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Math Help - bisection root algorithm

  1. #1
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    please help

    find the root of f(x)=2x^3-x^2+x-1=0 to 3 decimal places using interval bisection.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by bobby77
    find the root of f(x)=2x^3-x^2+x-1=0 to 3 decimal places using interval bisection.
    First sketch the graph to locate the approximate position of the root.
    See attachment:

    Now we see the root is located at about x=0.7. Now we need
    a pair of values x_{lo}, x_{hi} such that x_{lo}<x_{hi} such that

    sgn(f(x_{lo}) \ne sgn(f(x_{hi}) .

    It looks as though 0.6, 0.8 will do for these (check).

    Now we need a table:

    Code:
        x_lo     x_hi     f(x_lo)     f(x_hi)     x_mid     f(x_mid)  
        0.6      0.8      -0.328      0.184      0.7         -0.104
    This is the first row, the next row is obtained by replacing x_{lo} or x_{hi} by x_{mid}, so that we
    still bracket the zero, thus:

    Code:
        x_lo     x_hi     f(x_lo)     f(x_hi)     x_mid     f(x_mid)  
        0.6      0.8      -0.328      0.184      0.7         -0.104
        0.7      0.8      -0.104      0.184      0.75        0.0312
        0.7      0.75     -0.104      0.0312     0.725      -0.0385
    and so on utill |x_{hi}-x_{lo}|<0.001.

    RonL
    Attached Thumbnails Attached Thumbnails bisection root algorithm-gash.jpg  
    Last edited by CaptainBlack; August 13th 2006 at 10:10 AM.
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