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Secant Method Minimum Estimation

  1. Apr 18, 2016 #1
    1. The problem statement, all variables and given/known data

    I am having some trouble understanding the following solved problem:

    Use two iterations of the Secant method to estimate where ##x^2 -x## has a local minimum. Start with ##x^{(0)} =1##, ##x^{(1)} =1/2.##

    Answer:

    2016_04_18_174819.jpg


    2. Relevant equations

    Secant iteration:

    $$x_k = x_{k-1} - \frac{x_{k-1} - x_{k-2}}{f'_{k-1} - f'_{k-2}}$$

    3. The attempt at a solution

    So I understand that for the minimum we need to solve at the derivative of the function, which is

    $$3x^2 -1$$

    Using the given initial values ##x^{(0)} =1##, ##x^{(1)} =1/2,## the first iteration of the Secant method becomes


    $$x^{(3)} = \frac{1}{2} - \frac{(1/2)- 1}{(3(1/2)^2 -1) - (3(1)^2 -1)} = 0.2778$$

    So, how did they get ##0.556## for the first iteration?

    Also, why did they use ##1/4## instead of ##1/2## as ##x_{k-1}##?

    I can't really follow the given answer, so I am wondering if there is an error in the solution. Any help is greatly appreciated.
     
  2. jcsd
  3. Apr 18, 2016 #2

    BvU

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    I have learned a different secant method. Please check it out.

    In your calculation, however, in fact you correctly use the function, not the derivative .

    [edit] my mistake. Not the function nor the derivative. The derivative of ##x^2-x## is ##2x-1## !

    But you do forget to multiply with f(1/2).
    And you confuse yourself by calling it x(3)
     
  4. Apr 18, 2016 #3
    Thank you so much for the explanation, it makes perfect sense now. I got the right answer, but I believe there is a typo in the solution in that ##x_{k-1}## in the first iteration should be ##1/2## instead of ##1/4.##
     
  5. Apr 18, 2016 #4

    BvU

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    I think you've got it right ! Well observed and no wonder you were a bit disoriented :smile:
     
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