Presentation on the bisection method.pptx

This presentation for bisection method for solving polynomial equation in different fields as mathematics, physics, engineering etc.

1. Bisection Method By : Ankit Choudhary Ankush Rathore Anuj Yadav

2. The B isection M ethod in mathematics is a root finding method which repeatedly bisects an interval and then selects a subinterval in which a root must lie for further processing. INTRODUCTION

3. It is a very simple method , but it is also relatively slow . Because of this, it is often used to obtain a rough approximation to a solution which is then used as a starting point for more rapidly converging methods . Bisection method is also called Interval halving method

4. G r aphical Representation

5. Procedure of Bisection Method Step 1: Choose two approximations a and b such that, f(a)*f(b)<0 Step 2: Evaluate the midpoint c of [ a,b ] given by, c=( a+b )/2

6. Step 3 : If f(c)*f(b)<0 then rename b & c as a & b. If not rename of c as b, Then apply the formula of Step 2. Step 4: Stop evolution when the different of two successive values of c obtained from Step 2 is numerically less than e, the prescribed accuracy

7. If the function f (x) changes sign between the two points, more than one root for the equation f (x) = may exist between the two points.

8. If the function f (x) does not change sign between two points, there may not be any roots for the equation f (x) = between the two points.

9. If the function f (x) does not change sign between the two points, roots of the equation f (x) = may still exist between the two points.

10. FLOWCHART

11. Start Input : a, b f(a)*f(b) < Try with another values of a, b f(a)* f( x ) < b = x a = x | f( x )| < tolr Yes x =( a+b)/2 Yes Yes Output : root = x Stop No No No

12. PROGRAM CODE AND OUTPUT

13. input the values of a and b such that f(a) and f(b) have opposite values 1 the Root of Function f(x)= 0.567143440 the Root of Function f(x)= 0.567142963 the Root of Function f(x)= 0.567143202 the Root of Function f(x)= 0.567143321 the Root of Function f(x)= 0.567143261 the Root of Function f(x)= 0.567143321 the Root of Function f(x)= 0.567143321 the Root of Function f(x)= 0.567143321 the Root of Function f(x)= 0.567143321 the Root of Function f(x)= 0.567143321 the Root of Function f(x)= 0.567143321

14. input the values of a and b such that f(a) and f(b) have opposite values 1 the Root of Function f(x)= 0.656620502 the Root of Function f(x)= 0.656620264 the Root of Function f(x)= 0.656620383 the Root of Function f(x)= 0.656620324 the Root of Function f(x)= 0.656620383 the Root of Function f(x)= 0.656620383 the Root of Function f(x)= 0.656620383 the Root of Function f(x)= 0.656620383 the Root of Function f(x)= 0.656620383 the Root of Function f(x)= 0.656620383

15. input the values of a and b such that f(a) and f(b) have opposite values 1 2 the Root of Function f(x)= 1.83424330 the Root of Function f(x)= 1.83424306 the Root of Function f(x)= 1.83424318 the Root of Function f(x)= 1.83424306 the Root of Function f(x)= 1.83424306 the Root of Function f(x)= 1.83424306 the Root of Function f(x)= 1.83424306 the Root of Function f(x)= 1.83424306 the Root of Function f(x)= 1.83424306 the Root of Function f(x)= 1.83424306

16. input the values of a and b such that f(a) and f(b) have opposite values -2 -3 the Root of Function f(x)= -2.49086356

17. ANALYTICAL SOLUTION OF ROOT S

18. (a) To find roots of equation with the help of calculator using bisection method

19. x * exp (x)-1=0 Let f(x )=x* exp (x)-1 f(0)=0-1=-1 f(1 ) = 1.e-1=1.7182 So, the root lies in [0,1] a = 0, b = 1 first iteration x1 = (0 + 1)/2 = 0.5 f(x1)=f(0.5) = (0 . 5) exp (0.5)-1= -0.175639 Here f(x1)*f(b)< 0 f(0.5)*f(1 )< So , x1 = a = 0.5 New interval [ 0.5,1]

20. Second iteration x2 = (0.5+1)/2 = 0.75 f(x2)=f(0.75) = 0.5877 Here f(x2)*f(a)< 0 So, x2 = b = 0.75 New interval [ 0.5, 0.75]

21. Similarly, x3=0.625 [ 0.5 , 0.625 ] x4=0.5625 [ 0.5625, 0.625] ……………………………………… …………………………………… ……………………………………… x12=0.56713855 here x12=0.56714 (to five decimal places ) so , the root of equation f(x) is 0.56714

22. x**3-5*x+3=0 Let f(x ) = x**3- 5x + 3 f (0 )= 0-5(0 )+3= + 3 f(1) = 1 - 5 + 3 = -1 So, the root lies in [ 0,1 ] a = 0, b = 1 first iteration x1 = (0 + 1)/2 = 0.5 f(x1)= f(0.5) = (0 . 5)**3 - 5(0.5 ) + 3 = 0.625 Here f(x1)*f(b)< 0 f(0.5)*f(1 ) < , So, x1 = a = 0.5 New interval [ 0.5,1]

23. Second iteration x2 = ( 0.5+1 )/2 = 0.75 f(x2)=f(0.75) = ( 0.75)** 3 - 5(0.75 ) + 3 = -0.328125 Here f(x2)*f(a)< f(0.75 )* f a ) < , So, x2 = b = 0.75 New interval [ 0.5, 0.75]

24. Similarly, x3=0.625 [0,625, 0.75] x4=0.6875 [0.625, 0.6875] ……………………………………… …………………………………… ……………………………………… x20=0.65662075 x21=0.65662025 Here x20 ~ x21 (to six decimal places) So, the root of equation f(x) is 0.656620

25. (c) GRAPHICAL VERIFICATION OF THE ROOTS

26. x * exp (x)-1=0

27. X**3-5*x+3=0

28. Result The root of the given equation by Fortran programming and by alytical solution is almost same. Equation (1) x**3-5*x+3=0 x=0.656620 Equation (2) x* exp (x)-1=0 x=0.56714

29. Error Analysis Error % % for, xexp (x)-1=0 Error % =0.000004% for, x**3-5*x+3=0 Error % = 0.00005%  

30. ADVANTAGES The bisection method is very simple and easy to program on the computer. Does not involve complex calculations. The bisection method guarantees converges to a root in an interval as long as the function is continuous

31. DISADVANTAGES This method is very slow i.e. we have to do many steps to get the root of desired accuracy (slow to convergence) It fails to determine complex roots. It can not be applied if there are discontinuities in the guess interval .

32. THANK YOU