I am studying for finals and am looking over previous problems that I missed significant points on. My instructor doesn't comment on why things are incorrect and he doesn't discuss homework, so I am on my own to figure out why I am wrong. Anyway, I would like some help figuring out where I went wrong.

Prove that if $\displaystyle r$ is a zero of multiplicity $\displaystyle k$ of the function $\displaystyle f$, then quadratic convergence in Newton's iteration will be restored by making this modification

$\displaystyle

x_{n+1} = x_n - k \frac{f(x_n)}{f'(x_n)}

$

Proof:

If $\displaystyle x^*$ is a zero of multiplicity $\displaystyle k$ then $\displaystyle f(x)$ can be rewritten as

$\displaystyle

f(x) = (x - x^*)^k q(x)

$

where $\displaystyle q(x*) \ne 0,$ and so we have $\displaystyle f^{(k)}(x^*) \ne 0.$ Writing $\displaystyle f(x) \,\text{and} \, f'(x)$ as its Taylor series expansion about $\displaystyle (x - x*)$

$\displaystyle

f(x) = \frac{f^{(k)}(\xi)}{k!}(x - x^*)^k, \quad

f'(x) = \frac{f^{(k)}(\eta)}{(k - 1)!}(x - x^*)^{k - 1}

$

where $\displaystyle \xi, \eta$ are between $\displaystyle x$ and $\displaystyle x^*.$ Thus we have

$\displaystyle g(x) = x - \frac{f^{(k)}(\xi)}{f^{(k)}(\eta)}(x - x^*)$

$\displaystyle \lim_{x \to x^*} g'(x) = 1 - \lim_{x \to x^*}\frac{d}{dx}\left[ \frac{f^{(k)}(\xi)}{f^{(k)}(\eta)}(x-x^*)\right]$

This is where I went wrong. I originally thought that the derivative was just $\displaystyle 1 - \frac{f^{(k)}(\xi)}{f^{(k)}(\eta)}$, but $\displaystyle \xi, \eta$ depend on $\displaystyle x$ so that is not true.

$\displaystyle \lim_{x \to x^*} g'(x) = 1 - \lim_{x \to x^*}\left[\frac{d}{dx} \left(\frac{f^{(k)}(\xi)}{f^{(k)}(\eta)}\right)(x-x^*) + \frac{f^{(k)}(\xi)}{f^{(k)}(\eta)}\right]$

$\displaystyle = 1 - \lim_{x \to x^*}\left[\frac{d}{dx} \left(\frac{f^{(k)}(\xi)}{f^{(k)}(\eta)}\right)(x-x^*)\right] + \lim_{x \to x^*}\frac{f^{(k)}(\xi)}{f^{(k)}(\eta)}$

$\displaystyle = 1 - \lim_{x \to x^*}\left[\frac{d}{dx} \left(\frac{f^{(k)}(\xi)}{f^{(k)}(\eta)}\right)(x-x^*)\right] + 1$

$\displaystyle = \lim_{x \to x^*}\frac{d}{dx} \left[\frac{f^{(k)}(\xi)}{f^{(k)}(\eta)}\right](x-x^*).$

$\displaystyle = \lim_{x \to x^*}\frac{d}{dx} \left[\frac{\frac{df^{(k)}}{dx}(\xi) f^{(k)}(\eta) - f^{(k)}(\xi)\frac{df^{(k)}}{dx}(\eta) }{\left[f^{(k)}(\eta)\right]^2}\right](x-x^*).$

This is suppose to go to zero, but I don't really see how. I suppose it has something to do with $\displaystyle \frac{df^{(k)}}{dx}(\xi) \to \frac{df^{(k)}}{dx}(\eta)\text{ as } x \to x^*\text{ and } f^{(k)}(\xi) \to f^{(k)}(\eta)\text{ as } x \to x^*$.

Any help is appreciated, also if you have a different way to do this I will be glad to see it. Thank you in advance.