Show that the Bisection Method converges to...

  1. 1. The function defined by f(x)=[tex]\sin(\pi*x)[/tex]has zeros at every integer x. Show that when -1<a<0 and 2<b<3, the Bisection method converges to

    a. 0, if a+b<2
    b. 2, if a+b>2
    c. 1, if a+b=2


    2. Bisection Method

    An interval [tex][a_{n+1},b_{n+1}][/tex]containing an approximation to a root of f(x)=0 is constructed from an interval [tex][a_{n},b_{n}][/tex] containing the root by first letting

    [tex]p_{n}=a_{n}+\frac{(b_{n}-a_{n})}{2}[/tex]

    Then set
    [tex]a_{n+1}=a_{n}[/tex] and [tex]b_{n+1}=p_{n}[/tex] if [tex]f(a_{n})*f(p_{n})<0[/tex]

    and

    [tex]a_{n+1}=p_{n}[/tex] and [tex]b_{n+1}=b_{n}[/tex] otherwise.




    3. I attempted to randomly choose numbers for a and b to satisfy the relations a+b<2 , -1<a<0, and 2<b<3. I picked a=-0.5 and b=2.5 and went through many iterations of the Bisection method to see if p[n] converged to zero. It didn't. I'm not sure if I am approaching the problem correctly. Can someone give me a good head start with this?
     
    Last edited: Feb 1, 2007
  2. jcsd
  3. Ok, I thought that maybe I wasn't doing enough iterations. So to avoid hours of number crunching I wrote the bisection method in True Basic code and made it print out the results so I could graph it. It worked for parts (a) and (b) because when I graph p vs n, p converges to 0 for part (a) and 2 for part (b). Strangely enough, when I do the same for part (c) it converges to 0 instead of 1. Can anyone tell me why?

    I did the same as before:
    For part (a) I used a=- 0.75 and b= 2.5
    part (b) a= -0.25 and b= 2.5
    part (c) a=-0.5 and b=2.5

    Here is the code and graphs of all 3 parts:

    Code
    program Bisection

    option nolet
    input prompt "a: ":aa
    input prompt "b: ":bb
    input prompt "TOL: ":tol
    print "Enter file name for saving data (enter it as filename.dat)"
    input prompt "(AND make sure there isn't already a file with that name): ": file$
    open #23: name file$, access output, create newold, org text
    nn=log((bb-aa)/tol)/log(2)
    kk=1
    print "a", "b", "f*f", "p", "k"
    print #23: "a", "b", "f*f", "p", "k"
    do while kk<nn
    pp=aa+(bb-aa)/2
    ff=sgn(sin(Pi*pp))*sgn(sin(Pi*aa))
    print aa, bb, ff, pp, kk
    print #23: aa, bb, ff, pp, kk
    if ff<0 then
    bb=pp
    else
    aa=pp
    end if
    kk=kk+1
    loop
    pp=aa+(bb-aa)/2
    print aa, bb, ff, pp, kk
    print #23: aa, bb, ff, pp, kk
    end

    Graphs

    [​IMG]
    [​IMG]
    [​IMG]
     
  4. Dick

    Dick 25,810
    Science Advisor
    Homework Helper

    Your general problem here is that floating point numbers are not exact. In the a=-0.5, b=2.5 case your first midpoint is at 1.0. In an ideal world the value of the function SHOULD be exactly zero there. But it is not. So it tries for another midpoint either to the right or left of 1.0 - which one depends on roundoff error. It should have actually STOPPED at 1.0. How can you fix it?
     
  5. Oh ok. Thanks. Maybe if I set a and b so that it doesn't equal 1 exactly, it will converge like it's supposed to. I'll try it and let you know. Thanks
     
  6. Dick

    Dick 25,810
    Science Advisor
    Homework Helper

    That will work - but it's not a complete solution. Your current algorithm doesn't have any way to terminate. Since you can't generally expect to find an x value such that f(x)=0.0 EXACTLY, one easy way out is to pick a small number d and stop if f(x)<d. Also you probably want to terminate if |a-b|<e for some other small number e.
     
    Last edited: Feb 4, 2007
  7. Well I have it set so that the while loop terminates within a tolerance:

    nn=log((bb-aa)/tol)/log(2)
    kk=1
    .
    .
    do while kk<nn


    You input tolerance at the beginning and it gives you nn, an approximate number of iterations. I have it set so that once kk > nn, the loop stops. The problem usually gives a tolerance, but since this problem didn't, I chose a really small tolerance so I could be as accurate as possible.
     
  8. Dick

    Dick 25,810
    Science Advisor
    Homework Helper

    Ah, ok. As long as you see what the problem is in this particular case.
     
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