# Multiple root polynomial

Let $$c$$ be a fixed real number.Show that a root of the equation $x(x+1)(x+2).....(x+2014) = c$ can have multiplicity at most 2.Also determine the number of values of $$c$$ for which the equation has a root of multiplicity $$2$$.

4 years, 4 months ago

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We can clearly assume $$c\neq 0$$ - otherwise, there is nothing to prove, since all the roots are simple. If we suppose that $$x$$ is a root of $$f(x)=x(x+1)\cdot\ldots\cdot(x+2014)-c$$ with multiplicity three or more, we have that $$f(x)=0,f'(x)=0$$ and $$f''(x)=0$$. Since: $f'(x) = \left(\frac{1}{x}+\frac{1}{x+1}+\ldots+\frac{1}{x+2014}\right)\cdot x(x+1)\cdot\ldots(x+2014),$ $f''(x) = -\left(\frac{1}{x^2}+\frac{1}{(x+1)^2}+\ldots+\frac{1}{(x+2014)^2}\right)\cdot x(x+1)\cdot\ldots(x+2014)+\left(\frac{1}{x}+\frac{1}{x+1}+\ldots+\frac{1}{x+2014}\right)^2\cdot x(x+1)\cdot\ldots(x+2014),$ the three conditions give: $\sum_{j=0}^{2014}\frac{1}{(x+j)^2} = 0,$ that is impossible since all the terms are positive numbers.

The possible values of $$c$$ for which a double root exists are given by $$y(y+1)\cdot\ldots\cdot(y+2014)$$ where $$y$$ is a root of $$\sum_{j=0}^{2014}\frac{1}{y+j}=0.$$

- 4 years, 4 months ago

Let the left side be $$f(x)$$. First note that $$c = 0$$ doesn't work, as all the roots have multiplicity 1. Next, look at $$\frac{f'(x)}{f(x)} = \frac1{x} + \frac1{x+1} + \cdots + \frac1{x+2014}$$. The right side is a decreasing function with $$2015$$ vertical asymptotes; considering the graph, you can see that it crosses the $$x$$-axis once in every interval $$(-N-1,-N)$$, where $$N$$ is a nonnegative integer $$\le 2013$$. Since $$f'(x)$$ is a polynomial of degree $$2014$$, this shows that it has $$2014$$ distinct roots.

Since $$f'(x)$$ has no roots of multiplicity $$\ge 2$$, $$f(x)$$ cannot have any roots of multiplicity $$\ge 3$$. So we're done with the first half.

This also shows that there are $$2014$$ values $$a$$ such that $$f(x) = f(a)$$ has a root $$a$$ of multiplicity $$2$$. This might lead to fewer than $$2014$$ values of $$c$$, though. I don't think it does (but only because $$2014$$ is even!). But my reasoning is probably better confined to a second post, mostly because I haven't fully worked it out yet.

- 4 years, 4 months ago

It will take a lots of time to type the solution so I am writing the keypoints In order to have a multiplicity of 3 or more than that, the second derivative of the polynomial P(x) = x(x+1)....(x+2014) - c should be equal to zero. We can find the second derivative and apply Cauchy schwarz to know that it cannot be equal to zero.

- 4 years, 4 months ago