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Math Help - Error-bound on Regula Falsi

  1. #1
    Member Mollier's Avatar
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    Error-bound on Regula Falsi

    Hi, in the lack of a Numerical Analysis sub forum, I will post this question here as it is quite "calculus-ish".

    Does there exist a error-bound on Regula Falsi (Method of False Position) similar to the one that the Bisection method has? I would like to have an expression for the number of iterations necessary as a function of some prespecified error.

    Thanks.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Mollier View Post
    Hi, in the lack of a Numerical Analysis sub forum, I will post this question here as it is quite "calculus-ish".

    Does there exist a error-bound on Regula Falsi (Method of False Position) similar to the one that the Bisection method has? I would like to have an expression for the number of iterations necessary as a function of some prespecified error.

    Thanks.
    Probably not, or if it does it is not very widely reported.

    CB
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  3. #3
    Member Mollier's Avatar
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    Say that we have a function f and we know that a root r exists on [a_1,b_1].
    If we use the method of false position, our first approximation to this root will be,

     c_1 = b_1 - \frac{f(b_1)}{\frac{f(b_1)-f(a_1)}{b_1-a_1}}.

    The next one is,
    <br />
c_2 = b_1 - \frac{f(b_1)}{\frac{f(b_1)-f(c_1)}{b_1-c_1}}

    and so on..

    What if I use a similar bound to the one in the bisection method, and say that
    |r-c_n|\geq\frac{c_1}{2^n} such that

    n \geq log_2(\frac{c_1}{\epsilon_{step}})

    where \epsilon_{step}=|r-c_n| ?
    Last edited by Mollier; September 6th 2010 at 03:55 AM.
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by Mollier View Post
    Say that we have a function f and we know that a root r exists on [a_1,b_1].
    If we use the method of false position, our first approximation to this root will be,

    c_1 = b_1 - \frac{f(b_1)}{\frac{f(b_1)-f(a_1)}{b_1-a_1}}.

    The next one is,
    <br />
c_2 = b_1 - \frac{f(b_1)}{\frac{f(b_1)-f(c_1)}{b_1-c_1}}

    and so on..

    What if I use a similar bound to the one in the bisection method, and say that
    |r-c_n|\geq\frac{c_1}{2^n} such that

    n \geq log_2(\frac{c_1}{\epsilon_{step}})

    where \epsilon_{step}=|r-c_n| ?
    Which depends of the function in question, unlike the bounds for bisection.

    It would be interesting to see the error at the n-th step say in terms of bounds on the derivatives etc

    (I don't think this impossible, just not worth the candle given false-positions other limitations and the nice properties of bisection - which does have its own problems when you have a double root)

    CB
    Last edited by CaptainBlack; September 6th 2010 at 05:33 AM.
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  5. #5
    Member Mollier's Avatar
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    You mean we could perhaps bound the secant line from point (a,f(a)) to (b,f(b))?
    I do not see what sort of bound we could use to do this though, other than

    tan^{-1}\left(\frac{f(b)-f(a)}{b-a}\right)\leq 90deg.
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