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The question reads,

Suppose $p$ is a zero of multiplicity $m$ of $f$ , where $f^m$ is continuous on an open interval containing $p$. Show that the following fixed-point method has $g^\prime(p)=0$:

$g(x)=x-\frac{mf(x)}{f^\prime(x)}$

My attempt:

There are two cases, when $m=1$ and when $m\gt1$.

  • For $m=1$,

    The steps are,

    1. Find $g(x)$ with $m=1$
    2. Find $g^\prime(x)$
    3. Use theorem (showing zero of multiplicity) that the function $g$ has a zero of multiplicity $p$.

    This will give, $g\prime(p)=(f(p)f^{\prime\prime}(p))/(f^\prime(p))^2$,

    By definition, since $m=1, f(p)=0$ and $f\prime(p)\ne0$ it gives, $g\prime(p)$ = 0

  • For $m\gt1$,

    1. Find $g(x)$ with $m$
    2. Find $g^\prime(x)$,

    This will give us Equation1, $g^\prime(x)=1-m+m\cdot (f(x)f^{\prime\prime}(x))/(f^\prime(x))^2$

    We use the definition stating, $f(x) = (x-p)^m\cdot q(x)$ And find the first and second derivative too. Then we find $f^\prime(x)$ and $f^{\prime\prime}(x)$ 1. That's where I'm stuck. I thought substituting the derivatives in Eq1 would simplify to give what I aimed to prove, that $g^\prime(p)=0$

Is there a simpler way to approach this? Maybe I make a mistake? I'm not sure at this point! Thanks for even reading if you made it this far!

Lutz Lehmann
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John MacPherson
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    See for similar tasks https://math.stackexchange.com/questions/2362853/quadratic-rate-of-convergence-of-modified-nr-method, https://math.stackexchange.com/questions/4035640/derivation-of-modified-newton-method, https://math.stackexchange.com/questions/2656545/taylor-expansion-series-to-prove-modified-newtons-methods-convergence – Lutz Lehmann Oct 08 '21 at 08:25
  • It would be helpful if you gave details on where you are stuck. You could give the formulas for the first and second derivative, or at least the expression for $g'(x)$ that is cancelled of the common factors $(x-p)$. – Lutz Lehmann Oct 08 '21 at 08:32

1 Answers1

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As you said, since $p$ is a zero of $f(x)$ with multiplicity $m > 1$, there is a function $q(x)$ such that $f(x) = (x-p)^m q(x)$. Note that this implies that $f'(p), f''(p), ..., f^{m-1}(p)$ will also all equal zero. The algebra is nasty, but we can find $f'(x)$ and $f''(x)$ and plug them into $g'(x)$. After some simplifying of the resulting scary looking fraction, you eventually get that $g'(p) = 1 - m + m \frac{m-1}{m} = 0$.

homie o'morphic
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  • It's that scary looking fraction I'm stuck on. Unless things cancel out, which I can't find a way to do, it looks like a differential equation – John MacPherson Oct 08 '21 at 16:23
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    Sorry for the delay in getting back to you. Try expressing $f'(x)$ and $f''(x)$ in the form $f'(x)=m(x-p)^{m-1}+ q_1(x)$ and $f''(x)=m(m-1)(x-p)^{m-2}+ q_2(x)$ before plugging into the formula for $g'(x)$ and then at the end worry about evaluating $q_{1,2}(p)$ if needed after simplification. Ill have a go at this by hand later today, but the result worked when I put it into mathematica at the time of my original reply to you. – homie o'morphic Oct 09 '21 at 22:16