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Problem: Suppose $p \in \mathcal{P}(\mathbf{C})$ has degree $m$. Prove that $p$ has $m$ distinct roots if and only if $p$ and its derivative $p'$ have no roots in common.

My proof so far: If $m=0$, then $p(z)=a_0\neq 0$ and $p$ has no roots. Then $p'(z)=0$ and has no roots. If $m=1$, then $p(z)=a_0+a_1z$ with $a_1 \neq 0$ so $p$ has exactly one root, namely $-a_0/a_1$ and $p'(z)=a_1$ and has no roots. In both cases, $p$ and $p'$ have no roots in common.

Now suppose $m > 1$. We use induction on $m$, assuming that for every polynomial $r$ with $m-1$ distinct roots, $r$ and $r'$ have no roots in common. Let $p$ be a polynomial of degree $m$ with distinct roots. There exists $q$ such that \begin{align*} p(z)=(z-\lambda)q(z) \end{align*} for all $z \in \mathbf{F}$. Since q has $m-1$ distinct roots, $q$ and $q'$ have no roots in common. By the chain rule, \begin{align*} p'(z)=(z-\lambda)q'(z)+q(z) \end{align*} We know that $\lambda$ is not a root of $p'$ since $p'(\lambda)=(\lambda-\lambda)q'(z)+q(\lambda)=0+q(\lambda)\neq 0$ as lambda is not a root of $q$. All other roots of $p$ are roots of $q$. For these roots $\lambda_q$, $p'(\lambda_q)=(\lambda-\lambda_q)q'(\lambda_q)+q(\lambda_q)=(\lambda-\lambda_q)q'(\lambda_q) + 0 \neq 0$.

Now suppose $m > 1$. We use induction on $m$, this time assuming that for every polynomial $r$ of degree $m-1$ such that $r$ and $r'$ have no roots in common, $r$ has $m-1$ distinct roots.

I'm stuck here, since I don't know how to how to manipulate the derivatives. I'm not sure if proof by induction is the best approach here as well and I'd appreciate your help!

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4 Answers 4

up vote 5 down vote accepted

Not sure that induction is really the way to go. Here is a sketch of a possible argument.

Firstly, if $p$ does not have $m$ distinct roots then it has a double (or higher order) root $a$ and so $$p(x)=(x-a)^2q(x)$$ for some polynomial $q$. You can now easily calculate $p'$ and show that it has a root in common with $p$.

Conversely, if $p$ and $p'$ have a common root $a$ we can write $$p(x)=(x-a)q(x)\quad\hbox{and}\quad p'(x)=(x-a)r(x)\ .$$ Differentiating the first equation and equating it with the second, $$(x-a)q'(x)+q(x)=(x-a)r(x)\ ,$$ which shows that $x-a$ is a factor of $q(x)$. So $a$ is a double (at least) root of $p$.

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Try writting $p(x)=(x-\lambda)^kq(x)$ where $q(\lambda)\neq 0$ then $$p^{\prime}(x)=k(x-\lambda)^{k-1}q(x) +(x-\lambda)^kq^{\prime}(x)$$ so $$(x-\lambda)|(p(x),p^{\prime}(x))$$ of and only if $k>1$. So the root $\lambda $ is a multiple root if and only if $p$ and $p^{\prime}$ have $\lambda$ as common root. Thus if $p$ and $p^{\prime}$ have no common roots, all roots of $p$ are simple and there must be $m$ of them.

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Thanks! This is a very simple and elegant proof! –  Srinivas Vasudevan Jun 4 at 7:13
    
So would it be fine to combine this argument with the induction argument for the converse (I'm sure the induction is not efficient but it seems to work) to have a valid proof? –  Srinivas Vasudevan Jun 4 at 7:17

If $p(z)$ has $m$ distinct roots $\rho_j$, $1 \le j \le m$, then we may write

$p(z) = a \prod_1^m (z - \rho_j), \tag{1}$

where $0 \ne a \in \Bbb C$ is a complex constant. Then the derivative of $p(z)$ is

$p'(z) = a \sum_{k = 1}^m \prod_{j = 1, j \ne k}^m (z - \rho_j), \tag{2}$

and successively evaluating $p'(z)$ at $\rho_1$, $\rho_2$, $\ldots$, $\rho_m$ we see that

$p'(\rho_l) = a \prod_{j = 1, j \ne l}^m (\rho_l - \rho_j) \ne 0 \tag{3}$

since the $\rho_j$ are distinct; thus $p(z)$ and $p'(z)$ have no common zero.

If on the other hand we assume that $p(z)$has a multiple zero $\rho$, then

$p(z) = (z - \rho)^kq(z) \tag{4}$

for some $k \ge 2$ and $q(z) \in \Bbb C[z]$. We thus have

$p'(z) = k(z - \rho)^{k - 1}q(z) + (z - \rho)^kq'(z), \tag{5}$

which, since $k \ge 2$, shows that $(z - \rho) \mid p'(z)$; by contraposition, this in turn implies that if no $\rho$ satisfying $p(\rho) = p'(\rho) = 0$ exists, that is, there is no $\rho$ such that $(z - \rho) \mid p(z)$ and $(z - \rho) \mid p'(z)$, the roots of $p(z)$ must be distinct from one another.

QED.

Note Added in Response to Comment by Srinivas Vasudevan, Wednesday 4 June 2014, 4:43 PM PST: The transition 'twixt (1) and (2) follows the ordinary rules of differentiation applied to polynomials. In a nutshell, we have the Leibniz Product Rule for the differentiation operation: for $f(z), g(z) \in \Bbb C[z]$:

$(f(z)g(z))' = f'(z)g(z) + f(z)g'(z). \tag{6}$

(6) in fact applies to any differentiable functions, and may be validated by a standard $\epsilon$-$\delta$ proof built around the ordinary definition of derivative as

$f'(z) = \lim_{\Delta z \to 0} \dfrac{f(z + \Delta z) - f(z)}{\Delta z}, \tag{7}$

or it may be had for polynomials in $\Bbb C[z]$ in a purely algebraic manner wherein the derivative $Df(z) = f'(z)$ is defined as a linear operator on $\Bbb C[z]$ which in addition satisfies $Dc = 0$ for all constant $c \in \Bbb C$ and $Dz^n = nz^{n - 1}$ for all integer $n \ge 1$; the ordinary formula for the derivative of a polynomial follows easily, algebraically, from these defining properties, as does (6). Granting that (6) holds for any two polynomials $f(z), g(z) \in \Bbb C[z]$, we see that for any three polynomials $f(z)$, $g(z)$, $h(z)$,

$(f(z)g(z)h(z))' = (f(z)(g(z)h(z)))' = f'(z)g(z)h(z) + f(z)(g(z)h(z))'$ $= f'(z)g(z)h(z) + f(z)(g'(z)h(z) + g(z)h'(z))$ $= f'(z)g(z)h(z) + f(z)g'(z)h(z) + f(z)g(z)h'(z). \tag{8}$

It is easy, given (6) and (7), to show inductively that for any integer $m \ge 2$ and any collection $f_j(z) \in \Bbb C[z]$, $1 \le j \le m$ of polynomials, that

$(\prod_1^m f_j(z))' = \sum_{k = 1}^m (f'_k(z) \prod_{j = 1, j \ne k}^m f_j(z)); \tag{9}$

I'll leave the details to the reader. If we set $f_j(z) = z - \rho_j$ in (9), we have $f'_j(z) = 1$ and the implication $(1) \Rightarrow (2)$ is immediately had. End of Note.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!

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Could you share what theorem/algebraic manipulation you used to go from step (1) to step (2)? –  Srinivas Vasudevan Jun 4 at 16:13
1  
@Srinivas Vasudevan: Of course, I'll be glad to; but it may take me a few hours to respond fully since I have business this morning. I'll message you via comment when I've added the remarks you request. Cheers! –  Robert Lewis Jun 4 at 16:33
    
@SrinivasVasudevan: OK, I've added a note to my post addressing your concerns. Please feel free to contact me via comment if you have any further questions! Cheers! –  Robert Lewis Jun 4 at 23:46
    
Thank you Robert! This makes sense now. This is a very interesting approach. –  Srinivas Vasudevan Jun 5 at 2:09
    
@SrinivasVasudevan: my pleasure, sir! Indeed, direct calculation of $p'(z)$ from (1), (2) is illuminating. The rest of my argument is pretty standard and similar to some of the other answers . . . –  Robert Lewis Jun 5 at 2:16

Any polynomial $p(x)$ can be written as $\prod_{i=1}^mx-r_i$ if $r_i$ is the $i$th root of $p(x)$

The constant polynomial has no roots, so it cannot have any roots in common with its derivative.

A polynomial $x-r_1$ has the derivative $1$, so it has no roots in common with its derivative.

A polynomial $(x-r_1)(x-r_2)$ has derivative $2x-r_1-r_2$, which has a single root at $\frac{r_1+r_2}{2}$, which only has a root shared with $(x-r_1)(x-r_2)$ if $r_1=r_2$. This means that if and only if $(x-r_1)(x-r_2)$ has a multiple root, the derivative will also have that root.

Iff $p(x)$ has a multiple root then the derivative shares that root, that can be shown for all $p(x)$ inductively, in which they either are a polynomial of degree 2, in which case it is shown above, or they're of a higher degree. And it is shown below.

$$\frac{\text{d}}{\text{d}x}p(x)(x-r)=p(x)+(x-r)p'(x)$$

This means that if $p(x)=0$ and $p'(x)=0$, then the equation above is also $0$. If only $p(x)=0$ then the other term will evaluate to a nonzero value unless $x=r$, in which case $p(x)(x-r)$ will have a multiple root. If $p'(x)$ but not $p(x)(x-r)$ evaluates to zero, then the above equation will not be zero and they do not share a root.

This is not completely properly formulated but I hope you get how I'm trying to show this.

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