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Homework Help: A proof in Numerical Analysis

  1. Sep 25, 2005 #1
    Hello there!
    yet another proof, that i need help on
    I am supposed to prove that the following statement holds for the secant method
    dk+1/ek -> -1 for k->Infinity
    where
    dk+1 is the next change and ek is the error.

    I have this idea, but i want to hear whether its a valid proof.

    i use the expression for the secant method

    xk+1 = xk - f(xk) * ( xk-xk-1/f(xk)-f(xk-1) )

    and derive that
    dk+1 = xk+1 - xk = - f(xk) * ( xk-xk-1/f(xk)-f(xk-1) ) (1)

    I then use an expression in the lecture book, saying that
    f(xk) = ek* ( f(xk)-f(xk-1)/xk-xk-1 ) - (ek-1*ek * f''(xa)/2 )

    My argument is then that for k->Infinity, i will get that - (ek-1*ek * f''(xa)/2 ) goes towards zero. xa is in the interval between the exact solution and the current x, xk.
    This is the part that im not sure if im right about, can i argue like this?

    I then get the following expression

    f(xk) = ek* ( f(xk)-f(xk-1)/xk-xk-1 )

    Where I use the expression (1) and get
    f(xk) = ek* (- f(xk) /dk+1)
    Ánd from this I get
    dk+1/ek = -1

    Cheers
    -Daniel

    PS: How do you make those javascript math expressions ive seen in some of the posts?
     
  2. jcsd
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