# Nicer expression for 2.1369288...?

In Drift Analysis and Evolutionary Algorithms Revisited by Johannes Lengler and Angelika Steger in Theorem 10, there is mention of a constant "$$2.2$$", and in the proof it becomes apparent that the theorem is actually true for all c, which are the solution to the following system: Define $$f \colon \mathbb{R}^2 \to \mathbb{R}$$ as

$$f(a, c) := ac - e^{(a - 1)c} + \frac{a}{a - 1}$$ and consider the system of equations \begin{align}f(a, c) &= 0\\\\ \frac{\partial f}{\partial a}(a, c) &=0\end{align}

We are looking for a solution of this system near $$(0.237, 2.13)$$. It is easy to calculate solutions to this system numerically to arbitrary precision: Here is some code to get $$c = 2.13692884344059837908651709517671705999047797307894493...$$

For $$0 = \frac{\partial f}{\partial a}(a, c) - c\cdot f(a, c)$$ the exponential terms cancel and we get the relation $$c = \frac{\sqrt{-4a + 1} - 1}{2(a^2 - a)} \label{Eq:star}\tag{\star}$$ Substituting this expression back in $$f(a, c)$$, gives a function $$g\colon \mathbb{R} \to \mathbb{R}$$ defined as $$g(a) := f(a, \tfrac{\sqrt{-4a + 1} - 1}{2(a^2 - a)}).$$ One way of defining $$c$$ is to define $$a$$ as the root of $$g$$ (near $$0.237$$) and then use \eqref{Eq:star} to obtain $$c$$.

• Is there is a less implicit way of expressing $$c$$?
• Is $$c$$ related to some known other known constants?

Two observations:

1. It seems like $$a$$ is not so important here, in the sense that we can reparameterize $$a$$ and replace it by a function $$z(b)$$ (as long as $$z$$ is differntiable and goes through $$0.237$$ in the domain) and then have as the second equation the derivative with respect to $$b$$ instead of $$a$$. I played around with this a bit and for example $$z(b):= \frac{1 -b^2}{4}$$ then gives $$c = \frac{8}{b^3 + b^2 + 3b + 3}$$, but there might be an even better reparameterization.
2. Instead of expressing $$c$$ in terms of $$a$$ as the solution of a quadratic equation, one can also express $$a$$ in terms of $$c$$ as the solution of a cubic equation and then substitute back into $$f$$ to get a function only depending on $$c$$. Then the number I'm looking for is a root of that function, but this function is even more ugly than the description involving $$a$$.
• fyi: this constant is not (yet) in the OEIS... Feb 17 at 11:26
• Nor is it in the version of the Inverse Symbolic Calculator available at wayback.cecm.sfu.ca/projects/ISC/ISCmain.html Feb 17 at 13:04
• Wolfram Alpha also comes up empty. Feb 17 at 15:14

Apply Lagrange reversion to @TheSimpliFire’s equation:

$$\frac{k}{(k-1)^2}+1=e^k\iff k=1+\sqrt{\frac k{e^k-1}}\\\implies k=1+\sum_{n=1}^\infty\frac1{n!}\left.\frac{d^{n-1}}{dx^{n-1}}\left(\frac x{e^x-1}\right)^\frac n2\right|_1$$

Now use the Norlund polynomial, or generalized Bernoulli number, generating function. One notices the $$\frac1{(n-m+1)!}$$ truncates the inner sum:

$$\boxed{c=\frac{k^3}{k^2-k+1},k=\sum_{n=0}^\infty\sum_{m=0}^{n+1}\frac{B_n^{(\frac m2)}}{m! (n-m+1)!}}$$

shown here

• Note: the $n$ and $m$ indices were switched at the end Feb 17 at 23:00
• @PeterTaylor What does the sum become then; it seems to diverge? The closest formula for the inner sum was this one, but there is an extra $(-1)^k$ factor Feb 18 at 13:38
• Simple case of misunderstanding the notation: $B_n^{(\frac m2)} \neq \sqrt{B_n}^m$. Feb 18 at 23:04
• Are there any other formulas for $k$? The only other sum found was fairly long. Maybe there is an integral representation too. Feb 19 at 16:27

With a few substitutions, we find that $$c=\frac{k^3}{k^2-k+1}\quad\text{where}\quad1+\frac k{(1-k)^2}=e^k.$$ The solution for $$k$$ requires a more advanced function than Lambert $$W$$.

• Nice! That is already much shorter! Feb 17 at 19:15