17.1 - Equivalence Classes

In the previous section, we saw equivalence relation. We saw one solved example also. In this section , we will see a few more solved examples. Later in this section, we will see equivalence classes.

Solved example 17.3
Let L be the set of all lines in a plane. R is a relation in L.
R = {(L₁,L₂) : L₁ is perpendicular to L₂}. Show that R is symmetric but neither reflexive nor transitive.
Solution:
• Given that, L is the set of all lines in a plane. There will be infinite number of lines. Let us number them as: 1,2,3,4, . . .
All those lines will be present in L.
• R is a relation in L. So we must consider L × L
• (1,1), (1,2), (1,3), (1,4), . . . , (2,1), (2,2), (2,3), . . .  are all elements of L × L.
There will be infinite elements in L × L
1. First we check whether R is reflexive. It can be written in 3 steps:
(i) (1,1) is not eligible to be included in R. This is because, the first line is not perpendicular to itself. In fact, no line is perpendicular to itself.
(ii) In this way, (2,2), (3,3), (4,4), . . . are not eligible to be included in R.
(iii) Therefore, R is not a reflexive function.
2. Now we check whether R is symmetric. It can be written in 3 steps:
(i) Suppose that (3,9) is eligible to be included in R.
• Then it means that, the third line is perpendicular to the ninth triangle.
(ii) Now, (9,3) is an element of L × L.
• The element (9,3) is eligible to be included in R. This is because:
If third line is perpendicular to the ninth line, then ninth line will be perpendicular to the third.
(iii) So in general, if (L₁, L₂) is an element of R, then (L₂,L₁) will also be an element of R
• Therefore, R is a symmetric relation.
3. Now we check whether R is transitive. It can be written in 4 steps:
(i) Suppose that (4,7) is eligible to be included in R.
• Then it means that, the fourth line is perpendicular to the seventh line.
(ii) Also suppose that (7,10) is eligible to be included in R.
• Then it means that, the seventh line is perpendicular to the tenth line.
This situation is shown in fig.17.1 below:

Fig.17.1

(iii) It is clear that, fourth line is not perpendicular to the tenth line. In fact, fourth line is parallel to the tenth line.
So (4,10) is not eligible to be included in R.
(iv) In general, if both (L₁,L₂) and (L₂,L₃) are elements of R, then (L₁,L₃) will not be an element of R.
Therefore R is not a transitive relation.
4. We see that:
R is symmetric but neither reflexive nor transitive.
• So R is not an equivalence relation.

Solved example 17.4
Show that the relation R in the set {1,2,3} given by R = {(1,1), (2,2), (3,3), (1,2), (2,3)} is reflexive but neither symmetric not transitive.
Solution:
1. Let us check whether R is reflexive:
• There are three elements in the original set.
• All those three elements are present in reflexive form in the set R.
They are: (1,1), (2,2) and (3,3)
• Therefore, R is reflexive.
2. Let us check whether R is symmetric:
• (1,2) is present in R. But (2,1) is not present.
Therefore, R is not symmetric.
3. Let us check whether R is transitive:
• (1,2) and (2,3) are present in R. But (1,3) is not present.
Therefore, R is not transitive.

Solved example 17.5
Show that the relation R in the set Z of integers given by:
$\rm{R = \{(a,b) : 2 ~\text{divides}~ a-b\}}$
is an equivalence relation.
Solution:
• We have to consider the set Z. It is the set of integers. Integers are those numbers which do not have fraction or decimal parts. They can be +ve or -ve. Zero is also an integer. There are infinite integers. All of them will be present in Z.
• R is a relation in Z. So we must consider Z × Z.
• . . . (-1,0), (-1,1), (-1,2), (-1,3), (-1,4), . . . , (2,0), (2,1), (2,2), (2,3), . . .  are all elements of Z × Z.
There will be infinite elements in Z × Z

1. First we check whether R is reflexive. It can be written in 6 steps:
(i) (-1,0) is not eligible to be included in R. This is because, (-1 + 0) is not divisible by 2.
$\frac{-1 + 0}{2} = \frac{-1}{2}$. The quotient $\frac{-1}{2}$ is not an integer.
(ii) (-1,1) is eligible to be included in R. This is because, (-1 + 1) is divisible by 2.
$\frac{-1 + 1}{2} = 0$. The quotient zero is an integer.
(iii) (-1,2) is not eligible to be included in R. This is because, (-1 + 2) is not divisible by 2.
$\frac{-1 + 2}{2} = \frac{1}{2}$. The quotient $\frac{1}{2}$ is not an integer.
(iv) By doing a few calculations like these, we get to know some facts:
• For (a,b) to be eligible,
   ♦ Both a and b must be odd
   ♦ OR both a and b must be even.
   ♦ The sign of a and b does not matter.
(v) Based on this information, we can easily write some eligible pairs:
(-7, -3), (-4, -10), (-8, 12), (3, -7), (11,13), etc,.
(vi) It is clear that all elements of the form (a,a) are eligible. This is because, the difference (a-a) will be always zero.
• Some examples are: (-17,-17), (2,2) (5,5) etc.,
• Since all (a,a) are eligible, we can say that R is reflexive.

2. Now we check whether R is symmetric. It can be written in 3 steps:
(i) Consider an example: (-5,9).
• This pair is eligible.
(ii) Then (9,-5) is also eligible. We are just interchanging the positions. Both will be the same odd numbers. Or both will be the same even numbers. When (a-b) is calculated after the interchange, only the sign will change. Divisibility by 2 will not be affected.
(iii) We can write:
If (a,b) is an element of R, then (b,a) will also be an element of R.
• Therefore, R is a symmetric relation.

3. Now we check whether R is transitive. It can be written in 3 steps:
(i) Consider an example: (-5,9) and (9,11)
Both pairs are eligible.
(ii) (-5,11) is also eligible. This can be explained as follows:
The first pair consists of odd numbers.
‘9’ is common in both pairs.
So the second pair will also consist of odd numbers.
(iii) We can write:
If (a,b) and (b,c) are eligible, then (a,c) will be eligible.
• Therefore R is a transitive relation.

4. We see that:
R is reflexive, symmetric and transitive.
• So R is an equivalence relation.

5. In this problem, we must note an important point. It can be written in 2 steps:
(i) Consider the condition:
Both coordinates in the pair must be even numbers.
• For such pairs, one of the coordinates can be zero. So (0,2) (0,8), (0, -12), (16,0), (-24,0) etc., are acceptable.
• We can never use ‘1’ as one of the two coordinates. Because ‘1’ is an odd number.
• The smallest eligible pair is (0,2). We can go on adding 2. All numbers thus obtained can be used to form eligible pairs. We get:
0, 2, 4, 6, 8, . . .
• The smallest eligible pair is (0,2). We can go on subtracting 2. All numbers thus obtained can be used to form eligible pairs. We get:
0, -2, -4, -6, -8, . . .
• Combining the two, we get the set:
{. . . -8, -6, -4, -2, 0, 2, 4, . . . }
• This is the set of even numbers. We can denote it by the letter 'E'. Take any two elements from E. The ordered pair formed by those two numbers will be eligible.

(ii) Consider the condition:
Both coordinates in the pair must be odd numbers.
• For such pairs, one of the coordinates can be ‘1’. So (1,3) (1,9), (1, -11), (17,1), (-23,1) etc., are acceptable.
• We can never use ‘0’ as one of the two coordinates. Because ‘0’ is an even number.
• The smallest eligible pair is (1,3). We can go on adding 2. All numbers thus obtained can be used to form eligible pairs. We get:
1, 3, 5, 7, 9, . . .
• The smallest eligible pair is (1,3). We can go on subtracting 2. All numbers thus obtained can be used to form eligible pairs. We get:
1, -1, -3, -5, -7, -9, . . .
• Combining the two, we get the set:
{. . . -9, -7, -5, -3, -1, 1, 3, 5, 7, 9, . . . }
• This is the set of odd numbers. We can denote it by the letter 'O'. Take any two elements from O. The ordered pair formed by those two numbers will be eligible.

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Based on the above solved example 17.5, we can learn some new terms about relations. It can be written in steps:
1. Recall the basic definition about relations:
If (a,b) is an element in set R, then a will be related to b.
◼ For example, a is the sister of b.
We write: R = {(a,b) : a ∈ A, b ∈ B, a is the sister of b}
   ♦ A is the set of all students in class 11.
   ♦ B is the set of all students in class 12.
• Here, (a,b) pairs are picked from A × B
• The relation R is from A to B.
• However, R in this case is not an equivalence relation.

◼ Another example: a is the sister of b.
We write: R = {(a,b) : a, b ∈ A, a is the sister of b}
   ♦ A is the set of all students in the school.
• Here, (a,b) pairs are picked from A × A
• The relation R is on A.
• However, in this case also, R is not an equivalence relation.

◼ Another example: a is 3 greater than b
We write R = {(a,b) : a, b ∈ Z, a = b + 3}
   ♦ Z is the set of all integers.
• Here, (a,b) pairs are picked from Z × Z
• The relation R is on Z.
• However, in this case also, R  is not an equivalence relation.

2. In the solved example 17.5 that we saw above, R is an equivalence relation. Let us see some interesting points:
(i) Consider the condition: a and b must be even.
• We saw that, to satisfy this condition, we can make ordered pairs by taking any two elements from E.
• So any even number ‘a’ is related to any other even number ‘b’ by the relation: 2 divides (a-b)
(ii) Consider the condition: a and b must be odd.
• We saw that, to satisfy this condition, we can make ordered pairs by taking any two elements from O.
• So any odd number ‘a’ is related to any other odd number ‘b’ by the relation: 2 divides (a-b)

3. Based on the above step (2), we can write:
(i) The relation R partitions Z into E and O.
(ii) E can give ordered pairs which are eligible to be included in R.
(iii) O can also give ordered pairs which are eligible to be included in R.
(iv) E is said to be an equivalence class of Z. It is a subset of Z.
(v) O is also said to be an equivalence class of Z. It is also a subset of Z.

4. We can use this as a general result. It can be written in 5 steps:
(i) R is a relation in set X
(ii) R is an equivalence relation.
(iii) Then R will partition X into various sets: A₁, A₂, A₃, . . .
   ♦ These sets are called equivalence classes.
(iv) Union of all equivalence classes will give set X
(v) Take any two equivalence classes. There will not be any common elements between them. So intersection of any two equivalence classes will give a null set.

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Let us see an example:
Show that the relation R in the set Z of integers given by:
$\rm{R = \{(a,b) : 3 ~\text{divides}~ a-b\}}$
is an equivalence relation. Hence find the equivalence classes.
Solution:
• We have to consider the set Z. It is the set of integers. Integers are those numbers which do not have fraction or decimal parts. They can be +ve or -ve. Zero is also an integer. There are infinite integers. All of them will be present in Z.
• R is a relation in Z. So we must consider Z × Z.
• . . . (-1,0), (-1,1), (-1,2), (-1,3), (-1,4), . . . , (2,0), (2,1), (2,2), (2,3), . . .  are all elements of Z × Z.
There will be infinite elements in Z × Z

1. First we check whether R is reflexive. It can be written in 2 steps:
(i) Ordered pairs like (-1,-1), (0,0), (7,7) etc., are eligible because, $\frac{a-b}{3}$ will give zero.
(ii) Since all (a,a) are eligible, we can say that R is reflexive.

2. Now we check whether R is symmetric. It can be written in 2 steps:
(i) Pairs like (0,3), (3,6), (-6,-9) etc., are eligible.
(ii) Even if we interchange 'a' and 'b', they will be eligible.
(iii) We can write:
If (a,b) is eligible, then (b,a) is also eligible.
• Therefore, R is a symmetric relation.

3. Now we check whether R is transitive. It can be written in 3 steps:
(i) Consider an example: (-7,-1) and (-1,8)
Both pairs are eligible.
(ii) (-5,8) is also eligible.
We can check any two eligible pairs like this. The third pair formed will be eligible.
• The proof can be written as follows:
   ♦ (a-b) is divisible by 3. So (a-b) = 3k, where k is an integer.
   ♦ (b-c) is divisible by 3. So (b-c) = 3m, where k is an integer.
   ♦ Now, (a-c) = [(a-b) + (b-c)] = [3k + 3m] = 3(k+m).
   ♦ So (a-c) is divisible by 3.
(iii) We can write:
If (a,b) and (b,c) are eligible, then (a,c) will be eligible.
• Therefore R is a transitive relation.

4. We see that:
R is reflexive, symmetric and transitive.
• So R is an equivalence relation

5. Since R is an equivalence relation, we must find the equivalence classes. It can be done in four steps:

(i) The smallest eligible pair will contain zero. It is (0,0).
• So we put zero in the middle.
   ♦ Then we add 3 on the right side in a repeating manner.
   ♦ Also,we subtract 3 on the left side in a repeating manner.
• We get: . . . -12, -9, -6, -3, 0, 3, 6, 9, . . .

(ii) The next smallest eligible pair will contain 1. It is (1,-2).
• So we put 1 in the middle.
   ♦ Then we add 3 on the right side in a repeating manner.
   ♦ Also,we subtract 3 on the left side in a repeating manner.
• We get: . . . -11, -8, -5, -2, 1, 4, 7, 10, . . .

(iii) The next smallest eligible pair will contain -1. It is (-1,-4).
• So we put -1 in the middle.
   ♦ Then we add 3 on the right side in a repeating manner.
   ♦ Also,we subtract 3 on the left side in a repeating manner.
• We get: . . . -13, -10, -7, -4, -1, 2, 5, 8, . . .

(iv) So the equivalence classes are:
A₁ = {. . . -12, -9, -6, -3, 0, 3, 6, 9, . . .}
A₂ = {. . . -11, -8, -5, -2, 1, 4, 7, 10, . . .}
A₃ = {. . . -13, -10, -7, -4, -1, 2, 5, 8, . . .}
• There are no common elements between the above three sets. That means, they are disjoint sets.
• Also the union of the three sets will give Z.

Solved example 17.6
Show that the relation R in the set P={1,2,3,4,5,6,7} given by:
$\rm{R = \{(a,b) : ~\text{both a and b are either odd or even}\}}$
is an equivalence relation. Hence find the equivalence classes.
Solution:
• R is a relation in P. So we must consider P × P.
• (1,1), (1,2), 1,3), . . . (2,1), (2,2), (2,3), . . . are all elements of P × P.
There will be (7 × 7) = 49 elements in P × P

1. First we check whether R is reflexive. It can be written in 2 steps:
(i) Ordered pairs (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) and (7,7) are eligible to be included in R. This is because, both a and b are either odd or even.
(ii) Since all (a,a) are eligible, we can say that R is reflexive.

2. Now we check whether R is symmetric. It can be written in 2 steps:
(i) Pairs like (1,3), (2,6), (5,7) etc., are eligible.
(ii) Even if we interchange 'a' and 'b', they will be eligible. This is because, both a and b are either odd or even.
(iii) We can write:
If (a,b) is eligible, then (b,a) is also eligible.
• Therefore, R is a symmetric relation.

3. Now we check whether R is transitive. It can be written in 3 steps:
(i) Consider an example: (1,3) and (3,7)
Both pairs are eligible.
(ii) (1,7) is also eligible.
We can check any two eligible pairs like this. The third pair formed will be eligible.
• The proof can be written as follows:
If two pairs are eligible then it means that all four coordinates in them are either odd or even. So the new pair formed will also be either odd or even.
(iii) We can write:
If (a,b) and (b,c) are eligible, then (a,c) will be eligible.
• Therefore R is a transitive relation.

4. We see that:
R is reflexive, symmetric and transitive.
• So R is an equivalence relation

5. Since R is an equivalence relation, we must find the equivalence classes. It can be done in four steps:

(i) The set of all odd numbers in P is: {1,3,5,7}
(ii) The set of all even numbers in P is: {2,4,6}
(iii) So the equivalence classes are:
A₁ = {1,3,5,7}
A₂ = {2,4,6}
• There are no common elements between the above two sets. That means, they are disjoint sets.
• Also the union of the two sets will give P.

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Link to a few more solved examples is given below:

Exercise 17.1

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In the next section, we will see types of functions.

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Chapter 17 - Relations And Functions

In the previous section, we completed appendix B. With that, we have completed all topics in class 11. In this chapter, we will see Relations and Functions which is the first chapter in class 12.

• In class 11, we saw some basic details about relations and functions. See chapter 2.
• Let us recall five important points:
(i) A and B are two sets.
(ii) A × B is a new set formed from A and B. This new set will contain all possible ordered pairs between the elements of A and B.
(iii) The ordered pairs are written in the form (a,b).
   ♦ a is from set A
   ♦ b is from set B
(iv) We pick out some of those ordered pairs and form a new set R.
(v) Take any ordered pair from R. There will be a definite relation between the a and the b of that ordered pair.
• For example,
   ♦ a is the brother of b.
   ♦ a is 2 less than b.
• We write this as a R b.

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In this section, we will learn more details about relations and functions.

Types of relations

First we will see empty relation. It can be explained in 4 steps:
1. We have seen the relation from set A to itself. It is written as R in A.
2. Relation in A will be a subset of A × A
• But Φ (empty set) is a subset of any set. So there exists a relation R in A such that R is a empty set.
3. Let us see an example. It can be written in 4 steps:
(i) Let A = {1,2,3,4}
(ii) R in A in set builder form is:
R = {(a,b): a-b = 10}
(iii) a-b = 10 is same as b = a-10
• Take the first element 1. Ten subtracted from 1 is -9. This -9 is not present in A. So when a = 1, we cannot write b. In other words, no ordered pair in R can have a = 1.
• Take the second element 2. Ten subtracted from 2 is -8. This -8 is not present in A. So when a = 2, we cannot write b. In other words, no ordered pair in R can have a = 2.
• Similarly, we will find that:
   ♦ No ordered pair in R can have a = 3
   ♦ No ordered pair in R can have a = 4
(iv) None of the elements in A can become 'a' of the ordered pairs in R. So there will be no element in R. In other words, R is an empty set.
4. If R = Φ, then that relation is called an empty relation.

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Now we will see universal relation. It can be explained in 4 steps:
1. We have seen the relation from set A to itself. It is written as R in A.
2. Relation in A will be a subset of A × A
• But a set itself is a subset of any set. So there exists a relation R in A such that R is A × A.
3. Let us see an example. It can be written in 4 steps:
(i) Let A = {1,2,3,4}
(ii) R' in A in set builder form is:
R' = {(a,b): |a-b| ≥ 0}
(iii) A has four elements. So A × A will have 16 elements.
• Take the first element (1,1). Here a = 1 and b = 1
|a-b| = |1-1| = |0| = 0
So (1,1) is eligible to be included in the set R'.
• Take the second element (1,2). Here a = 1 and b = 2
|a-b| = |1-2| = |-1| = 1 ≥ 0
So (1,2) is eligible to be included in the set R'.
• Take the third element (1,3). Here a = 1 and b = 3
|a-b| = |1-3| = |-2| = 2 ≥ 0
So (1,3) is eligible to be included in the set R'.   
• Take any one of the 16 elements, say (4,1). Here a = 4 and b = 1
|a-b| = |4-1| = |3| = 3 ≥ 0
So (4,1) is eligible to be included in the set R'
• In this way we can check all the 16 elements. We will see that, all of them are eligible to be included in the set R'.
(iv) We can write: R' = A × A
4. If R' = A × A, then that relation is called a universal relation.
• Here every element in A has the relation R' with every other element of A.

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Both the empty relation and the universal relation are some times called trivial relations.

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Solved example 17.1
A is the set of all students of a boys school. Show that the relation R in A given by R={(a,b) : a is sister of b} is the empty relation and R’ = {(a,b) : the difference between heights of a and b is less than 3 meters} is the universal relation.
Solution:
Part (i):
1. We can try to write the elements in R. Those elements will be in the form (a,b)
‘a’ will be from set A. ‘b’ will also be from set A.
2. Set A contains only boys.
So ‘a’ can never be the sister of ‘b’.
3. As a consequence, we will not be able to write a single element of the form (a,b). That means, R is an empty set.
• So the relation R is an empty relation.
Part (ii):
1. Heights greater than 2 m are very rare. We can safely assume that, the tallest student has a height of 2.5 m.
2. Heights smaller than 1.5 are very rare. We can safely assume that, the shortest student has a height of 1 m.
3. So the maximum possible difference between the heights is (2.5 – 1) = 1.5 m.
• None of the “differences” will be greater than 1.5 m.
• Consequently, none of the “differences” will be greater than 3 m.
• In other words, all differences will be less than 3 m.
4. We can take any (a,b) from A × A.
• The difference in that pair will be less than 3 m.
• So all elements in A × A are eligible to be included in R`
• Thus we get R` = A × A.
• That means, R` is a universal relation.

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Next we will see reflexive relation. It can be explained in 3 steps:
1. Let A = {1,2,3,4}
• We know that there will be 16 elements in A × A.
• (1,1), (2,2), (3,3) and (4,4) will be among those 16 elements.
2. Suppose that, there is a relation R in A in such a way that, (1,1), (2,2), (3,3) and (4,4) are elements of R.
• Then that relation is called a reflexive relation.
3. We can write the definition in two steps:
(i) R is a relation in A
(ii) For every a ∈ A, if (a,a) ∈ R, then R is a reflexive relation.

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Next we will see symmetric relation. It can be explained in 3 steps:
1. Let A = {1,2,3,4}
• We know that there will be 16 elements in A × A.
• (1,2), (2,1), (1,3), (3,1) etc., will be among those 16 elements.
2. Suppose that, there is a relation R in A which satisfies the following conditions:
• If (1,2) is an element of R, then (2,1) is also an element of R.
• If (1,3) is an element of R, then (3,1) is also an element of R.

- - - - 

• If (2,4) is an element of R, then (4,2) is also an element of R.

so on . . .

• If this is true for all such pairs, then that relation is called a symmetric relation.
3. We can write the definition in two steps:
(i) R is a relation in A
(ii) For all a₁, a₂ ∈ A, if (a₁,a₂) ∈ R ⇒ (a₂,a₁) ∈ R, then R is a symmetric relation.

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Next we will see transitive relation. It can be explained in 3 steps:
1. Let A = {1,2,3,4}
• We know that there will be 16 elements in A × A.
• (1,2), (2,3), (1,3), (3,4) etc., will be among those 16 elements.
2. Suppose that, there is a relation R in A which satisfies the following conditions:
• If (1,2) and (2,3) are elements of R, then (1,3) is also an element of R.
• If (1,3) and (3,4) are elements of R, then (1,4) is also an element of R.

- - - - 

• If (2,4) and (4,3) are elements of R, then (2,3) is also an element of R.

so on . . .

• If this is true for all such pairs, then that relation is called a transitive relation.
3. We can write the definition in two steps:
(i) R is a relation in A
(ii) For all a₁, a₂, a₃∈ A,
if (a₁,a₂) ∈ R and (a₂,a₃) ∈ R ⇒ (a₁,a₃) ∈ R, then R is a transitive relation.

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Now we can write the definition of an equivalence relation. It can be written in 2 steps:
1. R is a relation in A.
2. This R is an equivalence relation if all three conditions below are satisfied.
(i) R is a reflexive relation.
(ii) R is a symmetric relation.
(iii) R is a transitive relation.

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Solved example 17.2
Let T be the set of all triangles in a plane. R is a relation in T.
R = {(T₁,T₂) : T₁ is congruent to T₂}. Show that R is an equivalence relation.
Solution:
• Congruent triangles are those triangles which have the same sides and same corresponding angles.
• Given that, T is the set of all triangles in a plane. There will be infinite number of triangles in that set. Let us number them as: 1, 2, 3, 4, . . .
All those triangles will be present in T.
• R is a relation in T. So we must consider T × T
• (1,1), (1,2), (1,3), (1,4), . . . , (2,1), (2,2), (2,3), . . .  are all elements of T × T.
There will be infinite elements in T × T
1. First we check whether R is reflexive. It can be written in 3 steps:
(i) (1,1) is eligible to be included in R. This is because, the first triangle is congruent to itself. In fact, any triangle is congruent to itself.
(ii) In this way, (2,2), (3,3), (4,4), . . . are eligible to be included in R.
(iii) Therefore, R is a reflexive function.
2. Now we check whether R is symmetric. It can be written in 3 steps:
(i) Suppose that (3,9) is eligible to be included in R.
• Then it means that, the third triangle is congruent to the ninth triangle.
(ii) Now, (9,3) is an element of T × T.
• The element (9,3) is eligible to be included in R. This is because:
If third triangle is congruent to the ninth triangle, then ninth triangle will be congruent to the third.
(iii) So in general, if (T₁, T₂) is an element of R, then (T₂,T₁) will also be an element of R
• Therefore, R is a symmetric relation.
3. Now we check whether R is transitive. It can be written in 4 steps:
(i) Suppose that (4,7) is eligible to be included in R.
• Then it means that, the fourth triangle is congruent to the seventh triangle.
(ii) Also suppose that (7,10) is eligible to be included in R.
• Then it means that, the seventh triangle is congruent to the tenth triangle.
(iii) It is clear that, fourth, seventh and tenth triangles are congruent.
Then fourth triangle is congruent to the tenth triangle.
So (4,10) is eligible to be included in R.
(iv) In general, if both (T₁,T₂) and (T₂,T₃) are elements of R, then (T₁,T₃) will also be an element of R.
Therefore R is a transitive relation.
4. We see that:
R is reflexive, symmetric and transitive. So R is an equivalence relation.   

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In the next section, we will see a few more solved examples.

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B.4 - Applications of Mathematical Modelling

In the previous section, we completed the examples of mathematical modelling. We saw four examples altogether. Those examples helped us to learn the basics about mathematical modelling. In this section, we will see some interesting situations where mathematical modelling can be effectively used.

1. In human beings and all other animals, blood flows through blood vessels and reach all parts of the body. Blood carries the oxygen and nutrients which must reach all parts. Any obstruction to the flow can cause serious health issues. The characteristics of the blood vessels are such that there is no obstruction to the flow. But due to some illness, those characteristics can change. In such a situation, a mathematical model can be prepared. The model will relate the new characteristics and the resulting new pattern of blood flow. Based on the results given by the mathematical model, the physiologist can make better judgements.

2. In cricket, the third umpire has to take decisions about LBW. He analyses the trajectory of the ball assuming that the batsman is not present. A mathematical model of the trajectory will help the umpire to decide whether an LBW will occur if the batsman was actually present.

3. Meteorology department has to make whether predictions. The parameters that they have to consider are: temperature, air pressure, humidity, wind speed etc., A mathematical model is prepared by considering all the essential parameters. Precision instruments are available to measure these parameters. Using the values obtained from the instruments, the mathematical model will give the required results. Interpretation of the results enable the department to make predictions about the whether conditions.

4. Department of Agriculture can predict the quantity of rice that can be harvested in a year. The quantity of rice available from a unit area is an essential parameter. Fertility of the soil in different parts of the country, availability of water for proper irrigation etc., are also essential parameters. The mathematical model is prepared based on the principles of statistics. Interpretation of the results given by the model, enable the department to make predictions about the quantity of rice.

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Given below is an actual situation which occurred in the 18th century. It was solved by mathematical modelling. It can be written in 3 steps:

1. The Pregel river flows through Konigsberg town. So the town is on the two river banks. River bank A and River bank B. This is shown in the fig.B.4 below:

Fig.B.4

• There are two islands C and D in the river. Seven bridges connect the islands and the banks.
2. The problem is to go around the town, visiting both the islands.
• But three conditions must be satisfied:
(i) The person can start the journey from any river bank or any island. But the journey must end at the exact starting point.
(ii) The person must use all the bridges.
(iii) Each bridge must be used only once.
3. Both residents and authorities could not solve the problem. So they requested the famous mathematician Leonhard Euler to find a solution.  
• The solution put forward by Euler can be written in 5 steps:
(i) In a pictorial representation of the problem, bridges are denoted by arcs or lines. The end points of bridges are denoted by dots. Those dots are called vertices (plural of vertex).

(ii) Let us see some examples:
In fig.B.5(I) below,
• Three arcs start from A. This is because, there are three bridges starting from the river bank A. Two of them to island C and the third to island D.

Fig.B.5

• Four arcs and one line start from C. This is because, there are five bridges starting from the island C. Two of them to river bank A, two of them to river bank B and the fifth to island D.
(iii) We see that:
    ♦ A has 3 arcs/lines. 3 is an odd number.
    ♦ B has 3 arcs/lines. 3 is an odd number.
    ♦ C has 5 arcs/lines. 5 is an odd number.
    ♦ D has 3 arcs/lines. 3 is an odd number.
• If a vertex has an odd number of arcs/lines, that vertex is called an odd vertex.
• If a vertex has an even number of arcs/lines, that vertex is called an even vertex.
• So in this problem, all are odd vertices.
(iv) Euler said that, in a journey which has to satisfy the conditions in (2), there must be only two odd vertices.
• In the present problem, there are four odd vertices. So will never be able to satisfy the conditions.
(v) Later, a new bridge was added between A and B. This is shown in fig.B.5(II). Now C and D are the required two odd vertices while A and B are even vertices.

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We have completed a discussion on mathematical modelling. With that, we have completed all the chapters of class 11. In the next chapter, we will see the first chapter of class 12.

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B.3 - Mathematical Modelling Involving Exponential Functions

In the previous section, we saw the mathematical model involving inequalities. In this section, we will see another example.

Example 4:
• The situation is:
A population control unit wants to find out the population of a country after 10 years.
• This situation can be converted into a mathematical problem. For that, we engage in the process of mathematical modeling. The final result can be obtained in 4 steps:
 

Step 1:
First we study the situation. We will write it in order:

(i) The population changes with time.
For example, population next year will be different from the present population.
(ii) Population increases with births.
(iii) Population decreases with deaths.
(iv) Based on the above three information, we can write:
time, number of births and number of deaths are the parameters.

• We can write:
Step 1 involves studying the situation and identifying the parameters. Essential parameters should be carefully identified.

Step 2:
This can be written in order:
(i) Let us denote time by the letter ‘t’. It is time in years.
    ♦ So when t = 0, it indicates the present year 2023.
    ♦ When t = 1, it indicates the next year 2024
    ♦ When t = 2, it indicates the year 2025 so on . . .

(ii) Let us denote the population at any time by p(t).
• So when t = 0, the population will be p(0).
    ♦ It is the population as on first of January 2023
• When t = 1, the population will be p(1).
    ♦ It is the population as on first of January 2024 so on . . .

(iii) We can find p(1) as follows:
p(1) = population as on first of January 2024
= population as on first of January 2023 [this is p(0)]
+ Number of births in the year 2023
- Number of deaths in the year 2023

(iv) In general, we can write:
p(t+1) = Population as on the first of January of year t [this is p(t)]
+ Number of births in the year t
- Number of deaths in the year t
• Let us denote,
    ♦ Number of births in the year t as B(t)
    ♦ Number of deaths in the year t as D(t)
• Now the equation becomes:
p(t+1) = p(t) + B(t) - D(t)

(v) Now we consider the term birth rate.
• It is the number of births per 1000 of the population per year.
• For example, suppose that the birth rate is 7 and p(t) is 1500000.
• Number of “thousands” in p(t) = $\frac{1500000}{1000}$ = 1500.
• There will be 7 new births in each of those thousands.
• So the number of new births in the year t = 1500 × 7 = 10500.
• Let us denote birth rate by the letter ‘b’. Then for our present example, b = $\frac{7}{1000}$.
• We see that, number of births can be easily calculated. All we need to do is, multiply the population by b.
• That means, B(t) = p(t) × b

(vi) Similarly, if d is the death rate, then number of deaths can be calculated as:
D(t) = p(t) × d

(vii) Now the equation that we wrote in (iv) becomes:
p(t+1) = p(t) + b p(t) - d p(t)
⇒ p(t+1) = (1 + b - d) p(t)
• So if we put t = 9, we will get the population in the tenth year. But for that, we need to know the population in the ninth year.
• To know the population of the ninth year, we need to know the population of the eighth year. So on . . .
• This is a lengthy process. We do not have the populations in eighth or ninth years. All we have is p(0), which is the population of the present year.

(viii) Let us try to find an alternate method.
• Consider the equation: p(t+1) = (1 + b - d) p(t)
• Let us put t = 0. Then we get:
p(0+1) = (1 + b - d) p(0)
⇒ p(1) = (1 + b - d) p(0)
• Let us put t = 1. We get:
$\begin{array}{ll}{}    &{p(2)}    & {~=~}    &{(1+b-d) p(1)}    &{} \\
{}    &{}    & {~=~}    &{(1+b-d) (1+b-d)p(0)}    &{} \\
{}    &{}    & {~=~}    &{(1+b-d)^2 p(0)}    &{} \\
\end{array}$ 
• Let us put t = 2. We get:
$\begin{array}{ll}{}    &{p(3)}    & {~=~}    &{(1+b-d) p(2)}    &{} \\
{}    &{}    & {~=~}    &{(1+b-d) (1+b-d)^2p(0)}    &{} \\
{}    &{}    & {~=~}    &{(1+b-d)^3 p(0)}    &{} \\
\end{array}$

• We see a pattern. We can easily write:
p(t) = (1+b-d)^t p(0)

(ix) Consider the term (1+b-d). All items inside the brackets are constants. So the term as a whole will be a constant. We denote it by the letter ‘r’. It is called the growth rate. It is also known as Malthusian parameter, in honor of Robert Malthus who first brought this model to popular attention.
• So the equation in (viii) can be modified as:
p(t) = r^t p(0)

(x) Consider the equation p(t) = r^t p(0).
• With the derivation of this equation, we have completed step 2.
• We see that population is a function of t. This is because, t is the only variable. r and p(0) are constants.
• p(t) is an exponential function. Any function of the form c r^t, where c and r are constants is called an exponential function.

• We can write:
Step 2 involves drawing the necessary diagrams and writing the relevant mathematical equations/inequalities. In short, we write a mathematical problem in this step. Any mathematical problem will have a definite solution. So in this step, it is necessary to recheck all the works done thus far.

Step 3:
In this step, we put the equation into actual use. This can be written in order:

(i) Suppose that, the present population p(0) is 250000000 and the rates b and d are 0.02 and 0.01 respectively.

(ii) We get r = (1+b-d) = (1+0.02-0.01) = 1.01

(iii) So the population in the tenth year will be:
p(10) = 1.01¹⁰ × 250000000 = 276,155,531.25

• We can write:
Step 3 involves the actual application of the result obtained in step 2. Calculators or digital computers can be used for lengthy problems.

Step 4:
This can be written in order:
(i) First we do the validation.
We have obtained the population in step 3. But it has decimal values. Population cannot be in the decimal form. It must be a whole number. We can say that, the equation derived in step 2 is not a very accurate equation.
• However, we can make an approximation. We can state that, the population in the tenth year will be 276,155,531 approximately.

(ii) Next we do the interpretation.
We state that the equation derived in step 2 is not completely dependable. This is because, we assumed b and d to be constant for all the years. In actual practice, this is not possible. Improvements or decline in health management systems in the society can alter b and d. Also, if there is migration into or out of the society, population will change considerably. To get an accurate mathematical model, we must take such factors also into consideration.

• We can write:
In step 4, we do validation and interpretation.

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It is clear that, after validation and interpretation, we may need to go back to step 1 and look for additional factors and parameters that may be playing crucial roles. This is shown in the flow chart in fig.B.3 below:

Fig.B.3

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In the next section, we will see some interesting situations where mathematical modelling can be used effectively.

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B.2 - Mathematical Modelling Involving Inequalities

In the previous section, we saw the mathematical model for the motion of a simple pendulum. In this section, we will see another example.

Example 3:
• The situation is:
A farm house uses atleast 800 kg of special food daily. The special food is a mixture of corn and soyabean with the following compositions:
$\begin{array}{cc}{}    &{\textbf{Material}}    &{\textbf{Protein present per Kg}}    &{\textbf{Fibre present per Kg}}    &{\textbf{Cost per Kg}}\\
{}    &{\textbf{Corn}}    &{0.09}    &{0.02}    &{\text{Rs 10}}\\
{}    &{\textbf{Soyabean}}    &{0.60}    &{0.06}    &{\text{Rs 20}}\\
\end{array}$
The dietary requirements of the special food stipulate atleast 30% protein and at most 5% fibre. How can this dietary requirement be satisfied at the least possible cost?
• This situation can be converted into a mathematical problem. For that, we engage in the process of mathematical modeling. The final result can be obtained in 4 steps:
 

Step 1:
First we study the situation. We will write it in order:

(i) “Atleast 800 kg” means:
    ♦ The total quantity should not be less than 800 kg.
    ♦ The total quantity can be greater than 800 kg.
(ii) The available nutrients and costs can be listed:
• One kg corn will
    ♦ give 0.09 kg protein.
    ♦ give 0.02 kg fibre.
    ♦ cost Rs 10
• One kg soya bean will give
    ♦ give 0.60 kg protein.
    ♦ give 0.06 kg fibre.
    ♦ cost Rs 20
(iii) Our aim is:
• The 800 kg (or more) food must
    ♦ contain atleast 30% protein.
    ♦ contain at most 5% fibre.
    ♦ cost the least possible amount.

• We can write:
Step 1 involves studying the situation and identifying the parameters. Essential parameters should be carefully identified.

Step 2:
This can be written in order:
(i) Let the food be made up of x kg corn and y kg of soya bean.
• Then from (i) of step 1, we can write: x + y ≥ 800

(ii) Available protein and fibre:
• From the x kg corn, we will get some protein and fibre.
• From the y kg soya bean also, we will get some protein and fibre.
• Then from (ii) of step 1, we can write:
    ♦ Total protein than can be obtained from x kg corn and y kg soya bean is:
0.09x + 0.60y   
    ♦ Total fibre than can be obtained from x kg corn and y kg soya bean is:
0.02x + 0.06y

(iii) Checking whether the dietary requirements are fulfilled:
• We are taking (x+y) kg food. This will contain (0.09x + 0.60y) kg of protein.
Then from (iii) of step 1, we can write:
$\begin{array}{ll}{}    &{\frac{0.09x + 0.60y}{x+y} \times 100}    & {~\ge~}    &{30}    &{} \\
{\Rightarrow}    &{\frac{0.09x + 0.60y}{x+y}}    & {~\ge~}    &{0.30}    &{} \\
{\Rightarrow}    &{0.09x + 0.60y}    & {~\ge~}    &{0.30 x + 0.30y}    &{} \\
{\Rightarrow}    &{0.30y}    & {~\ge~}    &{0.21 x}    &{} \\
{\Rightarrow}    &{0}    & {~\ge~}    &{0.21 x – 0.30 y}    &{} \\
{\Rightarrow}    &{0.21 x – 0.30 y}    & {~\le~}    &{0}    &{} \\
\end{array}$

• We are taking (x+y) kg food. This will contain (0.02x + 0.06y) kg of fibre.
Then from (iii) of step 1, we can write:
$\begin{array}{ll}{}    &{\frac{0.02x + 0.06y}{x+y} \times 100}    & {~\le~}    &{5}    &{} \\
{\Rightarrow}    &{\frac{0.02x + 0.06y}{x+y}}    & {~\le~}    &{0.05}    &{} \\
{\Rightarrow}    &{0.02x + 0.06y}    & {~\le~}    &{0.05 x + 0.05y}    &{} \\
{\Rightarrow}    &{0.01y}    & {~\le~}    &{0.03 x}    &{} \\
{\Rightarrow}    &{0}    & {~\le~}    &{0.03 x – 0.01 y}    &{} \\
{\Rightarrow}    &{0.03 x – 0.01 y}    & {~\ge~}    &{0}    &{} \\
\end{array}$

(iv) Checking cost:
We are taking (x+y) kg food. That food will cost (10x + 20y)
Then from (iii) of step 1, we can write: (10x + 20y) should be as small as possible.

• We can write:
Step 2 involves drawing the necessary diagrams and writing the relevant mathematical equations/inequalities. In short, we write a mathematical problem in this step. Any mathematical problem will have a definite solution. So in this step, it is necessary to recheck all the works done thus far.

Step 3:
In this step, we put the equations and inequalities into actual use. This can be written in order:

(i) From (i) of step 2, we have:
x + y ≥ 800
• The red line in fig.B.2 below represents
x + y = 800

Fig.B.2

• So any (x,y) in the upper half plane II of the red line will satisfy the inequality. See section 6.3.

(ii) From (iii) of step 2, we have:
0.21x - 0.30y ≤ 0
• The magenta line in fig.B.2 above represents
0.21x - 0.30y = 0
• So any (x,y) in the upper half plane II of the magenta line will satisfy the inequality.

(iii) Again from (iii) of step 2, we have:
0.03x - 0.01y ≥ 0
• The green line in fig.B.2 above represents
0.03x - 0.01y = 0
• So any (x,y) in the lower half plane I of the green line will satisfy the inequality.

(iv) So we have three regions:
    ♦ upper half plane II of the red line.
    ♦ upper half plane II of the magenta line.
    ♦ lower half plane I of the green line.
• The three regions will over lap in the yellow region shown in fig.B.2.
• So any (x,y) in the yellow region will satisfy all the three inequalities.

(v) From (iv) of step 2, we have:
(10x + 20y) must be as small as possible.
• For that, x and y must be as small as possible. At the same time, x and y must satisfy all three inequalities.
• This point is marked by a cyan dot in fig.B.2 above.
• If we draw a vertical line through the cyan dot, then that line will meet the x-axis at 470.59
• If we draw a horizontal line through the cyan dot, then that line will meet the y-axis at 329.41
• So the required point is (470.6,329.4)
• We can write:
Step 3 involves the actual application of the result obtained in step 2. Calculators or digital computers can be used for lengthy problems.

Step 4:
This can be written in order:
(i) First we do the validation.
• We are going to buy 470.6 kg corn and 329.4 kg soya bean.
• So total quantity = (470.6 + 329.4) = 800.
Thus “atleast 800” is satisfied.
• In this way all the dietary requirements can be checked. This will complete the validation process.
(ii) Next we do the interpretation.
• We state that:
If we buy 470.6 kg corn and 329.4 kg soya bean, then all the dietary requirements will be satisfied and at the same time, the cost will be the least possible value. The least possible cost is (470.6 × 10 + 329.4 × 20) = Rs 11294.00
• We can write:
In step 4, we do validation and interpretation.

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In the next section, we will see one more example.

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B.1 Mathematical Model of a Simple Pendulum

In the previous section, we saw an example which demonstrated the basics of mathematical modeling. In this section, we will see another example.

Example 2:
• The situation is:
A physicist (scientist who specializes in the field of physics) wants to understand the motion of simple pendulum. A simple pendulum consists of a mass (called bob) attached to one end of a string. The other end of the string is fixed at a point.
• This situation can be converted into a mathematical problem. For that, we engage in the process of mathematical modeling. The final result can be obtained in 4 steps:
 

Step 1:
The physicist studies the situation. He realizes that, the factors that may be affecting the motion of a pendulum are:
(i) Period of oscillation (T)
(ii) Mass of the bob (m)
(iii) Effective length of the pendulum (l). This is the distance between the point of suspension to the center of mass of the bob.
(iv) Acceleration due to gravity (g)

• Now, the physicist tries to confirm that, the above four are indeed the parameters. For that, he performs simple experiments.
• Time period of the following two pendulums are determined experimentally:
(i) Pendulum with effective length l and mass of bob m₁.
(ii) Pendulum with same effective length l but mass of bob m₂.
• He finds that, there is no appreciable change in T.
• Again, T of following two pendulums are determined experimentally:
(i) Pendulum with effective length l₁ and mass of bob m.
(ii)Pendulum with a different effective length l₂ but same mass m.
• He finds that, there is appreciable change in T.
• So he concludes that:
    ♦ m is not an essential parameter.
    ♦ l is an essential parameter.
• We can write:
Step 1 involves studying the situation and identifying the parameters. Essential parameters should be carefully identified.

Step 2:
This can be written in order:
(i) The physicist repeats the experiment with same m and different l. He notes T in each trial.
(ii) Using the observations, he plots a graph with T along the y-axis and l along the x-axis.
• The graph thus obtained is a parabola.
(iii) Equation of a parabola is of the form y² = kx.
• So he can conclude that, the relation between T and L will be of the form T² = kl
(iv) From the values of T and l of the various trials, k can be obtained. He gets:
$k = \frac{4 \pi^2}{g}$
• Substituting in (iii), we get:
$\begin{array}{ll}{}    &{T^2}    & {~=~}    &{\frac{4 \pi^2}{g} l}    &{} \\
{\Rightarrow}    &{T}    & {~=~}    &{\sqrt{\frac{4 \pi^2}{g} l}}    &{} \\
{\Rightarrow}    &{T}    & {~=~}    &{2 \pi \sqrt{\frac{l}{g}}}    &{} \\
\end{array}$

• We can write:
Step 2 involves drawing the necessary diagrams and writing the relevant mathematical equations. In short, we write a mathematical problem in this step. Any mathematical problem will have a definite solution. So in this step, it is necessary to recheck all the works done thus far.

Step 3:
This can be written in order:
(i) The physicist puts the equation into actual use.
(ii) He assumes two values for l. They are 225 cm and 275 cm.
(iii) When l = 225 cm, he gets:
$\begin{array}{ll}{}    &{T}    & {~=~}    &{2 \pi \sqrt{\frac{l}{g}}}    &{} \\
{}    &{}    & {~=~}    &{2 \pi \sqrt{\frac{2.25}{9.81}}}    &{} \\
{}    &{}    & {~=~}    &{3.04 ~\text{sec}}    &{} \\
\end{array}$ 
(iv) When l = 275 cm, he gets:
$\begin{array}{ll}{}    &{T}    & {~=~}    &{2 \pi \sqrt{\frac{l}{g}}}    &{} \\
{}    &{}    & {~=~}    &{2 \pi \sqrt{\frac{2.75}{9.81}}}    &{} \\
{}    &{}    & {~=~}    &{3.36 ~\text{sec}}    &{} \\
\end{array}$
• We can write:
Step 3 involves the actual application of the result obtained in step 2. Calculators or digital computers can be used for lengthy problems.

Step 4:
This can be written in order:
(i) The physicist tries to confirm the results obtained in step 3. For that, he performs a number of trials of the actual experiment.

(ii) When mass (m) of the bob is 385 gms and l is 275 cm, T is 3.371 sec
• This T obtained experimentally, is comparable with the 3.36 sec that he got by the mathematical calculations in step 3.  

(iii) When m is 385 gms and l is 225 cm, T is 3.056 sec
• This T obtained experimentally, is comparable with the 3.04 sec that he got by the mathematical calculations in step 3.  

(iv) When m is 230 gms and l is 275 cm, T is 3.351 sec
• This T obtained experimentally, is comparable with the 3.36 sec that he got by the mathematical calculations in step 3.  

(v) When m is 230 gms and l is 225 cm, T is 3.042 sec
• This T obtained experimentally, is comparable with the 3.04 sec that he got by the mathematical calculations in step 3.

(vi) However, there are some small errors. For example, in (ii) above, we can find an error of (3.371 - 3.36) = 0.011.
• But such errors are small. So the mathematical model in step 3 is acceptable.
• We can conclude that:
    ♦ T is directly proportional to l.
    ♦ T is inversely proportional to g.
(vii) We see that, the mathematical model is in good agreement with the practical values. But small errors are also seen.
• These errors are due to:
    ♦ mass of the string, which we did not consider.
    ♦ resistance of the air, which also we did not consider.
• When the physicist sees the errors, he investigates further. Such an investigation will help to find the reasons for the errors.
• Once the reasons are found out, the mathematical model can be improved. The improved model may contain many complex equations.
• We can surely say that, a simple mathematical model is a good starting point.

(viii) We can write:
• In step 4, we do validation and interpretation.
◼ What is validation?
The answer can be written in 4 steps:
(a) We have two items:
    ♦ Results obtained from the mathematical model.
    ♦ Known facts about the real problem.
(b) We compare the two items
(c) If there is no appreciable difference, the mathematical model can be considered to be valid.     
(d) If there is appreciable difference, it cannot be considered to be valid.

◼ Suppose that a mathematical model is found to be valid. Once found valid, we have to do the interpretation. What is interpretation?
The answer can be written in 2 steps:
(a) We have the results obtained from the mathematical model. We state, why and how those results are obtained. In our present case, the physicist state that, T is directly proportional to l and inversely proportional to g.
(b) We also state why and how some errors (if any) occur. In our present case, the physicist state that, mass of the string and air resistance are the cause of the errors.

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So we have seen the mathematical model related to the oscillation of a simple pendulum. In the next section, we will see a mathematical model related to inequalities.

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Appendix B - Mathematical Modelling

In the previous section, we completed a discussion on infinite series. In this appendix B, we will see mathematical modelling.

• Many difficult situations that we face in science, engineering, economics, finance, management, etc., can be written in the form of mathematical problems.
• When we write such mathematical problems, there are two advantages:
   ♦ A better understanding of the situation can be obtained.
   ♦ Better solutions can be obtained to overcome the situation.
• Nowadays it has become a trend to translate difficult situations into mathematical problems. This is partly due to the fact that, advanced digital computers are now available to solve lengthy mathematical problems.

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• The process of translating a situation into a mathematical problem is called mathematical modeling. We can become experts in mathematical modeling by practicing different types of problems. Let us see a few examples:

Example 1:
• The situation is:
A surveyor wants to measure the height of a tower. Since the tower is very tall, a measuring tape cannot be used.
• This situation can be converted into a mathematical problem. For that, we engage in the process of mathematical modeling. The final result can be obtained in 4 steps:
 

Step 1:
The surveyor studies the situation. He realizes that, a triangle can be drawn and trigonometry can be applied. For applying trigonometry, he will need three parameters:
(i) Height of the tower (h)
(ii) Angle of elevation (𝜃)
(iii) Distance from the foot of the tower to the point at which angle of elevation is measured (d).
• We can write:
Step 1 involves studying the situation and identifying the parameters.
Once identified, the surveyor proceeds to obtain those parameters which can be measured.
   ♦ He measures d using a tape measure. It is 450 m
   ♦ He measures 𝜃 using a theodolite. It is 40^o.

Step 2:
The surveyor draws the triangle and writes the relevant mathematical equations. The triangle is shown in fig.B.1 below:

Fig.B.1

   ♦ Length AB is h
   ♦ Length OA is d
   ♦ Angle AOB is 𝜃.
• The equation connecting the parameters is:
$\begin{array}{ll}{}    &{\tan \theta}    & {~=~}    &{\frac{AB}{OA}}    &{} \\
{\Rightarrow}    &{\tan \theta}    & {~=~}    &{\frac{h}{d}}    &{} \\
{\Rightarrow}    &{h}    & {~=~}    &{d \tan \theta}    &{} \\
{\Rightarrow}    &{h}    & {~=~}    &{450 \times \tan 40}    &{} \\
\end{array}$
• We can write:
Step 2 involves drawing the necessary diagrams and writing the relevant mathematical equations. In short, we write a mathematical problem in this step. Any mathematical problem will have a definite solution. So in this step, it is necessary to recheck all the works done thus far.

Step 3:
The surveyor solves the equation. He gets h = 377.6 m
• We can write:
Step 3 involves solving the mathematical problem obtained in step 2. Calculators or digital computers can be used for lengthy problems.

Step 4:
The surveyor writes the height of the tower:
The height of the tower is 378 m.
• We can write:
In step 4, we write the result of the mathematical calculations. With this final step, the situation is overcome.

• This final step is called:
Interpreting the mathematical solution to the real solution.

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In the next section, we will see the case of simple pendulum.

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A.2 Exponential Series

In the previous section, we saw infinite geometric series. In this section, we will see exponential series.

• Consider the series:
$1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} +~.~.~$
• The sum of this series is denoted by the letter e.
• e is very important for calculus and other areas of mathematics, economics, financial studies, science and engineering.
• Note that all terms in the series, are +ve. So e will be +ve.

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Let us find the value of e. It can be done in 8 steps:

1. Two series are written below. They are written in such a way that, the terms in the second series come directly below the corresponding terms in the first series.

\begin{array}{ll}{}    &{\frac{1}{3!}}    & {~+~}    &{\frac{1}{4!}}    & {~+~}    &{\frac{1}{5!}}    & {~+~.~.~.~+~}    &{\frac{1}{n!}}    & {~+~.~.~.~\color{green}{\text{- - - (I)}}}    &{} \\
{}    &{\frac{1}{2^2}}    & {~+~}    &{\frac{1}{2^3}}    & {~+~}    &{\frac{1}{2^4}}    & {~+~.~.~.~+~}    &{\frac{1}{2^{n-1}}}    & {~+~.~.~.~\color{green}{\text{- - - (II)}}}    &{} \\
\end{array}

2. We see an interesting point. It can be written in 3 steps:
(i) Take any term in (II). Note down the power of 2 in that term.
(ii) Take the corresponding term in (I). Note down the denominator of that term.
(iii) The power will be one less than the denominator.

3. Another interesting point can also be observed. It can be written in 4 steps:
(i) Take the first terms: $\frac{1}{3!}~\text{and}~\frac{1}{2^2}$
• We have:
$\frac{1}{3!} = \frac{1}{6}~\text{and}~\frac{1}{2^2} = \frac{1}{4}$
• So the first term in (I) is less than the corresponding term in (II)

(ii) Take the second terms: $\frac{1}{4!}~\text{and}~\frac{1}{2^3}$
• We have:
$\frac{1}{4!} = \frac{1}{24}~\text{and}~\frac{1}{2^3} = \frac{1}{8}$
• So the second term in (I) is less than the corresponding term in (II)

(iii) Take the third terms: $\frac{1}{5!}~\text{and}~\frac{1}{2^4}$
• We have:
$\frac{1}{5!} = \frac{1}{102}~\text{and}~\frac{1}{2^4} = \frac{1}{16}$
• So the third term in (I) is less than the corresponding term in (II)

(iv) Based on the above 3 steps, we can write:
Whenever n is greater than 2,
$\frac{1}{n!}~<~\frac{1}{2^{n-1}}$

4. We saw that, each term in (I) is less than the corresponding term in (II). So we can write:

$\begin{array}{ll}{}    &{\frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!}~ +~.~.~.~+~\frac{1}{n!}~+~.~.~.}    & {~<~}    &{\frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4}~ +~.~.~.~+~\frac{1}{2^{n-1}}~+~.~.~.}    &{} \\
    {\Rightarrow}    &{\left(1 + \frac{1}{1!} + \frac{1}{2!} \right)~+~\left(\frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!}~ +~.~.~.~+~\frac{1}{n!}~+~.~.~.\right)}    & {~<~}    &{\left(1 + \frac{1}{1!} + \frac{1}{2!} \right)~+~\left(\frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4}~ +~.~.~.~+~\frac{1}{2^{n-1}}~+~.~.~.\right)~\color{green}{\text{- - - (A)}}}    &{} \\
    {\Rightarrow}    &{\left(1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!}~ +~.~.~.~+~\frac{1}{n!}~+~.~.~.\right)}    & {~<~}    &{\left(1 + 1 + \frac{1}{2^1} \right)~+~\left(\frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4}~ +~.~.~.~+~\frac{1}{2^{n-1}}~+~.~.~.\right)}    &{} \\
    {\Rightarrow}    &{1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!}~ +~.~.~.~+~\frac{1}{n!}~+~.~.~.}    & {~<~}    &{1 + \left(1 + \frac{1}{2^1} + \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4}~ +~.~.~.~+~\frac{1}{2^{n-1}}~+~.~.~.\right)~\color{green}{\text{- - - (B)}}}    &{} \\
    {\Rightarrow}    &{1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!}~ +~.~.~.~+~\frac{1}{n!}~+~.~.~.}    & {~<~}    &{1 + \frac{1}{1 - \frac{1}{2}}}    &{} \\
    {\Rightarrow}    &{1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!}~ +~.~.~.~+~\frac{1}{n!}~+~.~.~.}    & {~<~}    &{3}    &{} \\
    {\Rightarrow}    &{e}    & {~<~}    &{3}    &{} \\
    \end{array}$

◼ Remarks:
• Line marked as A:
Here we add $\left(1 + \frac{1}{1!} + \frac{1}{2!} \right)$ on both sides.
• Line marked as B:
Consider the RHS of this line. We have a portion enclosed in brackets. This portion is a infinite geometric series. It’s sum can be easily calculated. It works out to 2.

5. Earlier we said that:
$1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} +~.~.~=~e$
• Consider the LHS. The first two terms are 1 and 1.
• So we can write:
e will be always greater than 2.

6. Let us compare the results:
    ♦ From (4) we have: e < 3
    ♦ From (5) we have: e > 2
• So we can write: 2 < e < 3
7. The approximate value of e is 2.71828. The accuracy will increase when we increase the number of terms in the series.
• e is an irrational number like π.
8. Consider the equation that we use to evaluate e:
$e~=~1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} +~.~.~.$
    ♦ In the LHS, the power of e is 1.
    ♦ In the RHS, all numerators are 1.
        ✰ These ones are actually, powers of 1.
• So we can write:
$e^1~=~1 + \frac{1^1}{1!} + \frac{1^2}{2!} + \frac{1^3}{3!} + \frac{1^4}{4!} +~.~.~.$
• If the power of e is 2, then we can write:
$e^2~=~1 + \frac{2^1}{1!} + \frac{2^2}{2!} + \frac{2^3}{3!} + \frac{2^4}{4!} +~.~.~.$     
• If the power of e is 7, then we can write:
$e^7~=~1 + \frac{7^1}{1!} + \frac{7^2}{2!} + \frac{7^3}{3!} + \frac{7^4}{4!} +~.~.~.$
◼ We can write the general form:
• If the power of e is x, then:
$e^x~=~1 + \frac{x^1}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} +~.~.~.~+ \frac{x^n}{n!}~+~.~.~.$

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Now we will see a solved example:

Solved example A.3
Find the coefficient of x² in the expansion of e^2x+3 as a series in the powers of x.
Solution:
1. We have the general form:
$e^x~=~1 + \frac{x^1}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} +~.~.~.~+ \frac{x^n}{n!}~+~.~.~.$
2. In our present case, we have (2x+3) in the place of x. So we can write:
$e^{2x+3}~=~1 + \frac{(2x+3)^1}{1!} + \frac{(2x+3)^2}{2!} + \frac{(2x+3)^3}{3!} + \frac{(2x+3)^4}{4!} +~.~.~.~+ \frac{(2x+3)^n}{n!}~+~.~.~.$
3. The general term is: $\frac{(2x+3)^n}{n!}$
The numerator of this general term can be expanded using binomial theorem. We get:

$\begin{array}{ll}{}    &{\frac{(2x+3)^n}{n!}}    & {~=~}    &{\frac{(3+2x)^n}{n!}}    &{} \\
{}    &{}    & {~=~}    &{\frac{1}{n!}\left[3^n + \rm{{}^n C_1}3^{n-1}(2x) +\rm{{}^n C_2}3^{n-2}(2x)^2 + \rm{{}^n C_3}3^{n-3}(2x)^3~+~.~.~.~+~(2x)^n  \right]}    &{} \\
\end{array}$

4. Note the third term of the above expansion. It is the term with "x²".
• So we can write:
General form of the coefficient of x² is $\frac{\rm{{}^n C_2}3^{n-2}(2x)^2}{n!}$

5. In (3), we considered the general term. For each term, there will be a term with x². We want the sum of coefficients of all such terms. That sum will be:
$\sum_{n=2}^{n=\infty}{\frac{\rm{{}^n C_2}3^{n-2}2^2}{n!}}$
(We saw similar problems in section 9.5) 

• This can be simplified as shown below:

$\begin{array}{ll}{}    &{\sum_{n=2}^{n=\infty}{\frac{\rm{{}^n C_2}3^{n-2}2^2}{n!}}}    & {~=~}    &{\sum_{n=2}^{n=\infty}{\frac{\frac{n!}{2! \times (n-2)!}\times 3^{n-2} \times 4}{n!}}}    &{} \\
{}    &{}    & {~=~}    &{\sum_{n=2}^{n=\infty}{\frac{\frac{1}{1 \times (n-2)!}\times 3^{n-2} \times 2}{1}}}    &{} \\
{}    &{}    & {~=~}    &{2\sum_{n=2}^{n=\infty}{\frac{3^{n-2}}{(n-2)!}}}    &{} \\
{}    &{}    & {~=~}    &{2 \left[\frac{3^{2-2}}{(2-2)!} + \frac{3^{3-2}}{(3-2)!} + \frac{3^{4-2}}{(4-2)!} + \frac{3^{5-2}}{(5-2)!}~+~.~.~. \right]}    &{} \\
{}    &{}    & {~=~}    &{2 \left[\frac{3^{0}}{0!} + \frac{3^{1}}{1!} + \frac{3^{2}}{2!} + \frac{3^{3}}{3!}~+~.~.~. \right]}    &{} \\
{}    &{}    & {~=~}    &{2 \left[\frac{1}{1} + \frac{3^{1}}{1!} + \frac{3^{2}}{2!} + \frac{3^{3}}{3!}~+~.~.~. \right]}    &{} \\
{}    &{}    & {~=~}    &{2 \left[1 + \frac{3^{1}}{1!} + \frac{3^{2}}{2!} + \frac{3^{3}}{3!}~+~.~.~. \right]}    &{} \\
{}    &{}    & {~=~}    &{2e^3}    &{} \\
\end{array}$

6. So the coefficient of x² is 2e³.

Alternate method:

1. $\begin{array}{ll}{}    &{e^{2x+3}}    & {~=~}    &{e^3 \times e^{2x}}    &{} \\
{}    &{}    & {~=~}    &{e^3 \left[1 + \frac{2x}{1!} + \frac{(2x)^{2}}{2!} + \frac{(2x)^{3}}{3!}~+~.~.~. \right]}    &{} \\
\end{array}$
2. Consider the third term in the above result. It is the term with x².
• This term can be written as: $\frac{4x^2 e^3}{2!}~=~2x^2 e^3$.
3. So we can write:
The coefficient of x² is $2e^3$

Solved example A.4
Find the value of e² rounded off to one decimal place.
Solution:
1. Let us expand e²:
$\begin{array}{ll}{}    &{e^2}    & {~=~}    &{1 + \frac{2^1}{1!} + \frac{2^2}{2!} + \frac{2^3}{3!} + \frac{2^4}{4!} + \frac{2^5}{5!} + \frac{2^6}{6!} + \frac{2^7}{7!} + \frac{2^8}{8!}~+~.~.~.}    &{} \\
{}    &{}    & {~=~}    &{\left(1 + \frac{2^1}{1!} + \frac{2^2}{2!} + \frac{2^3}{3!} + \frac{2^4}{4!} \right) + \left(\frac{2^5}{5!} + \frac{2^6}{6!} + \frac{2^7}{7!} + \frac{2^8}{8!}~+~.~.~.\right)}    &{} \\
{}    &{}    & {~=~}    &{\left(1 + 2 + 2 + \frac{8}{6} + \frac{16}{24} \right) + \left(\frac{2^5}{5!} + \frac{2^6}{6!} + \frac{2^7}{7!} + \frac{2^8}{8!}~+~.~.~.\right)}    &{} \\
{}    &{}    & {~=~}    &{\left(5 + \frac{4}{3} + \frac{2}{3} \right) + \left(\frac{2^5}{5!} + \frac{2^6}{6!} + \frac{2^7}{7!} + \frac{2^8}{8!}~+~.~.~.\right)}    &{} \\
{}    &{}    & {~=~}    &{7 + \left(\frac{2^5}{5!} + \frac{2^6}{6!} + \frac{2^7}{7!} + \frac{2^8}{8!}~+~.~.~.\right)}    &{} \\
\end{array}$

2. In the last line of the above result, consider the portion inside brackets. We will modify that portion as shown below:
$\begin{array}{ll}{}    &{\frac{2^5}{5!} + \frac{2^6}{6!} + \frac{2^7}{7!} + \frac{2^8}{8!}~+~.~.~.}    & {~=~}    &{\frac{2^5}{5!} + \frac{2^6}{6 \times 5!} + \frac{2^7}{7 \times 6 \times 5!} + \frac{2^8}{8 \times 7 \times 6 \times 5!}~+~.~.~.}    &{} \\
{\Rightarrow}    &{\frac{2^5}{5!} + \frac{2^6}{6!} + \frac{2^7}{7!} + \frac{2^8}{8!}~+~.~.~.}    & {~=~}    &{\frac{1}{5!}\left(\frac{2^5}{1} + \frac{2^6}{6} + \frac{2^7}{7 \times 6} + \frac{2^8}{8 \times 7 \times 6 }~+~.~.~. \right)}    &{} \\
{\Rightarrow}    &{\frac{2^5}{5!} + \frac{2^6}{6!} + \frac{2^7}{7!} + \frac{2^8}{8!}~+~.~.~.}    & {~<~}    &{\frac{1}{5!}\left(\frac{2^5}{1} + \frac{2^6}{6} + \frac{2^7}{6 \times 6} + \frac{2^8}{6 \times 6 \times 6 }~+~.~.~. \right)~\color{green}{{\text{- - - (A)}}}}    &{} \\
{\Rightarrow}    &{\frac{2^5}{5!} + \frac{2^6}{6!} + \frac{2^7}{7!} + \frac{2^8}{8!}~+~.~.~.}    & {~<~}    &{\frac{1}{5!}\left(\frac{2^5}{1-\frac{2}{6}} \right)~\color{green}{\text{- - - (B)}}}    &{} \\
{\Rightarrow}    &{\frac{2^5}{5!} + \frac{2^6}{6!} + \frac{2^7}{7!} + \frac{2^8}{8!}~+~.~.~.}    & {~<~}    &{\frac{1}{5!}\left(\frac{2^5}{\frac{2}{3}} \right)}    &{} \\
{\Rightarrow}    &{\frac{2^5}{5!} + \frac{2^6}{6!} + \frac{2^7}{7!} + \frac{2^8}{8!}~+~.~.~.}    & {~<~}    &{\frac{1}{5 \times 4 \times 3 \times 2 \times 1}\left(\frac{2^5 \times 3}{2} \right)}    &{} \\
{\Rightarrow}    &{\frac{2^5}{5!} + \frac{2^6}{6!} + \frac{2^7}{7!} + \frac{2^8}{8!}~+~.~.~.}    & {~<~}    &{\frac{1}{5 \times 1 \times 1 \times 1 \times 1}\left(\frac{2 \times 1}{1} \right)}    &{} \\
{\Rightarrow}    &{\frac{2^5}{5!} + \frac{2^6}{6!} + \frac{2^7}{7!} + \frac{2^8}{8!}~+~.~.~.}    & {~<~}    &{\frac{2}{5}}    &{} \\
{\Rightarrow}    &{\frac{2^5}{5!} + \frac{2^6}{6!} + \frac{2^7}{7!} + \frac{2^8}{8!}~+~.~.~.}    & {~<~}    &{0.4}    &{} \\
\end{array}$

◼ Remarks:
• Line marked as (A):
In this line, consider the denominators of the RHS.
   ♦ We put 6 in the place of 7.
   ♦ We put 6 in the place of 8 also.
         ✰ So the denominators decrease. As a result, the fractions increase in values.
         ✰ Then the whole RHS becomes larger than the LHS. We change the "=" sign to "<" sign.
• Line marked as (B):
In the line line marked as (A), in the RHS, the portion inside brackets is an infinite geometric series. It's sum can be easily calculated.

3. Now consider the last line of the result in (1).
Imagine that we replace $\frac{2^5}{5!} + \frac{2^6}{6!} + \frac{2^7}{7!} + \frac{2^8}{8!}~+~.~.~.$
by $\frac{1}{5!}\left(\frac{2^5}{1} + \frac{2^6}{6} + \frac{2^7}{6 \times 6} + \frac{2^8}{6 \times 6 \times 6 }~+~.~.~. \right)$
• Then we can write: e² < 7.4

4. Consider again the expansion of e². We want the first seven terms:

$\begin{array}{ll}{}    &{e^2}    & {~=~}    &{1 + \frac{2^1}{1!} + \frac{2^2}{2!} + \frac{2^3}{3!} + \frac{2^4}{4!} + \frac{2^5}{5!} + \frac{2^6}{6!} + \frac{2^7}{7!} + \frac{2^8}{8!}~+~.~.~.}    &{} \\
{}    &{}    & {~=~}    &{\left(1 + \frac{2^1}{1!} + \frac{2^2}{2!} + \frac{2^3}{3!} + \frac{2^4}{4!} + \frac{2^5}{5!} + \frac{2^6}{6!} \right) +  \frac{2^7}{7!} + \frac{2^8}{8!} ~+~.~.~.}    &{} \\
{}    &{}    & {~=~}    &{\left(1 + 2 + 2 + \frac{2^3}{3 \times 2 \times 1} + \frac{2^4}{4 \times 3 \times 2 \times 1} + \frac{2^5}{5 \times 4 \times 3 \times 2 \times 1} + \frac{2^6}{6 \times 5 \times 4 \times 3 \times 2 \times 1} \right) +  \frac{2^7}{7!} + \frac{2^8}{8!} ~+~.~.~.}    &{} \\
{}    &{}    & {~=~}    &{\left(1 + 2 + 2 + \frac{4}{3} + \frac{2}{3} + \frac{4}{15} + \frac{4}{45} \right) +  \frac{2^7}{7!} + \frac{2^8}{8!} ~+~.~.~.}    &{} \\
{}    &{}    & {~=~}    &{\left(1 + 2 + 2 + \frac{60 + 30 + 12 + 4}{45} \right) +  \frac{2^7}{7!} + \frac{2^8}{8!} ~+~.~.~.}    &{} \\
{}    &{}    & {~=~}    &{\left(1 + 2 + 2 + \frac{106}{45} \right) +  \frac{2^7}{7!} + \frac{2^8}{8!} ~+~.~.~.}    &{} \\
{}    &{}    & {~=~}    &{\left(1 + 2 + 2 + 2.3555556 \right) +  \frac{2^7}{7!} + \frac{2^8}{8!} ~+~.~.~.}    &{} \\
{}    &{}    & {~=~}    &{\left(7.355 \right) +  \frac{2^7}{7!} + \frac{2^8}{8!} ~+~.~.~.}    &{} \\
\end{array}$

• It is clear that, e² is greater than 7.355556.
• This value when rounded off to one decimal place is: 7.4.

5. Let us compare the results:
   ♦ From (3), we have: e² < 7.4
   ♦ From (4), we have: e² > 7.4

6. So the value of e² rounded off to one decimal place is 7.4.

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In the next section, we will see mathematical modelling.

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A.1 Infinite Geometric Series

In the previous section, we saw infinite series related to binomial expansions. In this section, we will see infinite series related to geometric series.

• In chapter 9, we saw the general form of a geometric series. Let us write it again:
$a + ar + ar^2 + ar^3 ~+ ~.~.~.~+~ar^{n-1}$
• This is a finite series. There are n terms.
• We know that, for such a finite series, the sum is given by:
$S_n = \frac{a \left(1 - r^n \right)}{1-r}$

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• Now let us suppose that, there are infinite number of terms. Then it will be an infinite geometric series.
• For such series also, we can derive a formula to find the sum.
• First we will see an example to learn the general features of infinite geometric series. It can be written in 7 steps:

1. Consider the geometric series:
$1 + \frac{2}{3} + \frac{4}{9} + ~.~.~.$
2. We see that, the first term, a = 1
3. Let us calculate r:
• $\frac{2}{3} \div 1 = \frac{2}{3}$

• $\frac{4}{9} \div \frac{2}{3} = \frac{4}{9} \times\frac{3}{2} = \frac{2}{3}$

• So we can write: $r = \frac{2}{3}$

4. Now we can write the sum of n terms:

$\begin{array}{ll}{}    &{S_n}    & {~=~}    &{\frac{a \left(1 - r^n \right)}{1-r}}    &{} \\
{}    &{}    & {~=~}    &{\frac{1 \times \left[1 - \left(\frac{2}{3} \right)^n \right]}{1-\left(\frac{2}{3} \right)}}    &{} \\
{}    &{}    & {~=~}    &{\frac{\left[1 - \left(\frac{2}{3} \right)^n \right]}{\frac{1}{3}}}    &{} \\
{}    &{}    & {~=~}    &{3 \left[1 - \left(\frac{2}{3} \right)^n \right]}    &{} \\
\end{array}$

5. We know that, when the number of terms in the series increases, n increases.
• When n increases, what happens to the term $\left(\frac{2}{3} \right)^n$ ?
◼ Let us put some convenient values for n:
(calculations can be easily done using a spreadsheet program)
• When n = 1, we get:
$\left(\frac{2}{3} \right)^n = \left(\frac{2}{3} \right)^1 = 0.666666666666667$
• When n = 5, we get:
$\left(\frac{2}{3} \right)^n = \left(\frac{2}{3} \right)^5 = 0.131687242798354$
• When n = 10, we get:
$\left(\frac{2}{3} \right)^n = \left(\frac{2}{3} \right)^{10} = 0.017341529915833$
• When n = 20, we get:
$\left(\frac{2}{3} \right)^n = \left(\frac{2}{3} \right)^{20} = 0.000300728659822$
• When n = 30, we get:
$\left(\frac{2}{3} \right)^n = \left(\frac{2}{3} \right)^{30} = 0.00000521509505084655$

6. It is clear that, when n increases, $\left(\frac{2}{3} \right)^n$ decreases.
• When n approaches infinity, $\left(\frac{2}{3} \right)^n$ will approach zero. There will be more than a million zeros just to the right of the decimal point. For all practical purposes, it can be considered as zero.
7. Now consider the result in (4).
• Using that result, we can write:
$S_n = 3[1 - 0] = 3~\text{where n is infinity}$

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Let us see if we can write a general form of the above result. It can be done in 4 steps:
1. Consider the geometric series:
$a + ar + ar^2 + ar^3 ~+ ~.~.~.~+~ar^{n-1}~+ ~.~.~.$
• The common ratio r is a proper fraction. It can be +ve or -ve.
• Mathematically, we write this as: |r| < 1
(Proper fraction is a fraction in which denominator is larger than the numerator)
• We want the sum when n is infinity.
2. We have:
$S_n = \frac{a \left(1 - r^n \right)}{1-r}$
• This can be written as:
$S_n = \frac{a}{1-r}~-~\frac{a r^n}{1-r}$
3. Now we note an important point:
    ♦ Since |r| < 1,
    ♦ $r^n$ will approach zero
    ♦ when n approach infinity.
• So the result in (2) will become:

$\begin{array}{ll}{}    &{S_n}    & {~=~}    &{\frac{a}{1-r}~-~\frac{a r^n}{1-r}}    &{} \\
{}    &{}    & {~=~}    &{\frac{a}{1-r}~-~\frac{a \times 0}{1-r}}    &{} \\
{}    &{}    & {~=~}    &{\frac{a}{1-r}~-~0}    &{} \\
{}    &{}    & {~=~}    &{\frac{a}{1-r}}    &{} \\
\end{array}$

4. Symbolically, when sum of an infinite geometric series is written, we use "S" instead of "S_n".
• So we have: $S = \frac{a}{1-r}$

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Now we will see a solved example:

Solved example A.2
Find the sum to infinity of the following geometric progressions:
(i) $1, \frac{1}{2}, \frac{1}{2^2}, \frac{1}{2^3},~.~.~.$
(ii) $1, \frac{-1}{2}, \frac{1}{2^2}, \frac{-1}{2^3},  ~.~.~.$
(iii) $\frac{-5}{4}, \frac{5}{16}, \frac{-5}{64}, ~.~.~.$
Solution:
Part (i):
1. The infinite geometric series associated with the given G.P is:
$1 + \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + ~.~.~.$
2. First term a is 1 and the common ratio r is $\frac{1}{2}$.
3. Absolute value of r is less than 1. So we can use the formula: $S = \frac{a}{1-r}$

$\begin{array}{ll}{}    &{S}    & {~=~}    &{\frac{a}{1-r}}    &{} \\
{}    &{}    & {~=~}    &{\frac{1}{1-\frac{1}{2}}}    &{} \\
{}    &{}    & {~=~}    &{2}    &{} \\
\end{array}$

Part (ii):
1. The infinite geometric series associated with the given G.P is:
$1 - \frac{1}{2} + \frac{1}{2^2} - \frac{1}{2^3} + ~.~.~.$
2. First term a is 1 and the common ratio r is $-\frac{1}{2}$.
3. Absolute value of r is less than 1. So we can use the formula: $S = \frac{a}{1-r}$

$\begin{array}{ll}{}    &{S}    & {~=~}    &{\frac{a}{1-r}}    &{} \\
{}    &{}    & {~=~}    &{\frac{1}{1- \left(-\frac{1}{2} \right)}}    &{} \\
{}    &{}    & {~=~}    &{\frac{1}{1+ \frac{1}{2}}}    &{} \\
{}    &{}    & {~=~}    &{\frac{1}{\frac{3}{2}}}    &{} \\
{}    &{}    & {~=~}    &{\frac{2}{3}}    &{} \\
\end{array}$

Part (iii):
1. The infinite geometric series associated with the given G.P is:
$\frac{-5}{4} + \frac{5}{16} - \frac{5}{64} + ~.~.~.$
2. First term a is $\frac{-5}{4}$ and the common ratio r is $-\frac{1}{4}$.
3. Absolute value of r is less than 1. So we can use the formula: $S = \frac{a}{1-r}$

$\begin{array}{ll}{}    &{S}    & {~=~}    &{\frac{a}{1-r}}    &{} \\
{}    &{}    & {~=~}    &{\frac{\frac{-5}{4}}{1- \left(-\frac{1}{4} \right)}}    &{} \\
{}    &{}    & {~=~}    &{\frac{\frac{-5}{4}}{1 + \frac{1}{4}}}    &{} \\
{}    &{}    & {~=~}    &{\frac{\frac{-5}{4}}{\frac{5}{4}}}    &{} \\
{}    &{}    & {~=~}    &{-1}    &{} \\
\end{array}$

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In the next section, we will see exponential series.

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