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I'm reading a book and have encountered a relation which seems to me to be impossible to prove, I would like to be sure if this is the case. The author gives a probability function as $$p_n = \frac{e^{-c_1 n - c_2/n}}{Z},$$ where $c_1$ and $c_2$ are constants and Z is a normalization factor and $n \geq 3$. Then by considering $\langle n \rangle = 6$ and defining $\alpha$ (the second moment) as $\alpha = \sum_{n = 3}^{\infty} p_n (n - 6)^2$, the author claims one can show that

\begin{equation} \alpha + p_6 = 1, \quad \quad \quad 0.66 < p_6 < 1, \end{equation} \begin{equation} \alpha p_6^2 = 1 / 2 \pi, \quad \quad \quad 0.34 < p_6 < 0.66. \end{equation}

How is such a thing possible in the first place as these relations are not even dependent on $c_1$ and $c_2$?

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    $\begingroup$ Do you want to tell us the name of the secret book? $\endgroup$
    – Kung Yao
    Jul 12, 2021 at 21:58

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The equality $\alpha + p_6 = 1$ can be rewritten as $$\sum_{n=3}^\infty g(n)p_n=1,$$ where $$g(n):=(n-6)^2+1(n=6)\ge1(n\in\{5,6,7\})+4\times1(n\notin\{5,6,7\}).$$ Therefore and because $p_n>0$ for all $n\ge3$, we have $$\sum_{n=3}^\infty g(n)p_n>\sum_{n=3}^\infty p_n=1,$$ so that the equality $\alpha + p_6 = 1$ is always false.


As for the equality \begin{equation} \alpha p_6^2 = 1 / 2 \pi, \quad \quad \quad 0.34 < p_6 < 0.66, \end{equation} numerics suggest that it is also false in general. In particular, using Mathematica's numerical summation NSum[] command, for $c_1=12$ and $c_2=500$ we get $p_6=0.50995\ldots\in(0.34,0.66)$ but $2\pi\alpha p_6^2=0.90785\ldots\ne1$.

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  • $\begingroup$ @Loi : I do not see your identities in the paper linked in your comment. Anyhow, do you see any errors in the above disproof? $\endgroup$ Jul 12, 2021 at 23:48
  • $\begingroup$ @Loi : A question about corresponding approximate relations would be something quite different, and should be posted separately. $\endgroup$ Jul 12, 2021 at 23:54
  • $\begingroup$ @Loi : Apparently, they mean it in some unspecified approximate sense, with this justification: "asymptotic behaviour is manifested in fig. 5," Anyhow, the mathematical question that you posted is now resolved. $\endgroup$ Jul 13, 2021 at 0:06
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    $\begingroup$ @Loi : Such an edit would invalidate the valid answer to a clearly stated question. So, an "approximate" question should be posted separately. $\endgroup$ Jul 13, 2021 at 0:21

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