(adsbygoogle = window.adsbygoogle || []).push({}); 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?

**Physics Forums - The Fusion of Science and Community**

# Show that the Bisection Method converges to

Know someone interested in this topic? Share a link to this question via email,
Google+,
Twitter, or
Facebook

Have something to add?

- Similar discussions for: Show that the Bisection Method converges to

Loading...

**Physics Forums - The Fusion of Science and Community**