Since $e$ is transcendental, the function $f:\mathbb Z_{\geq 0}^{n+1}\to \mathbb R$ is injective,

$$ f(\underset{\text{Integers}\ \geq\ 0}{\underbrace{a_0,a_1,\ldots, a_n}}) = a_0 + a_1e+a_2e^2+\cdots a_ne^n$$

Since $f$ is injective, in principle, the numbers $a_i$ can be deduced given the output of $f$. And since the numbers $a_i$ are non-negative, any interval of the real numbers intersects the codomain of $f$ on only finitely many points.$^*$

I'm wondering whether, in practice, the integer coefficients $a_i$ can be recovered from the value of $f$. Does it suffice to have only polynomially many (in $n$) bits of $f$? (i.e., polynomial in $n$). And can this be done in polynomial time?

There is an exponential-time algorithm, namely: Simply enumerate all $a$ for which $f(a)\leq t$ for a given target value $t$. In fact, in my application I know that $a_0+a_1+\cdots +a_n = 2^n$, so the enumeration is quite simple.

I have the same question for

$$g(a_0,a_1,\ldots, a_n) = \sum_{k=0}^ne^{\frac{i\pi}{p}k}a_k$$

In my application, I can choose $p$, e.g., a prime with $2n<p$, so that the contributions all lie in the north-east region of the complex plane.

$^*$ (Thanks to Emil Jerabek for pointing out that, without that constraint, $f$ cannot be computed from finitely many bits).

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    $\begingroup$ The reason you give in your first sentence is insufficient for injectivity of $f$. The correct reason is that $e$ is transcedental, and therefore not a zero of a polynomial. Also, it is not clear from the question whether $n$ is fixed. $\endgroup$ – Andrej Bauer Dec 23 '20 at 10:34
  • $\begingroup$ Yes, thanks. We can think of $n$ as the size of the problem, in the sense that I wish to know whether we only need $poly(n)$ many bits of $f$ in order to determine the values of $a$. $\endgroup$ – Lieuwe Vinkhuijzen Dec 23 '20 at 11:22

This is impossible. No finite number of bits of $f(a_0,\dots,a_n)$ suffices to determine any of $a_0,\dots,a_n$; in fact, any nondegenerate real interval contains the values $f(a_0,\dots,a_n)$ for infinitely many vectors $(a_0,\dots,a_n)\in\mathbb Z^{n+1}$, and this holds even you fix all but two of the $a_0,\dots,a_n$ in advance.

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    $\begingroup$ Thanks; in hindsight, that is obvious. I should have mentioned that, in my application, all the $a_i$ are non-negative, so this degenerate case is avoided. In this case, any real interval only contains finitely many values of $f$. $\endgroup$ – Lieuwe Vinkhuijzen Dec 23 '20 at 9:35
  • $\begingroup$ I see. I’m not sure how to directly exploit nonnegativity, but bounds such as $|a_i|\le2^n$ might be enough to get an efficient solution; I’ll think about it. $\endgroup$ – Emil Jeřábek Dec 23 '20 at 10:09
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    $\begingroup$ It is easy to check that explicit versions of Baker's theorem indeed imply that an approximation of $f(a_0,\dots,a_n)$ to $\mathrm{poly}(n,\log B)$ bits of precision is enough to uniquely determine the input among $(a_0,\dots,a_n)\in\{-B,\dots,B\}^{n+1}$. Now, the question is if we can also compute it efficiently. The first place to look at is the LLL algorithm as already mentioned in the other answer, but I have to stop now. $\endgroup$ – Emil Jeřábek Dec 23 '20 at 12:26
  • $\begingroup$ Thanks a lot! This is a great reference. $\endgroup$ – Lieuwe Vinkhuijzen Dec 23 '20 at 12:34
  • $\begingroup$ Thanks again for the answer; we are having some trouble locating explicit versions of Baker's theorem. Therefore, if you would be so kind as to provide a reference, we would be much obliged. A modest statement would suffice, I believe, namely that $\left| \sum_{k}^na_ke^k \right|$ is bounded away from $0$ by at least $\text{polylog}\left(1/\max_k a_k\right) = \text{log}(\max_k a_k)^{-O(1)}$ whenever the $a_k$ are not all zero, i.e., the value is of the form $0.000...0x$ for only polynomially many zeroes. $\endgroup$ – Lieuwe Vinkhuijzen Feb 9 at 15:06

If you know that the $a_i$'s are all not too large, and you have a good approximation to $f(a_0,\dots,a_n)$, I think LLL lattice basis reduction could be applicable (I haven't tried to verify the details).

Algorithms for finding integer relations look very closely related, and might possibly be directly applicable.


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