Results 1 to 3 of 3

Math Help - Secant Method - Calculating the number of steps required

  1. #1
    Junior Member
    Joined
    Jan 2009
    Posts
    36

    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:

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


    Hope someone can help!

    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,412
    Thanks
    1328
    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:

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


    Hope someone can help!

    Thanks
    So |x_1- r|\le C_{sct}|x_0- r|^{1.618}= 3|0.01|^{1.618}
    Then |x_2- r|\le 3(3|0.01|^{1.618})^{1.618}= 3^{1.618}(0.01)^{2(1.618)}, 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 |x_n- r|\le 3^{(n-1)1.618}(0.01)^{n(1.618)}.

    Set that equal to 10^{-14} and solve for n, using logarithms, and round up to the next nearest integer, of course.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jan 2009
    Posts
    36
    Quote Originally Posted by HallsofIvy View Post
    So |x_1- r|\le C_{sct}|x_0- r|^{1.618}= 3|0.01|^{1.618}
    Then |x_2- r|\le 3(3|0.01|^{1.618})^{1.618}= 3^{1.618}(0.01)^{2(1.618)}, 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 |x_n- r|\le 3^{(n-1)1.618}(0.01)^{n(1.618)}.

    Set that equal to 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

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

    is this not

    |x_2- r|\le 3(3|0.01|^{1.618})^{1.618}= 3(3^{1.618}(0.01)^{(1.618^2)})
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. secant method
    Posted in the Advanced Math Topics Forum
    Replies: 1
    Last Post: May 28th 2010, 01:36 AM
  2. Secant method in MATLAB
    Posted in the Math Software Forum
    Replies: 7
    Last Post: April 24th 2010, 10:53 AM
  3. Secant Method + Interpolation
    Posted in the Advanced Math Topics Forum
    Replies: 2
    Last Post: October 18th 2009, 08:49 AM
  4. [SOLVED] Calculating required number of wins
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: July 31st 2009, 09:13 AM
  5. secant method
    Posted in the Math Software Forum
    Replies: 1
    Last Post: September 29th 2008, 11:03 PM

Search Tags


/mathhelpforum @mathhelpforum