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Problem:

Consider the continuous function $f$ which is $k$ times differentiable: $f(\alpha )=f'(\alpha )=\cdots=f^{(k-1)}(\alpha )=0$ and $f^{(k)}(\alpha )\neq 0$. Assume that $\alpha$ is a root to $f$ with multiplicity $k$, i.e: $f(x)=(x-\alpha )^k g(x)$ where $g(\alpha)\neq0$.

I need to prove that $g'(\alpha)\neq0$. Any ideas?

I tried to differentiate the expression $f(x)=(x-\alpha )^k g(x)$ one time, and then find the expression of $g'(x)$ in terms of $f(x)$ and $f'(x)$, and take the limit as $x$ tends to $\alpha$, but it doesn't work. Any help is appreciated.

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    Yes. Especially for polynomials, p has root of multiplicity$k$precisely when p = (x-alpha)^k*g(x) with g(x) a poly for which g(alpha) nonzero.2012-10-09

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What if $k=2$ and $\alpha = 0$ and $f(x) = x^2g(x)$ where $g(x)=x^2+1$?

Then multiplied out,

$f(x) = x^4+x^2 $

$f'(x)=4x^3+2x, $

$f''(x)=12x^2+2.$

So we have your conditions that $f(0)=f'(0)=0$ while $f''(0)$ not $0$ [it's $2$].

And we have that the function $g(x)$ satisfies $g(0)$ nonzero [it's $1$].

However in our example here, $g'(x)=2x$ so $g'(0)=0$.

Therefore I think yor problem must be misphrased somehow. In fact I can't see how one could derive anything about the derivative of g...

We can make similar examples for any $k$, by choosing $f(x)=x^kg(x)$ and $g(x)=x^r+1$. Then $f(x)=x^{k+r}+x^r$ and the same phenomenon occurs, where this time g(0) is not zero and several derivatives of g at alpha=0 after that are all zero.

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    Thanks for the site reference. I used to know a bit of LaTe$x$, and will try dusting it off for future.2012-10-09