There's a sheaf of smooth real-valued functions on $\mathbb{R}$, and its germ at $0$ is some vector space $V$. I would like to understand this space. There is a surjective linear map
$$ \phi \colon V \to \mathbb{R}^{\mathbb{N}} $$
sending the germ $[f]$ of any smooth function $f: \mathbb{R} \to \mathbb{R}$ to its list of derivatives:
$$ \phi([f]) = (f(0), f'(0), f''(0), \dots ) $$
This is the 'easily understandable aspect' of $V$. I want to understand the kernel $ \mathrm{ker}(\phi)$.
So:
Question. Can we explicitly construct any nonzero linear map $\ell : \mathrm{ker}(\phi) \to \mathbb{R}$?
In simple terms: can can we extract any real numbers from the germ of a smooth function $f \colon \mathbb{R} \to \mathbb{R}$ in a linear way, other than by taking derivatives of that function at $0$?
Since $\ker(\phi)$ is infinite-dimensional, there exist infinitely many linearly independent linear maps $\ell : \mathrm{ker}(\phi) \to \mathbb{R}$. But this does not imply that we can actually get our hands on any of them, because it's possible that my last sentence can only be proved using the axiom of choice (or some weaker nonconstructive principle). There are some well-known examples of this frustrating situation in analysis, like the concept of Banach limit.
Someone asked a less precise version of this question on Math Stackexchange:
but it did not receive helpful answers, other than one pointing out that the question originally asked could be phrased in this way. My question is much less ambitious: I'm not asking for enough linear functionals $\ell \colon \mathrm{ker}(\phi) \to \mathbb{R}$ to separate points in $\mathrm{ker}(\phi)$, although that would be great; I'm just asking if any explicit linear functionals of this form are known.