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I have the following family of polynomials:

$$p_n(x) = \sum_{k=0}^n {n \choose k} \frac{n+2}{n+2-k} x^k$$

I conjecture that it has non-real roots for all $n \geq 2$. This holds for all the small cases I have checked. However, I'm not sure how to prove it. I feel that I am lacking in methods. I tried to use the argument principle around a sufficiently large segment of the negative real axis, but that became too complicated. It looks to me like there might be a combinatorial method of proving this. Does anyone have any ideas?

I would very much appreciate any suggestions on how to approach this problem, or more generally, problems such as this. I would much prefer suggestions or hints to a full solution.

Thank you very much!

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You can prove this by a bit of clever algebra and calculus. First, note that clearly $p_n(x)>0$ for $x\geq 0$, so we only have to check for negative roots. Now the idea is to modify $p_n$ so that we can relate it to $\sum_{k=0}^n {n \choose k} x^k=(x+1)^n$. In order to do this, we need to get rid of the $n+2-k$ in the denominator, so we need to make the exponent on $x$ be something like $n+2-k$ and then differentiate.

So let's let $$q_n(x)=\frac{p_n(x)}{(n+2)x^{n+2}}=\sum_{k=0}^n {n \choose k} \frac{1}{n+2-k} x^{k-n-2}.$$ Then $q_n$ has the same roots as $p_n$ (since $0$ is not a root of $p_n$). Differentiating $q_n$ gives $$q_n'(x)=-\sum_{k=0}^n {n \choose k} x^{k-n-3}=-x^{-n-3}\sum_{k=0}^n {n \choose k} x^k=-x^{-n-3}(x+1)^n.$$

From here you can then use some elementary calculus to determine exactly how many negative roots $q_n(x)$ has. The details are hidden below.

Let us first suppose $n$ is even. Then $q_n'(x)$ is nonnegative for all $x<0$. Note that $q_n(x)$ is positive as $x$ approaches $-\infty$ (the $k=n$ term dominates), so $q_n(x)$ is positive for all negative $x$. Thus $q_n(x)$ has no negative roots. It follows that in this case all the roots of $p_n(x)$ are nonreal.

Now suppose $n$ is odd. Then $q_n'(x)$ is positive for $x<-1$ and negative for $-1<x<0$, so the maximum value of $q_n(x)$ for $x<0$ is achieved at $x=-1$. As in the even case, $q_n(x)$ is positive as $x$ approaches $-\infty$. We can also see that $q_n(x)$ approaches $-\infty$ as $x$ approaches $0$ from below (the $k=0$ term dominates). It follows that $q_n(x)$ has a single simple root for $x$ negative, which is between $-1$ and $0$. Thus in this case, $p_n(x)$ has exactly one simple real root which is between $-1$ and $0$, and all the other roots of $p_n(x)$ must be nonreal.

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  • $\begingroup$ This is perfect. Thank you very much! $\endgroup$ – solstafir Jun 18 '16 at 14:09

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