Problem: Solve equation
$$ \binom xn'\ =\ \log(n) $$
Here prime stands for the derivative with respect to $x$.
Observe that:
$\quad$ for integer $n$ large, the approximate solution is $\ x=n\ $ (it's extra off by Euler's $\gamma$).
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$\begingroup$ Not for this forum. Besides, the right hand side does not involve x, meaning the left hand side is linear in x, so n must be one. (Except that doesn't' work either.) Gerhard "One For A Different Forum" Paseman, 2017.06.23. $\endgroup$– Gerhard PasemanCommented Jun 23, 2017 at 20:46
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1$\begingroup$ The OP wants to consider, for each positive integer $n$, ${x \choose n}$ as a polynomial $f_n(x)$ of degree $n$, and solve $f_n'(x) = \log(n)$ (where $x$ is near $n$). Of course a closed-form solution is out of the question, but asymptotic solutions are possible. $\endgroup$– Robert IsraelCommented Jun 23, 2017 at 20:52
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5$\begingroup$ I hope the OP will at least rewrite the question so it is not subject to the same misinterpretation that misled both @GerhardPaseman and me. $\endgroup$– Steven LandsburgCommented Jun 23, 2017 at 21:06
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1$\begingroup$ @RobertIsrael, only the classical arithmetic form is "out of question". But there is plenty of potential opportunities for other neat closed form answers. $\endgroup$– Wlod AACommented Jun 24, 2017 at 0:36
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2$\begingroup$ @WlodAA: Like Gerhard, I thought that you were seeking an $n$ that would make the equation true for all $x$, not for an $x$ that would make the equation true for a given $n$. I'd have avoided this misinterpretation if I'd realized who you were (in which case I'd have realized you were unlikely to ask anything quite so crazy), but I might also have avoided it if I'd stopped to think for a moment. I apologize for not taking that moment, and I am retracting my close vote. $\endgroup$– Steven LandsburgCommented Jun 24, 2017 at 0:49
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1 Answer
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Surely you don't expect a closed-form formula for a root of a polynomial of degree $n-1$. Approximations are all you can hope for.
If $f_n(x) = {x \choose n}$ and $H_{n,m} = \sum_{k=1}^n 1/k^m$ the generalized harmonic numbers, we have
$$ \eqalign{f_n(n) &= 1\cr f_n'(n) &= H_{n,1} \sim \ln(n) + \gamma + \frac{1}{2n} + O(1/n^2)\cr f_n''(n) &= H_{n,1}^2 - H_{n,2} \sim (\ln(n)+\gamma)^2 - \frac{\pi^2}{6} + \frac{\ln(n)+1+\gamma}{n} + O(\ln(n)/n^2)\cr }$$ and thus the next approximation after $x=n$ might be $$x = n - \frac{\gamma}{(\ln(n)+\gamma)^2 - \pi^2/6}$$
answered Jun 23, 2017 at 22:12
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1$\begingroup$ "Surely you don't expect a closed-form formula for a root of a polynomial of degree n−1n−1. Approximations are all you can hope for." *** *** *** Not true. A meaningful, closed form answer does not have to be in classical terms of the $4$ arithmetic operations plus roots $x^{\frac 1k}$. $\endgroup$– Wlod AACommented Jun 24, 2017 at 0:15
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