If $\sum_{k=0}^{n-1}\frac{x^k}{k!}=\sum_{k=n}^{\infty}\frac{x^k}{k!}$ for large $n$, approximate $x$ in terms of $n$.

Question

Split the Maclaurin series for $e^x$ into two sub-series - the first $n$ terms, and the remainder - then equate the two sub-series. What is a good approximation for $x$, for large $n$?

Since $e^x=\sum_{k=0}^{\infty}\frac{x^k}{k!}$, we have $\sum_{k=0}^{n-1}\frac{x^k}{k!}=\frac{1}{2}e^x$.

Experimenting with desmos, it seems that $x\approx n-\frac{1}{3}$ for even or odd $n$, and $x\approx -0.28n-0.4$ for even $n$. Other than that, I do not know how to approach this question.

(Context: I was trying to answer this question, but I think the question in this post is interesting by itself.)

Answer 1

$$\sum_{k=0}^{n-1}\frac{x^k}{k!}=\frac{e^x \Gamma (n,x)}{\Gamma (n)}$$ $$\sum_{k=n}^{\infty}\frac{x^k}{k!}=\frac{e^x (\Gamma (n)-\Gamma (n,x))}{\Gamma (n)}$$ So, we need to solve for $x$ the equation $$2\Gamma (n,x)=\Gamma (n)$$ Numerically, it is better to look for the zero of function $$f(x)=\log (2 \Gamma (n,x))-\log (\Gamma (n))$$ As you noticed, the solution is close to $n$.

If we make one single iteration of Newton method with $x_0=n$, we have $$x_1=n+e^n n^{1-n} \Gamma (n,n) (\log (2 \Gamma (n,n))-\log (\Gamma (n)))$$ Looking here, there is nice approximation : to first order $$\Gamma (n,n)\sim n^{n-1}\, e^{-n}\Bigg[\sqrt{\frac{\pi }{2}} \sqrt{n}-\frac{1}{3}+\frac{1}{12} \sqrt{\frac{\pi }{2n}} +O\left(\frac{1}{n}\right) \Bigg]$$ and then, guess what, $$x_1 \sim n-\frac 13 +\frac{1}{9 \sqrt{2 \pi n}} +\frac{1}{81 \pi n}+O\left(\frac{1}{n^{\frac 32}}\right)$$

Just a few checks $$\left( \begin{array}{ccc} n & \text{estimate} & \text{solution} \\ 5 & 4.68728 & 4.67091 \\ 10 & 9.68108 & 9.66871 \\ 15 & 14.6784 & 14.6680 \\ 20 & 19.6768 & 19.6677 \\ 25 & 24.6757 & 24.6675 \\ 30 & 29.6749 & 29.6673 \\ 35 & 34.6743 & 34.6672 \\ 40 & 39.6738 & 39.6672 \\ 45 & 44.6734 & 44.6671 \\ 50 & 49.6730 & 49.6671 \end{array} \right)$$

Taking more terms, we have $$x_1 \sim n-\frac 13 +\frac{1}{9 \sqrt{2 \pi n}}+\frac{5-12 \pi }{405 \pi n}+\frac{20+51 \pi }{4860 \sqrt{2} \pi ^{3/2}n^{3/2}}+O\left(\frac{1}{n^2}\right)$$ which, for $n=5$ gives $x_1=4.68177$.

Edit

For the case where $n=2p$, we need to solve $$g(x)=\log (2 \Gamma (2p,x))-\log (\Gamma (2p))$$ As you noticed, the solution is close to $-\frac 1 2p$. Doing the same as above $$x_1=-\frac 1 2p+2^{2 p-1} e^{-p/2} p^{1-2 p} \Gamma \left(2 p,-\frac{p}{2}\right) \left(\log(\Gamma (2 p))-\log \left(2 \Gamma \left(2 p,-\frac{p}{2}\right)\right)\right)$$ Using another approximation given in 1, $$\Gamma \left(2 p,-\frac{p}{2}\right)\sim -\frac{1}{27} e^p p^{2 p-2} (9 p+1)\sim -\frac{1}{3} e^p p^{2 p-1}$$

At this point, at least for the time being, I am stuck.

Answer 2

This is equivalent to the asymptotic inversion with respect to $x$ and for large $n$ the normalised upper incomplete gamma function $Q$: $$ Q(n,x) = \frac{1}{2}. $$ From this result, we have $$ x \sim n-\frac{1}{3}+\frac{8}{405n}+\frac{184}{25515n^2}+\frac{2248}{3444525n^3}+\ldots, $$ as $n\to +\infty$. To find the negative root when $n$ is even, we can proceed as follows. For the lower incomplete gamma function $\gamma$, we have $$ \frac{{\gamma (n + 1,nx)}}{{(nx)^n e^{ - nx} }} = \frac{{n\gamma (n,nx)}}{{(nx)^n e^{ - nx} }} - 1. $$ Thus, by this result, $$ \frac{{n\gamma (n,nx)}}{{(nx)^n e^{ - nx} }} = \frac{1}{1-x}+\mathcal{O}\!\left(\frac{1}{n}\right) $$ for $x<0$ and $n\to +\infty$. For the normalised lower incomplete gamma function this means $$ P(n,nx)= \frac{{(nx)^n e^{ - nx} }}{{\Gamma (n + 1)}}\frac{1}{1-x}\left( 1+\mathcal{O}\!\left(\frac{1}{n}\right) \right), $$ for $x<0$ and $n\to +\infty$. Employing Stirling's formula yields $$ P(n,nx)= \frac{{x^n e^{n(1 - x)} }}{{\sqrt {2\pi n} }}\frac{1}{1-x}\left( 1+\mathcal{O}\!\left(\frac{1}{n}\right) \right), $$ for $x<0$ and $n\to +\infty$. Now the problem is equivalent to solving $P(n,nx)=\frac{1}{2}$ for even $n$ and negative $x$. If $n$ is even, the above asymptotics may be written $$ P(n,nx) = \frac{{e^{n(1 - x + \log \left| x \right|)} }}{{\sqrt {2\pi n} }}\frac{1}{1-x}\left( 1+\mathcal{O}\!\left(\frac{1}{n}\right)\right) $$ for $x<0$ and $n\to +\infty$. A first approximation to the solution of the original problem is thus $$ x_0 n=(-0.2784645438\ldots )n, $$ where $x_0=-0.2784645438\ldots$ is the unique negative root of $1-x+\log|x|=0$. A better approximation follows by solving $$ (1 - x + \log \left| x \right|) = \frac{1}{{2n}}\log \left( {\frac{{(1-x)^2\pi n}}{2}} \right). $$ Writing $x = x_0 + \xi$, using linear approximation, and taking $1-x\approx 1-x_0$ on the right-hand side, we find $$ \frac{{1 - x_0 }}{{x_0 }}\xi \approx \frac{1}{{2n}}\log \left( {\frac{{(1-x_0)^2\pi n}}{2}} \right) \Longrightarrow \xi \approx \frac{{x_0 }}{{2(1 - x_0 )n}}\log \left( {\frac{{(1-x_0)^2\pi n}}{2}} \right).$$ Hence, a second approximation to the solution of the original problem is \begin{align*} & x_0 n + \frac{{x_0 }}{{2(1 - x_0 )}}\log \left( {\frac{{(1-x_0)^2\pi n}}{2}} \right) \\ & = (-0.2784645438\ldots)n +(-0.1089058528\ldots)\log \left( (2.5674219652\ldots)n \right). \end{align*} It is possible to obtain higher approximations using more terms from the asymptotics of $P(n,nx)$, but the procedure becomes tedious.