Upcoming SlideShare
×

Thanks for flagging this SlideShare!

Oops! An error has occurred.

×
Saving this for later? Get the SlideShare app to save on your phone or tablet. Read anywhere, anytime – even offline.
Standard text messaging rates apply

# Bisection

2,183
views

Published on

1 Like
Statistics
Notes
• Full Name
Comment goes here.

Are you sure you want to Yes No
• Be the first to comment

Views
Total Views
2,183
On Slideshare
0
From Embeds
0
Number of Embeds
1
Actions
Shares
0
70
0
Likes
1
Embeds 0
No embeds

No notes for slide

### Transcript

• 1. Bisection Method http://numericalmethods.eng.usf.edu Transforming Numerical Methods Education for STEM Undergraduates 07/28/10 http://numericalmethods.eng.usf.edu
• 2. Bisection Method http://numericalmethods.eng.usf.edu
• 3. Basis of Bisection Metho d
• Theorem
http://numericalmethods.eng.usf.edu An equation f(x)=0, where f(x) is a real continuous function, has at least one root between x l and x u if f(x l ) f(x u ) < 0. Figure 1 At least one root exists between the two points if the function is real, continuous, and changes sign.
• 4. Basis of Bisection Method
• Figure 2 If function does not change sign between two points, roots of the equation may still exist between the two points.
http://numericalmethods.eng.usf.edu
• 5. Basis of Bisection Method
• Figure 3 If the function does not change sign between two points, there may not be any roots for the equation between the two points.
http://numericalmethods.eng.usf.edu
• 6. Basis of Bisection Method http://numericalmethods.eng.usf.edu Figure 4 If the function changes sign between two points, more than one root for the equation may exist between the two points.
• 7. Algorithm for Bisection Method http://numericalmethods.eng.usf.edu
• 8. Step 1
• Choose x  and x u as two guesses for the root such that f(x  ) f(x u ) < 0, or in other words, f(x) changes sign between x  and x u . This was demonstrated in Figure 1.
http://numericalmethods.eng.usf.edu Figure 1
• 9. Step 2
• Estimate the root, x m of the equation f (x) = 0 as the mid point between x  and x u as
http://numericalmethods.eng.usf.edu Figure 5 Estimate of x m
• 10. Step 3
• Now check the following
• If , then the root lies between x  and x m ; then x  = x  ; x u = x m .
• If , then the root lies between x m and x u ; then x  = x m ; x u = x u .
• If ; then the root is x m. Stop the algorithm if this is true.
http://numericalmethods.eng.usf.edu
• 11. Step 4 http://numericalmethods.eng.usf.edu Find the new estimate of the root Find the absolute relative approximate error where
• 12. Step 5 http://numericalmethods.eng.usf.edu Is ? Yes No Go to Step 2 using new upper and lower guesses. Stop the algorithm Compare the absolute relative approximate error with the pre-specified error tolerance . Note one should also check whether the number of iterations is more than the maximum number of iterations allowed. If so, one needs to terminate the algorithm and notify the user about it.
• 13. Example 1
• You are working for ‘DOWN THE TOILET COMPANY’ that makes floats for ABC commodes. The floating ball has a specific gravity of 0.6 and has a radius of 5.5 cm. You are asked to find the depth to which the ball is submerged when floating in water.
http://numericalmethods.eng.usf.edu Figure 6 Diagram of the floating ball
• 14. Example 1 Cont.
• The equation that gives the depth x to which the ball is submerged under water is given by
• a) Use the bisection method of finding roots of equations to find the depth x to which the ball is submerged under water. Conduct three iterations to estimate the root of the above equation.
• b) Find the absolute relative approximate error at the end of each iteration, and the number of significant digits at least correct at the end of each iteration.
http://numericalmethods.eng.usf.edu
• 15. Example 1 Cont.
• From the physics of the problem, the ball would be submerged between x = 0 and x = 2 R ,
• where R = radius of the ball,
• that is
http://numericalmethods.eng.usf.edu Figure 6 Diagram of the floating ball
• 16. Example 1 Cont. To aid in the understanding of how this method works to find the root of an equation, the graph of f(x) is shown to the right, where http://numericalmethods.eng.usf.edu Figure 7 Graph of the function f(x) Solution
• 17. Example 1 Cont. http://numericalmethods.eng.usf.edu Let us assume Check if the function changes sign between x  and x u . Hence So there is at least on root between x  and x u, that is between 0 and 0.11
• 18. Example 1 Cont. http://numericalmethods.eng.usf.edu Figure 8 Graph demonstrating sign change between initial limits
• 19. Example 1 Cont. http://numericalmethods.eng.usf.edu Iteration 1 The estimate of the root is Hence the root is bracketed between x m and x u , that is, between 0.055 and 0.11. So, the lower and upper limits of the new bracket are At this point, the absolute relative approximate error cannot be calculated as we do not have a previous approximation.
• 20. Example 1 Cont. http://numericalmethods.eng.usf.edu Figure 9 Estimate of the root for Iteration 1
• 21. Example 1 Cont. http://numericalmethods.eng.usf.edu Iteration 2 The estimate of the root is Hence the root is bracketed between x  and x m , that is, between 0.055 and 0.0825. So, the lower and upper limits of the new bracket are
• 22. Example 1 Cont. http://numericalmethods.eng.usf.edu Figure 10 Estimate of the root for Iteration 2
• 23. Example 1 Cont. http://numericalmethods.eng.usf.edu The absolute relative approximate error at the end of Iteration 2 is None of the significant digits are at least correct in the estimate root of x m = 0.0825 because the absolute relative approximate error is greater than 5%.
• 24. Example 1 Cont. http://numericalmethods.eng.usf.edu Iteration 3 The estimate of the root is Hence the root is bracketed between x  and x m , that is, between 0.055 and 0.06875. So, the lower and upper limits of the new bracket are
• 25. Example 1 Cont. http://numericalmethods.eng.usf.edu Figure 11 Estimate of the root for Iteration 3
• 26. Example 1 Cont. http://numericalmethods.eng.usf.edu The absolute relative approximate error at the end of Iteration 3 is Still none of the significant digits are at least correct in the estimated root of the equation as the absolute relative approximate error is greater than 5%. Seven more iterations were conducted and these iterations are shown in Table 1.
• 27. Table 1 Cont. http://numericalmethods.eng.usf.edu Table 1 Root of f(x)=0 as function of number of iterations for bisection method.
• 28. Table 1 Cont. http://numericalmethods.eng.usf.edu Hence the number of significant digits at least correct is given by the largest value or m for which So The number of significant digits at least correct in the estimated root of 0.06241 at the end of the 10 th iteration is 2.
• Always convergent
• The root bracket gets halved with each iteration - guaranteed.
http://numericalmethods.eng.usf.edu
• 30. Drawbacks http://numericalmethods.eng.usf.edu
• Slow convergence
• If one of the initial guesses is close to the root, the convergence is slower
• 31. Drawbacks (continued)
• If a function f(x) is such that it just touches the x-axis it will be unable to find the lower and upper guesses.
http://numericalmethods.eng.usf.edu
• 32. Drawbacks (continued ) http://numericalmethods.eng.usf.edu
• Function changes sign but root does not exist