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Math Help - Numerical Analysis Bisection Method Problem

  1. #1
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    Numerical Analysis Bisection Method Problem

    Use the bisection methos to find solutions accurate to within 10^-2 for x^3-7x^2+14x-6=0 for each interval [0,1],[1,3.2],[3.2,4].

    I don't know how to find the number of iterations necessary for each interval. Thanks!
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  2. #2
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    Quote Originally Posted by wvlilgurl View Post
    Use the bisection methos to find solutions accurate to within 10^-2 for x^3-7x^2+14x-6=0 for each interval [0,1],[1,3.2],[3.2,4].

    I don't know how to find the number of iterations necessary for each interval. Thanks!
    The interval length is halved at every iteration so after n steps the inteval length is a\times 2^{-n} where a is the initial interval length.

    So if we want the n-th interval to be less than or equal 10^{-2} we need:

    10^{-2}\ge a\times 2^{-n}

    .
    Last edited by Constatine11; January 19th 2009 at 12:27 PM.
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  3. #3
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    for the interval [0,1] they get 7
    for the interval [1,3.2} they get 8

    I got the first one at 7 but I cannot get 8 for the second one. DO you knwo how they got 8?
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  4. #4
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    Quote Originally Posted by wvlilgurl View Post
    for the interval [0,1] they get 7
    for the interval [1,3.2} they get 8

    I got the first one at 7 but I cannot get 8 for the second one. DO you knwo how they got 8?
    In case 2 we have a=2.2, so we seek an integer n such that:

    10^{-2} \ge 2.2\times 2^{-n}

    which we can use trial and error to find that the smallest integer n such that this holds is 8

    .
    Last edited by Constatine11; January 19th 2009 at 12:46 PM.
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  5. #5
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    still dont get it. I get 6
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  6. #6
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    Quote Originally Posted by Constatine11 View Post
    In case 2 we have a=2.2, so we seek an integer n such that:

    10^{-2} \ge 2.2\times 2^{-n}

    which we can use trial and error to find that the smallest integer n such that this holds is 8

    .
    Put n=7,\ 2^{-7}\approx 0.0078 so 2.2 \times 2^{-7}\approx 0.017>0.01

    Put n=8,\ 2^{-8}\approx 0.0039 so 2.2 \times 2^{-8}\approx0.0086 <0.01
    .
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