# Find all positive integral values of $x$ if $\prod_{m=0}^{1008} (x-{2m+1 \over 2})^{2m+1} \lt 0$

Lately, I have been taking multiple classes on such math problems. So while I was solving some math problems, I came over this question. The question originally says: How many positive integer solutions has the following inequality: $$\left(x-{1\over 2}\right)^1\left(x-{3\over 2}\right)^3\cdots\left(x-{2017\over 2}\right)^{2017} \lt 0.$$ I managed to change to: $$\prod_{m=0}^{1008} \left(x-{2m+1 \over 2}\right)^{2m+1} \lt 0.$$ I have been trying so hard to simplify more to get a solution but failed, and I was hoping if I could receive some help on this one. Thank you anyways.

• Since this is true for any negative integer, $x$, the answer is infinite. Perhaps you meant to ask how many positive integer solutions there are? – robjohn Sep 28 '18 at 16:48
• @robjohn Yes that is what I meant, I forgot to include it. – user587054 Sep 29 '18 at 11:26

Hint :

The inequality is equivalent to $$\left(x-{1\over 2}\right)\left(x-{3\over 2}\right)...\left(x-{2017\over 2}\right)\lt 0\tag1$$ (why?)

Let $$f(x)$$ be the LHS of $$(1)$$, and consider the graph of $$y=f(x)$$. The degree of $$y=f(x)$$ is $$\frac{2017+1}{2}=1009$$ which is odd.

So, positive integer solutions are

$$2,4,\cdots, 1008$$.

• OP ask positive integers. – Takahiro Waki Sep 29 '18 at 16:07
• @Takahiro Waki : Thanks. I've edited it. – mathlove Sep 29 '18 at 16:10

\begin{align} f(x)&=\text{sgn}\prod_{m=0}^{1008}\left(x-m-\frac12\right)^{2m+1} \\ &=\text{sgn}\prod_{m=0}^{1008}\left(x-m-\frac12\right)^{2m} \, \text{sgn}\prod_{m=0}^{1008}\left(x-m-\frac12\right) \\ &=\text{sgn}\prod_{m=0}^{1008}\left(x-m-\frac12\right) \\ &=\text{sgn}\prod_{m=0}^{x-1}\left(x-m-\frac12\right) \, \text{sgn}\prod_{m=x}^{1008}\left(x-m-\frac12\right) \\ &=\text{sgn}\prod_{m=x}^{1008}\left(x-m-\frac12\right) =\prod_{m=x}^{1008}(-1)=\color{red}{(-1)^{1009-x}\space\colon\,x\in[0,1008]} \end{align}

This answers the original version of the question.

Hint: $$\deg\left(\prod_{m=0}^{1008} \Big(x-{2m+1 \over 2}\Big)^{2m+1}\right) = \sum_{m=0}^{1008} (2m + 1) \equiv \sum_{m=0}^{1008} 1 \equiv 1\pmod2,$$ so the given polynomial has odd degree.

Since a polynomial of odd degree will "explode at infinity" (i.e. $$p(x)\to-\infty$$ as $$x\to-\infty$$), but a polynomial can only have finitely many roots, it will be "eventually" negative for sufficiently large $$-M$$ ($$M > 0$$), so the answer is infinitely many.