How to show that a root of the equation $x (x+1)(x+2) … (x+2009) = c $ can have multiplicity at most 2?

Question

How to show that a root of the equation $$x (x+1)(x+2) ....... (x+2009) = c $$ can have multiplicity at most 2 , and to find the value of $ c $ for which this is possible.

I proceeded by using the derivative method, but the number of terms involved is huge.I have been trying to find some common property for all these terms, but I've come up with nothing so far.

Answer 1

Let denote by $$P(x)=x (x+1)(x+2) ....... (x+2009) - c .$$

We have $$P(0)=P(-1)=\cdots P(-2009)=-c$$ and by Rolle's theorem there's $$0>t_0>-1>\cdots>t_{2008}>-2009$$ s.t. $$P'(t_0)=P'(t_1)=\cdots=P'(t_{2008})=0.$$ Since the polynomial $P$ has the degree $2010$ then the degree of $P'$ is $2009$ and then the $t_i$ are the roots of $P'$.

If $c=0$ then the roots of $P$ are different from the roots of $P'$ and there is not a multiple root.

If $c\neq 0$, we apply the Rolle's theoem on $P'$ and we find the roots $s_i$ of $P''$ s.t $$t_0>s_0>t_1>\cdots>s_{2007}>t_{2008}$$ and if we have $P(\alpha)=P'(\alpha)=0$ then $\alpha=t_{i_0}$ ($\alpha$ is a root of $P'$) and then $\alpha\neq s_i\forall i=0,\ldots,2007$ and hence $P''(\alpha)\neq 0$.

Answer 2

For a more pedestrian approach involving explicit derivative computations, observe that if $c=0,$ then all roots have multiplicity $1.$ So we only consider $c\ne0.$ Let $f(x)=x(x+1)(x+2)\ldots(x+2009)$ and $g(x)=f(x)-c.$ There is a root with multiplicity greater than $2$ only if there is an $\alpha$ such that $g(\alpha)=g'(\alpha)=g''(\alpha)=0.$ But $$g'(x)=\frac{f(x)}{x}+\frac{f(x)}{x+1}+\frac{f(x)}{x+2}+\ldots\frac{f(x)}{x+2009}=\sum_{j=0}^{2009}\frac{f(x)}{x+j}$$ and $$g''(x)=\sum_{0\le j<k\le2009}\frac{2f(x)}{(x+j)(x+k)}=\frac{[g'(x)]^2}{f(x)}-\sum_{j=0}^{2009}\frac{f(x)}{(x+j)^2}.$$

If $g(\alpha)=g'(\alpha)=g''(\alpha)=0,$ then $$f(\alpha)=c,\qquad g'(\alpha)=\sum_{j=0}^{2009}\frac{c}{\alpha+j}=0,\qquad g''(\alpha)=-\sum_{j=0}^{2009}\frac{c}{(\alpha+j)^2}=0.$$ But this is impossible since all terms in the expression for $g''(\alpha)$ are non-zero of the same sign.

We have double roots for solutions $\alpha$ of the equation $$\sum_{j=0}^{2009}\frac{1}{\alpha+j}=0,$$ and the values of $c$ for which such a double root occurs are $f(\alpha).$ Note that if $\alpha$ is a double root, then so is $-2009-\alpha,$ and that the corresponding values of $c$ are equal. Hence those $c$ that give rise to double roots actually give rise to pairs of double roots.