Program: BE(Mechanical)
Class: TE
Course: Numerical Methods and
Optimization
Unit: Roots of Equation
Lecture 01: Types of Root
Topics 02
Introduction Types of Bisection
Numerical on
Numerical Method
Bisection
Method
Method
Flowchart of Problem for Closing
Bisection Homework
Method
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Title of the page 03
1. Prakash Madhukar Sutar
2. Years of Experience -09
3. M.E (CAD Cam & Automation)
SCHOOL OF ENGINEERING AND TECHNOLOGY www.sandipuniversity.edu.in
07
Numerical Methods to obtain root of Equations
Bracketing Method
Bisection Method
Open End Method
Newtons Raphson
Successive Approximation
SCHOOL OF ENGINEERING AND TECHNOLOGY
Bisection Method 07
If we consider initial guesses 𝒙𝟏 and 𝒙𝟐 such that
f(𝒙𝟏 ) and f(𝒙𝟐 ) are both positive or both negative
f(𝒙𝟏 ) X f(𝒙𝟐 ) > 0, then we are not closer to the
root or we will never get the root of an equation
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Bisection Method 07
• Now we find the value of f(x3) by substituting the value of x3 in f(x)
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Steps in Bisection Method 07
1.Let the curve be y = f(x)
2.Choose initial guess x1 and x2 such that
f(𝒙𝟏 ) X f(𝒙𝟐 ) < 0,
3.Estimate the root of 𝑿𝟑 by
𝒙𝟏 + 𝒙𝟐
𝑿𝟑 =
𝟐
4.Make following evaluation to determine
in which subinterval the root lies .
If f(𝒙𝟑 ) X f(𝒙𝟐 ) < 0, If f(𝒙𝟏 ) X f(𝒙𝟑 ) < 0,
Then 𝒙𝟏 = 𝒙𝟑 Then 𝒙𝟐 = 𝒙𝟑
SCHOOL OF ENGINEERING AND TECHNOLOGY
Bisection Method 07
Example 1.: Using three iterations of Bisection method determine root of equation
Initial guesses are 𝒙𝟏 = 2.8 and 𝒙𝟐 =3 .
f(𝑥) =−0.9𝑥 2 + 1.7𝑥 + 2.5
Solution : 4. Iteration 1
1. f(𝒙𝟏 )= −0.9 ∗ (2.8)2 + 1.7(2.8) + 2.5 𝒙𝟏 +𝒙𝟐 𝟐.𝟖+𝟑
5. 𝑿𝟑 = = = 2.9
𝟐 𝟐
f(𝒙𝟏 )= 0.204
5. Check f(𝒙𝟑 )
2. f(𝒙𝟐 )= −0.9 ∗ (3)2 + 1.7(3) + 2.5
f(𝒙𝟑 )= −0.9 ∗ (2.9)2 + 1.7 2.9 + 2.5
f(𝒙𝟐 )= −0.5
f(𝒙𝟑 )= − 0.139
3. Check f(𝒙𝟏 ) X f(𝒙𝟐 )
As this value is negative so we will replace it
as f(𝒙𝟏 ) X f(𝒙𝟐 ) < 0 ; initial guess are right
with negative value of f(𝑥)
Therefore replace 𝒙𝟐 by 𝒙𝟑 .
SCHOOL OF ENGINEERING AND TECHNOLOGY
Bisection Method 07
Example 1.: Using three iterations of Bisection method determine root of equation
Initial guesses are 𝒙𝟏 = 2.8 and 𝒙𝟐 =3 .
f(𝑥) =−0.9𝑥 2 + 1.7𝑥 + 2.5
Solution : Solution :
Iteration 2 Iteration 3
𝒙𝟏 +𝒙𝟐 𝟐.𝟖+𝟐.𝟗 𝒙𝟏 +𝒙𝟐 𝟐.𝟖𝟓+𝟐.𝟗
4. 𝑿𝟑 = = = 2.85 4. 𝑿𝟑 = = = 2.875
𝟐 𝟐 𝟐 𝟐
• Check f(𝒙𝟑 ) • Check f(𝒙𝟑 )
f(𝒙𝟑 )= −0.9 ∗ (2.85)2 + 1.7 2.85 + 2.5 f(𝒙𝟑 )= −0.9 ∗ (2.875)2 + 1.7 2.875 + 2.5
f(𝒙𝟑 )= 0.0345 f(𝒙𝟑 )= −0.051562
As this value is positive so we will replace it with positive As this value is negative so we will replace it with negative
value of f(𝑥) value of f(𝑥)
Therefore replace 𝒙𝟏 by 𝒙𝟑 . Therefore replace 𝒙𝟐 by 𝒙𝟑 .
SCHOOL OF ENGINEERING AND TECHNOLOGY
Bisection Method 07
Example 2.: Using five iterations of Bisection method determine root of equation
Initial guesses are 𝒙𝟏 = 1.7 and 𝒙𝟐 = 1.9 .
f(𝑥) = 𝑥 3 − 𝑥 2 − 𝑥 − 1
Solution : 4. Iteration 1
𝒙𝟏 +𝒙𝟐 𝟏.𝟕+𝟏.𝟗
1. f(𝒙𝟏 )= 1.73 − 1.72 − 1.7 − 1 5. 𝑿𝟑 = = = 1.8
𝟐 𝟐
f(𝒙𝟏 )= − 0.677 5. Check f(𝒙𝟑 )
2. f(𝒙𝟐 )= 1.93 − 1.92 − 1.9 − 1 f(𝒙𝟑 )= 1.83 − 1.82 − 1.8 − 1
f(𝒙𝟐 )= 0.349 f(𝒙𝟑 )= − 0.208
3. Check f(𝒙𝟏 ) X f(𝒙𝟐 ) As this value is negative so we will replace it
as f(𝒙𝟏 ) X f(𝒙𝟐 ) < 0 ; initial guess are right with negative value of f(𝑥)
Therefore replace 𝒙𝟏 by 𝒙𝟑 .
SCHOOL OF ENGINEERING AND TECHNOLOGY
Bisection Method 07
Example 2.: Using five iterations of Bisection method determine root of equation
Initial guesses are 𝒙𝟏 = 1.7 and 𝒙𝟐 = 1.9 .
f(𝑥) = 𝑥 3 − 𝑥 2 − 𝑥 − 1
Solution : Solution :
Iteration 2 𝒙𝟏 =1.8 and 𝒙𝟐 = 1.9 . Iteration 3 𝒙𝟏 =1.8 and 𝒙𝟐 = 1.85
𝒙𝟏 +𝒙𝟐 𝟏.𝟖+𝟏.𝟗 𝒙𝟏 +𝒙𝟐 𝟏.𝟖+𝟏.𝟖𝟓
4. 𝑿𝟑 = = = 1.85 4. 𝑿𝟑 = = = 1.825
𝟐 𝟐 𝟐 𝟐
• Check f(𝒙𝟑 ) • Check f(𝒙𝟑 )
f(𝒙𝟑 )= 1.853 − 1.852 − 1.85 − 1 f(𝒙𝟑 )= 1.8253 − 1.8252 − 1.825 − 1
f(𝒙𝟑 )= 0.05912 f(𝒙𝟑 )= −0.0772
As this value is positive so we will replace it with positive As this value is negative so we will replace it with negative
value of f(𝑥) value of f(𝑥)
Therefore replace 𝒙𝟐 by 𝒙𝟑 . Therefore replace 𝒙𝟏 by 𝒙𝟑 .
SCHOOL OF ENGINEERING AND TECHNOLOGY
Bisection Method 07
Example 2.: Using five iterations of Bisection method determine root of equation
Initial guesses are 𝒙𝟏 = 1.7 and 𝒙𝟐 = 1.9 .
f(𝑥) = 𝑥 3 − 𝑥 2 − 𝑥 − 1
Solution : Solution :
Iteration 4 𝒙𝟏 =1.825 and 𝒙𝟐 = 1.85 . Iteration 5 𝒙𝟏 =1.8375 and 𝒙𝟐 = 1.85
𝒙𝟏 +𝒙𝟐 𝟏.𝟖𝟐𝟓+𝟏.𝟖𝟓 𝒙𝟏 +𝒙𝟐 𝟏.𝟖𝟑𝟕𝟓+𝟏.𝟖𝟓
4. 𝑿𝟑 = = = 1.8375 4. 𝑿𝟑 = = = 1.83125
𝟐 𝟐 𝟐 𝟐
• Check f(𝒙𝟑 ) Ans :
f(𝒙𝟑 )= 1.83753 − 1.83752 − 1.8375 − 1 After 5th iteration the root of given function is 1.83125
f(𝒙𝟑 )= −0.00975
As this value is negative so we will replace it with negative
value of f(𝑥)
Therefore replace 𝒙𝟏 by 𝒙𝟑 .
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