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Thread: Secant Method - Calculating the number of steps required

  1. #1
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    Secant Method - Calculating the number of steps required

    I need to calculate the number of steps required to reach a specified accuracy by using the Secant method. Here is the question:

    $\displaystyle Secant: |x_{n+1} - r| \leq C_{sct}. |x_n - r|^{\alpha_{sct}} $ with $\displaystyle \alpha_{sct} \approx 1.618$
    If an initial guess, $\displaystyle x_0$, is supplied such that $\displaystyle |x_n - r| = 10^{-2}$, and we assume that $\displaystyle C_{sct} = 3$, how many steps of each method are necessary to ensure that $\displaystyle |x_n - r| \leq 10^{-14}$?


    Hope someone can help!

    Thanks
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  2. #2
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    Quote Originally Posted by bobby View Post
    I need to calculate the number of steps required to reach a specified accuracy by using the Secant method. Here is the question:

    $\displaystyle Secant: |x_{n+1} - r| \leq C_{sct}. |x_n - r|^{\alpha_{sct}} $ with $\displaystyle \alpha_{sct} \approx 1.618$
    If an initial guess, $\displaystyle x_0$, is supplied such that $\displaystyle |x_n - r| = 10^{-2}$, and we assume that $\displaystyle C_{sct} = 3$, how many steps of each method are necessary to ensure that $\displaystyle |x_n - r| \leq 10^{-14}$?


    Hope someone can help!

    Thanks
    So $\displaystyle |x_1- r|\le C_{sct}|x_0- r|^{1.618}= 3|0.01|^{1.618}$
    Then $\displaystyle |x_2- r|\le 3(3|0.01|^{1.618})^{1.618}= 3^{1.618}(0.01)^{2(1.618)}$, $\displaystyle x_3- r|\le 3(3^{1.618}(0.01)^{1.618})^{1.618}= 3^{2(1.618)}(0.01)^{3(1.618)}$.

    It looks to me like $\displaystyle |x_n- r|\le 3^{(n-1)1.618}(0.01)^{n(1.618)}$.

    Set that equal to $\displaystyle 10^{-14}$ and solve for n, using logarithms, and round up to the next nearest integer, of course.
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    So $\displaystyle |x_1- r|\le C_{sct}|x_0- r|^{1.618}= 3|0.01|^{1.618}$
    Then $\displaystyle |x_2- r|\le 3(3|0.01|^{1.618})^{1.618}= 3^{1.618}(0.01)^{2(1.618)}$, $\displaystyle x_3- r|\le 3(3^{1.618}(0.01)^{1.618})^{1.618}= 3^{2(1.618)}(0.01)^{3(1.618)}$.

    It looks to me like $\displaystyle |x_n- r|\le 3^{(n-1)1.618}(0.01)^{n(1.618)}$.

    Set that equal to $\displaystyle 10^{-14}$ and solve for n, using logarithms, and round up to the next nearest integer, of course.
    I'm not sure if some of your maths is correct there? When you wrote

    $\displaystyle |x_2- r|\le 3(3|0.01|^{1.618})^{1.618}= 3^{1.618}(0.01)^{2(1.618)}$

    is this not

    $\displaystyle |x_2- r|\le 3(3|0.01|^{1.618})^{1.618}= 3(3^{1.618}(0.01)^{(1.618^2)})$
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