Introduction to Relation

Relation:

A relation R from A to B is a subset of the Cartesian product A x B and is derived by describing a relationship between the first element (say x) and the other element (say y) of the ordered pairs in A & B.

Representation of Relation:

A relation is represented either by Roster method or by Set-builder method. Consider an example of two sets A = {9, 16, 25} and B = {5, 4, 3, -3, -4, -5}. The relation is that the elements of A are the square of the elements of B.

1. In set-builder form, R = {(x, y): x is the square of y, x ∈ A and y ∈ B}.
2. In roster form, R = {(9, 3), (9, -3), (16, 4), (16, -4), (25, 5), (25, -5)}.

Terminologies (Terms used):

1. Image:
   for any ordered pairs, in any Cartesian product (say A × B), the second element is called the image of the first element.
2. Domain:The set of all first elements of the ordered pairs in a relation R from a set A to a set B.
3. Range:The set of all second elements in a relation R from a set A to a set B is  called Range.
4. Co-domain:
   The whole set B is called Co-domain. Range ⊆ Co-domain.

Total Number of Relations:

For two non-empty set, A and B. If the number of elements in A is h i.e., |A| = h & that of B is k i.e., |B| = k, then the number of ordered pair in the Cartesian product will be |A × B| = hk. The total number of relations is 2^hk.

Types of Relations:

There are many types of relations.

Empty Relation:

If no element of set X is related or mapped to any element of Y, then the relation R in A is an empty relation, i.e, R = Φ.

Universal Relation:

A relation R, called universal relation if each element of A is related to every element of B.

Suppose A is a set of all natural numbers and B is a set of all whole numbers. The relation between A and B is universal as every element of A is in set B.

                    Empty relation and Universal relation are sometimes called trivial relation.

Inverse Relation:

Let R be a relation from set A to set B i.e., R ∈ A × B. The relation R^-1 is said to be an Inverse relation if R^-1 from set B to A is denoted by R^-1 = {(b, a): (a, b) ∈ R}.

Identity Relation:

In Identity relation, every element of set A is related to itself only. I = {(a, a), ∈ A}.

For example, If we throw two dice, we get 36 possible outcomes, (1, 1), (1, 2), … , (6, 6). If we define a relation as R: {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}, it is an identity relation.

Reflexive Relation:

If every element of set A maps to itself, the relation is Reflexive Relation. For every a ∈ A, (a, a) ∈ R.

Symmetric Relation:

A relation R on a set A is said to be symmetric if (a, b) ∈ R then (b, a) ∈ R, for all a & b ∈ A.

Transitive Relation:

A relation in a set A is transitive if, (a, b) ∈ R, (b, c) ∈ R, then (a, c) ∈ R, for all a, b, c ∈ A

Equivalence Relation:

A relation is said to be equivalence if and only if it is Reflexive, Symmetric, and Transitive.

Ordered Relation:

Partial Ordered Relation:
A relation R⊂X ✖️ X is called partial ordering if it is Reflexive,Anti-Symmetric, and Transitive, that is
   
             (x , x) ∈ R for all x ∈ X
             (x , y) ∈ R ,(y , x) ∈ R ⇒ x = y
             (x , y) ∈ R, (y , z) ∈ R ⇒ (x , z) ∈ R

Total Ordered Relation:
When all the elements of a partial order relation are comparable, the relation is called a total order Relation.
i.e;  a,b ∈ A and  a≤ b and b≤ a
[ ≤ is called relation ]

Linear Ordered:

That is, every element is related with every element one way or the other.
A total order is also called a linear order.

PreOrdered/Quasi Ordered:

A binary relation R on a set A is a quasi order if and only if it is
(1) irreflexive, { (x , x) ∈ R for all x ∉ X }
(2) transitive.  (x , y) ∈ R, (y , z) ∈ R ⇒ (x , z) ∈ R.


(minimal/maximal element):

Let < A, ≤ > be a poset, where ≤ represents an arbitrary partial order. Then an element   b ∈ A is a minimal element of A if there is no element  a ∈ A that satisfies a ≤ b. Similarly an element b ∈ A is a maximal element of A if there is no element a ∈ A that satisfies b ≤ a.
 
(least/greatest element):

Let < A, ≤ > be a poset. Then an element b ∈ A is the least element of A if for every element a ∈ A, b ≤ a.

(well order): A total order R on a set A is a well order if every non-empty subset of A has the least element.

THEOREMS ON PROBABILITY OF EVENTS( PART -2)

[If you are beginner in probability then you should read Probability Theorem (Part-1) before Read the below content]

THEOREM- 4:

                   For any 3 events A , B , C  P( A ⋃ B | C ) = P( A | C ) + P( B | C ) P( A ⋂ B | C )

 

PROOF:

      We have P( A ሀ B ) = P( A ) + P( B ) - P( A ⋂ B )

 ⇒P[( A ⋂ C) ⋃ ( B ⋂ C )] = P( A ⋂ C ) + P( B ⋂ C )     - P( A ⋂ B ⋂ C )

         Dividing both sides by P( C ), we get

 ⇒ P[( A ⋂ C) ⋃ ( B ⋂ C )] / P( C ) = P( A ⋂ C ) / P(         C ) + P( B ⋂ C ) / P( C )  - P( A ⋂ B ⋂ C ) / P( C )                                      , P( C ) > 0

                       

 ⇒  P[( A  ⋃ B ) ⋂ C ] / P( C ) = P( A | C ) + P( B | C ) -  P( A ⋂ B | C )

 ⇒   P[( A  ⋃ B ) | C ] = P( A | C ) + P( B | C ) - P( A ⋂ B | C ) 

                                                                        [Proved]

THEOREM - 5:

           For any 3 events A , B , C  P(A ⋂ B' | C ) + P( A ⋂ B | C ) = P( A | C ) , B' = complementary of event B.

PROOF :

          Given,

         P(A ⋂ B' | C ) + P( A ⋂ B | C )

     = P(A ⋂ B' ⋂ C ) / P( C ) + P( A ⋂ B ⋂ C ) / P( C )

     = [ P(A ⋂ B' ⋂ C ) + P( A ⋂ B ⋂ C ) ] / P( C ) 

     =  P(A ⋂ C ) / P( C ) 

     = P( A | C )                                  [Proved]

 

 THEOREM - 6:

         For any 3 events A , B , C defined on the sample space S such that B ⊂ C and P( A ) > 0 , P( C | A )  ≥ P( B | A ).

 PROOF:

      P( C | A ) = P( C ⋂ A ) / P( A )      [ By definition of conditional probability]

     = [ P( B ⋂ C ⋂ A ) ⋃ P( B' ⋂ C ⋂ A ) ] / P( A )

     =  [ P( B ⋂ C ⋂ A ) ] / P( A )  +  [ P( B' ⋂ C ⋂ A ) ] / P( A )

     =    P [ ( B ⋂ C | A ) + ( B' ⋂ C | A ) ]

    Since B ⊂ C ⇒ B ⋂ C = B

           ⇒ P( C | A ) = P( B | A ) + P( B' ⋂ C | A )

           ⇒ P( C | A ) ≥ P( B |  A ).                                         [Proved]

 Pair-wise Independent Events:

         Definition:  A set of events  A₁ , A₂  , ... , A_n  are said to be pair-wise independent  if  

          P(  A_i ⋂  A_j ) = P(  A_i ) . P(  A_j )  ∀ i ≠ j

 Conditions for Mutual Independence of ' n ' Events :

    Let S denote the sample space for a number of events . The events in S are said to be mutually independent if the probability of the simultaneous occurrence of (any) finite number of them is equal to the product of their separate probabilities.

               If  A₁ , A₂  , ... , A_n  are ' n ' events , then their mutual independence , we should have  

       (i)  P(  A_i ⋂  A_j ) = P(  A_i ) . P(  A_j ) ,    [ i ≠ j ; i , j = 1,2,...,n]

      (ii) P(  A_i  ⋂  A_j  ⋂  A_k ) = P( A_i ) . P( A_j ) . P( A_k ) ,  [ i ≠ j ≠ k ; i ,      j , k = 1,2,...,n]

                                    .                                     .

                                    .                                     .

                                    .                                     .

 P( A₁ ⋂  A₂ ⋂ ... ⋂ A_n ) = P( A₁ ) . P( A₂ ) . ... . P( A_n )

 

Remarks : 

       1. It may be observed that pair-wise or mutual independence of events A₁ , A₂  , ... , A_n  is defined only when P(  A₁ ) ≠ 0 , for i = 1,2,...,n .

       2. If the events A and B are such that P( A )  ≠ 0 , P( B ) ≠ 0 and A independent of B , then B is independent of A.

      Proof :  We are given that , P( A | B) = P( A )       

                   [∵ A is independent of B ]

                   ⇒ P( A ⋂ B ) / P( B ) = P( A )

                   ⇒  P( A ⋂ B ) = P( A ) . P( B )

         ⇒ P( B ⋂ A ) / P( A ) = P( B )    [∵ P( A ) ≠ 0 ,                    A ⋂ B =  B ⋂ A ]

         ⇒ P( B | A ) = P( B )  

A QUESTION FROM IMO 1995 | TITU'S LEMMA | CAUCHY-SCHWARZ INEQUALITY

INTRODUCTION:

In this session we will discuss about a question which had appeared in IMO 1995. 

So before staring the session i will discuss about ' Titu's Lemma ' and  ' Cauchy-Schwarz Inequality '.

Cauchy-Schwarz Inequality :

 The Cauchy-Schwarz inequality also known as Cauchy-Bunyakovsky-Schwarz inequality states that for all sequence of real number a_i and b_i,we have

Condition:

Equality holds if and only if  a_i = kb_i, for non-zero constant K € R.

TITU'S LEMMA:

Titu's lemma is a direct consequence of cauchy-Schwarz inequality .

It is also known as Engle's Form, or Sedrakyan's inequality.

      Titu's Lemma states that for positive real number a₁,a₂,a₃.....a_n and b_{1 ,}b_{2 ,}b_3....b_n is

This form is specially helpful when the inequality involvs fractions where the numeretor is a perfect square.

Then it becomes

           

QNs.

Let  a,b,c are positive real numbers such that a.b.c = 1.

Prove that

ANS.

There are two ways to solve the question 

1.By Titu's Lemma

2. By AM-GM inequality

Method-1:

We have already discussed the inequality at the top.

Substitute a= 1/x  , b= 1/y , c=1/z

by the use of Cauchy-Schwarz inequality 

And the Titu's lemma

we know that a.b.c =1 so ,the  result will be 3/2.

Method-2:

Consider (a+b) , (b+c) ,(c+a).Then by the AM - GM inequality, we have the first set of inequalities.

Multiplying altogether 

But  abc=1 ,then

       (a+b)(b+c)(c+a) >=   8

By the power mean inequality ( AP > GP >HP), and considering the arithmetic and harmonic means of a^3(b+c) and its permutations

This is how we prove.

Stay connected to see more blogs.

BOLZANO-WEIERSTRASS THEOREM FOR SETS | REAL ANALYSIS

INTRODUCTION:

Have you remembered  Bounded set ,Order completeness of Real numbers.
If not just Click to Read Real Analysis part-1 .

Bounded Set:
A set is said to be bounded if it is bounded above and below.
" I will  not explain this just read about it in the above link "

Bolzano-Weierstrass Theorem:

Statement: Every infinite bounded set of real numbers has a limit point.

Proof: Let ' S ' be any infinite bounded set of real numbers.
                 
                  m ≤ x ≤ M ,      ∀  x ∈S  ---(1)
                     
    m = infimum of S       M=supremum of S

Let us assume T is a set of Real numbers.
 
          T = { x : x is greater than only for finite number of elements } ------(2)

 x ∈ T { x ≥ only a finite number of elements of S } --(3)

{ x ≤ infinite number of elements of S         }---(4)

  [ Don't confuse with 'S' and 'T' .   S is a infinitely bounded set while T is a finitely bounded set and T is the subset of S also element 'x' is common in both sets]

From equation (1) and (4)
               
                      m ∈ T  ⇔  T ≠ Φ

From equation (1)
               
                       M ≥ x  ,     ∀  x ∈S
 
Let us assume another element 'y' from set T.
                  y ∈ T  ⇔  y ≥ only a finite number of elements of S  [From equation (3)]

    M ≥ y ,    ∀  y ∈ T  ⇔ T is bounded above.

 Now we have prove that T is bounded above

So it will follow Order Completeness Property and we know that " There exist a supremum of  a set of which set follows the completeness property ".

       Let  P be the Supremum of T .
 
Claim: ( P  is a limit point  of set S )

          Let є > 0 be arbitrary number.

 Since   P + є > P
     
          ⇒ P + є ≥ finite number of elements of S   -----(5)
   [ Because P is the Supremum of T and T is the subset of S]
 
Since     P - є < P
      
         ⇒ P - є ≤  infinite number of elements of S ----(6)

From equation (5) and (6) that  nbh
      
                ］  P - є , P + є [   of  P contains infinite number of elements of S.

Hence , P is a limit point.


Example:

1.  Z  has no limit point .{it has not bounded}

    2.   S = { -1 ⁿ(n/(n+1)  : n ∈ N}

Ans. 
           S = { -1/2 ,2/3 , -3/4 .....}
          infinite | Bounded   -1 < x < 1

hence lmit point will  be 0.

If any doubt , Comment below

THEOREMS ON PROBABILITY OF EVENTS (PART-1)

  Before you readout this blog I suggest you to read it's previous session i.e. " Introduction to Probability" , CLICK HERE to read. 

 THEOREM-1:

             Probability of the impossible event is zero i.e. p(Φ) = 0.

 

    PROOF:

              Impossible event contains no sample  point and hence the certain event S and   the impossible   event Φ are mutually     exclusive.  

     Hence      S U Φ = S  , [where S=    Sample space ]

             ∴      P( S U Φ ) = P( S )

             ⇒    P( S ) + P(Φ ) = P( S )

             ⇒    P( Φ ) = 0                                                                                           [Proved]

REMARK:

    It may be noted P( A )=0, does not imply  that  A is necessarily an empty set. In practice, probability ' 0 ' is assigned to the events which are so rare for example , let us consider the   random tossing of a coin. The event that the coin will stand erect on its edge,  is assigned the probability 0.

 

THEOREM -2:

           Probability of the complementary event Ā  of A is given by  P( Ā ) =1-P( A ).

    PROOF:

                 We know that A and Ā are dis-joint                          events.

             Moreover,          A U Ā = S 

                                    ⇒ P( A U Ā ) = P( S )

                                    ⇒ P( A ) + P( Ā ) = 1

                                    ⇒ P( Ā ) = 1 - P( A )                                                                               [Proved]

                                                

THEOREM - 3:

                      For any event A, 0 ≤ P( A ) ≤ 1.

PROOF:

         Clearly for any event A , 

               we have  Φ ⊆ A⊆ S

                         ⇒ | Φ | ⊆ | A | ⊆ | S |

                         ⇒ | Φ |/ | S | ≤ | A |/ | S | ≤ | S |/ | S |

                         ⇒  0  ≤ P( A ) ≤ 1        [Proved]

ADDITION RULE:

  ★   For any two events A and B                

            P( A ⋃ B ) = P( A ) + P( B ) - P( A ⋂ B ).

  ★    If A and B are two mutually exclusive  events then,

                      P( A ⋃ B ) = P( A ) + P( B )                                            [ ∵ P( A ⋂ B ) = 0 ]

 ★    For any three events A  , B , C ,

                        

                  P( A ⋃ B ⋃ C )=P( A ) + P( B ) + P( C ) -      P( A ⋂ B ) - P(B ⋂ C ) - P(C ⋂ A ) + P(A ⋂ B ⋂ C)

 

    PROOF:

           Let us take A ⋃ B = D

      P( A ⋃ B ⋃ C ) = P( D ⋃ C ).                                   

      = P( D ) + P( C ) - P ( D ∩ C )

     = P( A ⋃ B ) + P( C ) -  P(( A ⋃ B ) ∩ C ))                 

     = P( A ⋃ B ) + P( C ) -  [ P(( A ⋂ C )  ⋃ ( B ⋂ C )) ]

     = P( A ) + P( B ) + P( C ) -P( A ⋂ B ) - [ P( A ⋂ C )  + P( B ⋂ C ) - P((A ⋂ B ) ⋂ ( B ⋂ C)) ].  

     =  P( A ) + P( B ) + P( C ) -P( A ⋂ B ) - P(B ⋂ C ) -  P(C ⋂ A ) + P(A ⋂ B ⋂ C)

                                                                                                                                            [Proved]

 ★  If A , B , C are the pair-wise mutually exclusive events , then 

                            

             P( A ⋃ B ⋃ C )=P( A ) + P( B ) + P( C ) .

 Example:

          A dice is thrown is thrown twice , find the 

          i) P( getting total is 8)

         ii) P( getting total is 10)

        iii) P( total is 8 and 10 )

        iv) P ( total is 8 or 10 )

         v) P( total is an even number)

        vi) P( total is not 8)

       vii) P(  tota is an even number but not 8 )

Answer-  Since the dice is thrown twice , the sample space ( S ) = | 36 |

       i) Let A is the event of getting total is 8

         A={ (2,6) , (3,5) , (4,4) , (5,3) , (6,2) } , | A | = 5

                 so, P( A )= | A | / | S | = 5/36.

      ii) Let B is the event of getting total is 10

                   B={ (4,6) , (5,5) , (6,4) } , | B |=3

                 so, P( B )= | B | / | S | = 3/36= 1/12

     iii) P( getting total is 8 and 10)

           = P( A ⋂ B ) = P(Φ ) = 0

      iv)P( getting total is 8 pr 10)

          = P( A ⋃ B ) = P( A ) + P( B ) - P( A ⋂ B ) 

          = 5/36 + 3/36 - 0 = 8/36=2/9

       v) Let C be the event of total is an even      no. 

              P( C ) = 18/ 36 = 1/2

      vi) P( getting total is not 8)

            = 1 - P( A ) = 1- (5/36) = 31/ 36

   vii) P( getting total is an even no. but not 8)

             = P(getting total is an even no.) - 

                                                P(getting total is 8)

             = (18/36) - (5/36) = 13/36.

 

    

 

  

Introduction to Geometric Progression

GEOMETRIC PROGRESSION(G.P.):

⮚A progression is called a G.P. if the ratio of it's each term to it's previous term is always constant. If       'a'  be the first term and 'r' be the common ratio then a, ar, ar², ... ,ar^n-1 is a sequence of G.P . t_n =               ar^n-1. 

GENERAL TERM OF A G.P.:

•   n^th term of G.P. i.e.  t_n = ar^n-1. , where the common ratio (r) =  t₂ /  t₁ =  t₃ /  t₂ = ...
• If G.P.  consists of ' n ' terms , then  p^th term from the end = (n-p+1)^th term from the beginning = ar^n-p. Also , the   p^th   term from the end of the G.P. with the last term ' l ' and common ratio 'r'  is =  l(1/r)^n-1 . 
• 
  
  

SUM OF FIRST ' n ' TERMS OF A G.P.:

⮚ Sum of first ' n ' terms of a G.P. is given by

             S_n = ∑   t_n 

                  =   t₁ +   t₂  +   t₃ + ... +   t_n    

                  =   a + ar + ar²+ ... +ar^n-1

•  ^S_n =  a(1-rⁿ )/(1-r)  and  S_n = (a-lr)/(1-r) , [when | r | < 1. ] 
•  ^S_n =  a(rⁿ -1)/(r-1)  and  S_n = (lr-a)/(r-1) , [when | r | > 1. ]
•  S_n = na  [ when r=1 ]

SUM OF INFINITE TERMS OF G.P.:

⮚ When n→∞ and | r | < 1 ( -1 < r < 1),      S_∞  = a/(1-r).

⮚ When n→∞ and | r | ≥ 1 ,  S_∞  does not exist.

GEOMETRIC MEAN(G.M.):

⮚ If a , b , c are in G.P.  , then  b/a = c/b  ⇒ b²= ac ⇒ b= √(ac)  is the G.M. of a and c.

      Similarly  G.M. of a , b , c  is   (abc)^1/3 .

      G.M.  of   a₁ , a₂  , a₃ , ... , a_n  is     (a₁. a₂  . a₃  ...  .a_n )^1/n         

  ⮚ Let n G.M.s are inserted between a & b . Let    G₁ , G₂  , G₃ , ... , G_n   are n G.M.s ,

      then a ,  G₁ , G₂ , G₃ , ... , G_n , b are in G.P. .

       Then common ratio ( r ) =  (b/a )^1/n+1          

         ^So,   G₁  =   ar   =   a . (b/a )^1/n+1 

                  G₂  =   ar²  =   a . (b/a )^2/n+1 

                  ^{___________________________}

                 G_n  =   arⁿ =   a . (b/a )^n/n+1 

PROPERTIES OF G.P. :

✪  If all the terms of a G.P. be multiplied or divided by the same non-zero constant , then it                 remains a G.P.  , with the same common ratio ' r '.  

✪  The reciprocal of the terms of a given G.P.  form a G.P. with common ratio as reciprocal of the         common ratio(r) of the original G.P. i.e. ' 1/r '.

✪  If each term of a G.P.  with common ratio ' r ' be raised to the same power k , the resulting              sequence also forms a G.P. with common ratio  r^k  .

✪  In a finite G.P. , the product of terms equidistant from the beginning and the end is always the       same and is equal to the product of the first and last term i.e. if   a₁ , a₂  , a₃ , ... , a_n   be in G.P.        . Then                a₁. a_n  = a₂ . a_n-1  =  a₃ . a_n-2  = ... = a_r . a_n-r+1          

✪  If  a₁ , a₂  , a₃ , ... , a_n , ... is a G.P. of non-zero , non-negative terms , then  loga₁ , loga₂  ,log a₃          , ... ,  loga_n , ...  is in A.P. and vice-versa.

✪  Three non-zero numbers a , b , c are in G.P. iff  b²= ac .

✪  If    a^x1 , a^x2, a^x3, ... , a^xn   are in G.P. , then  x₁ , x₂  , x₃ , ... , x_n are in A.P. .

✪  If the first term of a G.P. of n terms is ' a ' and last term is ' l ' , then the product of all terms         of  the G.P.  is  (al )^n/2 .

  ⮚ So, in the below tables , these are terms taken , which should be used in some kinds of questions in        G.P. as same as in the  A.P. .

                  TABLE-1 : When the product is given.

            ^{No. of terms                        Terms taken} 

3	a/r , a , ar
4	a/ r3 , a/r , a/r , a/ r3
5	a/ r2 , a/r , a , a/r , a/ r2

                   TABLE -2 : When the product is not given. 

       

         ^{No. of terms                        Terms taken} 

3	a , ar , ar2
4	a , ar , ar2 ,  ar3
5	a , ar , ar2, ar3, ar4

       So guys , this is all about the introduction of G.P. .      And on later, we will learn about                         Harmonic  Progression and it's properties . 

       Thanks for reading this post .😇😇😇 

                                     And Stay tuned guys . ✌✌✌

Newton-Raphson Method ---Solution of Polynomial And Transcendental equation| Engineering Mathematics

Introduction:
≫In the previous session we read about Bisection method which is used to solve the equation for finding root.
if you are not read that blog i suggest you to read that blog first
Click Here to read.

Newton-Raphson Method:

Concept:

We will choose a +ve point much nearer to cut of the curve to X-axix . Let that point is x_n and it intersect the curve at (x_n,∱(x_n)) .

              If we draw a tagent at the point (x_n,∱(x_n)) then the slope will be ∱'(x_n) and cut the X-axis at x_n+1.

     Similarly,the point x_n+1 cuts the curve at (x_n+1,∱(x_n+1)) and if we draw a tangent at there then the slope will be ∱'(x_n+1) and the tangent cut the X-axix at x_n+2 .

         And The point x_n+2 cuts the curve at (x_n+2,∱(x_n+2)). 

We can see that at every step we move towards the root point of ∱(x)=0.

This method is very easy and less-effort than Bisection-Method.

Working Rule:

1. Given ∱(x)=0 ------(1)
   Find initial root x₀ such that ∱(x)~0
[ i.e; x₀ is near to the root ∱(x)=0]

2. Find ∱(x₀) and ∱'(x₀)

3. First approximate root by Newton-Raphson method:
            $x_{1=x_{0\mathrm{}}}- \frac{∱\left(x_{0\mathrm{}}\right)}{∱\text{'}\left(x_{0\mathrm{}}\right)}$      

Find ∱(x₁) and ∱'(x₁)

4. Similarly, Second approximate root is

$x_{2=x_{1\mathrm{}}}- \frac{∱\left(x_{1\mathrm{}}\right)}{∱\text{'}\left(x_1\right)}$              

5.Also Third approximate root is

          $x_{3=x_{2\mathrm{}}}- \frac{∱\left(x_{2\mathrm{}}\right)}{∱\text{'}\left(x_{2\mathrm{}}\right)}$                 

General Formula:


$x_{n =x_{n-1 }}- \frac{∱\left(x_{n-1}\right)}{∱\text{'}\left(x_{n-1}\right)}$

Example

QnS. ∱(x) = x - 2 + lnx has a root near x =1.5 .Use Newton-Raphson Method to obtain a better estimate 2 digit after decimal point.

Ans.

In this question we don't need to figure out the x₀, because it has given that the root is lies near x=1.5 .

Here x₀ =1.5 and ∱(1.5) = -0.5 + ln(1.5) = -0.0945

∱'(x)= 1+(1/x)

⇒ ∱'(1.5) = 5/3

Hence using the general formula

x₁=1.5567

The Newton-Raphson formula can be used again.

so ∱(x₁) = ∱(1.5567) = -0.0007

∱'(x₁) = ∱'(1.5567) = 1.6424

{ Calculation Part is skipped for Readers}

x₂= 1.5571

⇒We will note that the 2digits after decimal point of x₁ is same as x₂ .So

our answer will be x₂.