f(x) is differentiable, we need to show that its derivative exists at
every point in its domain. The derivative of f(x) is given by:
f ′ (x) = 3x2 − 4x + 1
We can see that f’(x) is a polynomial function that is continuous and defined for
all real values of x. Therefore, f(x) is differentiable for all real values of x.
To further demonstrate this, we can use the limit definition of the derivative to
find the derivative of f(x) at a specific point, say x = 2:
f’(2) = limh→ 0
f (2 + h) − f (2)
h
Substituting in the expression for f(x), we get:
f’(2) = limh→ 0
(2 + h)3 − 2(2 + h)2 + (2 + h) − (23 − 2(2)2 + 2)
h
Simplifying and factoring out an h in the numerator, we get:
f’(2) = {limh→ 0
h(3h + 6)
}
h
Canceling out the h terms, we get:
f’(2) = limh→ 0 (3h + 6) = 6
Therefore, f(x) is differentiable at x = 2 and its derivative at that point is f’(2) =
6. By extension, f(x) is differentiable for all real values of x, since its derivative is
a continuous function.
12
1.3.1
Real life applications of Differentiability
Optimization: Differentiability is used to find the maximum and minimum values
of a function, which is important in optimization problems. In such problems,
the goal is to find the optimal value of a variable that maximizes or minimizes a
given function. Differentiability helps in identifying the critical points of a function, which are the points where the function’s derivative is zero, and thus, finding
the optimal value.
Physics: Differentiability is used in the study of physics to analyze the motion of
objects. The derivative of position with respect to time gives the velocity of an
object, and the derivative of velocity gives acceleration. Differentiability is also
used in the study of electromagnetic fields, where it helps in understanding the
rate of change of electric and magnetic fields
.
Economics: Differentiability is used in economics to model and analyze production functions and utility functions. In production functions, the derivative of the
function represents the marginal product of a factor of production. In utility functions, the derivative of the function represents the marginal utility of a good or
service. These concepts are crucial in determining the optimal level of production
and consumption.
1.4
1.4.1
Practice problems
Problems on Continuity
1
Question 1: Evaluate the limit limx→ 0 xecos( x ) .
Solution:
We know that −1 ≤ cos x ≤ 1 for any x.
In the same way −1 ≤ cos( x1 ) ≤ 1 (when x ̸= 0)
Raising each side by ”e”,
1
e−1 ≤ ecos( x ) ≤ e1
Case 1: When x > 0
Multiply each side by x:
1
xe−1 ≤ xecos( x ) ≤ xe
0
= 0 and limx→ 0 xe1 = 0(e) = 0
e
1
cos( )
So by sandwich theorem, limx→ 0 xe x = 0
Now, limx→ 0 xe−1 =
13
Case 2: When x < 0
Multiply each side by x:
xe
−1
cos(
≥ xe
xe ≤ xe
cos(
1
)
x ≥ xe
1
)
x ≤ xe−1
Now, limx→ 0 xe =
0
= 0 and {limx→ 0 xe−1 } = 0(e) = 0
e
1
x
So by sandwich theorem, limx→ 0 xecos = 0.
Answer: In both the cases, the given limit = 0.
Question 2: Examine the function defined by
(
3
x2
, 0 ≤ x ≤ 1 ; 2x2 –3x + , x > 1
f(x) = f (x) =
2
2
for continuity at x=1.
Solution:
Here, f (1) =
12
1
=
2
2
12
1
x2
=
= and,
limx→ 1− f (x) = limx→ 1−
2
2
2
3
12
1
3
1
2
limx→ 1+ f (x) = limx→ 1+ 2x –3x + =
= = 2 × 12 –3 × 1 + =
2
2
2
2
2
⇒ limx→ 1− f (x) = limx→ 1+ f (x) = f (1)
Answer: f is continuous at x=1.
Question 3: If the function defined by
(
1 − cos cx
f (x) =
, x ̸= 0 ; 21 , x = 0
x sin x
is continuous at x=0, find the value(s) of c.
Solution: Given f(x) =
limx→ 0 f (x)
1
1
when x = 0 → f (0) =
2
2
1 − cos cx
x sin x
1 − cos cx
1 + cos cx
= {limx→ 0
}×
x sin x
1 + cos cx
= limx→ 0
14
sin2 cx
x sin x(1 + cos cx)
sin cx 2 x
1
= limx→ 0 (
).
.
cx.c
sin x cos cx
= limx→ 0
1
= (1.c2 ).1. 1+1
=
c2
2
Answer: For the function f to be continuous at x = 0,
limx→ 0 f (x) = f (0)
⇒
1
c2
=
2
2
⇒ c2 = 1
⇒ c = 1, −1.
√
2 cos x − 1
π
π
, x ̸= , find the value of f( ), so that the
cot(x − 1)
4
4
π
function f becomes continuous at x = .
4
Question 4: If f(x) =
Solution:
√
2 cos x − 1
π f (x) = lim
π
cot(x − 1)
x→
x→
4
4
√
( 2 cos x − 1) sin x
= lim
π
cos x − sin x
x→
4√
√
( 2 cos x − 1)( 2 cos x + 1)(cos x + sin x) sin x √
( 2 cos x+1)(cos x+sin x)
= limx→ π4
cos x − sin x
(cos x + sin x) sin x
2 cos2 x − 1
√
= lim
(
)(
)
π
2
cos2 x − sin x
x→
2 cos x + 1
4
cos 2x (cos x + sin x) sin x
= lim
)( √
)
π(
cos 2x
x→
2 cos x + 1
4
(cos x + sin x) sin x
√
= lim
π
x→
2 cos x + 1
4
1
1 1
(√ + √ )√
2
2 2
= √
1
2.( √ + 1)
2
1
=
2
π
1
π
Answer: Hence, f ( ) = will make function f continuous at x = ( )
4
2
4
lim
15
Question 5: Let f(x) = 3x - 2, x ≤ 1 ; 3x - 4, x > 1. Is it a continuous function?
Justify your answer.
Solution:
We note that domain of f = R, therefore, we have to examine f for continuity at x
ϵ R. Let c be any real number.
Three cases arise:
Case (i): If c < 1, then f(c) = 3c-2.
limx→ c f (x) = limx→ c 3x − 2 = 3c − 2 = f (c) ⇒ f is continuous for all c < 1.
Case (ii): If c > 1, then f(c)= 3c-4.
limx→ c f (x) = limx→ c (3x − 4) = 3c − 4 = f (c) ⇒ f is continuous for all c > 1.
Case (iii): If c=1, then f (1) = 3x1 − 2 = 1
Now limx→ 1− f (x) = limx→ 1− (3x − 2) = 3 × 1 − 2 = 1 = f (1)
⇒ f is continuous at x=1 from the left
But, limx→ 1+ f (x) = limx→ 1+ (3x − 4) = 3 × 1 − 4 = −1 ̸= f (1)
⇒ f is discontinuous at x=1 from the right.
Answer: Hence, f has a discontinuity at x=1. It follows that f is discontinuous at
x=1, and continuous at all other points. Therefore, f is not a continuous function
and its domain of continuity is R – {1}.
Question 6: Show that the function f (x) = 1 −
interval [-1, 1].
p
(1 − x)2 is continuous on the
Solution:
If -1 < a < 1 , then using the Limit Laws, we have
{limx→ a f (x)}
= {limx→ a 1 −
= 1 - {limx→ a
=1-
p
p
(1 − x2 )}
p
(1 − x2 )}
(1 − a2 )
= f(a)
Thus, f is continuous at a if -1 < a < 1. Similar calculations show that {limx→ −1+ f (x)} =
1 = f (−1) and {limx→ 1− f (x)} = 1 = f (1)
16
and so is f continuous from the right at -1 and continuous from the left at 1.
Answer: Therefore, f is continuous on [-1, 1].
Question 7: Evaluate {limx→ π
sin x
}.
2+cos x
Solution:
y = sin x is continuous.
The function in the denominator, y = 2 + cos x, is the sum of two continuous functions and is therefore continuous. This function is never 0, because cos x ≥ −1 for
all x and so 2 + cos x > 0 everywhere.
sin x
is continuous everywhere.
Thus the ratio f(x)= 2+cos
x
Answer: By definition of a continuous function,
{limx→ π
sin x
}
2+cos x
= {limx→ π f (x)} = f (π) =
sin π
2+cos π
=
0
2−1
=0
Question
) 8: Find the value of k, so that the function
f(x) =
1−cos 4x
,
8x2
x ̸= 0 ; k , if x = 0
is continuous at x = 0.
Solution: )
1−cos 4x
,
8x2
Let f(x) =
x ̸= 0 ; k , x = 0 is continuous at x = 0.
Then, LHLx=0 = RHLx=0 = f (0)
Now, LHL = {limx→ 0− f (x)}
= {limx→ 0−
1−cos 4x
}
8x2
= {limx→ 0−
1−cos 4(0−h)
}
8(0−h)2
[put x = 0 - h ; when x → 0− , then h → 0]
={limh→ 0
1−cos 4h)
}
8h2
={limh→ 0
2 sin2 2h)
}
8h2
= {limh→ 0
sin2 2h)
}
4h2
=limh→ 0 ( sin2h2h) )2 }
=1
17
On substituting this value in LHLx=0 = RHLx=0 = f (0) , we get 1 = f(0) ⇒ 1 =
k [f(0) = k ; given]
Hence, for k = 1, the given function f(x) is continuous at x = 0.
1.4.2
Problems on Differentibility
Question 1: If f is differentiable at x = a, find {limx→ a
x2 f (a)−a2 f (x)
}
x−a
Solution:
Let x = a + h, so that when x → a, h → 0.
{limx→ a
x2 f (a)−a2 f (x)
}
x−a
= {limh→ 0
(a+h)2 f (a)−a2 f (a+h)
}
h
2
2 f (a+h)−f (a)
= {limh→ 0 ( (2a+hh )f (a) ) − ( a
h
}
= {limh→ 0 (2a + h)f (a)} − a2 {limh→ 0
f (a+h)−f (a)
}
h
= (2a + 0)f (a)–a2 f ′ (a)
= 2af (a)–a2 f ′ (a).
Question 2: If f(a) = 2, f’ (a)= 1, g(a) = -1, g’(a) = 2, then find the value of
{limx→ a
g(x)f (a)−g(a)f (x)
}
x−a
Solution:
g(x)f(a) – g(a)f(x) = (g(x)f(a) – g(a)f(a)) – (g(a)f(x)- g(a)f(a)).
{limx→ a
g(x)f (a)−g(a)f (x)
}
x−a
= f (a){limx→ a
g(x)−g(a)
}
x−a
− g(a){limx→ a
f (x)−f (a)
}
x−a
Putting x= a+h, so that when x → a, h → 0.
= f (a){limh→ 0
g(a+h)−g(a)
}
h
− g(a){limh→ 0
= f (a)g ′ (a)–g(a)f ′ (a)
= 2 × 2 − (−1) × 1
18
f (a+h)−f (a)
}
h
= 5.
Question 3: Examine the function ‘f’ for derivability at x=0 where
(
f (x) =
1 − x2 , x ≤ 0
Solution: Given f (x) =
; 1 + x2 , x > 0
(
1 − x2 , x ≤ 0
; 1 + x2 , x > 0
⇒ Df = R
f−′ (0) = {limh→ 0−
f (0+h)−f (0)
}
h
= {limh→ 0−
f (h)−f (0)
}
h
= {limh→ 0−
(1−h2 )−1
}
h
= {limh→ 0− (−h)}
= 0.
And, f+′ (0) = {limh→ 0+
={limh→ 0+
= {limh→ 0+
f (0+h)−f (0)
}
h
f (h)−f (0)
}
h
(1+h2 )−1
}
h
= {limh→ 0+ (h)}
= 0.
f−′ (0) = f+′ (0) ⇒ f is derivable at x = 0.
Question 4: Show that the function f defined as follows, is continuous at x=2,
but not differentiable.
f(x) = 3x - 2, 0 < x ≤ 1
2x2 –x, 1 < x ≤ 2
5x–4, x ¿ 2
Solution: We note that Df = (0, ∞)
F is defined in the neighbourhood of 2 and f (2) = 2x22 –2 = 6.
{limx→ 2− f (x)} = {limx→ 2− (2x2 − x)} = 2 × 22 − 2 = 6
and ,
19
{limx→ 2+ f (x)} = {limx→ 2+ (5x − 4)} = 5 × 2 − 4 = 6
⇒ {limx→ 2− f (x)} = {limx→ 2+ f (x)} = f (2)
The given function f is continuous at x=2.
L f’(2) = {limh→ 0−
f (2+h)−f (2)
}
h
= {limh→ 0−
2(2+h)2 −(2+h)−6
}
h
= {limh→ 0−
2(4+4h+h2 )−2−h−6
}
h
= {limh→ 0−
7h+2h2
}
h
= {limh→ 0− (7 + 2h)}
=7+2×0
=7
And, R f’(2) = {limh→ 0+
= {limh→ 0+
5(2+h)−4−6
}
h
= {limh→ 0+
(10+5h+10)
}
h
= {limh→ 0+
5h
}
h
f (2+h)−f (2)
}
h
= {limh→ 0+ 5}
=5
⇒ Lf ′ (2) ̸= Rf ′ (2)
⇒ The given function is not differentiable at x = 2
Question 5: Show that the function f defined as follows is not differentiable at
x=2
f (x) = x2 , x ≤ 1
1
,x > 1
x
Solution:
L f’(1) = {limh→ 0−
f (1+h)−f (1)
}
h
20
= {limh→ 0−
(1+h)2 −1
}
h
= {limh→ 0−
2h+h2
}
h
= {limh→ 0− (2 + h)}
=2+0
=2
And, R f’(1) = {limh→ 0+
= {limh→ 0+
1
−1
1+h
h
}
= {limh→ 0+
(1−(1+h))
}
h(1+h)
= {limh→ 0+
−1
}
1+h
=
f (1+h)−f (1)
}
h
−1
1+0
= -1
⇒ Lf ′ (1) ̸= Rf ′ (1)
⇒ The given function is not differentiable at x = 1
Question 6: Where is the function f(x) = |x| differentiable?
Solution:
If x > 0, then |x| = x and we can choose h small enough that x + h > 0 and hence
|x + h| = x + h .
Therefore, for x > 0 , we have
f’(x) = {limh→ 0
= {limh→ 0
|x+h|−|x|
}
h
(x+h)−x
}
h
= {limh→ 0 hh }
= {limh→ 0 1}
= 1.
and so f is differentiable for any x > 0.
21
Similarly, for x < 0, we have |x| = −x and we can choose h small enough that
x + h < 0 and so |x + h| = −(x + h).
Therefore, for x < 0 ,
|x+h|−|x|
}
h
f’(x) = {limh→ 0
= {limh→ 0
−(x+h)−(−x)
}
h
= {limh→ 0
−h
}
h
= {limh→ 0 (−1)}
= -1.
and so f is differentiable for any x ¡ 0.
For x=0, we have to investigate
f’(0)= {limh→ 0
= {limh→ 0
f (0+h)−f (0)
}
h
|0+h|−|0|
h
} (if it exists)
Let’s compute the left and right limits separately:
{limh→ 0−
|0+h|−|0|
}
h
= {limh→ 0−
= {limh→ 0−
|h|
}
h
= {limh→ 0+
|h|
}
h
−h
}
h
= {limh→ 0− (−1)}
= -1
and
{limh→ 0+
|0+h|−|0|
}
h
= {limh→ 0+ hh }
= {limh→ 0+ 1}
=1
Since these limits are different, f’(0) does not exist. Thus, f is differentiable at all
x except 0.
22
Question 7: If f (x) =
√
x, find the derivative of f. State the domain of f’
Solution:
f’(x)= {limh→ 0
√
= {limh→ 0
f (x+h)−f (x)
}
h
√
x+h− x
}
h
= {limh→ 0
√
√
√
√
( x+h− x).( x+h+ x)
√
√
}
h.( x+h+ x)
= {limh→ 0
(x+h)−x
√
√ }
h.( x+h+ x)
= {limh→ 0
√
=
1 √
}
x+h+ x
√ 1√
x+ x
= 2√1 x
We see that f’(x) exists if x > 0 , so the domain of f’ is (0, ∞). This is smaller than
the domain of f, which is [0, ∞).
23
Lecture 2
Integral Calculus
23 january 2023 | G.Mathesvaran , Sairam.P.R.V , Aakash Nathan ,R.Arun Balaji
Lectures on Mathematical Method For Economics II by Dr.Rakesh Nigam
2.1
Integral calculus
Definition:-A Function F(x) is called an ”anti-derivative” of F(x) on some interval=I is ∈ ∀x ∈ I
F ′ (x) =
∂F (x)
= f (x)
∂x
Given f(x) and we have to find F(x).
Theorem 1:- The collection of anti-derivates of f(x) on I in of the form
{ F(x) + c | c ∈ R },where c is constant.
proof:- If F’(x) = G’(x) , where F(x) and G(x) belong to collection of anti-derivates
of f(x). Both F(x) and G(x) differs by a constant G(x) = F(x) + c ,where c is
constant.
Ex:1
Z
=
=
=
x4
dx.
1 + x2
x4 + 1 − 1
dx
x2 + 1
Z
Z
dx
2
(x − 1)dx +
2
x +1
Z
x3
− x + tan−1 +C
3
24
Ex:2
Z
tan xdx
=
Z
=
sin x
dx
cos x
Z
dt
−
t
=
− log(t)
Ex:-3
=
=
Z
tan−1 x.dx
Z
tan−1 .1.dx
x tan
=
Z
1
.x.dx
1 + x2
Z
1
dt
−1
x tan −
2
t
−1
−
=
x tan−1 (x) −
=
x tan−1 (x) −
1
ln(t) + c
2
1
ln(1 + x2 ) + c
2
Ex:-4
Z
=
=
=
=
Z
9x − 5
9x2 − 6x + 1
9x − 5
.dx
3x2 − 1
9x − 5
A
B
=
+
(3x − 1)2
3x − 1 (3x − 1)2
A(3X − 1) + B
(3x − 1)2
x(3A) + [B − A]
(3x − 1)2
25
=
3A = 9
so
A=3
...(1)
=
A−B =5
=
=⇒ B = −2
Z
Subs (1)
3
.dx .
(3x − 1)
Z
2
.dx
(3x − 1)2
3x − 1 be ′ t′
N ow let
=
Z
dt
−2
t
3
Z
=
3 log(3x − 1) +
1
.dt
t2
2
+c
3x − 1
Ex:-5
Z
x. sin x.dx
u=x
v = sin x.dx
du = 1dx
Z
dv = sin x.dx
Z
u.dv = uv −
=
v.du
Z
x(− cos x) +
cos x(1).dx
=
−x cos x + sin x + c
Ex:-6
Z
Z
=
(ax + b)2 dx
(a2 x2 + 2abx + b2 )dx
a2 3
x + abx2 + b2 x + C
3
where C is the constant of integration.
=
26
Ex:-7
Z
sin 2x − 4e3x dx
To solve this integral, we can use integration by parts for the second term:
Let u = −4e3x and dv = dx, then du/dx = −12e3x and v = x.
Using the formula for integration by parts, we have:
Z
Z
cos 2x
3x
− (−12e3x )(x)dx
sin 2x − 4e dx = −
2
To evaluate the second integral, we can use integration by parts again:
Let u = x and dv = −12e3x dx, then du/dx = 1 and v = −4e3x .
Using the formula for integration by parts, we have:
Z
Z
4
3x
3x
(−12e )(x)dx = −4xe + 4e3x dx = −4xe3x + e3x + C
3
where C is the constant of integration.
Substituting this into the original integral, we have:
Z
4
cos 2x
+ 4xe3x − e3x + C
sin 2x − 4e3x dx = −
2
3
Thus, the solution to the integral is:
Z
4
cos 2x
sin 2x − 4e3x dx = −
+ 4xe3x − e3x + C
2
3
Ex:-8
Z
3ax
dx
2
b + c2 x 2
we can use substitution. Let u = c2 x2 + b2 , then du/dx = 2cx and dx = du/(2cx)
Substituting this into the integral, we have:
Z
Z
3ax
3a du
3a
dx
=
=
ln |c2 x2 + b2 | + C
b2 + c 2 x 2
2c u
2c
where C is the constant of integration.
Thus, the solution to the integral is:
Z
3ax
3a
dx =
ln |c2 x2 + b2 | + C
2
2
2
b +c x
2c
27
2.2
Indefinite integrals
Z
f(x)dx
I=
Z
is called Indefinite integral, where
f(x)dx = G(x) + c
b
Z
f(x).dx is a real no.
D=
a
∴ N o constant required.
Ex:-1
f (x) = e(−|x|)
Z
I=
f (x) = e
(−|x|)
e
(−|x|)
=
Z
.dx =
f (x).dx
e−x , x ≥ 0 = −e−x + c1
e+x , x < 0 = ex + c2
Y
y = e−x
.1
Jump discontinuity of 2
0
X
.-1
28
(2.1)
(−|x|)
I=e
=
(−e−x + 2) + d; x ≥ 0
e+x + d; x < 0
(2.2)
where ’d’ is the constant of integretion.
Ex:-2
Z
e3x dx
To solve this integral, we can use u-substitution. Let u = 3x, then du/dx = 3 and
dx = du/3. Substituting these into the integral, we get:
Z
Z
Z
1
3x
u du
=
e dx = e
eu du
3
3
We can then integrate eu with respect to u to get:
Z
1
1
e3x dx = eu + C = e3x + C
3
3
where C is the constant of integration. Therefore, the solution to the integral is:
Z
1
e3x dx = e3x + C
3
Ex:-3
Z
4xe2x dx
we can use integration by parts. Let u = 4x and dv/dx = e( 2x) , then du/dx = 4
and v =(1/2)e( 2x). Using the formula for integration by parts, we have:
Z
Z
2x
4xe dx = uv − vdu
Substituting in our values for u and v, we get:
Z
Z
1 2x
1 2x
2x
4xe dx = 4x · e −
e · 4dx
2
2
Simplifying, we get:
Z
2x
4xe dx = 2xe
2x
Z
−
2e2x dx
Integrating the second term with respect to x, we get:
Z
4xe2x dx = 2xe2x − e2x + C
where C is the constant of integration. Therefore, the solution to the integral is:
Z
4xe2x dx = 2xe2x − e2x + C
29
Ex:-4
Z
1
dx
sin x cos2 x
2
=
Z
1
4
=
Z
=
Z
1
sin2 2x
1
( sin2 2x)−1
4
4cosec2 2x.dx
=
−2cot2x + c
30
Lecture 3
Indefinite Integrals
27 january 2023 | Snehaa , Anamika, Chetna
Lectures on Mathematical Methods for Economics II by Dr.Rakesh Nigam
3.1
Introduction to indefinite integrals:
An indefinite integral, also known as an anti-derivative, is a type of integral that
does not have limits of integration. It is used to find a function that, when differentiated, gives
R the original function. The indefinite integral of a function f(x) is
denoted by f (x)dx and represents the family of all functions that have f(x) as
their derivative.
Mathematically, if F(x) is an anti-derivative of f(x), then all functions of the form
F (x) + C, where C is a constant, are also anti-derivatives of f(x). The constant C
is called the constant of integration.
R
For example, the indefinite integral of the function f (x) = x2 is x2 dx = 13 x3 + C,
where C is a constant of integration. This means that any function of the form
1 3
x + C, where C is any constant, is an anti-derivative of x2
3
Indefinite integrals are important in calculus because they allow us to solve a wide
range of problems involving rates of change and accumulation. They are also used
extensively in physics, engineering, and other sciences to model and analyze a
wide range of phenomena.
3.2
Integration by Parts
This is a method of integration, that is found quite useful in integrating products
of functions.
31
If u and v are any two differentiable functions of a single variable x. Then, by
using the product rule of differentiation, we have
dv
du
d
(uv) = u + v
dx
dx
dx
Now by integrating both sides, we get
Z
Z
dv
du
uv = u dx + v dx
dx
dx
OR
Z
dv
u dx = uv −
dx
Z
du
vdx
dx
(1)
Equation (1) can also be written as,
Z
Z
′
uv dx = uv − u′ vdx
The choice of ’u’ matters, for example
Inverse Log
tan−1 logx
Algebraic
x, x2 , x3
Trigonometry
sinx
R
Example 1 : R xex dx
= xex − ex
= xex − ex + c
R
Example 2 : ex cosx dx
R
= cos ex + (sinx) ex dx
R
= ex cosx + ex sinx dx
R x
R
e cosx = ex + ex sinx − ex cosx dx
...(I)
I = ex cosx + ex sinx − I
2I = ex cosx + ex sinx
I = 12 ex (cosx + sinx) + c
R
Example 3: tan−1 (x) dx
R
= tan−1 (x) 1dx
R
1
= x tan−1 (x) − 12 2 (1+x
2 ) x dx
R
= xtan−1 (x) − 21 2t dt2
32
Exponent
ex
= xtan−1 (x) −
= xtan−1 (x) −
1
2
1
2
ln|t| + c
ln |1 + x2 | + c
Practice Questions
1) ex (tan−1 x +
1
)dx
1+x2
2) e2 x sinx
3) (sin−1 x)2
4) ex (sinx + cosx)
5) (x2 + 1) ln x
3.3
Integration by Partial Fractions
Integration by Partial Fraction Decomposition is a procedure where we can “decompose” a proper rational function into simpler rational functions that are more
easily integrated. Basically, we are breaking up one “complicated” fraction into
several different “less complicated” fractions. You may have learned how to use
this technique in your Algebra class, and it’s quite useful in Calculus!
The basic idea is to factor the denominator (if it isn’t already factored) of the
complicated factor, and then break it up into different fractions with denominators
of those factors. If an integral is improper (the degree of the numerator is greater
than or equal to the degree of the denominator), use polynomial long division to
get a term that’s not a rational function, and then decompose the remaining terms
33
BASIC PARTIAL FRACTION DECOMPOSITION RULES
34
PROBLEMS
Example 1:
R
dx
2
2x + 9x − 5
q(x) = 2x2 + 9x − 5
= 2x(x + 5) − 1(x + 5)
= R(2x − 1)(x + 5)
dx
= (2x−1)(x+5)
1
(2x−1)(x+5)
A
2x−1
=
+
B
x+5
= A(x + 5) + B(2x − 1) = 1
2
11
A=
=
1
11
2dx
2x−1
R
−1
11
B=
−
1
11
dx
x+5
R
let 2x − 1 = t implies 2dx = dt
=
1
11
dt
t
=
1
log|2x
11
=
1
log |2x−1|
11
|x+5|
R
1
log|x
11
−
− 1| −
+ 5| + C
1
log|x
11
+ 5| + C
+C
Example 2
I=
=
R
(9x−5)dx
9x2 −6x+1
R
(9x−5)dx
(3x−1)2
9x−5
(3x−1)2
=
A
3x−1
+
B
(3x−1)2
9x − 5 = A(3x − 1) + B
Implies A = 3 andB = −2
I=3
R
dx
3x−1
−2
R
dx
(3x−1)2
let 3x − 1 = t implies 3dx = dt
35
=
dt
t
R
−
2
3
R
dt
t2
= log|3t − 1| +
2
3(3x−1)
+C
Example 3
I=
R
2x4 −x3 −x2 +3x−4dx
(x2 =1)
I=
R
2x2 − x − 3 +
4x−1
x2 +1
dx
−
=
2x3
3
−
x2
2
− 3x −
R
4x
x2 +1
=
2x3
3
−
x2
2
− 3x +
R
2dt
t
=
2x2
2
−
x2
2
− 3x + 2 ln|x2 + 1| − tan−1 x + c
−
R
R
dx
x2 +1
dx
1+x2
Since the degree of the numerator is greater than the degree of the denominator,
by using long division method we get to this conclusion
PRACTICE PROBLEMS
1.
R
dx
(x+1)(x+2)
2.
R
(x2 +1)dx
x2 −5x+6
3.
R
(3x−2)dx
(x+1)2 (x+3)
4.
R
(x2 )dx
(x2 +1)(x2 +4)
5.
R
(3x−1)dx
(x−1)(x−2)(x−3)
36
3.4
Integration in quadratic forms:
3.4.1
Type 1
Z
1
dx(OR)
2
ax + bx + c
Z
√
1
dx
ax2 + bx + c
1.Make the co-efficient of x2 as 1.
2.Use completing squares method and write in the form (x ± a)2 ± b2
3. Apply formula from table.
3.4.2
Type 2
Z
px + q
dx(OR)
2
ax + bx + c
px + q = A
Z
px + q
√
ax2 + bx + c
dy
(ax2 + bx + c) + B
dx
px + q = A(2ax + b) + B
1.Equate co-efficient of x and constant term to find A and B.
2.Substitute for px+q in the numerator and split into two integrals and integrate.
37
Example 1:
R x4
dx
1+x2
=
R
(x4 −1)+1
dx
1+x2
=
R
(x4 −1)
dx
1+x2
=
R
(x2 −1)(x2 +1)
dx
(1+x2 )
+
R
1
dx
1+x2
+
R
R
= (x2 − 1)dx +
R
dx
1+x2
dx
1+x2
R
R
R
= (x2 )dx − dx +
=
x3
3
dx
1+x2
− x + tan− 1x + c
Example 2:
R x+2
2x2 +6x+5
x + 2 = A(4x + 6) + B
= 4Ax + 6A + B
38
A=
1
4
B=
1
2
=
1
4
R
4x+6
2x2 +6x+5
+
1
2
R
dx
2x2 +6x+5
= 14 log|2x2 + 6x + 5| + 12 tan− 1(2x + 3) + C
Example 3:
R
1
dx
9x2 +6x+5
=
1
9
R
1
x2 + 2x
+ 95 dx
3
=
1
9
R
1
x2 +2x 13 +( 13 )2 −( 31 )2 + 59
=
1
9
R
1
(x+ 31 )2 +( 23 )2 dx
= 61 tan− 1( 3x+1
)+c
2
3.4.3
Practice questions
R
1)
√ 5x+3
x2 +4x+10
R
√
3)
R
√ x+3
5−4x+x2
4)
R
1
3x2 +13x−10
R
5) √
3.5
1
7−6x−x2
2)
6x+7
(x−5)(x−4)
Application of integrals:
1. Lorenz Curves and the Gini Index A question that arises in economics looks at
the equity of income or wealth distribution in a country. In standard economic
theories either too much or too little equity indicates a lack of opportunity and
is a hindrance to growth. However, before being able to address the advantages
or disadvantages of a level of inequity we need to be able to quantify the level of
equity or inequity. The standard method is to use the Lorenz curve and the Gini
index.
39
G=
R1
0 (x−L(X))dx
R1
0 xdx
=2
R1
0
(x − L(X))dx
2. Volume of revolution: Integration can also be used to find the volume of a
solid of revolution. Suppose we have a curve y=f(x) and we revolve it around
the x-axis to create a solid. The volume of this solid can be found by integrating
the cross-sectional area of the solid over the interval of interest. Specifically, if the
solid is generated by rotating the curve y=f(x) about the x-axis over the interval
[a,b], then the volume can be computed as follows:
b
Z
π[f (x)]2dx
V =
a
3. Work: Integration can be used to find the amount of work done by a force.
Suppose we have a force F that acts on an object as it moves along a path given
by a function r(t). Then the work done by the force over the interval [a,b] c
Z
b
F · dr
W =
a
where F is the force vector and dr is the differential of the position vector r.
4. Center of mass: Integration can be used to find the center of mass of an object.
Suppose we have an object with density function p( x,y,z) and volume V. Then the
center of mass can be found as follows:
1
x̄ =
M
Z
1
ȳ =
M
Z
1
z̄ =
M
Z
xρ(x, y, z)dV
V
yρ(x, y, z)dV
V
zρ(x, y, z)dV
V
where M is the mass of the object.
40
Lecture 4
Area Under Curves
30 january 23 | Harinarayanan, Saikeerthi, Sukhil Aditya
Based on
Lectures on Mathematical Methods for Economics II by Dr.Rakesh Nigam
4.1
Introduction
Figure 4.1: Sample graph
Curves are an ”Integral” part of our daily lives. Everything from toys to massive passenger aeroplanes, is made with knowledge of integrals and mensuration!
They beautify our surroundings with neat, smooth and sometimes symmetrical appearances. Could you imagine a world where there are no curves and everything
is straight with sharp edges and corners?
But to design these curves, we need to get a measure of how much area is supporting the curve from beneath, so as to give important information on the measure
of the state variable that the curve is on top of - giving the boundary for the area.
This can help us calculate important information across many different fields of
study. Examples are: • Consumer Surplus under Demand Curve (Economics)
• Determining the area of Land to construct Buildings (Civil Engineering)
• Designing domes and buildings (Architecture)
41
The calculation of area under curves is actually a subset of the applications of
Integral Calculus. This topic is conventionally taught in 12th std as ”Application
of Integrals”. Hence, basic knowledge of integrals is required for this chapter...
4.2
Meaning and Concept
We shall now look at some of the important types of graphical area calculations
and their corresponding integral area solutions... Area under the curve is calculated by different methods, of which the anti derivative method of finding the area
under the curve is more popular. The area under the curve can be found by knowing the equation of the curve, the boundaries of the curve and the axis enclosing
the curve. Generally we have formulas for finding the area of regular figures such
as rectangle, square, circle, polygon, etc. But there is no definite formula to calculate the area under the curve. The process of integration helps solve the equation
and find the required area.
4.2.1
Area with respect to the X axis
Figure 2 shows the area enclosed by the curve and the X axis. The
R 10bounding2 values
for the curve are 10 and 20 on the X axis. the formula being = 0 10x − x + 2 dx
Figure 4.2: Shaded area is the Graphical solution
Thus, if the integral is taken, then the answer would come in square units. Since
42
the integral is taken in ’dx’, we need to calculate the area from the X axis. On
calculation, the area turns out to be 560/3 sq. units.
How to find area under the curve?
• We need to know the equation of the curve (y=f(x)), the limit curve which
the area is the calculated, and the axis enclosing the area.
• Must find the integration (anti-derivative) of the curve.
• We must apply the upper limit and lower limit to the integral and must take
the difference.
4.2.2
Area with respect to Y axis
The figure below shows the area enclosed by the curve, y=20 and the Y axis (In
the first Quadrant).
Figure 4.3: Shaded area is the graphical solution
Here, the area is
4.2.3
R 20
0
sqrt(y) dy Note: ’sqrt’ - square root
Area between 2 curves
The area between 2 curves can be calculated by taking the difference of the areas
of the higher curve and the lower curve. We take the integral of the difference
of the 2 curves and apply the boundaries. Here an example is given (for the first
43
R
R
quadrant) 2x − x2 + 4 dx and x2 dx . Calculate the intersection points first, by
equating both the functions
Figure 4.4: Shaded area is the graphical solution
Here,
the area comes tho the answer - 20/3 sq.units. Try calculating yourself :) R
f (x)–g(x) dx
Important note no.1: If the area of the region beneath the curve and/or axis is
negative, then take the modulus of the solution arrived at.
Important note no.2: For and area which is partially above the axis and partially
below the axis - Take the modulus of the area below the axis and calculate the
area above the axis as normal.
4.2.4
Area between a curve and a line
Similarly as was the case of the area between 2 curves, the area can be calculated
by the difference of integrals of the line and the curve. (Subtract which
R 2 ever is
Rlower from the onw which is higher). Here, we have the example of: x dx and
|x| dx
R
Here, the area turns out to be: 1/3 sq.units. try Calculating yourself :) - f (x)–g(x) dx
44
Figure 4.5: The enveloped area on both sides of the Y axis is the solution
4.3
Applications of This Concept
Now let’s look at some of the practical real world applications of how Area under
Curves is being used...
4.3.1
Applications in Architecture
The concept of ”Area Under Curves” is widely used in architecture for various
purposes. Here are some real-life applications of this concept in architecture:
• Calculation of building’s energy consumption: The area under the curve of
a building’s energy consumption can be used to estimate the building’s total
energy usage. This can help architects and engineers to design buildings that
are energy-efficient and sustainable.
• Design of building facades: The area under the curve of solar radiation and
daylight can be used to design building facades that optimize natural lighting
and reduce energy consumption for artificial lighting.
• Analysis of acoustic properties: The area under the curve of sound intensity
can be used to analyze the acoustic properties of a building and design appropriate sound insulation systems.
• Calculation of building materials: The area under the curve of the building’s
floor plan can be used to estimate the required amount of building materials,
such as tiles, carpets, and paint.
45
• Determination of building loads: The area under the curve of a building’s
wind loads and seismic loads can be used to determine the required structural strength of the building’s foundation, columns, and beams.
• Analysis of thermal comfort: The area under the curve of indoor air temperature and humidity can be used to analyze the thermal comfort of a building
and design appropriate heating, ventilation, and air conditioning systems.
Overall, the concept of ”Area Under Curves” is a useful tool for architects and
engineers in designing and analyzing buildings that are efficient, sustainable, and
comfortable.
4.3.2
Applications in AI
AUC (Area under the ROC Curve)
A receiver operating characteristic curve, or ROC curve, is a graphical plot that
illustrates the diagnostic ability of a binary classifier system as its discrimination
threshold is varied. The method was originally developed for operators of military
radar receivers starting in 1941, which led to its name.
• AUC is one of the most important for measuring the performance of any
classification model. It is a performance measurement for a classification
problem at various thresholds settings.
• The ROC Curve measures how accurately the model can distinguish between
two things (e.g. determine if the subject of an image is a dog or a cat). AUC
measures the entire two-dimensional area underneath the ROC curve. This
score gives us a good idea of how well the classifier will perform.
• AUC is related to another evaluation metric called the CONFUSION MATRIX
• AUC provides an aggregate measure of performance across all possible classification thresholds. One way of interpreting
• AUC is as the probability that the model ranks a random positive example
more highly than a random negative example
4.3.3
Applications in Economics
• Investment and the Stock of Capital Let net investment I is the rate of
change of the stock of capital K. If time is treated as a continuous variable,
we can express this as I(t) Thus, if the rate of investment I(t) is known, the
capital stock K(t) can be estimated through the formula, K(t) = (integral)
I(t) dt
• Obtaining the Total from the Margin Integration helps us recover the total
function from the marginal function if the concerned variable varies continuously. Thus, it will be possible to derive the total functions such as cost,
revenue, production and saving from their marginal functions.
46
• Capital Accumulation Over a Specified Period We may use the definite
integral to find the total capital accumulation during the time interval [a, b].
• Consumer Surplus Consumer surplus is measured as the area below the
downward-sloping demand curve, or the amount a consumer is willing to
spend for given quantities of a good, and above the actual market price
of the good, depicted with a horizontal line drawn between the y-axis and
demand curve.
4.4
Practice Questions
1. Using integration, find the area of the region bounded between the line x = 4
and the parabola y² = 16x.
2. Sketch the region bounded by y = 2x − x2 and x-axis and find its area using
integration.
3. In the area above x-axis, bounded by the curves y = 2kx and x = 0 and x = 2
3
then find the value of k.
is log2
4. Find the area of the region bounded by the line y = 3x + 2, the x-axis and the
ordinates x = -1 and x = 1.
5. Using integration, find the area of the region bounded by the following curves
after making a rough sketch: y = 1+—x+1—, x=-3, x=3, y = 0.
6. Calculate the area under the curve of a function, f(x) = 7-x2, the limit is given
as x = -1 to 2.
7. Find the area under the curve, for the region bounded by the circle x² + y² =
16 in the first quadrant
8. Find the area under the curve, for the region enclosed by the ellipse x²/36 +
y²/25 = 1.
9. Calculate the area enclosed by the curve A = 3x²
10. Find the area of the circle x2 + y 2 = 36 in the first quadrant of the coordinate
axis.
11. Find the area of the region where y²-4x and x²+y²-1 overlap.
12. Calculate with respect to Y axis-Area of the region between x²/9 + y²/4=1;
y=1 and y=0.5.
13. Calculate the area of the region enclosed by 5x²+27x+10; x=5 and the Y axis.
14. An architect wants to design an opening to a room with the ceiling of the door
as a curve. The area the door covers is in between x=5; x=10 and y= x² +10.
Will there be enough space for 5 units tall, 3 units wide person to enter?
15. Formulate a non-linear equation of your choice and calculate the area beneath
the curve with respect to the Y axis.
47
Lecture 5
Riemann Integral
30 january 23 | AKSHAMALIGHA G, VAISHNAVI S, ABHINAYAA K SUBRAMANIAN
Based on
Lectures on Mathematical Methods for Economics II by Dr.Rakesh Nigam
5.1
Definition
The Riemann integral created by Bernhard Riemann, is the simplest integral to
define and it allows one to integrate every continuous function as well as some nottoo-badly discontinuous functions.For many functions and practical applications,
the Riemann integral can be evaluated by the fundamental theorem of calculus or
approximated by numerical integration.
Definition 1
: Called the partition of closed intervals [a, b] denoted by P.
Def1 : a = x0 < x1 < x2 < ... < xi+1 < xn−1 < xn=b , Length of ith = ∆xi
∆x1 = (x1 − x0 ) = (x1 − a)
∆x2 = xn − xn − 1 = (b − xn − 1)
Definition 2
: Length of the widest interval in partition P.
Def2 : λ(P ) = max{∆x1 , ∆x2 ..∆xn }
Definition 3
n
X
i=1
n→∞
Z
x=b
f (i) ∆xi −−−−→ I =
λ(p)→0
Z
y=b
f (x)dx =
x=a
z=b
f (y)dy =
y=a
48
Z
f (z)dz
z=a
y
f(a).................
............................
f(x1 )
..................................
f(x2 )
1
x0
a
a =x0
2
x1
∆x = b − a
n
y = f(x)
3
...
a =x0
x1 =x0 +∆x
x2=x1+∆x
..
.
b = xn =xn-1 + ∆x
n
x2
xn-1
x
b
∆x
Fig 1.1 Definition for Riemann Integral
Approx Area= f(x0 )∆x + f (x1 )∆x + f (x2 )∆x + ... + f (xn-1)∆x
rect 1
rect 2
rect 3
Xn
i=1
rect n
q∆f (xi)
For the above given arbitrary function where, we use n rectangles in the limits ’a’
and ’b’, with an equal width. The width of each rectangle can also be varied but
it makes the calculation of the area more difficult. And use the function evaluated
at the left boundary of the rectangle to determine it’s height. So to approximate
the area we just have to add the area of all the rectangles.
49
5.2
Properties
There are basically three major properties:
• Linearity
The integral linearity is then a measure of the fidelity of the conversion that
is performed by the measuring system.
It is the relation of the output to the input over a range expressed as a percentage of the full-scale measurements.
• Monotonicity
A Riemann-integrable function on [a, b] (with a ¡ b) is continuous almost
everywhere, so in particular it’s continuous at one point;
if this ensures that if the function is strictly positive, then its integral is positive.
• Additivity
The additivity of the Riemann integral, i.e. the property that the integral is
an additive function of the interval of integration, is among the most important properties of Riemann integration.
As it is widely used in the manipulation and simplification of integrals that
appear not only in specific calculations, but also in cornerstone theorems,
such as the fundamental theorem of calculus.
5.2.1
Linearity
If f : [a, b] ⇒ R is integrable and C ∈ ER, then cf is integrable
Z b
Z b
f
cf = c
a
a
And,
Z b
Z b
Z b
f+
g
(f + g) =
a
5.2.2
a
a
Monotonicity
As per this property, if f ⩽ g, then;
Rb
Rb
f⩽ ag
a
5.2.3
Additivity
If a < c < b, then;
Z c
Z b
Z b
f+
f=
f
a
c
50
a
5.3
Questions
Q.1
y = f (x) = α f or x ∈ [a, b]
P
P
A = 4i=1 f (ci )∆xi = α 4i=1 ∆xi
A = Approximation
= α [∆x1 + ∆x2 + ∆x3 + ∆x4]
= α [x1 − a + x2 − x1 + x3 − x2 + b − x3 ]
The function is Riemann integrable.
= α [b − a] (N o error in approximation)
Rb
= a f (x)dx = I
Q.2
Let function f(x) be bounded on a finite interval [a,b]
51
Area in aRf unction of x
x
F (x) = a f (z) dz
Q.3
Though the function is jumping the area is increasing slowly.
Q.4
Show
that the constant function f( x) =1 on [ 0,1] is Riemann integrable , and
R1
1dx
= 1.
0
Solution: Let P = (I1 , I2 , ..., In )
be any partition of [0,1] with endpoints
(0, x1 , x2 , ..., xn−1 , 1)
As f is constant, Mk = supf = 1, mk = inf = 1 f or k = 1, ..., n
and therefore,
Pn
U (f : P ) = L(f : P ) =
K=1 [xk − xk−1 ] = xn − x0 = 1
Thus we see that the sum of the areas of the rectangles under the graph of a
constant function is equal to the area under the graph. So, all upper and lower
sums of f on[0,1] are equal to 1, which implies that the upper and lower integrals
are equal.
Hence it is also proved that the integral of f is 1.
5.3.1
Theorem
Rx
If f is integrable in [a,b], then F( x) =[ a f (y) dy, is a continuous function on [
a,b].
52
5.4
Application
The Riemann Integral finds application in many fields, such as:
• It is used in integration and also used in differential calculus.
• It is used in partial differential equations and in representing functions by
trigonometric series.
• It is used in measuring distance travelled by some body, as we can easily
calculate the average velocity of the journey and total time by the velocity
versus time graph. The distance so travelled is expressed by the area under
the given curve.
Though having major applications, the Riemann integral is quite challenging to
cope with as the definition is a bit sophisticated. So, it is considered a bit inconvenient to make use of the Riemann integral in practical life.
5.5
Summary
The Riemann integrals can be computed only for proper integrals. There are two
types of integrals: definite integral also known as Riemann integral and indefinite
integral also known as anti-derivative. The definite integral (Riemann Integral)
is used in solving many interesting problems in the fields like economics, finance,
probability and more.
53
Lecture 6
Intermediate Value Theorem
03 february 2023 | Vrisha Shah, Nilay Singh ,Rwitaban Guha,Adithi Arunkumar,
Rahul Akshith, Rahul Ramanathan, Jameela Suha
Lectures on Mathematical Methods for Economics II by Dr.Rakesh Nigam
6.1
Definition:
Intermediate value theorem states that if “f” be a continuous function over a closed
interval [a, b] with its domain having values f(a) and f(b) at the endpoints of the
interval, then the function takes any value between the values f(a) and f(b) at a
point inside the interval. This theorem is explained in two different ways
54
6.2
Application 1
We can use the Intermediate Value Theorem (IVT) to show that certain equations
have solutions, or that certain polynomials have roots. For instance, the polynomial
f (x) = x4 + x − 3
is complicated, and finding its roots is very complicated. However, it’s easy to
check that
f (-1) = -3 and
f (2) = 15
Since -3¡0¡15 there has to be a point c between -1 and 2 with
f (c) = 0
In other words, f(x) has a root somewhere between -1 and 2 We don’t know where,
but we know it exists.
6.3
Application 2
The intermediate value theorem has many applications. Mathematically, it is used
in many areas. This theorem is utilized to prove that there exists a point below or
above a given particular line. It is also used to analyze the continuity of a function
that is continuous or not.
This theorem has many implications in Physics and Chemistry problems too. It
is applicable whenever there is a continuously varying scalar quantity with endpoints sharing the same value for a variable. These quantities may be – pressure,
temperature, elevation, carbon dioxide gas concentration, etc.
Intermediate value theorem has some significant real-life applications too. Let us
take an example of a wobbly table due to the uneven ground. In order to fix this,
rotate the table, provided that the ground is continuous; i.e. no ups and downs
due to poorly-fitted tiles.
The wobbly table will have three of its legs touching the ground, while its fourth
leg will be the problem. While rotating the table at a point, the fourth leg will be
below the ground, and at some other point, it will lie above the ground. According
to the intermediate value theorem, there will be a point at which the fourth leg
will perfectly touch the ground, and the table is fixed.
6.4
Application 3
Use IVT to show that there is a root of the given equation in the given interval
f (x) = x2 − x − 12
in the interval of
[3, 5]
f (3) = 32 − 3 − 12 = −6
f (5) = 52 − 5 − 12 = 8
55
f(c)=0
0 = x2 − 2 − 12
0=(x-4)(x+3)
x=4, -3
(Since -3 doesn’t lie between 3 and 5 we consider only 4)
c=4
f (3) < f (4) < f (5) and 3 < 4 < 5
6.5
proof
We need to show F’(x) = f(x) ∀x.
For x = x0 , F ′ (x) = f (x0 ) ∀ x0 ∃[a, b]
(x0 )
F ′ (x0 ) ≡ limx→x0 F (x)−F
x−x0
Rb
F (x) ≡ a f (z)dz
[
+
a
x0
x
x0
= limx→x+0
b
x
R
f (z)dz− ax0 f (z)dz]
x−x0
R
R
[ ax f (z)dz]
[ ax0 f (z)dz]
− x−x0
x−x0
Rx
[ x 0 f (z)dz]
x−x0
F ′ (x0 ) = limx→x+0
= limx→x+0
]
[
Rx
a
limx→x+0 f (cx) → f (x0 )
x0
x+
0
cx
x
x
+
NOTE:- As x → x+
0 , then cx → x0
−
As x → x−
0 , then cx → x0
F’ (x0 ) → f (x+
0)
F’ (x0 ) → f (x−
0)
Given: F is continuous in [a,b]
F is continuous at x0
−
F’(x) = f(x+
0 ) = f (x0 ) = f (x)
⇒ F’(x) = f (x0 ) ∀ x0 ∃ [a,b]
⇒ F’(x) = f (x) ∀ x
56
NOTE:Every
continuous function f(x) has an antiderivative (a.k.a indefinite integral) –
R
f (x)dx
Their antiderivate may not be an elementary function or a closed-form formula.
Examples:
R 2
3
x dx = x3 + C
R
sinxdx = −cosx + C
57
Lecture 7
Newton Leibnitz Formula
03 february 2023 | Adithi Arunkumar, Rahul Akshith, Rahul Ramanathan, Jameela
Suha
Lectures on Mathematical Methods for Economics II by Dr.Rakesh Nigam
Let f (a, b] → R be a continuousf unction
Let F(x) be an indefinite integral of f(x)
R x=b
Then x=a f (x)dx = [F (b) − F (a)] = F (x)|x=b
x=a
Rb 2
3
For example, a x dx = [ x3 ]ba = 31 (b3 − a3 )
7.1
Proof of NLF:
Define G(x) =
Rx
a
f (y)dy
By Fundamental Theorem of Calculus, G’(x) = f(x)
F C → constant
FC subject to F(x) = [G(x)+ C]
F(b)= G(b)+ C
F(a)= G(a) + C
Subtracting both,
F(b) - F(a) = G(b) - G(a)
Rb
Rb
Ra
F
(y)dy
=
f
(y)dy
−
f (y)dy
a
a
a
Example: limx→x0
1
x3
R t=x
t=0
sin2 (3t)dt = 00 (indeterminatef orm)
[LH rule]
R
d
[ t=x sin2 (3t)dt]
dx t=0
x→x0
d 3
x
dx
⇒ lim
⇒ limx→x0
3.sin2 (3x)
9x2
⇒ 3 limx→x0 [ sin(3x)
]2
3x
⇒3
58
Now, one might wonder: why does ’3t’ become ’3x’?
f (t) = t2
Rx
Rx
G(x) = a f (t)dt = a t2 dt = 13 (x3 − a3 )
G(x) = 13 x3 − ( 13 a3 ) → constant
Therefore, G(x) = x2 = f (x)
⇒ Remark : If x is a Reimann integrable function, it doesn’t imply the function f
has an antiderivative.
⇒ Counter − argument :
f (x) =
0
1
-1
0
x 0
=x 1
y
Area = 0
-1
Area = 1
Jump
0
x
1
f (x) → piecewise continuous
But f has no indefinite integral at x = 0.
Since we cannot differentiate at x=0, but f is integrable in the Reimann sense. But
f has no antiderivative (-1, +1).
f (x) = x2 → is both a Reimann integrable and has an antiderivative.
R x=π
Example: x=b xsinxdx = π
Rπ
= [xcosx]πx=o + 0 1(cosx)dx
= (πcosπ − 0cos0) + [−sinx]π0
59
Lecture 8
Improper Integrals
10 February 2023 | Archana T, Bhoshanaa N, Samyuktha N E
Lectures on Mathematical Methods for Economics II by Dr Rakesh Nigam
8.1
Introduction
We have learnt that definite integrals,
Rb
a
f (x), dx, satisfy the following conditions:
– Finite domain of integration [a, b].
– Finite integrand f (x) < ±∞.
For improper integrals, the following issues can arise:
• Infinite limits of integration.
• Infinite discontinuities.
Examples:
R∞
a) 1 x13 , dx
R1
b) 0 x13 , dx
R∞ 1
c) −∞ 4+x
2 , dx
Explanation:
Since in cases a) and c) the intervals are infinite, the Fundamental Theorem of
Calculus (FTC) does not apply.
For case b), a discontinuity exists at x = 0, so FTC also does not apply.
These are Improper Integrals
8.2
8.2.1
If
Rt
a
Types of Improper Integrals
Type 1
f (x), dx exists for every t ≥ a, then
R∞
a
f (x), dx = limt→∞
Rt
a
f (x), dx
provided that the limit exists and is finite.
Rb
If t f (x), dx exists for every t ≤ b, then
Rb
−∞
f (x), dx = limt→−∞
Rb
t
f (x)dx
provided that the limit exists and is finite.
R∞
Rb
The improper integrals a f (x), dx and −∞ f (x), dx are called convergent if the
corresponding limit exists and
if the limit does not exist.
R ∞ is finite, and Rdivergent
a
If for any value of a both a f (x), dx and −∞ f (x), dx are convergent, then we
define
R∞
Ra
R∞
f
(x),
dx
=
f
(x),
dx
+
f (x), dx
−∞
−∞
a
R∞
a
f (x), dx = limt→∞
Rb
−∞
Rt
a
f (x), dx
f (x), dx = limt→−∞
Rb
t
R∞
provided that the limit exists and is finite. The improper integrals a f(x)dx and
Rb
f(x)dx are called convergent if the corresponding limit exists and is finite,
−∞
and divergent if the limit does
R ∞ not exist. R a
If for any value of a both a f (x)dx and −∞ f (x)dx are convergent, then we
define
R∞
Ra
R∞
f (x)dx = −∞ f (x)dx + a f (x)dx
−∞
provided that limit exists and it is finite.
If f (x) ≥ 0, we can give the definite integral above an area interpretation; namely
that if the improper integral converges, the area under the curve on the infinite
interval is finite.
Example:
Determine whether the following integrals converge or diverge:
R∞
R∞ 1
dx, 1 x13 dx.
1 x
By definition,
Rt
limt→∞ 1 x1 dx = limt→∞ [ln |x|]1t = lim t → ∞ ln |t| − ln |1| = limt→∞ ln |t| = ∞.
This integral diverges.
Figure 8.1: Area =
8.2.2
R∞
a
f (x)dx = limN →∞
RN
a
f (x)dx
Type 2
a) If f is continuous on [a, b) and is discontinuous at b, then
Z b
Z t
f (x)dx = lim−
f (x)dx
a
t→b
a
if that limit exists and is finite.
Example: Determine whether the following integral converges or diverges:
Z 2
1
dx
0 x−2
Solution: This integral has a vertical asymptote at x = 2, which means that f (x) =
1
is discontinuous at b = 2. However, f (x) is continuous on the interval [0, 2).
x−2
62
Using the above theorem, we have:
Z 2
Z t
1
1
dx = lim−
dx
t→2
0 x−2
0 x−2
Now, we evaluate the limit of the integral:
Z t
1
lim−
dx = lim− [ln |x − 2|]0t = lim t → 2− [ln |t − 2| − ln 2]t0
t→2
t→2
x
−
2
0
The limit does not exist because the difference of the natural logarithms of two
values that approach 2 from opposite sides is infinite. Therefore, the integral
diverges.
The integral
R2
1
dx
0 x−2
diverges.
Figure 8.2: Discontinuous at b
Example Determine whether the following integral converges or diverges.
Z 2
1
dx
0 x−2
1
is continuous on [0, 2]. f (x) is discontinuous at
• The function f (x) = x−2
R2 1
x = 2. Therefore, we can calculate the integral 0 x−2
dx.
Z 2
1
•
dx = lim ln | x − 2 ||t0
t→−2
x
−
2
0
• lim (ln |t − 2| − ln | − 2|) which does not exist since ln |t − 2| → −∞ as t → 2−
t→−2
Z
2
• Therefore, the improper integral
0
63
1
dx diverges.
x−2
b) If f is continuous on (a; b] and is discontinuous at a, then
Z b
Z b
f (x)dx = lim+
f (x)dx
t→a
a
t
if that limit exists and is finite.
Figure 8.3: Discontinuous at a
Example Determine whether the following integral converges or diverges
Z 1
1
dx
2
0 x
• The function f (x) = x12 is continuous on (0, 1] and discontinuous at 0. Therefore, we can calculate the integral.
R1 1
R1 1
−1 1
• 0 x2 dx = limt→0+ t x2 dx = limt→0+ x t
• lim t → 0+ −1 −
t → 0+
−1
t
= limt→0+
• Therefore the improper integral
1
t
− 1 which doesn’t exist since
R1
1
0 x2
→ ∞ as
diverges
c) If f has a discontinuity at c, where a < c < b and both
dx are convergent, then we define,
64
1
t
Rc
a
f (x) dx and
Rb
c
f (x)
Figure 8.4: Discontinuous at c
Rb
a
f (x)dx =
Rc
a
f (x) dx +
Rb
c
f (x)
Example: Determine if the following integral converges or diverges and if it converges, find its value.
R4 1
, dx
0 (x−2)2
• The function
1
(x−2)2
has a discontinuity at x = 2. Therefore, we must check if
R2 1
R4 1
both improper integrals 0 (x−2)
, dx converge or diverge.
2 , dx and
2 (x−2)2
−1 t
R2 1
Rt 1
=
limt→2− 0 (x−2)
=
limt→2− x−2
• 0 (x−2)
0
=
2 , dx
2 , dx
−1
1
lim t → 2− t−2 − 2 , which doesn’t exist.
R4 1
R2 1
R4 1
• Therefore we can conclude that 0 (x−2)
, dx + 2 (x−2)
3 , dx =
3 , dx
0 (x−2)3
diverges,
since Rthis integral converges only if both improper integrals
R2 1
4
1
, dx and 2 (x−2)
3 , dx converge.
0 (x−2)3
8.3
Applications of Improper Integrals
The following applications of the Improper Integral generalize applications of definite integral that you saw earlier.The strategy is to use the characterization of the
definite integral as the limit of a sum to construct an approximate definite integral
and then to let the upper limit of integration increase without bound.
65
8.3.1
Present Value
Present value of an investment that generates income over a finite period of time
is given by a definite integrals.The present value of an investment that generates
income in perpetuity is given by an improper integral.
Figure 8.5: Present value
Example:
A donor wishes to make a gift to a private college from which the college will
draw Rs 7,000 per year in perpetuity to support the operation of its computer
center. Assuming that the prevailing annual interest rate will remain fixed at 10
percent compounded continuously, how much should the donor give the college?
That is, what is the present value of the endowment?
To find the present value of a gift that generates Rs 7,000 per year for N years,
divide the N-year time interval 0≤t≤N into n subintervals of length ∆n t years
and let tj denote the beginning of the jth subinterval. Then,
Amount generated during the jth subinterval = 7000 ∆n t
Present value of the amount generated during the jth subinterval = 7000
e− 0.1tj ∆n t
Pn
−
Present Value of N-year gift= lim
j=1 7000 e 0.1tj ∆n t
nf i→+∞
RN
= 0 7000e− 0.1tdt
To find the present value of the total gift, take the limit of this Rintegral as N apN
proaches infinity. That is, Present Value of the total gift= lim
70000e− 0.1tdt
0
N f i→+∞
=
lim
N f i→+∞
= lim
N f i→+∞
(−7000e− 0.1t |N
0 )
−70000(e− 0.1N − 1)
= Rs 70,000
8.3.2
Radioactive waste
It is estimated that t years from now a certain nuclear power plant will be producing radioactive waste at the rate of f(t) = 400t pounds per year. The waste
66
decays exponentially at the rate of 2 percent per year. What will happen to the
accumulation of radioactive waste from the plant in the long run?
To solve this problem, we need to use the formula for exponential decay, which is:
A(t) = A0 e−rt
where:
A(t) is the amount of radioactive waste at time t.
A0 is the initial amount of radioactive waste.
r is the decay rate (in this case, 2t is the time elapsed since the waste was
produced.
In this problem, we know that the waste is being produced at a rate of f (t) = 400t
pounds per year. So the amount of waste produced in a time interval of dt is:
df = f (t)dt = 400tdt
To find the total amount of waste produced over a time interval from t to t + dt,
we can integrate df over that interval:
∆A =
R t+dt
t
df =
R t+dt
t
400tdt = 400
R t+dt
t
tdt = 400
h 2 it+dt
t
2
t
= 400 t +
dt 2
2
− 400t2
Now we can use the formula for exponential decay to find the change in the
amount of radioactive waste due to decay over the same time interval:
∆A′ = A(t + dt) − A(t) = A0 e−rt − A0 e−r(t+dt) = A0 e−rt (1 − e−rdt )
In the long run, we are interested in the accumulation of radioactive waste, which
is the integral of ∆A − ∆A′ over all time intervals. We can simplify this expression
using the fact that in the long run, dt becomes very small and we can approximate
∆A and ∆A′ as the rates of change of A and A′ , respectively. This gives us:
dA
dt
−
dA′
dt
= f (t) − rA(t)
Using the formula for f (t) and A(t) from above, we have:
dA
dt
−
dA′
dt
= 400t − 0.02A(t)
This is a first-order linear differential equation that we can solve using standard
techniques. The general solution is:
A(t) = 2r (200t2 + Ce−0.02t )
where C is a constant determined by the initial condition A(0) = A0 . To find the
long-term behavior of A(t), we can ignore the Ce−0.02t term, since it decays to
zero as t goes to infinity. This gives us:
A(t) ≈
400 2
t
r
= 10000t2
This tells us that in the long run, the accumulation of radioactive waste from the
67
plant will continue to increase quadratically with time, despite the exponential
decay of the waste
.
8.3.3
Capital Value
When the notion of the present value is extended to an infinite time interval, the
result is called the capital value of the income stream and is given by,
Capital Value =
R +∞
0
f (t)e− rtdt
where f(t) is the annual rate of flow at the time t, r is the annual interest rate,
compounded continuosly.
Example:
Suppose that an organization wants to establish a trust fund that will provide
a continuous income stream with an annual rate of flow at time t given by
f(t)=10,000. If the interest rate remains at 10 percent, compounded continuously, find the capital value of the fund.
The present value of the fund is the integral of the annual rate of flow, discounted
at theR continuous interest rate, over all time. That is,
∞
V = 0 fe(t)
rt dt
In this case,Rwe have f (t) = 10, 000 for all t, so we can simplify the integral as:
∞
V = 10, 000 0 e−rt dt
This is an improper integral, since the upper limit of integration is infinity. To
evaluate
use the formula for an exponential integral:
R −ax it, we1 can
e dx = − a e−ax + C
where C is a constant of integration. Substituting a = r and integrating from 0 to
infinity, we get:
∞
V = 10, 000 − 1r e−rt 0 =10, 000 1r = 10,000
=100, 000
0.1
The capital value of the fund is 100,000 dollars.
8.4
Practice Problems
1) Find if
R∞
0
e−x dx converges or diverges.
Solution:
Z
∞
Z
−x
e dx = lim
0
b→∞
0
b
e−x dx
= lim [e−x ]0b = lim b → ∞(−e−b + e0 )
b→∞
So the integral converges and equals 1.
68
=0+1 =1
2) Find if
R∞
1
−∞ 1+x2
dx converges or diverges
Solution:
By definition,
Z
∞
−∞
1
dx =
1 + x2
Z
c
−∞
1
dx +
1 + x2
Z
c
∞
1
dx
1 + x2
where we get to pick whatever c we want .Let’s pick c=0.
R0
1
dx
−∞ 1+x2
= lim [arctan(x)]0b
b→−∞
= lim [arctan(0) − arctan(b)]
b→−∞
=0 − lim arctan(b)
b→−∞
= π2
R∞ 1
Similiarly, 0 1+x
dx = π/2
R ∞ 21
Therefore, −∞ 1+x2 dx=π/2 + π/2
=π.
Therefore , it converges.
3) Find if
R2
2x
0 x2 −4
Solution:
The denominator of
x=2.So
R2
2x
dx
0 x2 −4
= lim−
c→2
dx converges or diverges
2x
x2 −4
Rc
is zero when x=2.So the function is not defined when
2x
dx
0 x2 −4
= lim− [ln(x2 − 4)]c0
c→2
(This is solved by substitution of x2 − 4 = t )
lim ln(x2 − 4) − ln(4)
= −∞
c→2−
So the integral diverges.
4) Check the convergence of
R1
1
−1 x2
dx
Solution:
The function f(x)=
compute,
R0
1
dx
−1 x2
+
1
x2
has a discontinuity at x=0,which lies in [-1,1].We must
R1
1
dx
0 x2
69
R1
1
0 x2
dx equals to ∞ which is divergent .
Similarly other is also divergent hence this function is divergent
5) Check the convergence of
R∞
0
(1 + 2x)e−x dx
Solution:
Z
∞
Z
−x
(1 + 2x)e dx = lim
t→∞
0
u = 1 + 2x → du = 2dx,
Z
−x
−x
(1 + 2x)e dx = −(1+2x)e +2
Z
t
(1 + 2x)e−x dx
0
dv = e−x → v = −e−x
e−x dx = −(1+2x)e−x −2e−x +c = −(3+2x)e−x +c
Now we use limits,
Z ∞
(1 + 2x)e−x dx = lim [−(3 + 2x)e−x ]0t = lim t → ∞[3 − (3 + 2t)e−t ]
t→∞
0
= lim 3 − lim [(3 + 2t)/et ] = 3 − lim 2/et = 3 − 0 = 3
t→∞
t→∞
t→∞
Therefore the integral converges at 3.
R1
6) Check the convergence of −5 1/(10 + 2y) dy
Solution:
There exists dicontinuity at y=-5. Now we will have to eliminate the discontinuity
. To do this we see which interval we are interested in that is [-5,1] , these are the
values that are greater than -5 so we use the right hand limit.
R1
R1 1
1
dy = lim+ t (10+2y)
dy
−5 (10+2y)
t→5
R
1
= (10+2y)
dy = 12 ln|10 + 2z| + c
Now taking limits,
= lim+ [ 21 ln | 10 + 2| + c]1t
t→5
= lim+ ( 21 ln | 12 | − 12 ln | 10 + 2t |)
t→5
= 21 ln | 12 | +∞
=∞
Therefore this integral diverges.
70
7) Check the convergence of
R1 √
−∞
6 − y dy
Solution:
Since the question has −∞ we try to eliminate it by,
Z 1p
Z 1 p
6 − y dy = lim
6 − y dy
t→−∞ t
−∞
3
3
2
2 3 2
1
= lim − (6 − y) 2 t
= lim t → −∞ − (5) 2 + (6 − t) 2
t→−∞
3
3
3
2 3
= (5) 2
3 √
10 5
=
3
√
Therefore, the integral converges to 203 5 .
R∞ 9
8) Check the convergence of 2 (1−3z)
4 dz. If yes determine its value.
Solution:
Z
2
∞
9
dz = lim
t→∞
(1 − 3z)4
Z
t
9
dz
4
2 (1 − 3z)
t
1
= lim
t→∞ 3(1 − 3z)3
2
1
1
= lim
−
t→∞ 3(1 − 3t)3
3(1 − 6)3
1
=
3(5)3
1
=
375
Therefore, the integral converges and its value is
9) Check the convergence of
R4
x
0 (x2 −9)
1
.
375
dx. If yes determine its value.
Solution:
For this discontinuity exists at x=2. To eliminate the discontinuity we break the
integral at x=2.
Z 4
Z 2
Z 4
1
1
1
dx =
dx +
dx
2
1 x +x−6
1 (x + 3)(x − 2)
2 (x + 3)(x − 2)
Z t
Z 4
1
1
= lim−
dx + lim+
dx
t→2
s→2
1 (x + 3)(x − 2)
s (x + 3)(x − 2)
Now we integrate this using partial fractions:
71
1
A
B
=
+
(x + 3)(x − 2)
x+3 x−2
1 = A(x − 2) + B(x + 3)
x = 2 : 1 = 5B → A = −1/5
x = −3 : 1 = −5A → B = 1/5
Z =
1
1
−
dx
5(x − 2) 5(x + 3)
x−2
1
ln
+C
5
x+3
1
1
1
t−2
1
2−2
s−2
1
ln
− ln
+ lim+
ln
− ln
= lim−
s→2
t→2
5
t+3
5
4
5
2+3
5
s+3
1
1
1
1
t−2
1
s−2
= lim−
ln
+ ln 4 + lim+ − ln
+ ln 2 − ln 7
t→2
s→2
5
t+3
5
5
s+3
5
5
=
= −∞ + ∞
Hence, these integrals diverge.
10) Check the convergence of
R∞
−∞
6w3 /(w4 + 1)2 dw. If yes determine its value.
Solution:
3
3 1
The antiderivative of (w6w
4 +1)2 is not − 2 w 4 +1 + c. Instead, we can make the substitution u = w4 + 1 to get:
Z
Z
Z
6w3
3
2w2
3
d(w4 + 1)
3 1
dw
=
dw
=
=−
+C
4
2
4
2
4
2
(w + 1)
2
(w + 1)
2
(w + 1)
2 w4 + 1
where C is the constant of integration.
Using this antiderivative, the solution proceeds as follows:
72
0
Z s
6w3
6w3
= lim
dw + lim
dw
t→−∞ t (w 4 + 1)2
s→∞ 0 (w 4 + 1)2
3 1
3 1
0
t + lim s → ∞ −
0s
= lim −
4
4
t→−∞
2w +1
2w +1
3 3 1
3 1
3
= lim t → −∞ − + 4
+ lim − 4
+
s→∞
2 2t +1
2s +1 2
3 3 3
=− + +
2 2 2
3
=
2
Z
Therefore, the integral converges and its value is 23 .
11) Check the convergence of
R4
1
1 x2 +x−6
dx. If yes, determine its value
x2 + x − 6 = (x + 3)(x − 2).
Next, note that the integrand is continuous on the interval [1, 4] except at the
points x = −3 and x = 2.
To check the convergence, we can use the Comparison Test. For x ∈ (1, 4], we have
1
< x12 . Thus,
x2 + x − 6 > x2 , which implies x2 +x−6
4
Z 4
Z 4
1
1
1
3
0≤
dx ≤
dx = −
= .
2
2
x 1 4
1 x +x−6
1 x
R4
Since the integral 1 x12 dx is a convergent p-series with p = 2, we conclude that
R4 1
1
the integral 1 x2 +x−6
dx converges. f (x) = x2 +x−6
and g(x) = x12 . Then, we have
f (x)
x2
= lim 2
= 1,
x→∞ g(x)
x→∞ x + x − 6
lim
since the leading terms in the numerator andR denominator are both x2R. Therefore,
∞
∞
1
by the Limit Comparison Test, if the integral 1 x12 dx converges, then 1 x2 +x−6
dx
also converges.
However, we have
∞
Z ∞
1
1
dx = −
= 1,
x2
x 1
1
which is a finite Rnumber. Therefore, by the Limit Comparison Test, we conclude
4
1
that the integral 1 x2 +x−6
dx diverges.
12) Check the convergence of
R0
−∞
e1/x /x2 dx. If yes determine its value.
73
Solution:
In this question there also exists a discontinuity at x=0 along with infinite limit.
Let us split this integral at x= -1.
Doing this we get ,
Z
0
−∞
−1
Z 0 1/x
e1/x
e
dx +
dx
2
2
−∞ x
−1 x
Z −1 1/x
Z s 1/x
e
e
= lim
dx + lim−
dx
2
2
t→−∞ t
s→0
x
−1 x
1/x 1/x e
e
−1
−
t + lim s → 0 −
−1s
= lim −
t→−∞
x
x
1
1/t
= lim t → −∞ − + e
+ lim− e−1 − e1/s
s→0
e
1
= − + 1 + e−1
e
=1
e1/x
dx =
x2
Z
Therefore, the integral converges and its value is 1.
R0
1
dx. If yes determine its value.
13) Check the convergence of −∞ √3−x
Solution:
= lim
R0
t→−∞ t
√ 1 dx
3−x
√
= lim [−2 3 − x]0t
t→−∞
√
√
= lim [−2 3 + 2 3 − t]
t→−∞
√
=[−2 3 + ∞]
=∞
R0 1
dx= lim t √3−x
dx
t→−∞
√
0
= lim t → −∞[−2 √3 − x]√
t
= lim√
t → −∞(−2 3 + 2 3 − t)
=[−2 3 + ∞]
=∞
So the limit is infinite and so this integral is divergent.
R0
−∞
√1
3−x
14) Check the convergence of
R∞
−∞
2
xe−x dx. If yes determine its value.
74
Solution:
R∞
R0
2
2
2
xe−x dx= −∞ xe−x dx + 0 xe−x dx
R0
R∞
2
2
= lim −∞ xe−x dx + lim 0 xe−x dx
R∞
−∞
t→−∞
t→∞
= lim [−1/2e
t→−∞
−x2 ]0t
+ lim [−1/2e−x
2 ]t
0
t→∞
− 2
= lim (−1/2 + 1/2e t + lim (−1/2e− t2 + 1/2) =-1/2+1/2 =0
t→−∞
t→∞
The integral is convergent and equals 0.
15) Check the convergence of
R∞
−2
sin(x) dx. If yes determine its value.
Solution:
R∞
Rt
sin(x)dx = lim −2 sin(x) dx
−2
t→∞
= lim [−cos(x)]t−2
t→∞
= lim [cos(2) − cos(t)]
t→∞
This limit does not exist and so the integral is divergent.
8.5
Conclusion
Improper Integrals is an integral over an infinite interval or an integral of a function containing an infinite discontinuity on the interval; An improper integral is
defined in terms of a limit. The improper integral converges if this limit is a finite
real number; otherwise, the improper integral diverges.
• Integrals of functions over infinite intervals are defined in terms of limits.
• Integrals of functions over an interval for which the function has a discontinuity at an endpoint may be defined in terms of limits.
• The convergence or divergence of an improper integral may be determined
by comparing it with the value of an improper integral for which the convergence or divergence is known.
R∞
Rt
f
(x),
dx
=
lim
f (x), dx
t→∞
aR
Rab
b
f (x), dx = limt→−∞ t f (x), dx
R−∞
R
R∞
∞
a
f (x), dx = −∞ f (x), dx + a f (x), dx
−∞
75
Lecture 9
Theorem
10 february 2023 | Vrisha Shah, Nilay Singh ,Rwitaban Guha,Rishabh Papnoi
Lectures on Mathematical Methods for Economics II by Dr.Rakesh Nigam
9.1
9.1.1
Mean Value Theorem
Proof:
Consider a line passing through the points, (a, f(a)) and (b, f(b)).
Equation of line is, y – f(a) = f(b) – f(a)/(b-a) . (x – a)
or y = f(a)+ f(b) – f(a)/(b-a) . (x – a)
Let h be a function define difference between any function f and the above line.
h(x) = f(x) – f(a) – f(b)-f(a)/(b-a) . (x – a)
using “Rolle’s theorem”, we have
h’(x) = f’(x) – f(b)-f(a)/(b-a)
Or f(b) – f(a) = f’(x) (b – a). Hence Proved.
76
77
9.2
Applications of Mean Value Theorem
1.|Cosx − Cosy| ≤ |x − y|
By M.V.T
Cosx-Cosy
x-y
wherex ≥ 0, y ≥ 0.
= −sin(c)
Cosx-Cosy
x-y
≥ −1 [The least value Sin(c) can take is -1]
= Cosx − Cosy ≤ x − y
= |Cosx − Cosy| ≤ |x − y| [+ve, if x¿y -ve, if x¡y]
2.Letf (x) = x3 + 7x2 + 2x + 3 on the interval (0,2). Find a value for the number(s)
that satisfies the mean value theorem for this function and interval.
Explanation: The mean value theorem states that for a planar arc passing
through a starting and endpoint (a,b);a¡b, there exists at a minimum one point,
c, within the interval (a,b) for which a line tangent to the curve at this point is
parallel to the secant passing through the starting and end points. In other words,
if one were to draw a straight line through these start and end points, one could
find a point on the curve where the tangent would have the same slope as this
f (b) − f (a)
line. f ′ (c) =
b−a
f (0) = 03 + 7(0)2 + 2(0) + 3 = 3
f (2) = 23 + 7(2)2 + 2(2) + 3 = 43
Then take the difference of the two and divide it by the interval
43 − 3
= 20
2−0
Now find the derivative of the function; this will be solved for the value(s)
found above.
f ′ (x) = 3x2 + 14x + 2
= 3x2 + 14x + 2 = 20
x = −5.72, 1.05
Since the interval is (0,2), x=1.05 satisfies the MVT.
78
9.3
Fundamental Theorem of Calculus
Introduction: The Fundamental Theorem of Calculus is an extremely powerful
theorem that establishes the relationship between differentiation and integration,
and gives us a way to evaluate definite integrals without using Riemann sums or
calculating areas.
9.3.1
First Fundamental Theorem of Integral Calculus
The first part of the calculus theorem is sometimes called the first fundamental
theorem of calculus. It affirms that one of the anti-derivatives (may also be called
indefinite integral) say F, of some function f, may be obtained as integral of f with
a variable bound of integration. From this, we can say that there can be antiderivatives for a continuous function.
Statement: Let f be a continuous function on the closed interval [a, b] and let
A(x) be the area function. Then
A′(x) = f (x), ∀x ∈ [a, b].
Or Let f be a continuous real-valued function defined on a closed interval [a, b].
Let F be the function defined, for all x in [a, b], by:
Rb
F(X)= a f (t) dt
Then F is uniformly continuous on [a, b] and differentiable on the open interval
(a, b), and
F ′ (x) = f (x)∀, x ∈ (a, b)
Here, the F’(x) is a derivative function of F(x).
9.3.2
Second Fundamental Theorem of Integral Calculus
The second fundamental theorem of calculus states that, if the function “f” is
continuous on the closed interval [a, b], and F is an indefinite integral of a function
“f” on [a, b], Rthen the second fundamental theorem of calculus is defined as:
b
f (b) − f (a) = a f (x) dx
Here R.H.S. of the equation indicates the integral of f(x) with respect to x.
f(x) is the integrand.
dx is the integrating agent. ‘a’ indicates the upper limit of the integral and ‘b’
indicates a lower limit of the integral.
The function of a definite integral has a unique value. The definite integral of a
function can be described as a limit of a sum. If there is an anti-derivative F of
the function in the interval [a, b], then the definite integral of the function is the
difference between the values of F, i.e., F(b) – F(a).
Remarks on the Second Fundamental Theorem of Calculus
79
Rb
The second part of the fundamental theorem of calculus tells us that a f (x) dx
= (value of the anti-derivative F of “f” at the upper limit b) – (the same antiderivative value at the lower limit a).
This theorem is very beneficial because it provides us with a method of estimating
the definite integral more quickly, without determining the sum’s limit.
In estimating a definite integral, the essential operation is finding a function whose
derivative is equal to the inte-grand.
However, this process will reinforce the relationship between differentiation and
integration.
Rb
In the expression a f (x) dx, the function f(x) or say “f” should be well defined
and continuous in the interval [a, b].
9.4
Applications
Real-world Examples Calculus’ fundamental theorem can be used to calculate the
area under a continuous graph and the tangent line at every point on the graph.
Calculus is used extensively in physics, chemistry, biology, economics, and a variety
of other subjects. It’s important to realise that the calculus theorem can be used in
a variety of fields, including engineering, mechanics, physics, statistics, and even
medicine. This theorem can be used to investigate a bacterial culture in medicine,
for example. A lot of factors influence the growth of substances. Differential
calculus aids in combining temperature and food sources and determining their
relationship to growth rate (Monet).
Modern business people and researchers can use calculus to increase their profit
margins. Statisticians assess a wide range of data from a variety of sources. Calculus is useful for predicting possible answers to various queries and taking the
necessary steps. Chemistry is another sector where calculus is used and has an
impact on scientists’ work. One common example is the prediction of chemical reaction functions and rates. Information concerning the radioactive decay reaction
was an important milestone in modern existence, according to Monet. Finally, as
80
the field of graphic design expands, the application of calculus rises. Working with
three-dimensional models takes a lot of time and effort.
Rx
1.Find the derivative of g(x)= 1
g’(X)= ( x31+1 )
9.5
1
t3 +1
dt
Questions
Rx
1.Evaluate f’(4) when f(x)= 0 (2t2 + 5t − 4) dx
Rx
2. Evaluate f’(7) when f(x)= 0 f (t2 − 2t + 5) dt
3. Evaluate
Rx
−π/2
f
√
2
sin t + 2
dt
4. Find the derivative of the function
f (x) =
R x2
3
dt
t
R x3
5. Find the derivative of the function f(x)= x2 tdt
81
Lecture 10
Logarithmic Function
17 February 2023 | PRIDIKSHA.S,YUVABUVANI.A,MADHU SREE.NK
Lectures on Mathematical Methods for Economics -2 by Dr. Rakesh Nigam
The logarithmic function is one of the important medium in math calculations.The
method of logarithms was publicly propounded by John Naiper in 1614, in a book
titled ” Mirifici Logarithmorum Canonis Descriptio”.Naiper coined the term of logarithm in Middle Latin ”logarithmus” derived from the Greek, liteally meaning
ratio number. It has wide range of applications in scientific ,astronoical and economics calculations.
10.1
Introduction
In genreal, logarithm of a positive number with respect to a given base is the index
or the power to which the given base must be raised in order to be equal to the
given number. Logarithmic functions are closely related to exponential functions
and are considered as an inverse of the exponential function.The exponent says
how many times to use the number in a multiplication.A logarithm answers the
question”How many of one number multiply together to make another number?”
10.1.1
Definition
A logarithm is defined as the power to which a number must be raised to get some
other values.
logb (x) = c
bc = x
Both the equations describe the same relationship between b,c,x where b is the
base, c is the exponent and x is the argument.
82
EXAMPLES:
• Log2 8 = 3 ⇒ 23 = 8
• log8 4 = 2/3 ⇒ 82/3 = 4
• logb 1 = 0 ⇒ 00 = 1
• log3 1/9 = −2 ⇒ 3−2 = 1/9
• logb ∞ = c ⇒ bc = ∞
10.2
Properties of Logarithm
• logb (xa ) = alogb (x)
Proof:
LHS= logb (xa ) = l
bl = x a
x = (bl )1/a = bl/a
RHS= alogb (x) = r
logb (x) = r/a
br/a = x
br/a = bl/a = x
l=r
logb (xa ) = alogb (x)
Hence proved
EXAMPLE:
log4 53 = 3log4 5
• logb (xy) = logb (x) + logb (y) proof:
RHS:
letlogb (x) = z and logb (y) = w
bz = xandbw = y
LHS:
logb xy = logb (bz bw ) = logb bz+w
= z + wlogb b
=z+w
RHS=LHS=z+w
Hence proved
EXAMPLE
log2 (4)(6) = log2 4 + log2 (2 ∗ 3)
4 = 2 + log2 2 + log2 3
= 2 + 1 + log2 3
83
• logb (x/y) = logb (x) − logb (y) Proof;
LHS=logb (x/y) = logb (xy −1 )
= logb (x) + logb y −1
= logb (x) − logb (y) = RHS
Hence proved
EXAMPLE;
log2 (3/4) = log2 (3 ∗ 4−1 ) = log2 (3) + log2 (4−1 )
= log2 (3) − log2 (4)
• logb (x) = loga (x)/loga (b) (change of base formula for log) for any
a>0 a ̸= 1 Proof;
logb (x) ∗ loga (b) = loga (x)
where logb (x) = z, loga (b) = c and loga (x) = r
l=zc=r (zc=LHS, r=RHS)
bz = x
ac = b
ar = x
RHS:
loga x = loga (bz )
= zloga b
=zc = l
LHS=RHS
Hence proved
EXAMPLE;
log4 (16) = log2 (16)/log2 (4)
=log2 (24 )/log2 (22 )
=4log2 (2)/2log2 (2)
=2
84
10.3
Types of logarithm
• Common logarithm
• Natural logarithm
Common logarithm: The logarithm to the base 10 are called common logarithm.
Usually the base 10 is not written but it is understood that if no base is mentioned
the base is to be taken as 10.
EXAMPLE :
log 1 = 0
log 100 = 2
Natural logarithm: The logarithm of any number to the base e is called natural
logarithm.
loge (X) = ln(X)
ln(X) = c implies x = eC where e = 2.718
NOTE;
eln(x) = x
Proof :
Lety = eln(X)
ln(y) = lneln(x)
ln(y) = ln(x)ln(e)
ln(y) = ln(x)
y=x
Therefore, eln(x) = x
Figure 10.1: exponential and logarithm function
85
10.4
Power Law Behaviour
A power law is a functional relationship between two quantities, where a relative
change in one quantity results in a proportional relative change in the other quantity, independent of the initial size of those quantities: one quantity varies as a
power of another
10.4.1
Log-Log Plot
Log–log plot is a two-dimensional graph of numerical data that uses logarithmic
scales on both the horizontal and vertical axes
Example:
y = Axn Logb y = logb Axn
logb y = logb A + logb xn
logb y = logb a + nlogb x
Taking,
logb y = Y
logb a = C
Y = C + n ,Where n is the slope
Figure 10.2: log-log plot
86
10.4.2
semi-log plot
Semi-logarithmic plot/graph has one axis on a logarithmic scale, the other on a
linear scale. Example:
y = cem t loge (y) = loge (c) + loge emt
Y=C+mt
Figure 10.3: semi-log plot
10.5
Applications of Logarithm
Logarithm has a wide range of applications in economics. They enable economists
to better understand complex economic relationships and make informed economic decisions.The following are some of the applications of logarithm in economics:
10.5.1
COMPOUND INTEREST :
Logarithm is used for determining the unknowns in compound interest calculation.
The formula for compound interest is given as
A = P (1 + r/n)( nt),
87
where A is the final amount, P is the principal amount, r is the annual interest
rate, n is the number of times interest is compounded per year, and t is the time in
years. By taking the logarithm of both sides of this equation, we can solve for one
of the variables.
Suppose compound interest rate is 6 percent, the sum of money will amount to
1000 pounds in 12 years is given by :
A = P (1 + R/100)n
1000 = P (1 + 0.0.6)( 12) = P (1.06)( 12)
⇒ 3 = logP + 12log1.06
⇒ logP = 3 − 12 × 0.0253059
= 2.6963292
⇒ P = 496.97
10.5.2
Economic growth
Logarithmic functions are used to measure economic growth. By using logarithmic
scales, economists can plot economic indicators over time and measure growth
rates accurately.
The rate of growth of a function f (t) over time t is the marginal over the total
function, or
r = f ′(t)/f (t)
But from the rules of derivatives of logarithmic functions, we also obtained that
the ratio f ′(t)/f (t)is the derivative of the natural log of f (t) with respect to time t
. Thus, we obtain for the instantaneous rate of growth r
r = f ′(t)/f (t) = dLnf (t)/f (t)
We can use our knowledge of logarithmic functions to find the rate of growth of
a combination of functions. Let us assume that the function y is a product of two
other functions of time u and v , respectively, such that
y=uv where u =f (t) and v= g(t)
The rules of logarithms allow us to find a relationship for the rates of growth of a
combination of functions. For example, if we take the natural log of both sides of
the equation for the y function, we obtain
lny = lnu + lnv
Differentiating both sides of this equation with respect to time t yields a relationship for the three rates of growth:
dlny/dt = dlnu/dt + dlnv/dt
or ry = ru + rv
where by definition each rate of growth is the derivative of the natural log of the
given function with
respect to time
88
10.5.3
Financial analysis
Logarithms are used extensively in financial analysis to help understand exponential growth rates, compounding periods, and interest rate calculations.
For example, the formula for the present value of a cash flow is given as
P V = F V /(1 + r)n ,
where PV is the present value, FV is the future value, r is the discount rate, and n
is the number of compounding periods. By taking the logarithm of both sides of
this equation we can solve for any of these variables.
To solve for the number of years given the future value, present value, and interest
rate, you can use the following formula:
n = log(F V /P V )/log(1 + r)
Where:
n is the number of years
FV is the future value
PV is the present value
r is the interest rate
To use this formula, you will need to take the natural logarithm
(ln) of both sides of the following equation:
F V = P V ∗ (1 + r)n
This will give you:
ln(F V ) = ln(P V ∗ (1 + r)n )
Using logarithm rules, you can simplify this to:
ln(F V ) = ln(P V ) + n ∗ ln(1 + r)
n = (ln(F V ) − ln(P V ))/ln(1 + r)
So to find the number of years, simply plug in the values of FV, PV, and r into this
formula and solve for n.
10.5.4
Elasticity of demand
Elasticity of demand measures the responsiveness of consumers to changes in the
price of a good. The formula for price elasticity of demand is given as
% change in quantity demanded /% change in price.
By taking the logarithmic derivative of this function, we can obtain a more precise
measure of elasticity that takes into account the nonlinear nature of demand. By
definition, the point elasticity of y with respect to x is the ratio of the marginal
over the average function:
89
Exy] =
dy/dx
y/x
=
dy
dx
×
x
y
where dy/dx × 1/y gives the marginal over the total function.
We know that this result can be expressed in log terms in the form
dy
dx
×
1
y
=
dlny
dx
Substituting in the point elasticity formula, we get
dy
dx
×
x
y
dlny
dx
=
×x
Again, by rules of logarithmic functions, the derivative of a natural log of a function x with respect to x gives dnx/dx=1/x where we substitute for x in the point
elasticity, and we obtain
dy
dlny
dx
Exy] = dy/dx
= dx
× xy = dlny
× x = dlny
× dlnx
= dlnx
y/x
dx
dx
10.6
Practice Questions
QUESTION-1
1/2
Differentiate y = xx
SOLUTION
Take log on both sides
1/2
dy
dx
dy
dx
logy = logxx
× y1 = x1/2 ×
1
= y( x1/2
+
1
x
logx
)
2x1/2
= x1/2 logx
+ logx × 2x11/2
= xx
1/2
(2 +
logx
)
2x1/2
QUESTION-2
√
Differentiate y = x3/4 x2 + 1/(3x + 2)5
SOLUTION
We take logarithm of both sides of the equation and use the laws of logarithms to
simply:
logy = 3/4logx + 1/2log(x2 + 1) − 5log(3x + 2)
differentiating implicitly with respect to x gives
1/y × dy/dx = 3/4 × 1/x + 1/2 × 2x/x2 + 1 − 5 × 3/3x + 2
solving for dy/dx, we get
dy/dx = y(3/4x + x/x2 + 1 − 15/3x + 2)
because we have an explicit expression for y,we can substitute and write
√
dy/dx = x3/4 x2 + 1/(3x + 2)5 (3/4x + x/x2 + 1 − 15/3x + 2)
90
QUESTION-3
The population of the European Union after the last enlargement in 2007 is approximately 300 million people. It is increasing at an annual rate of 1.5%. Express
the function of the European population at time t . What is the doubling time for
it?
SOLUTION
To express the time function of European population we start from the initial period:
P0 = 300, P (1) = 300(1 + 0.015), P (2) = 300(1 + 0.015)2
Thus, t years from the enlargement, European population will grow according to
the function
P (t) = 300(1.015)t
In order for the population to double,
600 = 300(1.015)t
2 = 1.015t
ln2 = tln1.015
t ≈ 46.56
QUESTION-4
Eddie has deposited Rs.4,000 in a bank account that pays 8% annual interest
compounded continuously. He hopes that in 4 years, when he graduates from
college, the value of the account will be Rs.10,000 and he can buy a new car.
When will Eddie’s account be worth Rs.10,000?
SOLUTION
The future value of Eddie’s account is
F (t) = Rs.4000e0.08t
10, 000 = 4000e0.0t
e0.08t = 2.5
Taking the natural logarithm of both sides yields
0.08t = ln2.5
t = ln2.5/0.08 = 0.916/0.08 ≈ 11.45
years is the approximate time when Eddie’s account will reach Rs.10,000.
QUESTION-5
How long will it take money to double in value if invested in a bank that pays 12%
compounded continuously?
SOLUTION
If we assume the value of the deposit to be A , we should have
91
2A = Ae0.12t
2 = e0.12t
ln2 = 0.12t
t = ln2/0.12 ≈ 5
2A = Ae0.12t
2 = e0.12t
ln2 = 0.12t
t = ln2/0.12 ≈ 5.8
years is the approximate time when the value of the deposit will double.
QUESTON-6
Suppose you deposited Rs.100 in a bank that compounds interest continuously,
but you forgot to ask the annual interest rate. All you know is that your deposit of
Rs.100 today will be worth Rs.150 in 5 years.What is the annual rate of interest
that this bank pays?
SOLUTION
Using the future-value formula, we have:
150 = 100E 5r
1.5 = e5r
ln1.5 = 5r
r = ln1.5/5 ≈ 0.08
Therefore, the interest rate that will increase our deposit of RS.100 to RS.150 in 5
years is 8%
QUESTION-7
If a Rs.100 deposit is made at a bank that pays 12% per year, compounded annually, determine how long it will take for the investment to reach Rs.2000.
SOLUTION
P = 100, A = 2000, i = 0.12
A = P (1 + i)n
2000 = 100(1 + 0.12)n
2000/100 = 100(1.12)n /100
20 = 1.12n
log20 = log1.12n
log20/log1.12 = nlog1.12/log1.12
26.4 = n
It will take 26.4 years for Rs.100 investment to be worth Rs.2000
QUESTION-8 kelly invests RS.5000 with a bank. The value of her investment
can be determined using the formula y = 5000(1.06)t ,where y is the value of the
invesetment at time t, in years. Approximately how many years will it take for
Kelly’s investment to reach rs.20 000?
SOLUTION
92
20000/5000 = 5000(1.060t /5000)
4 = 1.06t
log4 = log1.06t
log4/log1.06 = tlog1.06/log1.06
t=23.8 years
QUESTION-9
RS.2000 is invested at 5% per year, compounded semi-annually. How long in
months, will it take for the investment to triple in value?
SOLUTION
P = 2000, A = 3(2000) = 6000, n =?
6000 = 2000(1 + 0.025)n
6000/2000 = 1.025n
3 = 1.025n
log3 = log1.025n = nlog1.025
log3/log1.25 = n
n = 44.5
time=44.5/2=22.25years=267months
QUESTION-10
A national park has a population of 5000 deer in the year 2016. Conservationists
are concerned because the deer population is decreasing at the rate of 7% per year.
If the population continues to decrease at this rate, how long will it take until the
population is only 3000 deer?
SOLUTION
Let y be the number of deer in the national park t years after the year 2016: y = abt
r = −0.07and b = 1 + r = 1 − 0.07 = 0.93 and the initial population is a=5000
The exponential decay function is y = 5000(0.93)t
To find when the population will be 3000, substitute y = 3000
3000 = 5000(0.93)t
Next, divide both sides by 5000 to isolate the exponential expression
3000/5000 = 5000/5000(0.93)2
0.6 = 0.93t
Rewrite the equation in logarithmic form; then use the change of base formula to
evaluate.
t = log0 .93(0.6)
t = ln(0.6)/ln(0.93) = 7.039 years;
After 7.039 years, there are 3000 deers.
QUESTION-11
A video posted on YouTube initially had 80 views as soon as it was posted. The
total number of views to date has been increasing exponentially according to the
93
exponential growth function y = 80e0.2t , where t represents time measured in
days since the video was posted. How many days does it take until 2500 people
have viewed this video?
SOLUTION
Let y be the total number of views t days after the video is initially posted. We
are given that the exponential growth function is y = 80e0.2t and we want to find
the value of t for which y = 2500. Substitute y = 2500 into the equation and use
natural log to solve for t .
500 = 80e0 .12t
31.25 = e0.12t
0.12t = ln(31.25)
t = ln(31.25)/0.12
t = 3.442/0.12
t ≈ 28.7days
QUESTION-12
Suppose that Vinh invests rs.10000 in an investment earning 5% per year. He
wants to know how long it would take his investment to accumulate to rs.12000,
and how long it would take to accumulate to rs.15000.
SOLUTION
y = 10000(1.05)t
y/10000 = 1.05t
t = g(y) = log1.05 (y/10000)
Use the change of base formula to express t as a function of y using natural logarithm
t = g(y) = ln(y/10000)/ln(1.05)
To find the number of years until the value of this investment is RS.12,000, we
substitute y = RS.12,000 into function g and evaluate t:
t = g(12000) =
ln(12000/10000)
ln(1.05)
=
ln(1.2)
ln(1.05)
= 3.74years
To find the number of years until the value of this investment is rs15,000, we
substitute y = rs15,000 into function g and evaluate t :
t = g(15000) =
ln(15000/10000)
ln(1.05)
=
ln(1.5)
ln(1.05)
QUESTION-13
Solve: logx = 2 − log(x − 21)
SOLUTION
logx + log(x − 21) = 2
log(x(x − 21)) = 2
base is assumed to be 10
94
= 8.31years
x(x − 21) = 102 = 100
x(x − 21) = 100
2
x − 21x − 100 = 0
x = 25, −4
QUESTION-14
Find log to the base 10 if Log8 (64) = 2
SOLUTION
10 (64)
log8 (64) = log
log10 (8)
log10 (64) = log8 (64) × log10 (8) = log8 (8)2 × log10 (8)
= 2log8 (8) × log10 (8) = 2 × 0.9031 = 1.8062
QUESTION-15
Find the number of digits in 360
SOLUTION
no of digits in 360 =integral part of (60log3) + 1
(60 × 0.4771) + 1 = 28.6260 + 1 = 29
10.7
Conclusion
In this chapter we saw how to use the natural exponential and logarithmic functions to solve growth problems.We also Understood the concept and calculating
the present value of a future payment or receipt when continuously discounted.We
also came across how plotting graphs of economic and other variables with a log
scale on the vertical axis reveals the growth rate, and how to plot these graphs.We
now thus have understanding of the continuous exponential and logarithmic functions and their relevance to analysing various problems of growth and present
value.
95
Lecture 11
Trigonometry
20 February , 2023 | Ananya , Bhuvi ,Nishitaa
lecture on Mathematical Methods for Economics II by Dr. Rakesh Nigam
11.1
Degrees and Radians
Def: A radian is an angle Q subtended at the centre O of circle when arc length of
circle = s= r.
Circumference of circle = 2π r
π
2πradius − 360 = 1 = 180
π radius = 180
radius = 180
π
π
1◦ = 180
11.2
Trigonometric Functions
a2 + b2 = h2
sinθ = ha
cosθ = hb
tanθ = ab
1
cosecθ = sinθ
1
cotθ = tanθ
1
secθ = cosθ
11.3
Trigonometric Properties
sin2 θ + cos2 θ = 1
tan2 θ + 1 = sec2 θ
1 + cot2 θ = cosec2 θ
96
11.4
Odd and Even functions
y = f (x) If y = f (−x) , f is an even function. Eg: y = x2 = (−x)2
y = g (x) If g (x) = −g (−x) , gisanoddf unction. Eg: y = x3 = (−x)3
97
y = sinx = sinθ
sinθandcosθhaveperiod2π
sinθ = sin (θ + 2πk)
cosθ = cos (θ + 2πk)
If θ = 0, a = 0, sob = h
sin (0) = ha = 0
cos (0) = hb = bb = 1
Positive θ - Anticlockwise
Negative θ - Clockwise
cosθ = hb , cos(−θ) = hb = cosθ − evenf unction
sinθ = ha , sin(−θ) = ha = −sinθ − oddf unction
tanθ = sinθ
= sin(−θ)
= − sinθ
= −tanθ − oddf unction
cosθ
cos(−θ)
cosθ
11.5
Sum Formulae in Trigonometry
sin(x + y) = sinxcosy + cosxsiny
sin(x − y) = sinxcosy − cosxsiny
cos(x + y) = cosxcosy + sinxsiny
cos(x − y) = cosxcosy + sinxsiny
If θ = 0, a = 0, sob = h sin (0) = ha = 0 cos (0) = hb =
Positive θ - Anticlockwise Negative θ - Clockwise
cosθ = hb , cos(−θ) = hb = cosθ − evenf unction
sinθ = ha , sin(−θ) = ha = −sinθ − oddf unction
tanθ = sinθ
= sin(−θ)
= − sinθ
= −tanθ − oddf unction
cosθ
cos(−θ)
cosθ
sin2x = 2sinxcosx
Which can also be written as
98
b
b
=1
sinx = 2sin
x
2
cos
x
2
or
cos2x = cos2 − sin2 = 1
or
1 − 2sin2 x = 2cos2 x − 1
Therefore ,
sin2 x
−
cosx =
2
x
cosx = 1 − 2sin2
2
x
cosx = 2cos2
−1
2
11.6
cos2 x
2
Product Formulae in Trigonometry
(sin(x +y) + sin(x- y) = 2 sinx.cosy)
(sin(x +y) - sin(x- y) = 2 cosx.siny)
(cos(x +y) + cos(x- y) = 2 cosx.cosy)
(cos(x +y) - cos(x- y) = -2 sinx.siny
11.7
Sum Product Formulae in Trigonometry
sinC + sinD = 2sin
sinC − sinD = 2cos
cosC + cosD = 2cos
cosC − cosD = 2sin
C +D
2
C +D
2
C +D
2
C +D
2
99
C −D
2
C −D
2
C −D
2
C −D
2
cos
sin
cos
cos
11.8
Polar Co-ordinates
In mathematics, the polar coordinate system is a two-dimensional coordinate system in which each point on a plane is determined by a distance from a reference
point and an angle from a reference direction. The reference point is called the
pole, and the ray from the pole in the reference direction is the polar axis. Circle
based problems are associated with Polar Co-ordinates.
√
r = [ x2 + 1]
tanθ = xy θ = tan− 1 xy
11.9
Cartesian Co-ordinates
Cartesian coordinates allow one to specify the location of a point in the plane,
or in three-dimensional space. The Cartesian coordinates of a point are a pair of
numbers (in two-dimensions) or a triplet of numbers (in three-dimensions) that
specified signed distances from the coordinate axis.
100
11.10
Applications of Trigonometry
The applications of trigonometry are seen in different fields in real life. Let us
look at all these fields briefly to understand the application of trigonometry better.
11.10.1
Measuring Heights
Trigonometry is used to calculate the height of a mountain or a building. The
height can be determined by measuring the horizontal distance from the base and
finding the angle of elevation to the top. This formation will represent a rightangle triangle with perpendicular, base, and hypotenuse.
11.10.2
Aviation
Aviation technology has evolved in the last few years. It now considers the speed,
direction and distance, as well as the speed and direction of the wind. The wind
plays a vital role in when and how a flight will travel. This equation can be solved
using trigonometry.
For example, if an aeroplane is travelling at 250 miles per hour, 55° of the north
of east, and the wind is blowing southwards at 19 miles per hour. This calculation
can be solved using trigonometry, finding the third side of the triangle that will
lead the aircraft in the right direction.
101
11.10.3
Criminology
Trigonometry is even used in the investigation of a crime scene. The functions of
trigonometry are helpful to calculate a trajectory of a projectile and estimate the
causes of a collision in a car accident. Further, it is used to identify how an object
falls or at what angle the gun is shot.
11.10.4
Video Game Development
Trigonometry gets widely used in game development.
• The sin function can be used to move any object in an oscillating manner.
For example, a character can be programmed to move back and forth.
• A direction vector is made from the user’s input, which can be used for the
character to turn and face towards the direction. This is used a lot in Third
Person Perspective games. To find the angle from vector, we use a function
called Inverse Tangent or arctan, which takes in y/x as input and gives an
angle as output.
11.10.5
Economics and Finance
• The sine and cosine functions are needed in modelling cyclical trends such as
seasonal variation in demand for certain goods. Here, the cyclical behaviour
of trigonometry is essential.
• A natural application of trigonometric functions is in the analysis of spatial
data.
102
• The Fourier series is used in Finance and Econometrics. A Fourier series is
an expansion of a periodic function into a sum of trigonometric functions.
Phases of a Business Cycle
103
Lecture 12
Binomial Theorem
24 february 2023 | Mayoukh Dutta, Mithalia S, Moksha Bhangria
Lectures on Mathematical Methods for Economics II by Dr. Rakesh Nigam
12.1
Introduction
The Binomial Theorem is a special way to multiply two numbers that are being
raised to a power (like squaring or cubing them). It helps us to quickly find out
what the answer will be without having to multiply everything out by hand.
Let’s say we have two numbers, a and b and we want to raise them to a power,
like (a + b)2 . The Binomial Theorem tells us that we can expand this power into a
sum of simpler terms using a special formula.
For example, (a + b)2 = a2 + 2ab + b2 .
This means that when we multiply (a + b) by itself, we get the first term a2 plus
twice the product of a and b, plus the last term b2 .
We can use this formula for any power, like (a + b)3 or (a + b)4 , and it will help us
quickly find the answer without having to do a lot of multiplication.
The following material intends to delve into an in-depth excursion of the idea of
Binomial Theorem and its applications.
12.2
What is the special formula?
Suppose we have two elements a and b raised to the power of n.
Then, (a + b)n = (a + b)(a + b)(a + b) . . . (a + b)
The special formula in question shall help us find (a + b)n with ease. But to get
there, we need some prerequisites – knowing the methods of counting.
12.3
The Art of Counting – Permutation and
Combination
Suppose we have the word “CAT”. In how many ways can we arrange the letters
to form words?
104
We can actually arrange the letters in six ways to form words – namely CAT, CTA,
ACT, TAC, ATC, TCA. It can also be inferred that the word can be arranged in 3!
ways.
These are called the permutations of the word. Here, each arrangement differs
from one another and the order is important.
So, in permutations, the order of the elements matter.
Similarly, we have something called combinations.
It is the selection of items in which the order of the elements does not matter
while arranging.
So, the number of combinations for “CAT”? It is actually 1.
Since the order does not matter, CAT=CTA=ACT=TAC=ATC=TCA while
considering combinations of the three letters.
NOTE: Number of Permutations are always higher than the Number of
Combinations.
Now, suppose there are three places - P, T and S in which to get to T, the routes
are:
(P T1 , T S1 ) → P S1
(P T1 , T S2 ) → P S2
(P T1 , T S3 ) → P S3
(P T2 , T S1 ) → P S4
(P T2 , T S2 ) → P S5
(P T2 , T S3 ) → P S6
Therefore, 3 ∗ 2 = 6 ways.
This is also called the multiplication rule. If an event A can occur in x different
ways and another event B can occur in y different ways, then there are x × y
ways of occurrence of both the events simultaneously.
Now let us take another case of counting. Suppose there are ‘n’ number of chairs
and ‘n’ number of people.
Therefore, chair number 1 can be occupied in ‘n’ ways.
When chair number 2 is occupied, it can be occupied in ‘n-1’ ways and so on, so
forth.
105
Chair number ‘n’ can be occupied in only one way.
Therefore, permutation of the ways the chairs can be occupied = n! =
n(n-1)(n-2). . . 1.
Now, suppose we have ‘r’ number of chairs and ‘n’ number of people, where n > r.
And in that case, chair number ‘r’ can be occupied in ‘n-(r-1)’ ways.
Thus, number of permutations are n(n − 1)(n − 2) . . . (n − r + 1)
=
[n(n − 1)(n − 2) . . . (n − r + 1)] [(n − r)(n − r − 1) . . . 1]
(n − r)(n − r − 1) . . . 1
=
n!
(n − r)!
Therefore, if we have ‘n’ number of objects and have to select ‘r’ number of
n!
objects, the number of permutations is: n Pr = (n−r)!
Suppose we are given ‘n’ number of objects.
Therefore, permutation of all the objects = n!
n!
When we select ‘r’ terms from the objects, the permutation becomes (n−r)!
Now, suppose there are ‘x’ combinations having ‘r’ terms.
Therefore,
x ∗ r! =
=> x =
n!
(n − r)!
n!
r!(n − r)!
Thus, if you have ‘n’ number of objects and
to select ‘r’ number of objects,
have
n
n!
n
the number of combinations is: Cr = r = r!(n−r)! = nPr! r
n
is also known as the binomial coefficient.
r
12.3.1
Deriving the Special Formula
Now, we come back to our pursuit of finding the value of (a + b)n .
When n=1, it becomes (a + b)1 = (a + b)
Similarly, when n=2, it becomes
(a + b)2 = (a + b)(a + b) = 1a2 b0 + 2a1 b1 + 1a0 b2 = a2 + 2ab + b2
(a + b) = F1 & (a + b) = F2
Ways of selecting a.a from F1 & F2 = 1
Ways of selecting a.b from F1 & F2 = 2
Ways of selecting b.b from F1 & F2 = 1
The order doesn’t matter in the binomial theorem as it is a combination.
106
Here, nr and n → total number of a and b in the set (depending on the context)
r → number of a and b selected (depending on the context)
To do the same in the case of (a + b)n , we generalise it using ‘r’ since it is the
varying element.
Thus, we get:
n X
n r n−r
(a + b) =
ab
r
r=0
n
12.4
Pascal’s Triangle
Pascal’s triangle is a triangular array of numbers. It starts with a 1 at the top and
then each row below has more numbers than the row above. Each number in the
triangle is made by adding the two numbers above it. So, for example, the third
row has a 1, then the two numbers above it added together (1+1=2), and then
another 1.
The first few rows of the Pascal’s Triangle look like this:
Now, what’s interesting about Pascal’s Triangle? We can actually use it to find the
coefficients of the binomial expansions. In simpler words, we can use it to
expand our special formula, (a + b)n with even more ease.
The values of the elements in the rows signify the coefficients after expansion and
each row signifies the value of “n” - the first row stands for the expansion when
n=1. And so on, so forth.
For example, (a + b)3 = (a + b)(a + b)(a + b)
k = number of a
(a + b)3 = 1a3 b0 + 3a2 b1 + 3a1 b2 + 1a0 b3
107
n = 3 f actors
n
(a + b) =
n X
n
k=0
k
ak bn−k − > f or a
n X
n
an−k bk − > f or b
n
−
k
k=0
Now, we need to prove
n
n
=
k
n−k
Thus, proved.
108
Problems on Binomial Theorem
1. Use binomial formula to prove that (𝒂 + 𝒃)𝟔 = 𝒂𝟔 + 𝟔𝒂𝟓 𝒃𝟏 + 𝟏𝟓𝒂𝟒 𝒃𝟐 + 𝟐𝟎𝒂𝟑 𝒃𝟑 +
𝟏𝟓𝒂𝟐 𝒃𝟒 + 𝟔𝒂𝟏 𝒃𝟓 + 𝟏𝒂𝟎 𝒃𝟔
Using the Binomial Coefficient formula [ (𝑎 + 𝑏)𝑛 = ∑
𝑛
(𝑛𝑟)𝑎𝑟 𝑏 𝑛−𝑟 ] on LHS, we get:
𝑟=0
(𝑎 + 𝑏)6
6
6
6
6
6
6
= ( ) 𝑎 (6−0) 𝑏 0 + ( ) 𝑎(6−1) 𝑏1 + ( ) 𝑎(6−2) 𝑏 2 + ( ) 𝑎(6−3) 𝑏 3 + ( ) 𝑎(6−4) 𝑏 4 + ( ) 𝑎 (6−5) 𝑏 5
0
1
2
3
4
5
6 (6−6) 6
+ ( )𝑎
𝑏
6
= (60)𝑎6 𝑏 0 + (61)𝑎5 𝑏1 + (62)𝑎4 𝑏 2 + (63)𝑎3 𝑏 3 + (64)𝑎2 𝑏 4 + (65)𝑎1 𝑏 5 + (66)𝑎0 𝑏 6
=(
6!
)𝑎6 𝑏 0 + (
0! 6!
6!
)𝑎5 𝑏1 + (
1! 5!
𝑛!
6!
)𝑎4 𝑏 2 + (
2! 4!
6!
)𝑎3 𝑏 3 + (
3! 3!
6!
)𝑎2 𝑏 4 + (
4! 2!
6!
)𝑎1 𝑏 5 + (
5! 1!
6!
)𝑎0 𝑏 6
6! 0!
[using the 𝑟!(𝑛−𝑟)! formula, substituting the values of ‘n’ and ‘r’]
= 𝑎6 + 6𝑎5 𝑏1 + 15𝑎4 𝑏 2 + 20𝑎3 𝑏 3 + 15𝑎2 𝑏 4 + 6𝑎1 𝑏 5 + 1𝑎0 𝑏 6
Therefore, RHS = LHS.
2. Verify by direct computation that
8
a) (83) = (8−3
)
8
b) (8+1
) = (83) + (3+1
)
3+1
Solution
a) Taking LHS
𝑛!
[using the 𝑟!(𝑛−𝑟)! formula, substituting the values of ‘n’ and ‘r’]
8!
(83) = (3! (8−3)!)
8!
= (3! 5!)
=(
8 𝑥 7 𝑥 6 𝑥 5!
3! 5!
8𝑥7𝑥6
= (3 𝑥 2 𝑥 1)
)
=8x7
= 56
Now taking RHS
[using the
𝑛!
𝑟!(𝑛−𝑟)!
formula, substituting the values of ‘n’ and ‘r’]
8
8!
(
) = (
)
8−3
(8 − 3)! (8 − (8 − 3))!
8!
= (5! (8−8+3)!)
8!
)
= (
5! 3!
= (
8 𝑥 7 𝑥 6 𝑥5!
)
5! 3!
8𝑥7𝑥6
)
= (
3 𝑥 2 𝑥1
= 8𝑥7
= 56
Here LHS = RHS, hence verified
8
b) (8+1
) = (83) + (3+1
)
3+1
Taking RHS
𝑛!
[using the 𝑟!(𝑛−𝑟)! formula, substituting the values of ‘n’ and ‘r’]
8+1
(8 + 1)!
(
) = (
)
3+1
(3 + 1)! (8 + 1 − (3 + 1))!
9!
)
= (
4! (9 − 3 − 1)!
= (
9!
)
4! (9 − 4)!
9!
)
= (
4! 5!
9 𝑥 8 𝑥 7 𝑥 6 𝑥 5!
)
= (
4! 5!
9𝑥8𝑥7𝑥6
)
= (
4 𝑥 3 𝑥 2 𝑥1
=9x2x7
= 126
Taking LHS
𝑛!
[using the 𝑟!(𝑛−𝑟)! formula, substituting the values of ‘n’ and ‘r’]
8
8
8!
8!
( )+(
) = (
)+(
)
3
3+1
3! (8 − 3)!
4! (8 − 4)!
8!
8!
)+(
)
= (
3! 5!
4! 4!
8 𝑥 7 𝑥 6 𝑥5!
8 𝑥 7 𝑥 6 𝑥 5 𝑥 4!
5! 3!
4!4!
= (
)+(
8𝑥7𝑥6
= (
3 𝑥 2 𝑥1
8𝑥7𝑥6𝑥5
)+(
4𝑥3𝑥2𝑥1
)
)
= (8 𝑥 7) + (2 x 7 x 5)
= 56 + 70
= 126
Here LHS = RHS. Hence, verified.
3. Find all the coefficients of the expansion of the polynomial (𝒂 + 𝒃)𝟖 using the Pascals
triangle.
1st row
1 1
2nd row
1 2 1
3rd row
1 3 3 1
4th row
1 4 6 4 1
5th row
1 5 10 10 5 1
6th row
1 6 15 20 15 6 1
7th row
1 7 21 35 35 21 7 1
8th row
1 8 28 56 70 56 28 8 1
According to the Pascals triangle, the coefficients of the expansion of (𝑎 + 𝑏)8 will be the
elements in row 8 of the Pascals triangle.
Answer: The coefficients of the expansion of (𝑎 + 𝑏)8 are 1, 8, 28, 56, 70, 56, 28, 8, 1.
4. Sum of binomial coefficient in a particular expansion is 2048. What is the number of terms in
the expansion?
Given : Sum of binomial coefficient = 2048
From the formula we know that,
2𝑛 = 2048
2𝑛 = 2048 = 211
n = 11
The number of terms in the expansion is 11.
5. Find the coefficient of 𝒙𝟑 in the expansion (2x + 1) (9x + 2) (2x + 3) (x + 1) algebraically.
Algebraically,
(2x + 1) (9x + 2) (2x + 3) (x + 1)
= (18𝑥 2 + 4x +9x + 2) (2x + 3) (x + 1)
= (36𝑥 3 + 54𝑥 2 + 8𝑥 2 + 12x + 18𝑥 2 + 27x + 4x + 6) (x+1)
= (36𝑥 3 + 80𝑥 2 + 43x + 6)(x+1)
= 36𝑥 4 + 36𝑥 3 + 80𝑥 3 + 80𝑥 2 + 43𝑥 2 + 43x+ 6x+ 6
= 36𝑥 4 + 116𝑥 3 + 123𝑥 2 + 49x+ 6
The coefficient of 𝑥 3 in the expansion (2x + 1) (9x + 2) (2x + 3) (x + 1) is 116.
6. Find the coefficient of 𝒙𝟑 in the expansion (2x + 1) (9x + 2) (2x + 3) (x + 1) binomially.
Using binomial theorem we assume the given factors as factor 1, factor 2, factor 3, and factor 4
respectively.
As we need to find the coefficient of 𝑥 3
(2x + 1) - FACTOR 1
(9x + 2) - FACTOR 2
(2x + 3) - FACTOR 3
(x + 1) - FACTOR 4
We multiply these factors in such a way that we’d get 𝑥 3 in all of it.
So, multiplying these factors like
Eq.1 - x-terms in factors 1,2,3 and constant from factor 4
[(2)(9)(2)]1 = 36
Eq.2 - x-terms in factors 4,1,2 and constant from factor 3
[(1)(2)(9)]3 = 54
Eq. 3 - x-terms in factors 3,4,1 and constant from factor 2
[(2)(1)(2)]2 = 8
Eq. 4 - x-terms in factors 2,3,4 and constant from factor 1
[(9)(2)(1)]1 = 18
Adding the equations, we get:
𝑥 3 = 116
7. Find out the fourth member of (𝒂 + 𝒃)𝟓after expansion.
𝑟
Using the Binomial Coefficient formula: [(𝑛𝑟)𝑎(𝑛−𝑟) 𝑏 ]
5
5
5
5
5
5
0
1
2
3
4
5
( ) 𝑎(5−0) 𝑏 + ( ) 𝑎(5−1) 𝑏 + ( ) 𝑎(5−2) 𝑏 + ( ) 𝑎(5−3) 𝑏 + ( ) 𝑎(5−4) 𝑏 + ( ) 𝑎(5−5) 𝑏
0
1
2
3
4
5
5
5
5
5
5
5
0
1
2
3
4
5
= ( ) 𝑎5 𝑏 + ( ) 𝑎4 𝑏 + ( ) 𝑎3 𝑏 + ( ) 𝑎2 𝑏 + ( ) 𝑎 1 𝑏 + ( ) 𝑎0 𝑏
0
1
2
3
4
5
[using the
=(
5!
) 𝑎5 𝑏 0 + (
0! 5!
5!
)𝑎4 𝑏1 + (
1! 4!
5!
𝑛!
𝑟!(𝑛−𝑟)!
)𝑎3 𝑏 2 + (
2! 3!
5!
formula, substituting the values of ‘n’ and ‘r’]
)𝑎2 𝑏 3 + (
3! 2!
5!
)𝑎1 𝑏 4 + (
4! 1!
5!
)𝑎0 𝑏 5
5! 0!
= 1 𝑎5 𝑏 0 + 5 𝑎4 𝑏1 + 10 𝑎3 𝑏 2 + 10 𝑎2 𝑏 3 + 5 𝑎1 𝑏 4 + 1𝑎0 𝑏 5
The fourth member of (𝑎 + 𝑏)5 after expansion is 10 𝑎2 𝑏 3 .
8. Expand (𝟏 − 𝟐𝐱)𝟓
We know that, (𝑎 + 𝑏)𝑛 = nC0𝑎𝑛 +nC1𝑎𝑛−1 𝑏1+ nC2𝑎𝑛−2 𝑏 2 +………+ nCn-1𝑎1 𝑏 𝑛−1 +nCn𝑏 𝑛
Thus, (𝑎 + 𝑏)6 = 5C0𝑎5 +5C1𝑎4 𝑏1 + 5C2𝑎3 𝑏 2 +………+5C5𝑏 5
=
5!
0!(5−0)!
5!
𝑎5 +
5!
1!(5−1)!
5!
𝑎 4 𝑏1 +
5!
2!(5−2)!
5!
𝑎3 𝑏 2 +…+
5!
5!(5−5)!
𝑎0 𝑏 5
5!
= 5!0! 𝑎5 +1!4! 𝑎4 𝑏1 +2!3! 𝑎3 𝑏 2 +…..+5!0! 𝑎0 𝑏 5
5!
5×4!
= 5! 𝑎5 +
4!
𝑎 4 𝑏1 +
5×4×3!
2!3!
5!
𝑎3 𝑏 2 +…..+5! 𝑎0 𝑏 5
= 𝑎5 + 5𝑎4 𝑏1 + 10𝑎3 𝑏 2 +…..+𝑏 5
= 𝑎5 + 5𝑎4 𝑏1 + 10𝑎3 𝑏 2 + 10𝑎2 𝑏 3 +5𝑎1 𝑏 4+𝑏 5 =(𝑎 + 𝑏)5
Now put a=1, b=−2𝑥
We get,
15 +5(1)4 (−2𝑥)+ 10(1)3 (−2𝑥)2+ 10(1)2 (−2𝑥)3 +5(1)1 (−2𝑥)4+(−2𝑥)5
= 1 − 10𝑥 + 40𝑥 2 − 80𝑥 3 + 80𝑥 4 − 32𝑥 5 .
9. Evaluate (𝟗𝟔)𝟑
We know that, (𝑎 + 𝑏)𝑛 = nC0𝑎𝑛 +nC1𝑎𝑛−1 𝑏1+ nC2𝑎𝑛−2 𝑏 2 +………+ nCn-1𝑎1 𝑏 𝑛−1+nCn𝑏 𝑛
Thus, (𝑎 + 𝑏)3 = C0𝑎3 + C2𝑎2 𝑏1 + C3𝑎𝑏 3
3
=
3!
0!(3−0)!
𝑎3 +
3!
1!(3−1)!
3
3
3!
𝑎 2 𝑏1 +
2!(3−2)!
𝑎𝑏 3
= 𝑎3 +3𝑎2 𝑏1 +3𝑎𝑏 2 +𝑏 3
𝑎 = 100, 𝑏 = −4
(100)3+3(100)2 (−4)1 +3(100)(−4)2 +(−4)3
= 1000000 − 120000 + 4800 − 64
= 884736
Thus, 963 = 884736
10. Evaluate (𝟏𝟎𝟐)𝟑
Thus, (𝑎 + 𝑏)3 = 3C0𝑎3 +3C2𝑎2 𝑏1 + 3C3𝑎𝑏 3
=
3!
0!(3−0)!
3!
𝑎3 +
1!(3−1)!
𝑎 2 𝑏1 +
3!
2!(3−2)!
𝑎𝑏 3
= 𝑎3 +3𝑎2 𝑏1 +3𝑎𝑏 2 +𝑏 3
𝑎 = 100, 𝑏 = 2
(100)3+3(100)2 (2)1+3(100)(2)2+(2)3
= 1000000 + 60000 + 1200 + 8
= 1061208
11. Using Binomial Theorem, indicate which number would be larger: (𝟏. 𝟏)𝟏𝟎𝟎𝟎𝟎 or 1000.
(𝑎 + 𝑏)𝑛 = nC0𝑎𝑛 +nC1𝑎𝑛−1 𝑏1+ nC2𝑎𝑛−2 𝑏 2 +………+ nCn-1𝑎1 𝑏 𝑛−1+nCn𝑏 𝑛
(1 + 0.1)10000 = 10000C0𝑎10000 +10000C1𝑎9999 𝑏1 +……….(1 + 0.1)10000
10000!
10000!
= 0!(10000−0)! 𝑎10000 +1!(10000−1)! 𝑎9999 𝑏+………
𝑎 = 1, 𝑏 = 0.1
(1)10000+1000(1)9999 + ⋯ = 1 + 1000 > 1000
Therefore, (1.1)10000 > 1000.
12. Show that 𝟗𝐧+𝟏 − 𝟖𝐧 − 𝟗 is divisible by 64 whenever n is a positive integer.
Numbers divisible by 64 are
64 = 64 × 1
128 = 64 × 2
640 = 64 × 10
Any number divisible by 64 = 64 × 𝑁𝑎𝑡𝑢𝑟𝑎𝑙 𝑁𝑢𝑚𝑏𝑒𝑟
Hence, in order to show that 9𝑛+1 − 8𝑛 − 9 𝑖𝑠 𝑑𝑖𝑣𝑖𝑠𝑖𝑏𝑙𝑒 𝑏𝑦 64,
We must prove that,
9𝑛+1 − 8𝑛 − 9 = 64𝑘, where k is some natural number
Writing (9)𝑛+1 = (8 + 1)𝑛+1
We know that, (𝑎 + 𝑏)𝑛 = C0𝑎𝑛 + C1𝑎𝑛−1 𝑏1+ C2𝑎𝑛−2 𝑏 2 +………+ Cn-1𝑎1 𝑏 𝑛−1+ Cn𝑏 𝑛
n
n
n
n
n
Putting 𝑎 = 1, 𝑏 = 8, 𝑛 = 𝑛 + 1
Thus, (9)𝑛+1= n+1C0 +nC1(8)1+ nC2(8)2 +………+n+1Cn+1(8)𝑛+1…..9𝑛+1 − 8𝑛 − 9 = 64𝑘
Therefore, it is divisible by 64.
13. Prove that ∑𝐧𝐫=𝟎 𝟑𝐫 Cr=𝟒𝐧
n
𝑛
𝑛
(𝑥 + 𝑎) = ∑ 𝑛𝐶𝑟 𝑥 𝑘 𝑎𝑛−𝑘
𝑘=0
Putting x = 3, a = 1 in the above equation:
n
n
(1 + 3) = ∑ nCr 3k 1n−k
n
k=0
(4)n = ∑ nCr 3r
k=0
Therefore, hence proved.
14. Find the coefficient of 𝐱 𝟓 in (𝐱 + 𝟑)𝟖
We know that the general term of expansion (𝑎 + 𝑏)𝑛 =Tr+1= Cr a b
For the general term of expansion (𝑥 + 3)8
Putting a=x, b=3, n=8
Tr+1=8Cr a8-r3r------(1)
We need to find coefficient of 𝑥 5
So, 𝑥 8−𝑟 =𝑥 5
Comparing powers,
8-r=5
r=3
putting r=3 in (1)
8
8-3 3
T3+1= C3 a 3
n
8!
= 5!3! 𝑥 5 (3×3×3)
8×7×6×5!
=
5!3!
𝑥 5 (27)
=1512 𝑥 5
Therefore, the coefficient of 𝑥 5 =1512.
n-r
r
12.5
Applications of Binomial Theorem
Now that we did a brief excursion in Binomial Theorem, it’d be even more
enriching to know how Binomial Theorem is applied in various fields.
12.5.1
In the Distribution of The Internet Protocol Address
The Internet is accessed via a protocol and each device which accesses the
Internet is actually assigned a unique address which is in turn known as the
Internet Protocol Address. Binomial Theorem is used to distribute the IP Address
via a method called variable subnetting.
12.5.2
Used to forecast the economy
Economists use the Binomial Theorem to calculate the probabilities that depend
on numerous and distributed variables to predict the economy in the future.
12.5.3
For ranking
The binomial theorem is used to determine scores and ranks when we take an
exam and wait for the results so you can get into the college of your choosing or
obtain a scholarship for our higher studies. The binomial theorem is also used to
compute the various national rankings we get based on various indices.
118
Lecture 13
PERMUTATION and COMBINATION
24 february2023 | SHIRSHA, HARI, SRISHTI, RAMITA
Lectures on Mathematical Methods for Economics II by Dr.Rakesh Nigam
13.1
introduction
Permutations and Combinations involve counting the number of different
selections possible from a set of objects given certain restrictions and
conditions.It refers to the different ways of arranging a specific group of data.
13.2
Fundamental Principle of Counting
13.2.1
Multiplication Principle
When considering two events, both mutually exclusive (independent), then the
total number of ways to do both events is the number of ways to do the first(m),
multiplied by the number of ways to do the second(n).
13.2.2
Addition
If two events are performed both independent of each other. Then either of the
two operations can be performed in (m+n) ways.
For OR = +
13.3
Permutation
A permutation is an arrangement in a definite order of a number of objects taken
some or all at a time, which suggests permutation is the arrangement of r things
from a set of n things without replacement. Order matters in the permutation.
119
13.3.1
Permutations when all the objects are distinct
Theorem The number of permutations of n different objects taken r at a time,
where 0 ≤ r ≤ n and the objects do not repeat n ( n – 1) ( n – 2). . .( n – r + 1),
which is denoted by n Pr
Proof There will be as many permutations as there are ways of filling in r vacant
places by the n objects. The first place can be filled in n ways; following which,
the second place can be filled in (n – 1) ways, following which the third place can
be filled in (n – 2) ways,..., the rth place can be filled in (n – (r – 1)) ways.
Therefore, the number of ways of filling in r vacant places in succession is n(n –
1) (n – 2) . . . (n – (r – 1)) or n ( n – 1) (n – 2) ... (n – r + 1)
Factorial notation The notation n! represents the product of first n natural
numbers, i.e., the product 1 × 2 × 3 × . . . × (n – 1) × n is denoted as n!. We
read this symbol as ‘n factorial’. Thus, 1 × 2 × 3 × 4 . . . × (n – 1) × n = n !
Some important Properties
(i) n! = n(n-1)!
(ii) 0! = 1! = 1
(iii) factorials of negative integers are not defined.
Derivation of the formula for n Pr
n!
n
,0≤r≤n
Pr = (n−r)
n
Pr =
n(n − 1)(n − 2)...(n − r + 1)(n − r)(n − r − 1)...3 × 2 × 1
n!
==
(n − r)(n − r − 1)...3 × 2 × 1
(n − r)
n!
Thus, n Pr = (n−r)
,0≤r≤n
The number of permutations of n objects, where p objects are of the same kind
and rest are all different =n!/p!
13.4
Combination
A combination is the choice of r things from a set of n things without
replacement. The order does not matter in combination.
Derivation of the formula: n Cr
n!
C(n,r)= n Cr = (n−r)r!
n!
n
Pr = (n−r)
n
Let Cr = a
r!
P (r, r) = (r−r)
= r!
a combination will give rise to a ×r!
P (n, r) = a × r!
(n,r)
a = P (r!)
n!
n
Cr = (n−r)r!
Some important Properties
(i) n Cr is a natural number.
(ii) n C0 =n Cn = 1,n C1 = n
(iii) n Cr =n Cn − r
120
(iv) n Cr +n Cr − 1 =n +1Cr
13.5
Some solved Questions
Question 1. 56 Pr+6 :54 Pr+3 = 30800 : 1, f indr p2
solution. we have,
56
Pr+6
54 P
=
r+3
30800
1
56!
(51 − R)!
30800
×
=
(50 − R)!
54!
1
56.55(51 − R) = 30800 → r = 41
therefore...
r
P2 =41 P2 = 41.40 = 1640.
Question 2. find the number of permutations of the letters of the words ’DADDY
DID A DEADLY DEED’
Solution. here total number of letters is 19, of which there are 9D’s, 3A’s, 2Y’s,
3E’s and rest occur only once.
19!
therefore the required number of permutations = 9!×3!×2!×3!
Question 3. How many words can be formed with the letters of word
’PATLIPUTRA’ without changing the relative positions of the vowels and
consonants?
solution. the word ’PATLIPUTRA’has eleven letters, of which there are 2P’s, 3A’s,
2T’s,1L,1U,1R,1I.Vowels are AAIUA
5!
these vowels can be arranged in 3!
= 20ways.
6!
the consonants are PTLPTR which can be arranged in 2!×2!
= 180ways
therefore the required number of words=20 × 180 = 3600ways
Question 4. Find the Numbers of permutations that can be had from the letters
of the words ’OMEGA’
i) O and A occupying end places ii) E being always in the middle iii) Vowel
occupying odd places iv) Vowels being always apart.
solution. there are five letters in the word ’OMEGA’
i)when O and A occupy end places i.e., M,E,G,(OA)
The first three letters (M,E,G) can be arranged by 3! = 6 ways and last two letters
(O,A) can be arranged by2!= 2 ways
therefore total number of such words = 6 × 2 = 12ways
ii) When E is fixed in the middles then there four places left to be filled by four
remaining letters O,M,G and A, this can be done in 4! ways.
therefore total number of such words = 4!=24 ways.
three vowels (O,E,A) can be arranged in the odd places in 3! ways (1st, 3rd and
5th) and the two consonants (M,G) can be arranged in the even places in 2! ways
(2nd and 4th)
121
therefore the total numbers of such words = 3! × 2! = 12ways
iv) total number of words 5!=120 combine the vowels into one bracket as (OEA)
and treating them as one letter we have three letters viz (OEA), M, G and these
can be arranged in 3! ways and three vowels within the bracket can be arranged
in 3! ways too.
therefore the number of ways where the vowels come together
= 3! × 3! = 36ways.
hence number of ways in which vowels remain apart from each other = 120-36=
84 ways.
Question 5. 20 persons were invited to a party. In how many ways can they and
the host be seated at a circular table ? In how many of these ways will two
particular persons be seated on either side of the host?
Solution. 1st part.: total persons on the circular table = 20 guest + 1 host = 21
they can be seated in (21-1)!= 20! ways.
2nd part: after fixing the places of three persons (1 host + 2 persons). treating
(1 host + 2 persons)=1 unit, so we have now [(remaining 18 persons +1
unit)=19] and the number of arrangement will be (19-1)!=18! also these two
particular persons can be seated on either side of the host in 2! ways.
hence the number of ways of seating 21 persons on the circular table such that
two particular persons are seated on either side of the host = 18! × 2! = 2 × 18!
Question 6. find the number of ways in which 12 different beads can be
arranged to form a necklace.
solution. 12 different beads can be arranged among themselves in a circular
order in (12-1)!=11! ways. Now in the case of necklace there is no distinction
between clockwise and anticlockwise arrangements. so the required
arrangements = 21 × 11!
Question 7
Prove that product of r consecutive positive integers is divisible by r!
Solution
Let r consecutive positive integers be
(m), (m + 1), (m + 2), ...., (m + r − 1)wherem ∈ N
∴ P roduct = m(m + 1)(m + 2)...(m + r − 1)
=
(m − 1)!m(m + 1)(m + 2)...(m + r − 1)
(m − 1)!
=
(m + r − 1)!
(m − 1)!
=
(m + r − 1)!
r!.
(r!(m − 1)!
=
r!.m+r−1 Cr
122
Question 8
If n Cr = 36, n Cr = 84 and n Cr = 126, then find r.
Solution
Here
n
84
Cr
=
nC − 1
36
r
n−r+1
7
=
r
3
3n − 3r + 3 = 7r
10r − 3n = 3.............................(1)
and
n
n − (r + 1) + 1
Cr + 1
126
=
=
nC
(r + 1)
84
r
n−r
3
=
r+1
2
2n − 2r = 3r + 3
5r − 2n = −30
10r − 4n = −6.............................(2)
Subtracting (2) from (1) we get
n=9
10r − 27 = 3
10r = 30
r=3
Question 9
In how many ways can a cricket team of eleven players be chosen out of a batch
of 15 players if
• A particular player is always chosen
• A particular player is never chosen?
Solution (i)Since a particular player is always chosen it means that 11 − 1 = 10
players are chosen out of remaining 15 − 1 = 14 players Therefore required
number of ways = 14 C1 0= 14 C4
=
14.13.12.11
= 1365
1.2.3.4
(ii)Since a particular player is never chosen. It means that 11 players are selected
out of remaining 15 − 1 = 14 players. therefore required number of ways =
14
C1 1=14 C3
14.13.12
= 364
1.2.3
123
Question 10
in how many ways can a pack of 52 cards be formed into 4b groups of 13 cards
each ?
Solution
Here order of groups is not important, then the number of ways in which 52
different cards can be divided equally into 4 groups is
52!
4!(13!)4
Question 11
A bag contains 23 balls out of which 7 are identical. Then find the number of
ways of selecting 12 balls from the bag.
Solution
Here n = 23, p = 7, r = 12(r > p) Therefore required number of selections = Σ12
r=5
16
Cr
=16 C5 +16 C6 +16 C7 +16 C8 +16 C9 +16 C1 0 +16 C11 +16 C12
= (16 C5 +16 C6 ) + (16 C7 +16 C8 ) + (16 C9 +16 C1 0) + (16 C11 +16 C12 )
=17 C6 +17 C817 C10 +17 C12
17
C11 +17 C917 C10 +17 C12
(17 C11 +17 C12 )(17 C9 +17 C10 )
18
C1 2 +18 C10
18
C6 +18 C8
Question 12
Find the number of non negative integral solutions of x1 + x2 + x3 + 4x4 = 20
Solution
Number of non negative integral solutions integral solutions integral solutions of
the given equation coefficient of
x20 in(1 − x)−1 (1 − x)−1 (1 − x)−1 (1 − x4 )−1 )
= coefficient of
x20 in(1 − x)−3 (1 − x4 )−1 )
= 1 +6 C4 +10 C8 +14 C12 +18 C16 +22 C20
= 1 +6 C2 +10 C2 +14 C2 +2 C2 +22 C2
6.5 10.9 14.13 18.17 22.21
+
+
+
+
=1+
1.2
1.2
1.2
1.2
1.2
= 1 + 15 + 45 + 91 + 153 + 231
= 536
Question 13
124
Find the exponent of 3 in 100!.
Solution
In terms of prime factors 100! can be written as 2a .3b .5c .7d ...
Now,
100 100 100 100
+ 2 + 3 + 4
E2 (100!) =
3
3
3
3
= 33 + 11 + 3 + 1 = 48
Question 14. find the sum of all five digit numbers that can be formed using the
digits 1,2,3,4 and 5( repetitions of digits not allowed)
5
solution. required sum + (5 − 1)!(1 + 2 + 3 + 4 + 5)( 10 9−1 )
=24.15.11111
=3999960
Question 15.six x’s have to be placed in the squared of the figure below,such that
each row contains at least one x.In how many different ways can this be done?
solution. the required no. of ways
= coefficient of x6 in(2 C1 x +2 C2 X 2 )(4 C1 X +4 C22 +4 C3 X 3 +4 C4 X 4 )(2 C1 x +2 C2 X 2 )
=Coefficient of X 3 in(2 = x)2 (4 = 6x + 4x2 + x3 )
= coefficient of x3 in(4 + 4x + x2 )(4 + 6x + 4x2 + x3 )
=4 + 16 + 6
=26
13.6
Applications
13.6.1
Probability
Combinations have an important implication in probability theory.There are
many such cases where combinations are considered and used for finding the
125
probability of a certain event. Let’s understand this with an example.Suppose ,a
bag contains 5 green and 7 red balls . If 2 balls are drawn,what is the probability
that one ball is green and the other is red? In order to solve this,we must have a
basic idea of combinations. Since it is given that there are 5 green balls and 7 red
balls,As we know that probability of an event is favourable outcome divided by
total outcome,the probability of getting one green and one red will be
= 5 C1 ×7 C1 /12 C2 = 35/66.
13.6.2
Binomial theorem
Binomial theorem pops up from the idea of permutations and combinations,
which says (a + b)n =n Σnr=0 Cr an bn−r
13.6.3
Business
In business decision-making, combinations are used in the selection of the
available objects. In contrast, permutations are used in the arrangement of such
objects that are opted through combinations. In such a way, permutations and
combinations help the management of a company make decisions by minimizing
the variables in selecting the profitable option from the available objects to
proceed with business activities.
126
Lecture 14
Hyperbolic Functions
3 March 2023 | Aastha Agarwal, Chirag Sridhar, Nandan Nair, and Sumana
Alladi
Lectures on Mathematical Methods in Economics II by Dr Rakesh Nigam
14.1
Introduction
Hyperbolic functions are a family of functions which are analogous to the
traditional trigonometric functions. What do we mean by this? Well, the
traditional trigonometric functions sin(x) and cos(x) are such that if we
parameterize x and y to be dependent on them, or, in mathematical terms, such
that x = cos(t) and y = sin(t), and plot this on the Cartesian plane varying t, a
unit circle is formed. Using this same principle, if we parameterize the functions
cosh(x) and sinh(x), where
cosh(x) =
ex − e−x
ex + e−x
, sinh(x) =
2
2
and draw a curve on the Cartesian plane such that x = cosh(t) and y = sinh(t), it
would result in a unit hyperbola. Therefore, hyperbolic functions defined in this
way are the parameters that satisfy the condition that varying the independent
variable and plotting the resulting dependent variables would form a unit
hyperbola. The three basic hyperbolic functions are the two defined above and a
third, tanh(x), defined as
tanh(x) =
ex − e−x
sinh(x)
= x
cosh(x)
e + e−x
These functions are used in a wide variety of fields for multiple purposes. In this
chapter, you will learn about the functions’ various characteristics, identities and
rules that apply to them, their applications, and how to solve problems related to
them. Also, there will be an exercise available at the end of this chapter
containing problems that you can solve.
127
14.2
Features of Hyperbolic Functions
The two functions that form all three basic hyperbolic functions are y =
−x
y = e 2 . This is how they look:
ex
2
and
Figure 14.1: the 2 base function
The blue line Figure 1 is the graph
e−x
y=
2
and the red line is
ex
2
These are the two ’base functions’ which make up the three basic hyperbolic
functions. It is essential that we understand the behavior of these functions, so
here are a few things to note about them. (1) As x → ∞, y1 = e−x → 0 and
y2 = ex → ∞, (2) As x → −∞, y1 = e−(−∞) → ∞ and y2 = e−∞ → 0, (3) They
meet at the point (0,0.5), and (4) They are both positive ∀x.
Now that we have sufficiently defined and described these base functions, let us
look at the graphs of each of the three basic hyperbolic functions, sinhx, coshx,
and tanhx, and some of their other characteristics including their asymptotic
properties.
y=
128
14.2.1
Cosh(x)
Figure 14.2: graph of cosh(x)
The addition of the amplitudes of the two curves in Figure 1 gives us the line
y = cosh(x), as defined in the introduction. This is the first basic hyperbolic
function, and Figure 2 is its graph. Some things to note about this function are
(1) it is an even function or in mathematical terms, f (−x) = f (x), which is
visually apparent because of its reflect-ability, (2) it is concave upward as its
slope is continually increasing, or in simple words, it takes the shape of a U, (3) it
dy
= 0, which is at (0,1) because at x = 0,
has a singular stationary point, where dx
x
−x
e
e
y for both y = 2 and y = 2 are y1 = y2 = 0.5, and cosh(x) is the addition of
amplitudes of these two functions at any given point and (4) the function is
positive ∀x and therefore the point (0,1) is a global minimum. Let us examine
what happens to this function when x → ∞ and x → −∞
x → ∞,
x → −∞,
e−x
e−∞
ex + e−x
ex
e∞
=
→ 0, y = cosh(x) =
→
=
→∞
2
2
2
2
2
ex
e−∞
ex + e−x
e−x
e−−∞
e∞
=
→ 0, y = cosh(x) =
→
=
=
→∞
2
2
2
2
2
2
The above facts (1) reinforce the previous fact that cosh(x) seems to be positive
∀x in a more reliable way than visual, and (2) tell us that this function is
129
unbounded as y = cosh(x) stretches to ∞ in both directions and does not have an
endpoint.
14.2.2
Sinh(x)
Figure 14.3: graph of sinh(x)
We will now move on to the next hyperbolic function, sinh(x). This is made up of
x
−x
the two base functions y = e2 and y = −e2 , so we can use the same principle of
superposition of these two curves to obtain sinh(x) graphically.
x
−x
The difference between the amplitudes of the functions y = e2 and y = e 2 gives
us the line y = sinh(x), as defined in the introduction. This is the second basic
hyperbolic function, and Figure 3 is its graph. Some things to note about this
function are (1) it is an odd function or in mathematical terms, f (−x) = −f (x),
x
−x
(2) its stationary point is at (0,0) because at x = 0, y for both y = e2 and y = e 2
are y1 = y2 = 0.5, and sinh(x) is the difference of amplitudes between these two
functions at any given point and (3) the function stretches both above and below
the x-axis, so it is neither fully positive or negative ∀x, therefore the point (0,0) is
a point of inflection. Let us examine what happens to this function when x → ∞
and x → −∞
x → ∞,
−e−x
−e−∞
ex − e−x
ex
e∞
=
→ 0, y = sinh(x) =
→
=
→∞
2
2
2
2
2
130
x → −∞,
ex
e−∞
ex − e−x
−e−x
−e−−∞
−e∞
=
→ 0, y = sinh(x) =
→
=
=
→ −∞
2
2
2
2
2
2
The above facts (1) reinforce the previous fact that sinh(x) stretches infinitely
both above and below y = 0 in a more reliable way than visual, and (2) tell us
that this function is unbounded as y = sinh(x) stretches to ∞ on the right and
−∞ on the left and does not have an endpoint.
14.2.3
tanh(x)
We will now move on to the third and last basic hyperbolic function, tanh(x).
Figure 14.4: graph of tanh(x)
Let us first derive the simplified form of y = tanh(x), using the first definition in
the introduction, which was that y = tanh(x) = sinh(x)
cosh(x)
x
−x
( e −e
)
sinh(x)
ex − e−x
2
= x
= ex +e
−x
cosh(x)
e + e−x
( 2 )
This is the third basic hyperbolic function, and Figure 4 is its graph. Some things
to note about this function are (1) it is an odd function or in mathematical terms,
131
f (−x) = −f (x), (2) it is neither fully positive nor negative ∀x, as is visually
apparent, (3) it seems to be approaching one non-infinite value as x → ∞ and a
different non-infinite value when x → −∞, and (4) because of the previous
observation, it seems to have only one real root at the point (0,0). Let us examine
what value this function approaches when x → ∞ and x → −∞
x → ∞, tanh(x) =
e∞
e∞ − 0
e∞ − e−∞ −∞
=
,
e
→
0,
tanh(x)
→
=1
e∞ + e−∞
e∞ + 0
e∞
e−∞ − e−−∞
e−∞ − e∞ −∞
=
,e
→ 0,
e−∞ + e−−∞
e−∞ + e∞
0 − e∞
−e∞
tanh(x) →
=
= −1
0 + e∞
e∞
From this we see that as x → ∞, tanh(x) tends to 1 and as x → −∞, tanh(x)
tends to -1. This shows us (1) that tanh(x) is bounded unlike sinh(x) and
cosh(x), (2) that |tanh(x)| < 1, or that tanh(x) is a fraction ∀x. Why is this true?
We can prove this through two methods.
, we can prove the above fact
Method 1: Because we defined y = tanh(x) as sinh(x)
cosh(x)
if we prove that cosh(x) > sinh(x)∀x. This is easily done:
x → −∞, tanh(x) =
cosh(x) − sinh(x) =
ex + e−x ex − e−x
e−x
−
=
2
2
2
We know that ez > 0 ∀z, and if t > 0 then
t
2
> 0 ∀t, therefore
e−x
> 0∀x
2
Because the difference between cosh(x) and sinh(x) is always positive, sinh(x)
has to be smaller than cosh(x).
Method 2:
ex − e−x
tanh(x) = x
e + e−x
All terms in the above identity are positive, proven in the same way as proven
earlier. Therefore, the fact that the numerator is subtracting e−x from ex , but the
denominator is adding e−x to ex proves that numerator is less than denominator
∀x.
Note also about y = tanh(x) that a small part of the curve which lies close to the
origin on both sides looks like a linear function, or a straight line if you zoom into
the origin enough.
14.3
Hyperbolic Identities and Rules
There exist a set of hyperbolic identities which we can use to prove/disprove any
claim. They may seem familiar, because their overall structure mimics that of
common trigonometric identities, but there are some slight differences which we
will examine. In this section, we will discuss 3 of these many identities and one
set of rules.
132
14.3.1
Identities
Identity 1
The first and arguably simplest identity is cosh2 (x) − sinh2 (x) = 1. Let us prove
this result using the definitions of cosh(x) and sinh(x).
cosh2 x − sinh2 x = (
ex − e−x 2
ex + e−x 2
) −(
)
2
2
1
= ((ex + e−x )2 − (ex − e−x )2 )
4
1
= (e2x + 2ex e−x + e−2x − (e2x − 2ex e−x + e−2x )
4
1
1
= (2ex e−x + 2ex e−x ) = (2 + 2) = 1
4
4
Identity 2
The second identity is sinh(2x) = 2sinh(x)cosh(x), which we will prove this by
simply substituting 2x for x into y = sinh(x)
1
1
sinh(2x) = (e2x − e−2x ) = (ex + e−x )(ex − e−x )
2
2
1
= (2cosh(x))(2sinh(x)) = 2cosh(x)sinh(x)
2
Identity 3
The third identity is cosh(2x) = cosh2 (x) + sinh2 (x). We will prove this from the
RHS to LHS, as it is simpler to illustrate that way
cosh2 (x) + sinh2 (x) = (
ex − e−x 2
ex + e−x 2
) +(
)
2
2
1
= (e2x + 2ex e−x + e−2x ) + (e2x − 2ex e−x + e−2x )
4
1
1
= (2e2x + 2e−2x ) = (e2x + e−2x ) = cosh(2x)
4
2
Reciprocals
Just like trigonometric functions, there exist a set of 3 more hyperbolic functions
which are the reciprocal of the basic hyperbolic functions, named in much the
same way as the trigonometric functions
1
1
1
cosh(x)
= cosech(x),
= sec(x),
=
= coth(x)
sinh(x)
cosh(x)
tanh(x)
sinh(x)
133
14.3.2
Osborn’s Rule
As the identities above indicate, there are many similarities between hyperbolic
and trigonometric identities. But there are key differences that do not seem
intuitive. To make it simpler for us to convert proven trigonometric identities to
hyperbolic ones, we use Osborn’s rule. This rule acts as a type of legend, telling
us what some basic trigonometric terms can be converted to in order to simply
and quickly obtain the hyperbolic counterpart of the identity. The set of rules:
+cos(x) → +cosh(x)
+sin(x) → +sinh(x)
+sin2 (x) → −sinh2 (x)
This means that given any arbitrary trigonometric identity with any of the LHS
terms, you can obtain the corresponding hyperbolic identity by substituting the
corresponding RHS term in its place. This is apparent in the identities above.
Note: the third rule applies to any term that has a product of sines, even if it is
not sin2 (x). This means that sin(a)sin(b) → −sinh(a)sinh(b). Let us exemplify
this. We know from trigonometry that
cos(x + y) = cos(x)cos(y) − sin(x)sin(y)
Using Osborn’s rule by substituting the RHS with LHS terms in the above identity,
we get
cosh(x + y) = cosh(x)cosh(y) + sinh(x)sinh(y)
Let us prove this backwards, starting with expanding the RHS:
cosh(x)cosh(y) + sinh(x)sinh(y) = (
=
ex − e−x ey − e−y
ex + e−x ey + e−y
)+(
)
2
2
2
2
ex+y + ex−y + ey−x + e−x−y ex+y − ex−y − ey−x + e−x−y
1
+
= (2ex+y + 2e−(x+y) )
4
4
4
1
= (ex+y + e−(x+y) ) = cosh(x + y)
2
14.4
Hyperbolic Equations
The questions you may need to solve are often in the form of hyperbolic
equations. If you keep all the rules and definitions in mind, these should be very
simple for you to solve. Because of this, we will only work out one example here
before moving on.
Ex: Solve the equation sinh(x) = 43 and find x
Answer - There are two methods to solve this:
Method 1 - by converting the question into an identity (using identity 1)
3
9
cosh2 x − sinh2 x = 1 = cosh2 x − ( )2 = cosh2 x −
4
16
134
9
25
5
3+8
= cosh2 x = , cosh(x) = , sinh(x) + cosh(x) =
=2
16
16
4
4
ex + e−x + ex − e−x
= ex = 2, x = ln2
2
Method 2 - Through directly substituting sinh(x) into the equation
1+
3
3
ex − e−x
= , ex − e−x =
2
4
2
3ex
= 0 = 2e2x − 3ex − 2
2
2x
x
x
x x
= 2e − 4e + e − 2 = 2e (e − 2) + 1(ex − 2) = (2ex + 1)(ex − 2)
e2x − 1 −
ex = 2, x = ln2
Note: the other root of this quadratic equation is negative, therefore we disregard
it, as ex cannot be negative by nature. All hyperbolic equations can be solved
using the same principles given above.
14.5
Applications of Hyperbolic Functions
One physical application of hyperbolic functions involves hanging cables. If a
cable of uniform density is suspended between two supports without any load
other than its own weight, the cable forms a curve called a catenary. High-voltage
power lines, chains hanging between two posts, and strands of a spider’s web all
form catenaries. Figure 5 shows a chain hanging from two posts, forming a
hyperbolic curve.
Satellite systems make heavy use of hyperbolas and hyperbolic functions. When
scientists launch a satellite into space, they must first use mathematical equations
to predict its path. Because of the gravity influences of objects with heavy mass,
the path of the satellite is skewed even though it may initially launch in a straight
path. Using hyperbolas, astronomers can predict the path of the satellite to make
adjustments so that the satellite gets to its destination.
Objects designed for use with our eyes make use of hyperbolas. These objects
include microscopes, telescopes and televisions. Before you can see a clear image
of something, you need to focus on it. Your eyes have a natural focus point that
does not allow you to see things too far away or close up. To view such things as
planets or bacteria, scientists have designed objects that focus light into a single
point. The designs of these use hyperbolas to reflect light to the focal point.
When using a telescope or microscope, you are placing your eye in a
well-planned focal point that allows the light from unseen objects to be focused
in a way for you to view them.
An Activation Function decides whether a neuron should be activated or not.
This means that it will decide whether the neuron’s input to the network is
important or not in the process of prediction using simpler mathematical
operations. In neural networks, as an alternative to a sigmoid function, a
hyperbolic tangent function could be used as an activation function. When you
135
Figure 14.5: A Catenary
backpropage, the derivative of the activation function would be involved in the
calculation of error effects on weights. The derivative of a hyperbolic tangent
function has a simple form just like a sigmoid function. This explains why
hyperbolic tangents are common in neural networks.
14.6
Practice Problems
The following problems will help you revise and reinforce the concepts discussed
earlier in this chapter. They also draw upon concepts discussed in previous
chapters, so make sure to be thorough with integration, differentiation, and
limits before attempting this exercise.
1. Consider the function limx→0 tanh(x). What happens if L’Hospital’s rule is
followed?
2. Show that cosh(x) ≥ 1 + 21 x2 ∀x
3. If tanh(x) = − 35 then 83 sinh(x) =?
136
4. Evaluate limx→0 sinh(ln(x))
5. Consider the function y = sech−1 (x + 1). Find
6. Given sinh(x) =
5
,
12
dy
dx
and simplify if possible.
find (a) cosh(x), (b) tanh(x), (c) sech(x), (d) coth(x)
7. Given that cosh(x) = 45 , determine the 2 possible values of sinh(x)
8. Solve the following equations for x
(a) 3cosh2 (x) − 7sinh(x) + 1 = 0
(b) 20cosh(2x) − 21sinh(x) = 200
(c) 4cosh(x) − e−x = 3
9. For f (x) = sinh−1 (x), find
e− 1
(a) f ( 2 e )
(b) find the domain and range of f (x)
(c) is f (x) an odd or even function?
10. For f (x) = tanh−1 (x), find f ( 45 )
11. Evaluate (a)
d
tanh−1 (sin(x))
dx
and (b)
d
csch−1 ( x1 )
dx
12. Determine the equation of the tangent of y = cosh−1 x at x =
13. For f (x) = tanhx, evaluate f (−ln2) for x > 0
14. Evaluate
d
sinh(x
dx
+ a)
R
R
2
15. Evaluate (a) sech(lnx)dx and (b) xcosh xa2 dx
137
5
4
Lecture 15
Maxima and Minima and L’Hopital
Rule
06 march 2023 | Puvan Karthik, Aryambikhaa, Shivani
Lecture on MathematicAL Method for Economics II by Dr.Rakesh Nigam
15.1
maxima and minima
Figure 15.1: Maxima and Minima
y = f (x)
At, x=a
Max; f ′ (x)x=a = 0f ′′ (x)x=a < 0
At,x=b
Min;f ′ (x)x=b = 0f ′′ (x)x=b > 0
The first order derivative is zero because the slope of the tangent is zero at x=a
and x=b
138
Figure 15.2: Increasing Function
For an increasing function the first derivative is positive
Figure 15.3: Integration
For an decreasing function the first derivative is negative
Let,
g(x) = f ′ (x)
Asx from a− to a+ g(x) is decreasing, so the property of a decreasing function
For a, it goes from +ve to -ve
For b, it goes from -ve to +ve.
139
g(x) is increasing, so the property of an increasing function g’(x) > 0
Figure 15.4: No Derivative
Some functions don’t have a derivative of zero so you need to look at the end
points to find maxima and minima
15.2
APPLICATIONS OF MAXIMA AND MINIMA
There are numerous practical applications in which it is desired to find the
maximum or minimum value of a particular quantity. Such applications exist in
economics, business, and engineering. Many can be solved using the methods of
differential calculus described above. For example, in any manufacturing
business it is usually possible to express profit as a function of the number of
units sold. Finding a maximum for this function represents a straightforward way
of maximizing profits.In other cases, the shape of a container may be determined
by minimizing the amount of material required to manufacture it. The design of
piping systems is often based on minimizing pressure drop which in turn
minimizes required pump sizes and reduces cost. The shapes of steel beams are
based on maximizing strength.
140
15.3
L’Hopital’s Rule
According to LH rule,
f (x)
is of intermediate
x→a g(x)
Suppose f(x) and y(x) are differentiable at x=a , if lim
form
∞/∞
Then we can write
it as
f ′ (x)
f (x)
= lim ′
lim
x→a g (x)
x→a g(x)
15.3.1
Proof of L’Hopital’s Rule
suppose lim f (x) = 0, lim g(x) = 0
x→a
x→a
f (x)
f ′ (x)
lim
= 0 ⇒ lim ′
x→a g(x)
x→a g (x)
x−a
f ′ (x)
= lim [f (x) − f (a) ÷
− g(a) ÷ (x − a)]
lim ′
x→a
x→a g (x)
g(x)
f (a)
lim [f (x) −
− g(a)]
x→a
g(x)
f (x)
lim
x→a g(x)
LHS = RHS
Ex-1
lim x2 + 1÷(2x − 1 ) =3
x→1
Ex-2
lim sin x ÷ log x
x→0
Ex-3
lim log x ÷ (x − 1) (intermediate form)
x→1
15.4
Taylor’s series
f(x) is a smooth function i.e there is derivatives of all orders.
Let,
a0 , a1 , a2 , a3 , ........................., an depend of f(x)
f (0) = a0 + a1 x + a2 x2 + a3 x3 + ..................... + an xn + ...................
a0 = f (0)
df /dx = a1 + 2a2 x + 3a3 x2 + ................... + nan xn−1 ......
a1 = f ′ (0)
d2 f (x)/dx2 = f ′′ (x) = 2a2 + (3)(2)a3 x + ............. + n(n − 1)an xn−2
f ′′ (x)x=0 = 2a2
a2 = 1/2f ′′ (0)
d3 f (x)/dx3x=0 = f ′′′ (x)x=0 = f 3 (x) = (3)(2)a3 + ........... + n(n − 1)(n − 2)xn−3
f 3 (x) = 6a3
141
a3 = 1/6f 3 (0)
an = 1/n!f (n)
f (x) = f (0) + f ′ (0) + 1/2f ′′ (0) + 1/6f ′′′ (0) + ................ + 1/n!f n (0)
This is the taylor series expansion of f(x) around at x=0
Figure 15.5: Graphical representation of Taylor series
15.5
Error in Approximation
e(x) = f (x) − y(x)
e(x) = f (x) − xf ′ (0) − f (0)
lim e(x) → 0
x→0
lim f (x) → f (0)
x→0
Consider,
f (x) − y(x)
e(x)
=
x
x
e(x)
f (x)
=
− f ′ (0) − f (0)
x
x
e(x)
f (x) − f (0)
=
− f ′ (0)
x
x−0
e(x)
f (x) − y(x)
f (x) − f (0)
lim
= lim
= lim
− lim f ′ (0)
x→0 x
x→0
x→0
x→0
x
x−0
f (x) − f (0)
lim
→ f ′ (0)
x→0
x−0
lim f ′ (0) → f ′ (0)
x→0
′
f (0) − f ′ (0) = 0
limx→0 ex /x → 0 (strong)
142
15.6
Exercise
f (x) = sinx
= sin0 + xcos0 −
3
= 0 + x − x6
3
= x − x6
x2
sin0
2
−
x3
cos0
6
+
x4
sin0
24
143
Lecture 16
taylor series
13 March 2023 | Kavinaya K
Lectures on Mathematical Methods For Economics II by Dr. Rakesh Nigam
16.1
Introduction
Taylor series is the polynomial or a function of an infinite sum of terms. Each
successive term will have a larger exponent or higher degree than the preceding
term. The above Taylor series expansion is given for a real values function f(x)
where f’(a), f”(a), f”’(a), etc., denotes the derivative of the function at point a. If
the value of point ‘a’ is zero, then the Taylor series is also called the Maclaurin
series.
16.2
application of taylor series
The uses of the Taylor series are:
Taylor series is used to evaluate the value of a whole function in each point if the
functional values and derivatives are identified at a single point. The
representation of Taylor series reduces many mathematical proofs. The sum of
partial series can be used as an approximation of the whole series. Multivariate
Taylor series is used in many optimization techniques. This series is used in the
power flow analysis of electrical power systems.
16.2.1
Taylor’s series about x = 0
f (x) = f (0) + xf ′ (0) +
16.2.2
x2 ′′
2! f (0)
+ ........ +
Taylor’s series about x = a
z = (x − a) ⇒ Z = 0
144
xn n
n! f (0)
+ ......
16.2.3
Taylor’s series about z = 0
2
n
f (z) = f (0) + z 1 f ′ (0) + z2! f ′′ (0) + ........ + zn! f n (0) + ......
2
(x−a)n ′′′
′′
f (x) = f (a) + (x − a)f ′ (a) + (x−a)
f
(a)
+
.....
2!
n! f (a) + ....
example:1
F (X) = x3 − 4x + 5, n = 3anda = 2
f ′ (x) = 3x2 − 4 ⇒ 3(2)2 − 4 = 8
f ′′ (x) = 6x ⇒ 6(2) = 12
f ′′′ (x) = 6
2
′′
= f (a) + (x − a)f ′ (a) + (x−a)
2! f (a) +
16.3
(x−a)3 ′′′
3! f (a)
inverse function
Is a 2nd function that undoes the work of the 1st function
if f (x) = y −→ 1st function
then g(x) = f −1 (y) = x∀x −→ 2nd f unction
note:
f −1 of anything reverse the operation f(.) does on x
EXAMPLES:
1. f (x) = 4x −→ f multipliesxby4
f ′ (x) = 41 x −→ f −1 x by 4
2. f (X) = 3x + 2
x → 3x → 3x + 2 = f (x)
(3x + 2) → 3x → x = f ′ (x) = x−2
3
145
3. The graph of
16.4
f (x)
f ′ (x)
is basically a reflection or interchanged y and x
Tangent equation
y − y1 = m(x − x1 )
y − f (a) = f ′ (a)(x − a)
yt (x) = f (a) + f ′ (a)(x − a)
note: if x in very near a then
dy ⋍ ∆y
∆y ⋍ dy
146
16.5
Maxima and Minima
At maxima/minima f’(x) changes sign
147
maxima/minima⇒ F ′ (c) = 0(or)F ′ (d) = 0
iff ′ (c) = 0 ⇏ max/min
16.5.1
Other scenarioes
f(x) is defined on interval D=[a,b]
f’( )→ need not exist for it to be maxima,minima
148
16.6
critical point
It is a point such that either f’(c)=0 or f’(c) doesn’t exist or if D =
[a,b] includes the end points a,b or critical point
16.6.1
example:
y=f(x)=—x— defined on D=[-1,2] find all critical point
critical point : x=-1,0,2 ( f’(0) doesn’t exist )
max : x=1
min : x= 1
149
dxdy → r(drdθ)
(x, y)p→ (r, θ)
r = x2 + y 2
θ = tan−1 ( xy )
x = rcosθ
y = rsinθ
δ(x,y)
δ(r,θ) = J(jacobian matrix)
J = | δ(x,y)
δ(r,θ) |
δx δx
δr δθ = cosθ −rsinθ = rcos2 θ
= δy
δy
sinθ rcosθ
δr δθ
dxdy
drdθ
=
=r
150
+ rsin2θ
Lecture 17
taylor series in 2D
14 March 2023 | Adithi Arunkumar, Rahul Akshith,
Rahul Ramanathan, Jameela Suha
Lectures on Mathematical Methods for Economics II by
Dr.Rakesh Nigam
17.1
Taylor Series in 2D (x,y)
17.1.1
Taylor series about t = 0
2
n
g(t) = g(0) + 1!t g ′(0) + t2! g ′′(0) + ... + tn! g n0 + ...
Idea - map 2D → 1D
y
1
t= (a, b)
t 1
0
(x,y)
(x(t), y(t))
(x0, y0)
x
x(t) = Q0x0 + t(a − x0) = dx
dt = (a − x0 )
dy
y(t) = y0 + t(b − y0) = dt = (b − y0)
151
g(t) = f(x,y) = f(x(t) , y(t))
g(t) =f [x0 + t(a − x0), y0 + t(b − y0)]
df dx
df dy
g’(t) = dg
dt = dz dt + dy dt
g’(t) = fxxt + fy yt
= (a − x0)f x + (b − y0)f x
g’(t) = (a − x0)f x (x(t) y(t)) + (bx − y0)f g(x(t), y(t))
g’(0) = (a − x0)f x (x(0) y(0)) + (b − y0)f (y)(x(0)y(0))
g’(0) =(a − x0)f x (x0, y0) = (b − y0)f g(x(0), y(0))
Since f(x,y) is continuous, fxy = fyx
g”(t) =
d2 y
dt2
dy
dy
dx
= (a − x0)[fxx dx
+
f
]
+
(b
−
y
)[f
+
f
xy
0
yx
yy
dt
dt
dt
dt ]
= (a − x0)2f xx + (a − x0)(b − y0)f xy + (b − y0)(a −
x0)f yx + (b − y0)2f yy
2
g”(t) = ddt2y =
(a − x0)2f xx + 2(a − x0)(b − y0)f xy + (b − y0)f yy
g”(0) = (a − x0)2f xx(x0y0) + 2(a − x0)(b − y0)f xy(x0y0) +
(b − y0)2f yy(x0y0)
Coming back to the equation:
g(t) = f (x, y) =
f (x0, y0) + [(x − x0)f x (y0) + (y − y0)f y (x0y0)]+
− x0)2fxx (x0y0) + 2(x − x0)(y − y0)fxy (x0y0) +
(y − y0)2f yy (x0y0)]
→
−
x →
x − →
−
X =
; −
x0 0 →
g (−
x0 ) = →
g (x0, y0)
y
y0
x
−
p(x, y) = →
r =
y
1
2 [(x
152
p
→
−
r = | r | = x2 + y 2
θ = tan−1( xy )
1
0
=x
+y
0
1
x
0
x+0
x
=
+
=
=
0
y
0+y
y
17.2
Gradient Vector
→
−
→
−
−
−
g (→
x0 ) = →
g
(x
,
y
)
=
∀ f (x0, y0)
0
0→
−
f x( x0 )
→
−
−
−
g (→
x0 ) =
= gradient vector of f at →
x0
→
−
f y( x0 )
−
−
f (→
x ) = f (→
x0 )
→
(x
−
x
)
T
0
−
−
−
−
−
−
x −→
x0 ) = [f x(→
x0 ), f y(→
x0 )]
g (→
x0 ) (→
(y − y0)
→
−
−
= (x − x0) f x( x0 ) + (y − y0) f y(→
x0 )
T
→
− ) = f (→
−
−
−
−
−
x0 ) + (→
g (→
x0 )) (→
x −→
x0 )
f( x
0 a b x
= ax2 + bxy + xyb + xy 2
x y
b c y
T −
−
−
−
−
−
−
−
f (→
x ) = f (→
x0 ) + [ →
g (→
x 0)] (→
x −→
x0 ) + 12 (→
x −→
x0 )
−
−
−
H(→
x 0)(→
x −→
x 0)
......
−
−
Multidimensional Taylor around →
x =→
x0
f (x, y) = (x2 + y 2)
1
→
−
x0 =
1
df /dx
2x
x
→
−
−
g (→
x0 ) =
=
=2
df /dy →
2y
y
−
→
x =−
x
0
153
=
−
2→
x 
→
−
→=1
x =−
x
0
1
2
=2
=
1
2
1
f xx f xy
α 0
−
H(→
x 0) =
=
Change in f at
f xy f yy →
0
α
−
−
x =→
x0
→
−
→
−
→
−
−
x 0 = ∆f ( x 0) = f ( x ) − f (→
x 0)
T
−
−
−
−
−
−
−
= [→
g (→
x 0)](→
x −→
x 0) + (→
x −→
x 0) H(→
x 0)
→
−
→
−
1. if x 0 is maxima: ∆ is negative, then ∆f ( x 0) < 0
−
−
since f (→
x ) < f (→
x 0)
→
−
−
2. if x 0 is minima: ∆ is positive, then ∆f (→
x 0) > 0 since
−
−
f (→
x ) > f (→
x 0) →
− 0
→
−
−
g (→
x 0) = 0
0
→
−
−
W LOG =⇒ →
x0 = 0
−
−
2
x H→
x = x2( f xx
∆f = 12 →
2 ) + xy(f xy) + y (f yy)
−
A = fxx (→
x 0)= 2a
−
B = fxy (→
x 0)= b
→
−
C = fyy ( x
0) = 2c f
xx
f
xy
A
B
−
H(→
x 0) =
=
f xy f yy
B C
→
−
|H( x 0)| = (AC − B 2) = (2a)(2c) − b2 = (4ac − b2)
∆f = ax2 + bxy + cy 2 = y 2[a( xy )2 + b( xy ) + c]
t = x/y
∆f = y 2[at2 + bt + c]
Quadratic equation at2 + bt + c = 0 = (b2 − 4ac)
Saddle point =⇒ both maxima and minima
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