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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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    Differentiate $k$ times. We get $f^{(k)}(x)=k!g(x)+H(x)$ where $H(\alpha)=0$.2012-10-08
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    @André Nicolas: How did you get that expression for $f^{(k)}(x )$. Aren't we supposed to apply the product rule? I I think you are using the product rule when you differentiated.2012-10-08
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    Yes, I used the product rule many times, but did not pay any attention to the parts that after all the differentiations would still have an $x-\alpha$ in them.2012-10-08
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    @André Nicolas: Ok, I checked your formula. I now I see why it is true, but I can't see how does this formula can help us to prove that $g'(\alpha)\neq0$. Any further hints please?2012-10-09
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    Sorry, can't type out answer, dinner, then commitment. Will write answer (quite a bit) later, unless someone has done it already.2012-10-09
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    All what I can say is that: $g'(\alpha)=\frac{1}{k!}f^{(k+1)}(\alpha)$, but we don't know anything about $f^{(k+1)}(\alpha)$?2012-10-09
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    I think the question means to ask for the standard result that if the first $k-1$ derivatives of $f$ are $0$, and the $k$-th derivative isn't $0$, then $f(x)=(x-\alpha)^kg(x)$ where $g(\alpha)\ne 0$.2012-10-09
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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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    One site that offers easy use of LaTex is:http://www.codecogs.com/latex/eqneditor.php, it would help the equations look cool.2012-10-09
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    Thanks for the site reference. I used to know a bit of LaTex, and will try dusting it off for future.2012-10-09