Introduction to Sequences and Series (Chapter-1)

PROGRESSION:

» A progression is a sequence whose terms a certain pattern i.e. the terms are arranged  under a definite     rule. 

ARITHMETIC PROGRESSION:

»  A sequence of numbers < t_{n  >}is said to be in arithmetic progression (A.P) when the difference                t_n -  t_n-1  is a constant ∀ n є N. This constant is called the  common difference and is denoted by 'd '.

»  If 'a' is the first term and 'd' is the common difference then the A.P can be represented as 

                                         a ,a+d ,a+2d ,a+3d ,...

GENERAL TERM OF AN A.P:

» nth term of an A.P i.e.  t_n =a+(n-1)d.

SUM OF 'n' TERMS OF AN A.P:

» The sum of 'n' terms of the series is given by 

           S_n   =∑  t_n   =  t₁ + t₂    + t₃  +... +t_n 

                             = n/2 [ 2a+(n-1)d]    

       or,

           S_n   =n/2 [a+l].    [where a=first term  and l= last term]

  » NOTE:        t_n   =   S_n -  S_n-1

  ARITHMETIC MEAN:

 » If three terms are in A.P then the middle term is called A.M between the other two.

    Ex.- a ,a+d ,a+2d are in A.P , then a+d is the A.M of a and (a+2d)

            i.e. a+d = [a+(a+2d)]/2

      🟇 A.M of a,b is = (a+b)/2.

     🟇 A.M of a,b,c is = (a+b+c)/3.

 » If  'n' terms are inserted between a & b i.e.  A₁ ,A₂ ,..., A_n  are A.Ms that means  

     a , A₁ ,A₂ ,..., A_n  , b are in A.P. Then the common difference is(d)=. (b-a)/(n+1). 

         Then  A₁ = a + d = a+(b-a)/(n+1).

                   A₂ = a + 2d = a+ 2[(b-a)/(n+1)].

                   --------------------------------------

                    A_n = a + nd = a+ n[ (b-a)/(n+1)].   

PROPERTIES of A.P:

✮ If a fixed number is added or subtracted i.e.a₁士k , a₂士k , a₃ 士k ,... of given A.P , then the resulting       sequence is also an A.P with the common difference as that of given A.P .

✮ If each term of an A.P is multiplied by a fixed number or divided by non-zero fixed number i.e

      ka₁ , ka₂ , ka₃ ,...  ; a₁/k , a₂/k , a₃ /k ,...[k≠0] , then the resulting sequence is also an A.P . Common          difference is multiplied by the fixed number.

✮ The sum of terms of an A.P equidistant from the beginning and the end is equal to sum of first and         last term .        i.e.   a₁  + a_n    = a₂ + a_n-1 =a₃ + a_n-2   and               so on.

✮ If a₁ , a₂ , a₃ ,... and b₁ , b₂ , b₃ ,... are two A.Ps with common differences d and d' , then 

      a₁+b₁ , a₂+b₂ , a₃+b₃  ,... is also an A.P with common difference d+d'.

✮ The nth term of any sequence is linear expression in n , then the sequence is an A.P with the common difference is the coefficient of "n".

      

         Ex.- Let                t_n= 2n+1.

                      put n=1,   t₁= 2(1)+1=3

                      put n=2,   t₂= 2(2)+1=5               { d =2}

                      put n=3,   t₃= 2(3)+1=7

       Here, the common difference (d)= coefficient of n=2.

✮ The sum of nth terms of any sequence is quadratic in n , then the sequence is an A.P with common difference twice the coefficient of n^2.

        Ex. - Let           S_n= 3n^2 + 2n+1.       

        As we know, t_n =   S_n -  S_n-1

                   _{= (3n^2 + 2n+1) - [3(n-1)^2 + 2(n-1)+1]}

               _{= 3n^2 +2n +1 -(3n^2+3-6n+2n-2+1)}

               =  3n^2 +2n +1 -3n^2 -3 +6n -2n +2-1

               = 6n -1.                                           

         here d = 6 = twice of the coefficient of n^2.

     

  ➤ So, in the below tables , these are the terms taken , which should be used in some kind of questions in A.P . 

 TABLE -1: When the sum is given.

                  

               No. of terms                   Term taken

3	a-d , a , a+d
4	a-3d , a-d , a+d , a+3d
5	a-2d , a-d , a , a+d , a+2d

           TABLE -2: When the sum is not given.

                  

     No. of terms                   Term taken

3	a , a+d , a+2d
4	a , a+d , a+2d , a+3d
5	a , a+d , a+2d , a+3d , a+4d

 

                   Here is the sample question :-                                                                                                 

               So guys, this is all about the introduction of sequence and series and about the A.P.

               On later, we will learn about G.P and it's properties.

                 ∴ Stay tuned, for next session.😇

Solution of Polynomial And Transcendental equation - Bisection Method | Engineering Mathematics

Introduction:

Before Expressing the Bisection Method.We should discuss about the equations and graphs.

            (Graph of an equation)

This is a graph of an equation. As we all know equations are of two types

1.Polynomial Equation

2.Transcendental Equation 

A polynomial equation of the form

is called an Algebraic equation. For Example

An equation which contains polynomials,trigonometry functions,logarithmic functions,exponential functions etc.,is called a Transcendental equation.For example

are Transcendental equations.
>>A polynomial equation of degree n will have exactly n roots,real or complex , simple or multiple. A transcendental equation may have one root  or no root or infinite numbers of roots depending on the form of  ∱(x) .

The Methods of finding the roots of ∱(x) =0 are classified as

1.Direct Method:

Direct methods give the exact values of all the roots in a finite number of steps.
                               In these methods we start with one or two initial approximation of the root and obtain a sequence of approximation X0 ,X1 ,...... ,Xk  which in the limit as K ￫  ∞  converge to the exact root X =a.
                                                       There are no direct methods for solving higher degree algebraic equation or transcendental equation.

2.Numerical Method:

Numerical methods are based are based on the idea of successive approximations.
                           In these method , we first find an interval in which the root lies.If 'a' and 'b' are two number such that ∱(a) and ∱(b) have opposite sign,then a root ∱(x)=0 lies in between 'a' and 'b' . We will discuss few important Numerical Methods to find root of ∱(x)=0 .

2.1 Bisection Method:
Bisection method is a time taking method which needs a much effort.
Working Rule:
(i) Let ∱(x)=0 be the given equation . Find 'a' and 'b' such that ∱(a).∱(b)<0
we consider ∱(a)<0 and ∱(b)>0.
{One of them must take +ve and other should -ve}

(ii)Find first approx. root using bisection method as
                                    X1= (a+b)/2
Calculate ∱(x1) and examine the sign.

(iii)If ∱(x1) <0 ➡️ root lies between X1 and 'b'.
Similarly 2nd approx root is given by
                                  X2=(X1 +b)/2
(iv)If ∱(x)>0 ➡️ root lies between 'a' and X1 then 2nd approx. root is given by
                                  X2=(X1 +a)/2
Calculate ∱(X2) and repeat step (iii) & (iv) until we get the required accuracy of the root.

QNs. Find The real roots of the equation x³ -x -4=0.using bisection method correct upto 3 decimal places?
ANS:
Here  ∱(x)=x³ -x -4=0

Find 'a' and 'b':
∱(0)=-4 < 0
∱(1) =-4 <0
∱(2)=2>0
{we go much closer to root thats why we can find the root with less effort}
∱(1.5)=-2.125<0
∱(1,6)=-1.504<0
∱(1.7)=-0.787<0
∱(1.8)=0.032>0
{We can notice that ∱(1.7) and ∱(1,8) are much closure to zero }

Let a=1.7 and b=1.8


First Approximate root using bisection method:

X1= (a+b)/2=1.75
➡️∱(1.75)=-0.34<0
Hence root lies between 1.75 and 1.8

Second Approximate root:
X2=(1.75 +1.8)/2=1.775
➡️∱(1.775)= -0.182<0
Hence root lies between 1.775 and 1.8.

Third Approximate root:
X3=(1.775+1.8)/2 = 1.7875
➡️∱(1.7875)= -0.076<0
Hence the root lies between 1.7875 and 1.8

Fourth Approximate root:
X4=(1.7875 +1.8)/2 = 1.79375
∱(1.79375)= -0.022<0
Hence the root lies between 1.79375 and 1.5

Fifth Approximate root:
X5=(1.79375 +1.8)/2=1.796875
∱(1.796875)=0.0048>0
Hence root lies between 1.79375 and 1.796875

Sixth Approximate root:
X6=(1.79375+1.796875)/2=1.795312
∱(1.795312)=-0.0087<0
Hence the root lies between 1.795312 and 1.796875

Seventh Approximate root:
X7=(1.795312+1.796875)/2=1.796093

If  new root's 3digit after decimal is matched with any of the Previous two root's 3digit after decimal then we concluded that we found the answer.

So X7=1.796093 matched with X5=1.796875.Now X7 is our approximate root.