Given an arbitrary sequence {$x_n$} in $\Bbb{R}$, find a test function $f$ with $f^{(n)}(0)=x_n$ Given an arbitrary sequence {$x_n$} in $\Bbb{R}$, can I find a test function having the $n$-derivative equal to $x_n$ at $0$?
Taking the $n^{th}$ derivative of an infinite polynomials gives:

$
f^{(n)}(z) = n!(a_n) +  ((n+1)!/1!)(a_{n+1})z + ((n+2)!/2!)(an_{n+2})z^2 + ...$

$f^{(n)}(z) = n!(a_n) + O{(z)}
$

So in general we need $a_n = x_n/n!$

Hence the function is $f(z) = x_0 + x_1z + (x_2/2)z^2 + ... + (x_n/n!)z^n + ...$
Taking the $n^{th}$ derivative of an infinite polynomials gives:

$
f^{(n)}(z) = n!(a_n) +  ((n+1)!/1!)(a_{n+1})z + ((n+2)!/2!)(an_{n+2})z^2 + ...$

$f^{(n)}(z) = n!(a_n) + O{(z)}
$

So in general we need $a_n = x_n/n!$

Hence the function is $f(z) = x_0 + x_1z + (x_2/2)z^2 + ... + (x_n/n!)z^n + ...$