I can see why this works for a root $p$ with multiplicity $k\geq 1$, when $f(x)=(x-p)^k$.

But, why is that true if $f(x)=(x-x_1)(x-x_2)\cdots(x-x_n)$ has distinct roots $x_1\neq x_2\neq \cdots \neq x_n$?

Is it something to do with Lagrange's Mean Value Theorem?

  • 1
    $\begingroup$ I guess that instead of "continuous functions" you mean "polynomials"? And by "non-zero derivations" you mean non-identically-null derivatives. If so, and if already know the factorization theorem, then it's almost trivial: just expand the product. $\endgroup$ – leonbloy Feb 13 '12 at 20:10
  • $\begingroup$ this question derives from analysis,i am not asking why the polynom of order n has n roots,it is clear as explained below,i am asking for a proof $f^(k^)(x)\neq0$ for every $1<=k<=n$ $\endgroup$ – user24927 Feb 13 '12 at 20:26
  • $\begingroup$ i know that the the n+1 derivation of $f(x)=a0+a1x+a2x^2+..+anx^n$ & $an\neq0$ will result in zero value,i am looking for a proof based on lagrange mid-value theorem,is there one? $\endgroup$ – user24927 Feb 13 '12 at 20:33

Throughout, suppose we're working over a field of characteristic $0$.

Let $$f(x) = \displaystyle \sum_{k=0}^n a_kx^k$$ be a polynomial of degree $n$, so that $a_n \ne 0$. Then its derivative $$f'(x) = \displaystyle \sum_{k=0}^{n-1} (k+1)a_{k+1}x^{k}$$ has leading coefficient $na_n$, which is nonzero if $n>0$. So if $n>0$ then this has degree $n-1$. Apply some sort of induction argument and you'll see that this implies that a polynomial of degree $n$ (over a field of characteristic $0$) has exactly $n$ nonzero derivatives, and that its $(n+1)^{\text{th}}$ derivative is $0$.

So if you're working over $\mathbb{C}$ then a consequence of the fundamental theorem of algebra is that $$\#\text{nonzero derivatives} = \deg f = \#\text{roots}$$

This is true over any algebraically closed field of characteristic $0$. In fact, it's true for a polynomial of degree $n$ over any algebraically closed field of characteristic $0$ or $p>n$.

| cite | improve this answer | |

The statement is not true over all functions and fields, but is certainly true of polynomials over the complex numbers. This is known as the fundamental theorem of algebra.

| cite | improve this answer | |
  • $\begingroup$ yes,i know it is true,i am looking for proof,the fundamental theorem of algebra is irrelevant for the proof as i already given the form $f(x)=(x-x_1)(x-x_2)..(x-x_n)$ $\endgroup$ – user24927 Feb 13 '12 at 20:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.