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Math Help - Using the Bisection Method

  1. #1
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    Using the Bisection Method

    let f(x)=(x-1)(x-2)(x-3)(x-4)(x-5). This function has five roots on the interval [0,7]. I have to use the bisection method on this interval and figure out what root is located.
    I'm not entirely sure I know how to use this method correctly..
    Here it goes,
    First, since f(0)<0 and f(7)>0, there exists a c\in\mathbb{R} in between those two points such that f(c) = 0 ..
    Now would you divide the interval in two?
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  2. #2
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    This is quite a strange function to use with this method as it is fully factored and gives you the roots.

    Anyhow you choose the interval [0,7] and find the signs of f(0) <0 and f(7)>0
    Now bisect the interval \frac{7-0}{2} = 3.5 to make 2 smaller intervals and find the sign of f(3.5)

    We know f(3.5)>0 so we choose the smaller intervals with opposite signs f(0) <0 and f(3.5)>0 , bisect again and so on until the interval is small enough.
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  3. #3
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    Quote Originally Posted by pickslides View Post
    This is quite a strange function to use with this method as it is fully factored and gives you the roots.

    Anyhow you choose the interval [0,7] and find the signs of f(0) <0 and f(7)>0
    Now bisect the interval \frac{7-0}{2} = 3.5 to make 2 smaller intervals and find the sign of f(3.5)

    We know f(3.5)>0 so we choose the smaller intervals with opposite signs f(0) <0 and f(3.5)>0 , bisect again and so on until the interval is small enough.
    Ok that's kind of what I thought... and yea it gives you all the roots but they want you to find the root that will be found after using that method, which apparently ends up being x=1
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  4. #4
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    Quote Originally Posted by frenchguy87 View Post
    let f(x)=(x-1)(x-2)(x-3)(x-4)(x-5). This function has five roots on the interval [0,7]. I have to use the bisection method on this interval and figure out what root is located.
    I'm not entirely sure I know how to use this method correctly..
    Here it goes,
    First, since f(0)<0 and f(7)>0, there exists a c\in\mathbb{R} in between those two points such that f(c) = 0 ..
    Now would you divide the interval in two?
    Wow thats crazy i just took an exam today with almost that same exact problem just a little difference in like one of the numbers. Anyway the way I did it was that I just kept applying the bisection method where a=0 b=7 and you have to find thier signs and all that, but then when you plug in p_n it either matches the sign of a or b, however you keep doing it until f(p)=0. However in your case you can kind of see where its going and it keeps getting closer and closer to 1 so your root that will be determined would be 1. I'm not sure if its all the correct but I mean it worked for me on both my hw's and the practice exam. Hope this helped
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  5. #5
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    Quote Originally Posted by tn11631 View Post
    Wow thats crazy i just took an exam today with almost that same exact problem just a little difference in like one of the numbers. Anyway the way I did it was that I just kept applying the bisection method where a=0 b=7 and you have to find thier signs and all that, but then when you plug in p_n it either matches the sign of a or b, however you keep doing it until f(p)=0. However in your case you can kind of see where its going and it keeps getting closer and closer to 1 so your root that will be determined would be 1. I'm not sure if its all the correct but I mean it worked for me on both my hw's and the practice exam. Hope this helped
    How would you formally write the end of the proof, whenever you have a small enough interval?
    I am down to the [0,1.75] interval, so how would I go one to say that the root must be 1?
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  6. #6
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    Quote Originally Posted by frenchguy87 View Post
    How would you formally write the end of the proof, whenever you have a small enough interval?
    I am down to the [0,1.75] interval, so how would I go one to say that the root must be 1?
    I don't understand. Are you to play dumb? Clearly f(x)=0,\text{ }x=1,2,3,4,5 and f(x)\ne 0\text{ }x\in \mathbb{R}-\{1,2,3,4,5\}. So clearly the zero given must be one.
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  7. #7
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    Quote Originally Posted by Drexel28 View Post
    I don't understand. Are you to play dumb? Clearly f(x)=0,\text{ }x=1,2,3,4,5 and f(x)\ne 0\text{ }x\in \mathbb{R}-\{1,2,3,4,5\}. So clearly the zero given must be one.
    lol I just wasn't sure how to write it properly... sorry
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