Fix an integer $N$, and consider the unique positive solution $z$ to the following equation:

$$\text{e}^{-(z+N)} \sum \limits_{k=0}^{N} \frac{(z+N)^k}{k!}=\frac{1}{2}$$

For $N = 0$, we find that $z = \ln(2) \approx 0.6931$. For $N=10$, the solution is $z \approx 0.6685$. It appears that as $N$ approaches infinity, $z$ is approaching a constant value (maybe $2/3$?). What constant is this approaching?


This is an equation I derived to find the median (or rather, the difference between the mean and median) of a generalized exponential distribution. The case $N=0$ corresponds to the usual exponential distribution, or finding the inter-arrival times $\Delta t = t_1 - t_0$. Higher values of $N$ represent the elapsed time $t_{N+1} - t_0$.

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    $\begingroup$ A related question. $\endgroup$ – Lucian Nov 28 '15 at 1:45
  • $\begingroup$ @Lucian Definitely related. Is there some kind of continuity argument to show that $z=0$ is the limit? Because I'm having trouble seeing why $z$ couldn't take on other values, especially when $N$ blows up. $\endgroup$ – Ryan Nov 28 '15 at 7:24
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    $\begingroup$ There are two routes I see. First we could use the Laplace method with the formula $$e^{-x} \sum_{k=0}^{n} \frac{x^k}{k!} = \frac{1}{n!} \int_x^\infty e^{-t} t^n\,dt$$ (after setting $x = z+n$). As mentioned by Lucian the leading order will be $1/2$, so it will take some work to calculate the higher-order terms in the asymptotic (which will hold the information about the desired solution). Or we could use some other known asymptotic formulas which are more complicated and whose derivation is too technical to describe in an answer, but which can be manipulated to give the desired result. $\endgroup$ – Antonio Vargas Nov 28 '15 at 15:50
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    $\begingroup$ (...) If you're interested in the latter method then I'd be happy to write an answer for it. In any case, your guess that the solution tends to $2/3$ is indeed correct. $\endgroup$ – Antonio Vargas Nov 28 '15 at 15:55
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    $\begingroup$ @Lucian right, as I mentioned the leading order is indeed $1/2$. The idea is that we can calculate the asymptotic $$e^{-(z+N)} \sum \limits_{k=0}^{N} \frac{(z+N)^k}{k!} = \frac{1}{2} + \frac{2-3z}{3\sqrt{2\pi n}} + o(n^{-1/2}) \qquad \text{as } n \to \infty,$$ where the error $o(n^{-1/2})$ holds uniformly with respect to $z$ for $z$ in compact subsets of $\mathbb C$. So the bounded solutions to the equation $$e^{-(z+N)} \sum \limits_{k=0}^{N} \frac{(z+N)^k}{k!} = \frac{1}{2}$$ satisfy $$0 = \frac{2-3z}{3\sqrt{2\pi n}} + o(n^{-1/2}),$$ from which it follows that $z = 2/3 + o(1)$. $\endgroup$ – Antonio Vargas Nov 29 '15 at 0:11

If we let

$$ p_n(z) = \sum_{k=1}^{n} \frac{z^k}{k!}, $$

then the limit linked by Lucian in the comments can be written as

$$ \frac{p_n(n)}{\exp(n)} = \frac{1}{2} + o(1) \qquad \text{as } n \to \infty, \tag{1} $$

and as Aryabhata notes this can be found in Donald J. Newman's book A Problem Seminar. This limit is a special case of the following, due to Donald J. Newman and Theodore J. Rivlin${}^1$:

$$\DeclareMathOperator{erfc}{erfc} \frac{p_n(n+w\sqrt{n})}{\exp(n+w\sqrt{n})} = \frac{1}{2}\erfc\!\left(\frac{w}{\sqrt{2}}\right) + o(1) \qquad \text{as } n \to \infty, \tag{2} $$

where $\erfc$ is the complementary error function and the error term $o(1)$ is uniform with respect to $w$ for $w$ in compact subsets of $\mathbb C$. Indeed, since $\erfc(0) = 1$, limit $(1)$ follows from limit $(2)$ by setting $w=0$.

Now, the expression in your question is obtained by setting $w = z/\sqrt{n}$, but the asymptotic

$$ \begin{align} \frac{p_n(z+n)}{\exp(z+n)} &= \frac{1}{2}\erfc\!\left(\frac{z}{\sqrt{2n}}\right) + o(1) \\ &= \frac{1}{2} - \frac{z}{\sqrt{2\pi n}} + O(n^{-1}) + o(1) \qquad \text{as } n \to \infty, \tag{3} \end{align} $$

where in the second line we've used the Maclaurin series for $\erfc$, doesn't given us enough information to determine the $o(\sqrt{n})$ roots of your equation

$$ \frac{p_n(z+n)}{\exp(z+n)} = \frac{1}{2}. $$

Indeed, after substituting this into $(3)$ we see that to first order they must satisfy

$$ \frac{z}{\sqrt{2\pi n}} = o(1) \qquad \text{as } n \to \infty, $$

and we evidently need more information how this $o(1)$ error term balances with $1/\sqrt{n}$. In other words, we need more accurate information about the rate of convergence to the limit in $(2)$.

This information was obtained by Kriecherbauer et al.${}^2$ using a version of the Riemann-Hilbert method. After some massaging their Theorem 3.3 gives us the asymptotic

$$ \frac{p_n(n\zeta)}{\exp(n\zeta)} = \frac{1}{2}\erfc\!\left(\sqrt{n\varphi(\zeta)}\right) + \frac{(n\zeta e^{-\zeta})^n}{n!} - \frac{\exp(-n\varphi(\zeta))}{3\sqrt{2\pi n}} + o(n^{-1/2}), \tag{4} $$


$$ \varphi(\zeta) = \zeta - 1 - \log\zeta $$

and the correct branch of $\sqrt{}$ must be chosen, which is valid in the limit $n \to \infty, \zeta \to 1$. One can check that this is a very good approximation even for small values of $n$. Here's a plot of the left-hand side of $(4)$ in blue versus the right-hand side in orange for $0 < \zeta < 2$ and $n=2$:

enter image description here

Substituting $\zeta = 1+z/n$ then expanding everything in series and collecting like terms yields

$$ \frac{p_n(z+n)}{\exp(z+n)} = \frac{1}{2} + \frac{2-3z}{3\sqrt{2\pi n}} + o(n^{-1/2}) \qquad \text{as } n \to \infty, \tag{5} $$

where the error holds uniformly for $z = o(\sqrt{n})$. Below is a plot of $e^{-z-n} p_n(z+n) - 1/2$ in blue versus $(2-3z)/3\sqrt{2\pi n}$ in orange for $0 < z < 1$ and $n=2$:

enter image description here

and for $n=100$:

enter image description here

It then follows that the $o(\sqrt{n})$ roots of the equation

$$ \frac{p_n(z+n)}{\exp(z+n)} = \frac{1}{2} $$

must satisfy

$$ 0 = \frac{2-3z}{3\sqrt{2\pi n}} + o(n^{-1/2}) \qquad \text{as } n \to \infty, $$

and solving for $z$ yields

$$ z = \frac{2}{3} + o(1) \qquad \text{as } n \to \infty. $$

It's worth noting that your equation has many complex roots which are $\Theta(\sqrt{n})$ and are picked up by limit $(2)$ above, and many which are $\Theta(n)$ which require a different asymptotic to detect.

  1. D. J. Newman and T. J. Rivlin
    The zeros of the partial sums of the exponential function
    J. Approx. Theory 5 (1972), 405–412.

  2. T. Kriecherbauer, A. B. J. Kuijlaars, K. D. T.-R. McLaughlin, and P. D. Miller
    Locating the zeros of partial sums of $e^z$ with Riemann-Hilbert methods
    Integrable Systems and Random Matrices, Contemp. Math., vol. 458, Amer. Math. Soc., Providence, RI, 2008, pp. 183–195.
    arXiv link

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    $\begingroup$ Good lord, hadn't expected that one needs this heavy machinery to answer this question (+1) $\endgroup$ – tired Nov 29 '15 at 17:30
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    $\begingroup$ @tired, definitely one can get the same asymptotic in $(5)$ using the Laplace method, but it can get a little tedious, plus I had been looking at this recently so it was fresh in my mind. $\endgroup$ – Antonio Vargas Nov 29 '15 at 23:43

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