Is there a ("continuous") base-$e$ number system?

Question

Let $b=10$. Then for each $x \in [0,1)$, there is a unique sequence of numbers $a_1,a_2,\ldots$ in $\{0,\ldots,9\}$ not ending in an infinite sequence of $9$'s such that $$x=a_1b^{-1}+a_2b^{-2}+a_3b^{-3}+\cdots.$$

For example: $$0.83 = 8\cdot b^{-1}+3\cdot b^{-2}+0\cdot b^{-3}+\cdots$$

I'm wondering if we there's a similar theorem involving $e$, but with everything made continuous. Something like this:

Theorem. For each $x \in [0,1)$, there is a unique function $f: (0,\infty) \rightarrow [0,\infty)$ satisfying [insert deep condition here] such that $x = \int_0^\infty f(\alpha) e^{-\alpha} d\alpha$.

(This is the Laplace transform of $f$ evaluated at $1$.)

The idea is that we then proceed to define that $f$ is the "base-$e$" representation of the number $x$. It would then probably be easy to extend the idea to all of $\mathbb{R}_{>0}$, and hectic insights would hopefully follow.

Answer 1

What you are looking for is the concept of $\beta$-expansion. This is the same as the usual base-$b$ number systems for integer $b$, except now $\beta>1$ can be an arbitrary real number. That is, you can still write all reals as sums $$x=\sum_{k=-N}^\infty a_k\beta^{-k}$$ with each $a_j$ being an integer strictly smaller than $\beta$. So the $e$-expansion would have coefficients $a_k\in\{0,1,2\}$. In general you don't have uniqueness of the coefficients, but this is also the case in the usual decimal expansion, since $0.999\dots = 1$.

The problem with the integral formulation is that there is too much freedom. You could take for instance just the constant functions to represent all of $[0,1)$ uniquely as $$x = \int_0^\infty xe^{-\alpha}d\alpha,$$ but this wouldn't be very interesting. Or you could take any function $f:(0,\infty)\to(0,\infty)$ such that the integral diverges. Then $$\int_0^xf(\alpha)e^{-\alpha}d\alpha$$ would give any positive real for some $0\leq x<\infty$. That is the unique representation in this version would just be $f$ multiplied by the characteristic function of some interval $[0,x]$.

Answer 2

You would need a rather restrictive condition on $f$. Already if you stick with the discrete generalization you would have a system that "works", so the restriction on $f$ would be that it basically is a sequence.

For the discrete case you have representations on the form $\sum^\infty c_ke^{-k}$, where $0\le c_k < e$, analogous to normal bases. You don't have uniqueness here either since $\sum_1^\infty 2e^{-k} = 2/(e-1) > 1$ (compare with decimal system where $\sum_1^\infty 9\times 10^{-k} = 9/(10-1) = 1$ which corresponds to the only ambiguity of of $1.000\dots = 1.\dots.999\dots$).

So you have two sources of ambiguity in the continuous case, one is inherited from the ambiguity in the discrete case and one is added to by the continuos function. You have:

$$\int f(x)e^{-x}dx = \sum \int_j^{j+1} f(x)e^{-x}dx = \sum e^{-j}\int f(x)e^{-(x-j)}dx$$

So $\int_j^{j+1} f(x)e^{-(x-j)}dx$ corresponds to $c_n$ in the discrete expansion. There's simply a lot of functions $f$ that have the same integral.

Note that you can handle the first source of ambiguity basically the same way as for integer-base systems, but instead of not allowing the number to end in an infinite series of $(b-1)$s you require that each digit has the maximal value. That is $x \le \sum^nc_kb^{-k} < x + b^{-n}$, that way the values of $c_k$ would be uniquely determined.

The second source of ambiguity could be handled by restricting $f$ to a certain class of functions (for example constant on each aligned unit interval, or continuous and linear on each aligned unit interval). An alternative is to introduce the equivalence relation $f\sim g$ as $\int_j^{j+1}(f(x)-g(x))e^{j-x}dx = 0$, then the uniqueness with respect to this equivalence relation is guaranteed.