# Bisection method

• Feb 16th 2010, 11:49 PM
JackRyder
Bisection method
The function f(x) = sin(pi*x) has zero at every integer. If -1 < a < 0 and 2 < b < 3, show that it converges to 0 if a + b < 2.

So p1 = (a1 + b1) / 2 for those restrictions which is (-1 + 2) / 2 = 1/2 and (-1 + 3) / 2 = 1 and 1/2 < p1 < 1.

1 > f(p1) > 0 so a2 = a1 = a and b2 = p1.

I found out that the solution is p = 0 because it is the only zero of f in [a2, b2] but how do I actually get it? What do I choose for b2 = p1 and a2 = a? Since both of them are a range. -1 < a2 < 0 and 1/2 < b2 < 1. I'd understand it if I started off with a = something and b = something

p = p2 = (a2 + b2) / 2 = ? = 0.
• Feb 17th 2010, 04:26 AM
CaptainBlack
Quote:

Originally Posted by JackRyder
The function f(x) = sin(pi*x) has zero at every integer. If -1 < a < 0 and 2 < b < 3, show that it converges to 0 if a + b < 2.

So p1 = (a1 + b1) / 2 for those restrictions which is (-1 + 2) / 2 = 1/2 and (-1 + 3) / 2 = 1 and 1/2 < p1 < 1.

1 > f(p1) > 0 so a2 = a1 = a and b2 = p1.

I found out that the solution is p = 0 because it is the only zero of f in [a2, b2] but how do I actually get it? What do I choose for b2 = p1 and a2 = a? Since both of them are a range. -1 < a2 < 0 and 1/2 < b2 < 1. I'd understand it if I started off with a = something and b = something

p = p2 = (a2 + b2) / 2 = ? = 0.

If a real function \$\displaystyle f(x)\$ is continuous on an interval \$\displaystyle [u,v]\$ and \$\displaystyle f(u)f(v)<0\$ then the bisection method converges to a zero of \$\displaystyle f(x)\$ in the interval \$\displaystyle [u,v].\$

So it is sufficient to show that \$\displaystyle f(a)f(p1)<0\$.

(By the way why do we end with \$\displaystyle [a,p1]\$ after the first step rather than \$\displaystyle [p1,b]\$ ? )

CB
• Feb 23rd 2010, 02:10 PM
JackRyder
But how do you choose which is a and which is b.

p = (a + b) / 2 so if both a and b aren't specific numbers but a range, how do you choose a and b to get that p.

Part b of the question is: a + b > 2 and p1 = (a + b) / 2 so 1 < p1 < 1. Thus f(p1) < 0 so a2 = p1 and b2 = b1 = b. The only zero in [a2, b2] is p = 2 so convergence will be to 2. So how do you get p2 = p = 2?
• Feb 24th 2010, 04:31 AM
CaptainBlack
Quote:

Originally Posted by JackRyder
But how do you choose which is a and which is b.

p = (a + b) / 2 so if both a and b aren't specific numbers but a range, how do you choose a and b to get that p.

Part b of the question is: a + b > 2 and p1 = (a + b) / 2 so 1 < p1 < 1. Thus f(p1) < 0 so a2 = p1 and b2 = b1 = b. The only zero in [a2, b2] is p = 2 so convergence will be to 2. So how do you get p2 = p = 2?

The conditions given for use of the bisection method are not satisfied by a and b, since the function is negative for both. This mean (a,b) contains 0 or an even non-zero number of roots (this is because it is continuous, the bisection method finds point at which the function changes sign and fortunatly this particular function does not have a root which kisses the x-axis).

CB