Even though the irregularities lying behind the Bernoulli's numbers are yet to be clarified, it seems that we already have algorithms to effectively compute them. [Source1, Source2]
In this question, I want to focus on the inverse function of Faulhaber's formula. (also known as Bernoulli's formula)
WLOG, setting an arbirary polynomial as $u(x)=2x^2+2x$ and plugging it into Faulhaber's formula, we obtain $v(x)=\frac{2x^3}{3}+2x^2+\frac{4x}{3}$. Let's denote this mapping, a functional procedure of plugging an univariate polynomial into Faulhaber's formula, as $f: u \rightarrow v$. Then my questions are as follows.
Q1. Does the inverse function $f^{-1}:v \rightarrow u$ exist? That is to say, is $f: u \rightarrow v$ bijective? If so, how can I prove it?
Q2. If the inverse function $f^{-1}:v \rightarrow u$ exist, is there any effective algorithm to calculate the outputs of the function?
Looking forward to any kind of advices from comments and answers. Recommending any useful sources related to this problem would be also grateful. Thanks.
Since $f$ sends $x^n$ into a polynomial of degree $n+1$ with nonzero leading coefficient (actually, $\tfrac 1{n+1}$) we can see that it is injective. On the other hand, its image is not the full space of polynomials: $f$ misses all polynomials of degree $0$, i.e. all nonzero constants. Therefore we can only define a left-inverse, i.e. a map $g$ on the space of polynomials such that $$ g\circ f\,[p(x)]=p(x) $$ for all polynomials $p(x)$. This left-inverse is moreover non-unique, it can send the constant polynomial 1 to an arbitrary polynomial. Moreover, $f$ does not have a right-inverse.
Let us fix this left-inverse $g$ by saying $g(1):=0$. Then you can check that $$ g[p(x)]=p(x)-p(x-1) $$ works. Note that $$ f\circ g\,[p(x)]=p(x)-p(-1). $$ For example, when $p(x)=\frac {2x^3}3+2x^2+\frac{4x}3$ we have $$ p(x)-p(x-1)=2x^2+2x, $$ in agreement with your example.