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First step to proof of convergance using Newton's root finding method

  1. Jan 20, 2010 #1
    Suppose I am looking for the root of a function of the form:
    [tex]f(x)=x ^{m}-c[/tex], where c,m>1.

    Suppose I take my first guess to be [tex]x _{0}=c[/tex].

    Then using newtons method my next guess will be given by:
    [tex]x _{n+1}=x _{n} - \frac {f(x _{n})}{f'(x _{n})}[/tex].

    From this, or thinking about this graphically it is obvoius that
    [tex]x _{0}>x _{1}>x _{n}>x _{n+1}>c^{1/m}[/tex].

    However I don't know how should I go about formally proving this.
  2. jcsd
  3. Jan 21, 2010 #2


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    To save writing, let x be current value and x* the next.

    x*=x - (xm-c)/mxm-1={(m-1)xm+c}/mxm-1
    < mxm/mxm-1=x, since c<xm.
  4. Jan 21, 2010 #3
    Hmm, I knew this, but is this a proof?

    Don't we need to prove that [tex]x^{m}>c [/tex] ?

    I get the impression that if I want to prove that
    x _{0}>x _{1}>x _{n}>x _{n+1}
    I need to use [tex]x^{m}>c [/tex]

    And it is easy to prove that [tex]x^{m}>c [/tex] but only if we are sure that:
    x _{0}>x _{1}>x _{n}>x _{n+1}

    I'm a physics student so I'm not used to proving things formally. Are you sure what you have written is enough?

    I also thought of the following proof:
    We know that:
    x _{0}>x _{1}>x _{n}>x _{n+1}
    Due to the fact that the function as well as its derivative are >0.
    This is true only for [tex]x_{n}>c^{1/m} [/tex]
    But we can show that using Newtons method the function x* (what you had written) has a minimum at [tex]x_{n}=c^{1/m} [/tex] (I do this by taking (x*)' and equating it to 0).

    Hence we know that [tex]x_{n}>=c^{1/m} [/tex] Therefore the argument stated in the beginning is definitely true: using this fact we know that x* is strictly decreasing and from this it follows that x*>[tex]c^{1/m} [/tex]
  5. Jan 22, 2010 #4


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    Initially you have c > 1, so the initial guess (c) needs cm > c, which is certainly true for c > 1.
  6. Jan 24, 2010 #5


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    Hi trelek, you could also use the "mean value theorem" to prove it fairly easily. An outline would go something like this :

    Let "a" be the root and "b" (with b>a) be the inital guess. By the mean value theorem there exists some k : a < k < b, such that

    f(b) - f(a) = f'(k)(b-a).

    Since f' is monotonic increasing it follows that

    f(b) - f(a) < f'(b)(b-a)

    and since f(a)=0 it follows that f(b) / f'(b) < (b-a)


    b - f(b)/f'(b) > a
    Last edited: Jan 24, 2010
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