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Newton-Raphson Method for derivation of iteration formula

  1. Dec 22, 2004 #1

    Prolly a dumb question, but the general equation for finding Xn+1 is

    Xn+1 = Xn - f(Xn) / f`(Xn)

    What will it be for Xn-1?
  2. jcsd
  3. Dec 22, 2004 #2
    X(n - 1) = X(n - 2) - f(X(n - 2)) / f'(X(n - 2)) (obviously only valid for n > 2 or n >= 2).
  4. Dec 22, 2004 #3


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    Gold Member

    Or, if you want to "walk backwards" from xm to x(m-1), you can substitute n=m-1 in your equation and solve for x(m-1):

    x((m-1)+1) = x(m-1) - f(x(m-1))/f`(x(m-1))
    x(m) = x(m-1) - f(x(m-1))/f`(x(m-1))
    x(m-1) = x(m) + f(x(m-1))/f`(x(m-1))

    From here I'm not quite sure, since you don't have f(x(m-1)) and its derivative,... maybe we can approximate them using the value at m, like so:

    x(m-1) = x(m) + f(x(m))/f`(x(m))

    But I'm not certain that's a valid approx... is this reasonable, guys?
  5. Dec 22, 2004 #4
    Thanks for the help.

    The question I am doing is the following.

    f(x) = x^2 - a

    Show that for x1 > 0 is any initial estimate for sqrt(a) then the Newton-Raphson method gives the iteration

    Xn-1 = (1/2)(Xn + a/Xn) , n >= 1
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