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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.

user1851281
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    For someone who have missed a simple algebra like me, knowing $v(n)=\sum_{1}^{n} u(x)$, $;; \sum_{1}^{n} u(x) - \sum_{1}^{n-1} u(x) = u(n)$ gives us the output what we desired. This corresponds to the $p(n)−p(n−1)$ in the answer. – user1851281 Nov 14 '22 at 18:23

1 Answers1

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  1. 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.

  2. 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.

Giulio R
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  • Thank you very much! Your answer seems quite terse compared to my mathematical maturity so I will take some time to study your answer. – user1851281 Nov 14 '22 at 15:08
  • Since I haven't got straight of what you are saying, does your answer on my second question imply that there is an effective algorithm to compute the inverse function by some recurrence relation? – user1851281 Nov 14 '22 at 15:09
  • The left-inverse is simply the map that takes a polynomial $p(x)$ and gives back the polynomial $p(x)-p(x-1)$, as in the example at the end. – Giulio R Nov 14 '22 at 15:10
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    Indeed, my short numerical experiments have shown that the $p(x)-p(x-1)$ method works correctly. Thank you very much! – user1851281 Nov 14 '22 at 15:22