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Math Help - Bisection method

  1. #1
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    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.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by JackRyder View Post
    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 f(x) is continuous on an interval [u,v] and f(u)f(v)<0 then the bisection method converges to a zero of f(x) in the interval [u,v].

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

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

    CB
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  3. #3
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    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?
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by JackRyder View Post
    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
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