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,
- Find $g(x)$ with $m=1$
- Find $g^\prime(x)$
- 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 $m\gt1$,
- Find $g(x)$ with $m$
- 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!
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 $g'(p) = 1 - m + m \frac{m-1}{m} = 0$.