# What function satisfies $F'(x) = F(2x)$?

The exponential generating function counting the number of graphs on $n$ labeled vertices satisfies (and is defined by) the equations $$F'(x) = F(2x) \; \; ; \; \; F(0) = 1$$ Is there some closed form or other nice description of this function? Does it have a name?

Of course, the series itself does not converge for any nonzero $x$, but like the Lambert W function (counting trees) it has combinatorial meaning. And the Lambert W function has a nice description as the inverse of $x e^x$; maybe there is a similar description of $F$?

• Well, if the function is analytic at $0$, then its Maclaurin series is $$\sum \frac{2^{n(n-1)/2}}{n!},$$from repeated differentiation at $x = 0$, but I'm guessing you knew this. – Theo Bendit Jul 18 '15 at 7:28
• Yes I did. I've been making little errors like this all week :-( – Theo Bendit Jul 18 '15 at 7:44
• the power series's radius of convergence is zero though. – mercio Jul 18 '15 at 8:20
• Hmm.. if we set $H(t)=F(\frac{1}{\log 2}2^{-t})$ then $H$ satisfies the delay differential equation $H'(t)=-2^{-t}H(t-1)$, which looks like it has many different solutions with $\lim_{t\to\infty}H(t)=1$ – Henning Makholm Jul 18 '15 at 10:37
• @6005: I don't. I just have the heuristic argument that you can choose $H$ freely on, say, the interval $[1,2]$ making sure the $H'(2)=-\frac12H(1)$, and then integrate the values of $H(t)$ up to $t=\infty$ numerically. Because of the $2^{-t}$ factor it should end up with a horizontal asymptote, which will be nonzero unless you're pretty unlucky. Then scale everything such that the limit is $1$ instead (the equation is linear after all). – Henning Makholm Jul 18 '15 at 20:02

## 1 Answer

There are infinitely many differentiable functions $$F\colon [0,\infty)\to\mathbb{R}$$ that satisfy $$F'(x)=F(2x)\qquad\text{and}\qquad F(0)=1.$$ We will follow the argument outlined by @HenningMakholm in the comments. First, consider the substitution $$G(x) = F\left(\dfrac{2^x}{\log 2}\right)$$. Plugging this into the above equation gives $$G\,'(x) = 2^x G(x+1)\qquad\text{and}\qquad \lim_{x\to-\infty} G(x)=1.$$ This delay differential equation has a large number of solutions. Roughly speaking, one can choose any function for $$G$$ on $$[0,1]$$, and then define $$G(x) = 2^{1-x}G\,'(x-1)\tag*{(1)}$$ recursively for $$x> 1$$, and $$G(x) = G(0) - \int_{x+1}^1 2^{u-1} G(u)\,du\tag*{(2)}$$ recursively for $$x<0$$. (See this question for a simpler example of this kind of solution.)

### Obstacles and Boundary Conditions

There are two obstacles to this construction:

1. If $$G$$ is $$C^n$$ on the interval $$(0,1)$$, then $$G$$ will be $$C^{n-1}$$ on $$(1,2)$$ by equation (1), and then $$C^{n-2}$$ on $$(2,3)$$, and so forth. Thus, if we want $$G$$ to be everywhere defined, we must start with a $$C^\infty$$ function on $$[0,1]$$.

2. If we start with an arbitrary $$C^\infty$$ function on $$[0,1]$$ and then switch to definition (1) for $$x>1$$, we must make sure that $$G$$ is $$C^\infty$$ at $$x=1$$, i.e. the the left and right hand derivatives of every order match up at this point.

For the second obstacle, repeatedly differentiating the equation $$G(x+1)=2^{-x}G\,'(x)$$ gives the sequence of equations $$G^{(n)}(x+1) \;=\; \sum_{i=0}^n \binom{n}{i}(-\log 2)^{n-i}\,2^{-x}\, G^{(i+1)}(x)$$ and plugging in $$x=0$$ yields $$G^{(n)}(1) \;=\; \sum_{i=0}^n \binom{n}{i}(-\log 2)^{n-i}\, G^{(i+1)}(0).$$ These are the boundary conditions that $$G$$ must satisfy on the interval $$[0,1]$$, but it is not hard to find $$C^\infty$$ functions that satisfy these, e.g. any bump function that satisfies $$G^{(n)}(0) = G^{(n)}(1) = 0\;\;\text{ for all }n\geq 1 \qquad\text{and}\qquad G(1)=0.$$

### The Limit as $$x\to-\infty$$

For negative values of $$x$$, the function $$G(x)$$ is defined by the integral equation $$G(x) = G(0) - \int_{x+1}^1 2^{u-1} G(u)\,du\tag*{(2)}$$ If we assume that $$G(x)\geq 0$$ for $$x\in [0,1]$$ and, say, that $$\int_0^1 2^{u-1} G(u)\,du \;<\; \frac{G(0)}{10}$$ then $$\frac{9}{10}G(0) \leq G(x) \leq G(0)$$ for all $$x\in[-1,0]$$. It follows recursively that $$G(0) \;\geq\; G(x) \;\geq\; \left(\frac{9}{10} - \frac{1}{\log 4}\right)G(0) \;>\; 0$$ for all $$x<-1$$, since \begin{align*} G(x) \;&=\; G(0)-\int_{x+1}^1 2^{u-1}G(u)\,du \\ &=\; G(0) - \int_0^1 2^{u-1}G(u)\,du - \int_{x+1}^0 2^{u-1}G(u)\,du \\ &\geq\; G(0) - \frac{G(0)}{10} - \int_{-\infty}^0 2^{u-1}G(0)\,du \\ &=\; \left(\frac{9}{10} - \frac{1}{\log 4}\right)G(0). \end{align*} Then $$G$$ is monotone on $$(-\infty,0)$$, so $$\displaystyle\lim_{x\to-\infty} G(x)$$ exists and is positive. Scaling $$G$$ linearly, we can arrange for this limit to be $$1$$.