# Root of the polynomial $x(x-1)(x-2)\cdots(x-K)=C$

Is there an analytic way to obtain the highest root of the polynomial $x(x-1)(x-2)\cdots(x-K)=C$ in terms of $K$ and $C$? The integer $K \ll x$ and the constant $C$ are known.

The other way to ask the question is given $K$ and $C$ find $N$ such that $\binom{N}{K}=C$

• What do you mean by $K \ll \color{red}{x}$? – user49685 Jun 12 '14 at 5:42
• So you're assuming $C \in \Bbb Z$? – Robert Lewis Jun 12 '14 at 5:42
• To user49685, in my problem the value of the integer $K$ is much smaller than $x$ – Oliver Jun 12 '14 at 5:47
• To Mr. Robert Lewis, Yes in the second case (the other way to ask the question) $C \in Z$. However, in the original question, $C \in R$. – Oliver Jun 12 '14 at 5:49
• But $x$ is an indeterminant, how can $K \ll x$? Let's say $C = 0$, then the highest root is $K$. I still find the requirement $K \ll x$ a little bit confusing. – user49685 Jun 12 '14 at 6:37

Is there an analytic way to obtain the highest root of the polynomial $x(x-1)\cdots(x-K)\!=\!C$

Doubtful, since it is well-known that there is no general formula for the roots of a polynomial with degree greater than $4$. Also, the inverse of the factorial or falling factorial function is not known to possess a closed form.

• Thank you for your answer!!! – Oliver Jun 13 '14 at 2:08

Doesn't Stirling formula help? Doing some simplifications using $K \ll N$, the question becomes something like finding a solution to $$N \log N + (K-N) \log(N-K) = K + \log C$$ (plus a small error term like $\frac{1}{2} \log\frac{N}{N-K}$). Depending on how small $K$ is with respect to $N$, maybe you can also ignore the difference between $\log(N)$ and $\log(N-K)$ in the equation above and find an approximated solution explicitly.

• Thank you for your insight. But the approximate solution does not work for me! I tried it before. I need to find an exact expression or something very closer. Approximate solution according to your answer is far from the true solution. – Oliver Jun 12 '14 at 6:42
• I have tried some computations with $K = 10$ and $N = 100$ and already the error is less than $0.5$. Are you sure it is not good enough? What I find (for a solution of the first equation, i.e. $N!/(N-K)! = C$) is: $K \log N - \frac{K^2+K/2}{N} = \log C$. This should be right up to the second order in $\frac{K}{N}$. – Spi Jun 12 '14 at 7:13
• Thank you Mr. Spi. However, we do not know N. We know only $K$ and $C$. In such as case, what should I do? I mean I need an expression like $N=f(C,K)$ – Oliver Jun 12 '14 at 11:18
• Such a nice closed formula is problematic to obtain. The first approximation (i.e. $K/N = 0$) gives $K \log N = \log C$, hence $N = C^{\frac{1}{K}} e^\delta$, for some small $\delta$. This $\delta$ should satisfy $\delta e^\delta = \frac{K+1/2}{C^{\frac{1}{K}}}$, which can't be solved by analytic methods. Thus there is no hope to find an explicit solution for $N$, either, but approximation iterative methods are generally used... sorry I can't be more concrete than that! (also I could have had wrong an arbitrary number of constants) – Spi Jun 12 '14 at 12:02
• Thank you for your valuable discussion Mr. Spi.. I agree with your answer. – Oliver Jun 13 '14 at 2:06