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.