Recently, I wondered about the following problem: let $n\geq 5$ and let
$$ P_n(x)=(x-1)(x-2)\ldots (x-n)-1 $$
Is it true that $P_n(x)$ has $n$ distinct real roots for any $n\geq 5$ ? I checked it for up to $n=50$. I found nothing else so far.
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The same polynomial appears in this question. – Jyrki Lahtonen Jun 12 '14 at 19:19
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When thinking for a solution to this, a problem came up. How would I differentiate this for any n? – Cruncher Jun 12 '14 at 19:36
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1The problem is equivalent to find $x$ in $\Gamma(x)=\Gamma(x-n)$ because $P_n(x)=-1+\dfrac{\Gamma(x)}{\Gamma(x-n)}$. – Jika Jun 12 '14 at 19:58
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What happens when you evaluate this at $x=0, 1, 1.5, 2, 2.5 \dots$? Then use the intermediate value theorem.
At the integer points the polynomial evaluates to $-1$. At alternate intermediate points the value is positive, and you get a pair of roots.
If $n$ is even you have positive values at $x=0$ and $x=n+1$ so roots at "either end" to make the full complement.
If $n$ is odd you have a positive value at $x=n+1$ which does the same.
Mark Bennet
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Let $Q_n(x) = P_n(x) +1 $. Chect the values at $Q_n(x)$ at points $\frac{i}{2}$ for $i = 1,3,5...,2n-1$. The signs will alternate and the value of $|Q_n(x)|$ will be greater than $1$. So, the value of $P_n(x)$ will also alternate at these points. This means that there are $n$ roots, in between these points.
rajatsen91
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