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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 $H(\alpha)=0$. – André Nicolas Oct 8 '12 at 23:51
@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. – C. Lambda Oct 8 '12 at 23:55
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. – André Nicolas Oct 8 '12 at 23:58
@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? – C. Lambda Oct 9 '12 at 0:06
Sorry, can't type out answer, dinner, then commitment. Will write answer (quite a bit) later, unless someone has done it already. – André Nicolas Oct 9 '12 at 0:17
up vote 1 down vote accepted

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

Then multiplied out,

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

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


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, it would help the equations look cool. – NoChance Oct 9 '12 at 2:33
Thanks for the site reference. I used to know a bit of LaTex, and will try dusting it off for future. – coffeemath Oct 9 '12 at 2:51

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